To Pm, sine qua non

!Preface 🧱

Contents Preface 🧱 1 Introduction 1 What Are Manifolds? … … … … … … … … 1 Why Study Manifolds? … … … … … … … … 4 2 Topological Spaces 17 Topologies … … … … … … … … … … . 17 Bases … … … … … … … … … … … 27 Manifolds … … … … … … … … … … . 30 Problems … … … … … … … … … … . 36 3 New Spaces from Old 39 Subspaces … … … … … … … … … … . 39 Product Spaces … … … … … … … … … . 48 Quotient Spaces … … … … … … … … … 52 Group Actions … … … … … … … … … . 58 Problems … … … … … … … … … … . 62 4 Connectedness and Compactness 65 Connectedness … … … … … … … … … . 65 Compactness … … … … … … … … … . . 73 Locally Compact Hausdorff Spaces … … … … … . . 81 Problems … … … … … … … … … … . 88

xvi Contents 5 Simplicial Complexes 91 Euclidean Simplicial Complexes … … … … … … . 92 Abstract Simplicial Complexes … … … … … … . 96 Triangulation Theorems … … … … … … … . . 102 Orientations … … … … … … … … … … 105 Combinatorial Invariants… … … … … … … . . 109 Problems … … … … … … … … … … . 114 6 Curves and Surfaces 117 Classification of Curves … … … … … … … . . 118 Surfaces … … … … … … … … … … . . 119 Connected Sums … … … … … … … … … 126 Polygonal Presentations of Surfaces… … … … … . . 129 Classification of Surface Presentations … … … … … 137 Combinatorial Invariants… … … … … … … . . 142 Problems … … … … … … … … … … . 146 7 Homotopy and the Fundamental Group 147 Homotopy … … … … … … … … … … . 148 The Fundamental Group … … … … … … … . . 150 Homomorphisms Induced by Continuous Maps … … … . 158 Homotopy Equivalence … … … … … … … … 161 Higher Homotopy Groups … … … … … … … . 169 Categories and Functors … … … … … … … . . 170 Problems … … … … … … … … … … . 176 8 Circles and Spheres 179 The Fundamental Group of the Circle … … … … … 180 Proofs of the Lifting Lemmas … … … … … … . . 183 Fundamental Groups of Spheres… … … … … … . 187 Fundamental Groups of Product Spaces … … … … . . 188 Fundamental Groups of Manifolds … … … … … . . 189 Problems … … … … … … … … … … . 191 9 Some Group Theory 193 Free Products … … … … … … … … … . . 193 Free Groups … … … … … … … … … … 199 Presentations of Groups … … … … … … … . . 201 Free Abelian Groups … … … … … … … … . 203 Problems … … … … … … … … … … . 208 10 The Seifert–Van Kampen Theorem 209 Statement of the Theorem … … … … … … … . 210 Applications… … … … … … … … … … 212 Proof of the Theorem … … … … … … … … 221

Contents xvii Distinguishing Manifolds … … … … … … … . . 227 Problems … … … … … … … … … … . 230 11 Covering Spaces 233 Definitions and Basic Properties … … … … … … 234 Covering Maps and the Fundamental Group … … … … 239 The Covering Group … … … … … … … … . 247 Problems … … … … … … … … … … . 253 12 Classification of Coverings 257 Covering Homomorphisms … … … … … … … . 258 The Universal Covering Space … … … … … … . . 261 Proper Group Actions … … … … … … … … 266 The Classification Theorem … … … … … … … 283 Problems … … … … … … … … … … . 289 13 Homology 291 Singular Homology Groups … … … … … … … 292 Homotopy Invariance … … … … … … … … . 300 Homology and the Fundamental Group … … … … . . 304 The Mayer–Vietoris Theorem … … … … … … . . 308 Applications… … … … … … … … … … 318 The Homology of a Simplicial Complex … … … … . . 323 Cohomology … … … … … … … … … … 329 Problems … … … … … … … … … … . 334 Appendix: Review of Prerequisites 337 Set Theory … … … … … … … … … … 337 Metric Spaces … … … … … … … … … . . 347 Group Theory… … … … … … … … … . . 352 References 359 Index 362

Chapter 1-Introduction

Ac ourse on manifolds differs from most other introductory graduate mathematics courses in that the subject matter is often completely unfamiliar. Most beginning graduate students have had undergraduate courses in algebra and analysis, so that graduate courses in those areas are continuations of subjects they have already begun to study. But it is possible to get through an entire undergraduate mathematics education, at least in the United States, without ever hearing the word “manifold.” One reason for this anomaly is that even the definition of manifolds involves rather a large number of technical details—for example, in this book the formal definition will not come until the end of Chapter 2. Since it is disconcerting to embark on such an adventure without even knowing what it is about, we devote this introductory chapter to a nonrigorous definition of manifolds, an informal exploration of some examples, and a consideration of where and why they arise in various branches of mathematics.

What Are Manifolds?

Let us begin by describing informally how one should think about manifolds. The underlying idea is that manifolds are like curves and surfaces, except, perhaps, that they might be of higher dimension. Every manifold has a dimension, which is, roughly speaking, the number of independent numbers (or “parameters”) needed to specify a point. The prototype of

2 1. Introduction

FIGURE 1.1. Plane curves. FIGURE 1.2. Space curve. an n-dimensional manifold is n-dimensional Euclidean space Rn, in which each point is an n-tuple of real numbers. An n-dimensional manifold is an object modeled locally on Rn; this means that it takes exactly n numbers to specify a point, at least if we do not stray too far from a given starting point. A physicist would say that an n-dimensional manifold is an object with n “degrees of freedom.” Manifolds of dimension 1 are commonly called curves (although they need not be “curved” in the ordinary sense of the word). The simplest example is the real line; other examples are provided by familiar plane curves such as circles, parabolas, or the graph of any continuous function of the form $$y =f(x)$$ (Figure1.1). Still other familiar 1-dimensional manifolds are space curves, which are often described parametrically by equations such as $$ \begin{aligned} (x,y,z) &= (f(t),g(t),h(t)) \[6pt] \text{for some continuous functions }f,g,h. \end{aligned} $$ (Figure 1.2). In each of these examples, a point on the curve can be unambiguously specified by a single real number. For example, a point on the real line is a real number. We might specify a point on the circle by its angle, a point on a graph by its $x$ coordinate, and a point on a parametrized curve by its parameter t. Note that although a parameter value determines a point, different parameter values may correspond to the same point, as in the case of angles on the circle. But in every case, as long as we stay close to some initial point, there is a one-to-one correspondence between nearby real numbers and nearby points on the curve.

FIGURE 1.3. Doughnut surface. Manifolds of dimension 2 are surfaces. The two most common examples are planes and spheres. (When mathematicians speak of a sphere, we invariably mean a spherical surface, which is 2-dimensional, not a solid ball, which is 3-dimensional.) Other familiar surfaces include cylinders, ellipsoids, paraboloids, and the doughnut-shaped surface in R3 obtained by revolving a circle around the z-axis (Figure 1.3). (This doughnut-shaped surface is often called a torus, but we will reserve that name for a slightly different but closely related object, to be introduced in the next chapter.) In these cases two coordinates are needed to determine a point. For example, on the plane we typically use Cartesian or polarcoordinates; on the sphere we might use latitude and longitude; while on the doughnut surface we might use two angles. As in the 1-dimensional case, the correspondence between points and pairs of numbers is in general only local. The only higher-dimensional manifold that we can visualize is Euclidean 3-space. But it is not hard to construct subsets of higher-dimensional Euclidean spaces that might reasonably be called manifolds. First, any open subset of Rn is an n-manifold for obvious reasons. More interesting examples are obtained by using one or more equations to “cut out” lower-dimensional subsets. For example, the set of points $(x1,x2,x3,x4)$ in R4 satisfying the equation $$(x1)2+(x2)2+(x3)2+(x4)2 =1$$ (1.1) is called the (unit) 3-sphere. It is a 3-dimensional manifold because in a neighborhood of any given point it takes exactly three coordinates to specify a nearby point: Starting at, say, the “north pole” $$(0,0,0,1)$$, we can solve equation (1.1) for $x4$, and then each nearby point is uniquely determined by choosing appropriate (small) $(x1,x2,x3)$ coordinates and setting$x4 =(1−(x1)2−(x2)2−(x3)2)1/2$. Near other points, we may need to solve for different variables; but in each case three coordinates suffice.

4 1. Introduction The key feature of these examples is that an n-dimensional manifold “looks like” Rn locally. To make sense of the intuitive notion of “looks like,” we will say that two subsets of Euclidean spaces U ⊂ Rk, V ⊂ Rn are topologically equivalent or homeomorphic (Greek for “same form”) if there exists a one-to-one correspondence ϕ: U →V such that both ϕ and its inverse are continuous maps. (Such a correspondence is called a home- omorphism.) A subset M of some Euclidean space Rk is locally Euclidean of dimension n if every point of M has a neighborhood in M that is topo- logically equivalent to a ball in Rn. Nowwecangiveapreliminarydefinitionofmanifolds.Ann-dimensional manifold(n-manifoldforshort)isasubsetofsomeEuclideanspaceRk that is locally Euclidean of dimension n. Later, after we have developed more machinery,wewillgiveaconsiderablymoregeneraldefinition;butthisone will get us started. Why Study Manifolds? What follows is an incomplete survey of some of the fields of mathematics in which manifolds play an important role. Topology Roughlyspeaking,topologyisthebranchofmathematicsthatisconcerned with properties of sets that are unchanged by “continuous deformations.” More accurately, a topological property is one that is preserved by home- omorphisms. ThesubjectinitsmodernformwasinventedacenturyagobytheFrench mathematician Henri Poincar´e, as an outgrowth of his attempts to classify geometric objects that appear in analysis. In a seminal 1895 paper titled Analysis Situs (theoldnamefortopology,Latinfor“analysisofposition”) andaseriesofcompanionpapersin1899–1905,Poincar´elaidoutthemain problems of topology and introduced an astonishing array of new ideas for solvingthem.Asyoureadthisbook,youwillseethathisnameiswrittenall overthesubject.Intheinterveningcentury,topologyhastakenontherole of providing the foundations for just about every branch of mathematics that has any use for a concept of “space.” (An excellent historical account of the first six decades of the subject can be found in [Die89].) Hereisasimplebuttellingexampleofthekindofproblemthattopology wasinventedtosolve.Considertwosurfacesinspace:asphereandacube. Itshouldnotbehardtoconvinceyourselfthatthecubecanbecontinuously deformed into the sphere without tearing or collapsing it. It is not much harder to come up with an explicit formula for a homeomorphism between them (we will do so in Chapter 2). Similarly, with a little more work, you

Why Study Manifolds? 5 FIGURE 1.4. Deforming a doughnut into a coffee cup. shouldbeabletoseehowadoughnutsurfacecanbecontinuouslydeformed into the surface of a one-handled coffee cup, by stretching out one-half of the doughnut to become the cup, and shrinking the other half to become the handle (Figure 1.4). Once you decide on an explicit set of equations to define a “coffee-cup surface” in R3, you could in principle come up with a setofformulastodescribeahomeomorphismbetweenitandthedoughnut surface. On the other hand, a little reflection will probably convince you that there is no homeomorphism from the sphere to the doughnut surface: Any such map would have to tear open a “hole” in the sphere, and thus could not be continuous. Itisusuallyrelativelystraightforward(thoughnotalwayseasy!)toprove that two manifolds are topologically equivalent once you have convinced yourself intuitively that they are: Just write down an explicit homeomor- phism between them. What is much harder is to prove that two manifolds are not homeomorphic—even when it seems “obvious” that they are not as in the case of the sphere and the doughnut—because you would need to show that no one, no matter how clever, could find such a map. History abounds with examples of operations that mathematicians long believedtobeimpossible,onlytobeprovedwrong.Hereisanexamplefrom topology.Imagineasphericalsurfacecoloredwhiteontheoutsideandgray ontheinside,andimaginethatitcanmovefreelyinspace,includingpassing freely through itself. Under these conditions you could turn the sphere insideoutbycontinuouslydeformingit,sothatthegraysideendsupfacing out, but it seems obvious that in so doing you would have to introduce a creasesomewhere.(Thereareprecisemathematicaldefinitionsoftheterms “continuouslydeforming”and“creases,”butyoudonotneedtoknowthem to get the general idea.) The simplest way to proceed would be to push the northern hemisphere down and the southern hemisphere up, allowing them to pass through each other, until the two hemispheres had switched places (Figure 1.5); but this would introduce a crease along the equator. The topologist Stephen Smale stunned the mathematical community in 1958 [Sma58] when he proved it was possible to turn the sphere inside out without introducing any creases. Several ways to do this are beautifully illustrated in video recordings [Max77, LMM94, SFL98].

6 1. Introduction FIGURE 1.5. Turning a sphere inside out (with a crease). The usual way to prove that two manifolds are not topologically equiv- alent is by finding topological invariants: properties (which could be num- bers or other mathematical objects such as groups, matrices, polynomials, or vector spaces) that are preserved by homeomorphisms. If two manifolds have different invariants, they cannot be homeomorphic. It is evident from the examples above that geometric properties such as circumferenceandareaarenottopologicalinvariants,becausetheyarenot generallypreservedbyhomeomorphisms.Intuitively,thepropertythatdis- tinguishes the sphere from the doughnut surface is the fact that the latter has a “hole,” while the former does not. But it turns out that giving a precise definition of what is meant by a hole takes rather a lot of work. One invariant that is commonly used to count holes in a manifold is called the fundamental group of the manifold, which is a group (in the algebraic sense) attached to each manifold in such a way that homeomorphic man- ifolds have isomorphic groups. Then the “size” of the fundamental group is a measure of the number of holes possessed by the manifold. The study of the fundamental group will occupy a major portion of this book. It is the starting point for algebraic topology, which is the subject that studies topological properties of manifolds (or other geometric objects) by attach- ingalgebraicstructuressuchasgroupsandringstotheminatopologically invariant way. One of the most important problems of topology is the problem of clas- sifyingmanifolds.Ideally,onewouldliketoproducealistofn-dimensional manifolds, and a theorem that says every n-dimensional manifold is home- omorphic to exactly one on the list, together with a list of computable topological invariants that could be used to decide where on the list any given manifold belongs. Precisely such a theorem is known for surfaces: It says that every compact surface is homeomorphic to a sphere, or to a doughnut surface with a finite number of holes, or to a connected sum of projective planes. (We will define these terms and prove the theorem in Chapter 6.) For higher-dimensional manifolds, the situation is much more compli- cated.Forexample,Poincar´econjecturedaround1900thatanycompact3-

Why Study Manifolds? 7 manifoldwhosefundamentalgroupisthetrivial(one-element)groupmust be homeomorphic to the 3-sphere. For a long time, topologists thought of this as the simplest first step in a potential classification of 3-manifolds. Butalthoughanalogousconjectureshavebeenmadeforhigher-dimensional manifolds and were proved in the intervening years (for 5-manifolds and higher by Stephen Smale in 1961 [Sma61], and for 4-manifolds by Michael Freedman in 1982 [Fre82]), the original Poincar´e conjecture remains as of this writing a preeminent unsolved problem in topology. The best hope for a classification of 3-manifolds is the geometrization conjecture made in the 1970s by William Thurston (see [Sco83, Thu97] for an explanation), which says, roughly, that every compact 3-manifold can be cut into finitely many pieces each of which admits one of eight (mostly non-Euclidean) geometric structures. Since the manifolds with geometric structures are much better understood, a proof of this conjecture would go a long way toward provid- ing a complete classification of 3-manifolds; in particular, it would imply that the Poincar´e conjecture is true. Indimensions4andhigher,ontheotherhand,thereisnohopeforacom- pleteclassification:Itwasprovedin1958byA.A.Markovthatthereisno algorithmforclassifyingmanifoldsofdimensiongreaterthan3(see[Sti93]). Nonetheless, there is much that can be said using sophisticated combina- tions of techniques from algebraic topology, differential geometry, partial differentialequations,andalgebraicgeometry,andspectacularprogresswas made in the last half of the twentieth century in understanding the vari- ety of manifolds that exist. The topology of 4-manifolds, in particular, is currently a highly active field of research. Geometry TheprincipalobjectsofstudyinEuclideanplanegeometry,asyouencoun- tered it in secondary school, are figures constructed from portions of lines, circles, and other curves—in other words, 1-manifolds. Similarly, solid ge- ometry is concerned with figures made from portions of planes, spheres, and other 2-manifolds. The properties that are of interest are those that are invariant under rigid motions. These include simple properties such as lengths,angles,areas,andvolumes,aswellasmoresophisticatedproperties derivedfromthemsuchascurvature.Thecurvatureofacurveorsurfaceis aquantitativemeasureofhowitbendsandinwhatdirections;forexample, apositivelycurvedsurfaceis“bowl-shaped,”whileanegativelycurvedone is “saddle-shaped.” Geometric theorems involving curves and surfaces range from the trivial totheverydeep.Atypicaltheoremyouhaveundoubtedlyseenbeforeisthe angle-sumtheorem:ThesumoftheinterioranglesofanyEuclideantriangle is π radians. This seemingly trivial result has profound generalizations to thestudyofcurvedsurfaces,whereanglesmayadduptomoreorlessthan π depending on the curvature of the surface. The high point of surface

8 1. Introduction theory is the Gauss–Bonnet theorem: For a closed, bounded surface in R3, thistheoremexpressestherelationshipbetweenthetotalcurvature(i.e.,the integral of curvature with respect to area) and the number of holes it has. If the surface is topologically equivalent to an n-holed doughnut surface, the theorem says that the total curvature is exactly equal to 4π −4πn. In the case n = 1 this implies that no matter how a one-holed doughnut surface is bent or stretched, the regions of positive and negative curvature will always precisely cancel each other out so that the total curvature is zero. The introduction of manifolds has allowed the study of geometry to be carried into higher dimensions. The appropriate setting for studying geo- metricpropertiesinarbitrarydimensionsisthatofRiemannian manifolds, which are manifolds on which there is a rule for measuring distances and angles,subjecttocertainnaturalrestrictionstoensurethatthesequantities behave analogously to their Euclidean counterparts. The properties of in- terest are those that are invariant under isometries, or distance-preserving transformations. For example, one can study the relationship between the curvatureofann-dimensionalRiemannianmanifold(alocalproperty)and its global topological type. A typical theorem is that a complete Riemann- ian n-manifold whose curvature is everywhere larger than some fixed posi- tivenumbermustbecompactandhaveafinitefundamentalgroup(nottoo many holes). The search for such relationships is one of the principal ac- tivitiesinRiemanniangeometry,athrivingfieldofcontemporaryresearch. See Chapter 1 of [Lee97] for an informal introduction to the subject. Complex Analysis Complexanalysisisthestudyofholomorphic(i.e.,complexanalytic)func- tions. Some such functions are naturally “multiple-valued.” A typical ex- ample is the complex square root. Except for zero, every complex number hastwodistinctsquareroots.Butunlikethecaseofpositiverealnumbers, where we can always √unambiguously choose the positive square root to denote by the symbol x, it is not possible to define a global continuous square root function on the complex plane. To see why, write z in polar c√oordinates a√s z = reiθ. Then the two square roots of z can be written reiθ/2 and rei(θ/2+π).Asθ increasesfrom0to2π,thefirstsquareroot goes from the positive real axis through the upper half-plane to the neg- ative real axis, while the second goes from the negative real axis through the lower half-plane to the positive real axis. Thus whichever continuous square root function we start with on the positive real axis, we are forced to choose the other after having made one circuit around the origin. Eventhougha“two-valuedfunction”isproperlyconsideredasarelation and not really a function at all, we can define the graph of such a relation in an unambiguous way. To warm up with a simpler example, consider the two-valued square root “function” on the nonnegative real axis. Its graph

Why Study Manifolds? 9 u u2 =x x FIGURE 1.6. Graph of the two branches of the real square root. √ is defined to be the set of pairs (x,u) ∈ R×R such that u = ± x, or equivalently u2 =x. This is a parabola opening in the positive x direction (Figure 1.6), which we can think of as the two “branches” of the square root. Similarly, the graph of the two-valued complex square root “function” is the set of pairs (z,w) ∈ C×C such that w2 = z. Over each small disk U ⊂ C that does not contain 0, this graph has two branches or “sheets,” correspondingtothetwopossiblecontinuouschoicesofsquarerootfunction on U (Figure 1.7). If you start on one sheet above the positive real axis andpassoncearoundtheorigininthecounterclockwisedirection,youend upontheothersheet.Goingaroundoncemorebringsyoubacktothefirst sheet. ItturnsoutthatthisgraphinC2isa2-dimensionalmanifold,ofaspecial type called a Riemann surface—this is essentially a 2-manifold on which there is some way to define holomorphic functions. Riemann surfaces are of great importance in complex analysis, since any holomorphic function gives rise to a Riemann surface by a procedure analogous to the one we sketched above. The surface we constructed turns out to be topologically equivalent to a plane, but more complicated functions can give rise to m√orecomplicatedsurfaces.Forexample,thetwo-valued“function”f(z)= ± z3−z yields a Riemann surface that is homeomorphic to a plane with one “handle” attached. One of the fundamental tasks of complex analysis is to understand the topological type (number of “holes” or “handles”) of the Riemann surface

10 1. Introduction w √ w = reiθ/2 y U √ w = rei(θ/2+π) x FIGURE 1.7. Two branches of the complex square root. of a given function, and how it relates to the analytic properties of the function. Algebra Oneofthemostimportantobjectsstudiedinabstractalgebraisthegeneral linear group GL(n,R), which is the group of n×n invertible real matrices. As a set, it can be identified with a subset of n2-dimensional Euclidean space, simply by stringing all the matrix entries out in a row. Since a matrixisinvertibleifandonlyifitsdeterminantisnonzero,GL(n,R)isan opensubsetofRn2 ,andisthereforeann2-dimensionalmanifold.Similarly, the complex general linear group GL(n,C) is the group of n×n invertible complex matrices; it is a 2n2-manifold, because we can identify Cn2 with R2n2 . A Lie group is a group (in the algebraic sense) that is also a manifold, togetherwithsometechnicalconditionstoensurethatthegroupstructure and the manifold structure are compatible with each other. They play a central role in differential geometry, representation theory, and mathemat- ical physics, among many other fields. The most important Lie groups are subgroups of the real and complex general linear groups. Some commonly

Why Study Manifolds? 11 FIGURE 1.8. A plane curve with disconnected pieces. encountered examples are the special linear group SL(n,R) ⊂ GL(n,R), consisting of matrices with determinant 1; the orthogonal group O(n) ⊂ GL(n,R),consistingofmatriceswhosecolumnsareorthonormal;thespecial orthogonal group SO(n) = O(n)∩SL(n,R); and their complex analogues, the complex special linear group SL(n,C) ⊂ GL(n,C), the unitary group U(n)⊂GL(n,C), and the special unitary group SU(n)=U(n)∩SL(n,C). ItisimportanttounderstandthetopologicalstructureofaLiegroupand howitstopologicalstructurerelatestoitsalgebraicstructure.Forexample, it can be shown that SO(2) is topologically equivalent to a circle, SU(2) is topologically equivalent to the 3-sphere, and any connected abelian Lie groupistopologicallyequivalenttoaCartesianproductofcirclesandlines. Liegroupsprovidearichsourceofexamplesofmanifoldsinalldimensions. Algebraic Geometry Algebraic geometers study the geometric properties of solution sets to sys- tems of polynomial equations. Many of the basic questions of algebraic geometry can be posed very naturally in the elementary context of plane curves defined by polynomial equations. For example: How many intersec- tion points can one expect between two plane curves defined by polyno- mials of degrees k and l? (Not more than kl, but sometimes fewer.) How many disconnected “pieces” does the solution set to a particular polyno- mialequationhave(Figure1.8)?Doesaplanecurvehaveanyselfcrossings (Figure 1.9) or “cusps” (points where the tangent vector does not vary continuously—Figure 1.10)? Buttherealpowerofalgebraicgeometrybecomesevidentonlywhenone focusesonpolynomialswithcoefficientsinanalgebraically closedfield(one in which every polynomial decomposes into a product of linear factors), since polynomial equations always have the expected number of solutions (counted with multiplicity) in that case. The most deeply studied case is the complex field; in this context the solution set to a system of complex

12 1. Introduction FIGURE 1.9. A self crossing. FIGURE 1.10. A cusp. polynomials in n variables is a certain geometric object in Cn called an algebraic variety, which (except for a small subset where there might be self crossings, cusps, or more complicated kinds of behavior) is a manifold. The subject becomes even more interesting if one enlarges Cn by adding “ideal points at infinity” where parallel lines or asymptotic curves can be thoughtofasmeeting;theresultingspaceiscalledcomplexprojectivespace, and is an extremely important manifold in its own right. The properties of interest are those that are invariant under projective transformations (the natural changes of coordinates on projective space). One can ask such questions as these: Is a given variety a manifold or does it have singular points (points where it fails to be a manifold)? If it is a manifold, what is its topological type? If it is not a manifold, what is the geometric structure of its singular set, and how does that set change whenonevariesthecoefficientsofthepolynomialsslightly?Iftwovarieties are homeomorphic, are they equivalent under a projective transformation? How many times and in what way do two or more varieties intersect? Algebraic geometry has contributed a prodigious supply of examples of manifolds. In particular, much of the recent progress in understanding 4- dimensionalmanifoldshasbeendrivenbythewealthofexamplesthatarise as algebraic varieties. Classical Mechanics Classical mechanics is the study of systems that obey Newton’s laws of motion. The positions of all the objects in the system at any given time can be described by a set of numbers, or coordinates; typically, these are not independent of each other but instead must satisfy some relations. The relations can usually be interpreted as defining a manifold in some Euclidean space.

Why Study Manifolds? 13 Q θ R P d PQ FIGURE 1.11. A rigid body in space. For example, consider a rigid body moving through space under the influence of gravity. If we choose three noncollinear points P, Q, and R on thebody(Figure1.11),thepositionofthebodyiscompletelyspecifiedonce weknowthecoordinatesofthesethreepoints,whichcorrespondtoapoint in R9. However, the positions of the three points cannot all be specified arbitrarily: Because the body is rigid, they are subject to the constraint that the distances between pairs of points are fixed. Thus, to determine the position of the body, we can arbitrarily specify the coordinates of P in space (three parameters), and then we can specify the position of Q by giving, say, its latitude and longitude on the sphere of radius d , the PQ fixed distance between P and Q (two more parameters). Finally, having determined the position of the two points P and Q, the only remaining freedom is to rotate R around the line PQ; so we can specify the position of R by giving the angle θ that the plane PQR makes with some reference plane(onemoreparameter).Thusthesetofpossiblepositionsofthebody is a certain 6-dimensional manifold M ⊂R9. Newton’ssecondlawofmotionexpressestheaccelerationoftheobject— that is, the second derivatives of the coordinates of P, Q, R—in terms of the force of gravity, which is a certain function of the object’s position. This can be interpreted as a system of second-order ordinary differential equations for the position coordinates, whose solutions are all the possible paths the rigid body can take on the manifold M. The study of classical mechanics can thus be interpreted as the study of ordinarydifferentialequationsonmanifolds,alsoknownassmooth dynam- ical systems. A wealth of interesting questions arise in this subject: How

14 1. Introduction do solutions behave over the long term? Are there any equilibrium points or periodic trajectories? If so, are they stable, that is, do nearby trajecto- ries stay nearby? A good understanding of manifolds is necessary to fully answer these questions. General Relativity Manifolds play a decisive role in Einstein’s general theory of relativity, which describes the interactions among matter, energy, and gravitational forces. The central assertion of the theory is that spacetime (the collec- tion of all points in space at all times in history) can be modeled by a 4-dimensional manifold that carries a certain kind of geometric structure called a Lorentz metric; and this metric satisfies a system of partial differ- entialequationscalledtheEinsteinfieldequations.Gravitationaleffectsare then interpreted as manifestations of the curvature of the Lorentz metric. In order to describe the global structure of the universe, its history, and its possible futures, it is important to understand first of all what kinds of 4-manifolds exist and what kinds of Lorentz metrics they can carry. There are especially interesting relationships between the local geometry of spacetime (as reflected in the local distribution of matter and energy) and the global topological structure of the universe; these relationships are similar to those described above for Riemannian manifolds, but are morecomplicatedbecauseoftheintroductionofforcesandmotionintothe picture.Inparticular,ifweassumethatonacosmicscaletheuniverselooks approximatelythesameatallpointsandinalldirections(suchaspacetime issaidtobehomogeneousandisotropic),thenitturnsoutthereisacritical value for the average density of matter and energy in the universe: Above this density, the universe closes up on itself spatially and will collapse to a point singularity in a finite time (the “big crunch”); below it, the universe extendsinfinitelyfarinalldirectionsandwillexpandforever.Interestingly, physicists’bestcurrentestimatesplacetheaveragedensityrathernearthe critical value, and they have so far been unable to determine whether it is above or below it, so they do not know whether the universe will go on existing forever or not. Quantum Field Theory Thetheoryofelementaryparticleinteractions,calledquantumfieldtheory, has become increasingly geometric in recent decades. In particular, the latest attempts to unify quantum theory and gravitation have led to ever moreinterestingandexoticgeometricstructures.Theapproachtoquantum gravity that is currently considered most promising by many physicists is string theory, in which manifolds appear in several different starring roles. First, in order to obtain a consistent theory, it seems to be necessary to assumethatspacetimehasmorethanfourdimensions.Weexperienceonly

Why Study Manifolds? 15 four of them directly, because the dimensions beyond four are so tightly “curledup”thattheyarenotvisibleonamacroscopicscale,muchasalong but microscopically narrow cylinder would appear to be one-dimensional whenviewedfromfarenoughaway.Thetopologicalpropertiesoftheman- ifold that appears as the “cross section” of the curled-up dimensions have such a profound effect on the observable dynamics of the resulting quan- tum field theory that it is possible to rule out most cross sections a priori. It currently appears that a consistent theory can be constructed only if the cross section is a certain kind of 6-dimensional manifold known as a Calabi–Yau manifold. These developments in physics have stimulated pro- found developments in the mathematical understanding of 6-manifolds in general and Calabi-Yau manifolds in particular. Anotherrolethatmanifoldsplayinstringtheoryisindescribingthehis- tory of an elementary particle. One of the central tenets of string theory is thatparticlesshouldberepresentednotaspoints,butastiny1-dimensional objects(“strings”)movingthroughspacetime.Asaparticlemoves,ittraces out a 2-dimensional manifold called its world sheet. Physical phenomena arisefromtheinteractionsamongthesedifferenttopologicalandgeometric structures: the world sheet, the Calabi-Yau cross section, and the macro- scopic four-dimensional spacetime that we see. Manifolds are used in many more areas of mathematics than the ones listed here, but this brief survey should be enough to show you that mani- folds have a rich assortment of applications. It is time to get to work.

2 Topological Spaces In this chapter we begin our study in earnest. The first order of business is to build up enough machinery to give a proper definition of manifolds. The chief problem with the preliminary definition given in Chapter 1 is that it depends on having an “ambient Euclidean space” in which our n- manifold lives. This introduces a great deal of extraneous structure that is irrelevant to our purposes. Instead, we would like to view a manifold as a mathematical object in its own right, not as a subset of some larger space. The key concept that makes this possible is that of a “topological space,” which is the main topic of this chapter. Webeginbydefiningtopologicalspaces,motivatedbytheopensetcrite- rionforcontinuityinmetricspaces.Afterthedefinitionweintroducesome oftheimportantelementarynotionsassociatedwithtopologicalspacessuch as convergence, continuity, homeomorphisms, closures, interiors, and exte- riors, and then explore how to construct topologies from bases. At the end of the chapter we give the official definition of a manifold as a topological space with special properties. Topologies One of the most useful tools in analysis is the concept of a metric space. (See the Appendix for a brief review of metric space theory.) The most important examples, of course, are (subsets of) Euclidean spaces with the

18 2. Topological Spaces Euclideanmetric,butmanyothers,suchasfunctionspaces,arisefrequently in analysis. Our goal in this book is to study manifolds and those of their properties that are preserved by homeomorphisms (continuous maps with continuous inverses). To accomplish this, we could choose to view our manifolds as metric spaces. However, a metric still contains extraneous information. It isobviousthatahomeomorphismbetweenmetricspacesneednotpreserve distances (just think of the obvious homeomorphism between two spheres ofdifferentradii).Sowewillpushtheprocessofabstractionastepfurther, andcomeupwithakindof“space”withoutdistancesinwhichcontinuous functions still make sense. The key motivation behind the definition of this new kind of space is the open set criterion for continuity (Lemma A.5 in the Appendix), which shows that continuous functions between metric spaces can be detected knowing only the open sets. Motivated by this observation, we make the followingdefinition.AtopologyonasetX isacollectionT ofsubsetsofX, called open sets, satisfying the following properties: (i) X and ∅ are elements of T. (ii) T is closed under finite intersections: If U1,…,U n ∈ T, then their intersection U1 ∩···∩U n is in T. (iii) T is closed under arbitrary unions: If {U α } α∈A is any(cid:2)(finite or in- finite) collection of elements of T, then their union U is in α∈A α T. A pair (X,T) consisting of a set X and a topology T on X is called a topological space. The elements of a topological space are usually called its points. Since we will rarely have occasion to discuss any other type of space in this book, we will sometimes follow the common practice of callingatopologicalspacesimplyaspace.Asiscommoninmathematicsin discussingasetendowedwithaparticularkindofstructure,ifthetopology is understood from the context, we will typically omit it from the notation and simply say “X is a topological space” or “X is a space.” Asidefromthesimplicityoftheopensetcriterionforcontinuity,theother reason for choosing open sets as the primary objects in the definition of a topologicalspaceisthattheygiveusaqualitativewaytodetect“nearness” to a point without necessarily having a quantitative measure of nearness as we would in a metric space. If X is a topological space and q ∈ X, a neighborhood of q is just an open set containing q. More generally, a neighborhood of a subset K ⊂ X is an open set containing K. (In some books, the word neighborhood is used in the more general sense of a set containing an open set containing q; but for us neighborhoods will always be open.) We think of something being true “near q” if it is true in some (or every, depending on the context) neighborhood of q. The following exercises give some simple examples of topological spaces.

Topologies 19 {{1},{2,3},{1,2,3},∅} Discrete topology Trivial topology FIGURE 2.1. Topologies on {1,2,3}. Exercise 2.1. Showthateachofthefollowingisatopologicalspace.(See Figure 2.1.) (a) Let X denote the set {1,2,3}, and declare the open sets to be {1}, {2,3}, {1,2,3}, and the empty set. (b) AnysetX whatsoever,withT={all subsets of X}.Thisiscalledthe discrete topology on X, and (X,T) is called a discrete space. (c) Any set X, with T={∅,X}. This is called the trivial topology on X. (d) Any metric space (M,d), with T equal to the collection of all subsets of M that are open in the metric space sense. This topology is called the metric topology on M. Metric spaces provide a rich source of examples of topological spaces. In fact,alargepercentageofthetopologicalspaceswewillneedtoconsiderare actually subsets of Euclidean spaces Rn, with the metric topology induced by the Euclidean metric (which we call the Euclidean topology). Unless we specify otherwise, subsets of Rn will always be considered as topological spaces with this topology. Thus our intuition regarding topological spaces will rely heavily on our understanding of subsets of Euclidean space. Another important class of examples of topological spaces is obtained by taking open subsets of other spaces. If X is a topological space, and Y is any open subset of X, then we can define a topology on Y just by declaring the open sets of Y to be those open sets of X that are contained in Y. It is trivial to check that the three defining properties of a topology are satisfied. (In the next chapter, we will show how to put a topology on any subset of a topological space.) Convergence and Continuity Theprimaryreasontopologicalspaceswereinventedwasthattheyprovide the most general setting for studying the notions of convergence and con- tinuity. For this reason, it is appropriate to introduce these concepts next. We begin with convergence. The definition of convergence of a sequence of points in a metric space (see the Appendix) is really just a fancy way of saying that as we go far

20 2. Topological Spaces enoughoutinthesequence,thepointsofthesequencebecome“arbitrarily close” to q. Intopologicalspaces,weuseneighborhoodstoencodethenotionof“ar- bitrarily close.” Thus, if X is a topological space and {q } is any sequence i of points in X, we say that the sequence converges to q ∈ X, and q is the limit of the sequence, if for every neighborhood U of q there exists N such that q ∈U for all i≥N. Symbolically, this is denoted by either q →q or i i lim i→∞q i =q. Exercise 2.2. Show that in a metric space, this topological definition of convergence is equivalent to the metric space definition. Forthetypesoftopologicalspaceswewillbechieflyinterestedin(mostly manifolds), convergent sequences behave very much the same way we are used to from our experience with Euclidean space. Nevertheless, it is good to be aware that for some of the stranger examples of topological spaces, convergence can sometimes have an unintuitive meaning, as the following exercises show. Exercise 2.3. (a) Let X be a discrete topological space. Show that the only convergent sequences in X are the ones that are “eventually constant,” that is, sequences {qi } such that qi =q for all i greater than some N. (b) Let Y be a trivial topological space (that is, a set with the trivial topology {∅,Y}). Show that every sequence in Y converges to every point of Y. Attheendofthischapterwewilldescribearestrictedclassoftopological spaces(Hausdorffspaces)forwhichthepathologicalbehaviorof (b)cannot occur. Next we address the most important topological concept of all: continu- ous maps. If X and Y are topological spaces, a map f: X → Y is said to be continuous if for every open set U ⊂Y, f−1(U) is open in X. The open set criterion (Lemma A.5) for continuity in metric spaces says preciselythatamapbetweenmetricspacesiscontinuousinthissenseifand only if it is continuous in the usual ε-δ sense. Therefore, all the maps that youknowtobecontinuousfrommetricspacetheoryarealsocontinuousas mapsoftopologicalspaces.ExamplesincludepolynomialfunctionsfromR to R, linear transformations from Rn to Rk, and, more generally, any map from a subset of Rn to Rk whose component functions are continuous in the ordinary sense, such as polynomial, exponential, rational, logarithmic, absolute value, and trigonometric functions (where they are defined), and functions built up from these by composition. Thenextlemmagivessomeelementarybutimportantpropertiesofcon- tinuous maps. The ease with which properties like this can be proved is one of the virtues of defining continuity in terms of open sets.

Topologies 21 Lemma 2.1. Let X, Y, and Z be topological spaces. (a) Any constant map f: X →Y is continuous. (b) The identity map Id: X →X is continuous. (c) If f: X → Y is continuous, so is the restriction of f to any open subset of X. (d) If f: X → Y and g: Y → Z are continuous, so is their composition g◦f: X →Z. Proof. We will prove (d) and leave the other parts as exercises. We have to show that if U is any open subset of Z, then (g◦f)−1(U) is an open subset of X. By elementary set-theoretic considerations, (g ◦f)−1(U) = f−1(g−1(U)). Applying the definition of continuity to g, g−1(U) is open; and then doing the same for f shows that f−1(g−1(U)) is open. Exercise 2.4. Prove parts (a)–(c) of Lemma 2.1. Inmetricspacesitmakessensetotalkaboutamapbeing“continuousat apoint”(f: M1 →M2 iscontinuousatx∈M1 ifforallε>0,thereexists δ > 0 such that for each y ∈ M1, d1(y,x) < δ implies d2(f(y),f(x)) < ε), and a map is continuous if and only if it is continuous at every point. In topological spaces, continuity at a point is generally not a very useful con- cept.However,itisanimportantfactthatcontinuityisa“local”property, in the sense that a map is continuous if and only if it is continuous in a neighborhoodofeverypoint.Theprecisestatementisgiveninthefollowing important lemma. Lemma 2.2 (Local Criterion for Continuity). Amapf: X →Y be- tween topological spaces is continuous if and only if each point of X has a neighborhood on which (the restriction of) f is continuous. Proof. If f is continuous, we may simply take each neighborhood to be X itself.Conversely,supposef iscontinuousinaneighborhoodofeachpoint, andlet U ⊂Y beanyopenset;wehavetoshowthat f−1(U)isopen.Any point x∈f−1(U) has a neighborhood V on which f is continuous (Figure x 2.2). Continuity of f| implies, in particular, that (f| )−1(U) is open in Vx Vx V , and therefore also open in X. Unwinding the definitions, we see that x (f| )−1(U)={x∈V :f(x)∈U}=f−1(U)∩V , Vx x x which contains x and is contained in f−1(U). Since f−1(U) is the union of all such open sets as x ranges over f−1(U), it follows that f−1(U) is open, as desired.

22 2. Topological Spaces X f Y V x U f−1(U) x FIGURE 2.2. Local criterion for continuity. If X and Y are topological spaces, a homeomorphism from X to Y is defined to be a continuous bijective map ϕ: X → Y with continuous in- verse. If there exists a homeomorphism between X and Y, we say that X and Y are homeomorphic or topologically equivalent. Sometimes this is abbreviated X ≈Y. Exercise 2.5. Show that “homeomorphic” is an equivalence relation. The homeomorphism relation is the most fundamental relation in topol- ogy. In fact, as we mentioned in Chapter 1, “topological properties” are exactly those that are preserved by homeomorphisms. Here are a few explicit examples of homeomorphisms that you should keep in mind. Example 2.3. Any open ball in Rn is homeomorphic to any other open ball; the homeomorphism can easily be constructed as a composition of translations x(cid:10)→x+x0 and dilations x(cid:10)→cx. Similarly, all spheres in Rn are homeomorphic to each other. These examples illustrate that “size” is not a topological property. Example 2.4. Let Bn denote the open unit ball B1(0) ⊂ Rn, and define a map F: Bn →Rn by x y =F(x)= . 1−|x|2 Note that |F(x)|=|x|/(1−|x|2)→∞ as |x|→1. It is straightforward to check that the inverse of F is given by 2y x=F−1(y)= (cid:3) . 1+ 1+4|y|2 Thus F is a homeomorphism, so Rn is homeomorphic to Bn. This shows that “boundedness” is not a topological property. Example 2.5. Another illustrative example is the homeomorphism be- tween a cube and a sphere alluded to in Chapter 1. Let S2 = {x ∈

Topologies 23 FIGURE 2.3. Deforming a cube into a sphere. R3 : |x| = 1} denote the unit sphere in R3, and set C = {(x,y,z) : max(|x|,|y|,|z|) = 1}, which is the cubical surface of side 2 centered at the origin. Let ϕ: C →S2 be the map that projects each point on C radi- ally inward to the sphere (Figure 2.3). More precisely, given a pointq ∈C, ϕ(q)istheunitvectorinthedirectionofq.Thusϕisgivenbytheformula (x,y,z) ϕ(x,y,z)= (cid:3) , x2+y2+z2 which is continuous by the usual arguments of elementary analysis. The next exercise shows that ϕ is a homeomorphism. This example demon- strates that “corners” are not topological properties. Exercise 2.6. Show that the map ϕ: C → S2 is a homeomorphism by showing that its inverse can be written ϕ−1(x,y,z)= (x,y,z) . max(|x|,|y|,|z|) In the definition of a homeomorphism, it is important to note that al- though the assumption that ϕ is bijective guarantees that the inverse map ϕ−1 exists for set-theoretic reasons, continuity of ϕ−1 is not automatic. Thenextexercisegivesanexampleofacontinuousbijectionwhoseinverse is not continuous. Exercise 2.7. Let X denote the half-open interval [0,1) ⊂ R, and let S1 denote the unit circle in R2 (both with the Euclidean metric topologies, of course). Define a map a: X → S1 by a(t) = (cos2πt,sin2πt) (Figure 2.4). Show that a is continuous and bijective but not a homeomorphism.

24 2. Topological Spaces a FIGURE 2.4. A map that is bijective but not a homeomorphism. A map f: X → Y (continuous or not) is said to be an open map if for any open set U ⊂X, the image set f(U) is open in Y. A map can be open but not continuous, continuous but not open, both, or neither. There is a generalization of homeomorphisms that is often useful. We say that a continuous map f: X →Y between topological spaces is a local homeomorphismifeverypointx∈X hasaneighborhoodU ⊂X suchthat f(U) is an open subset of Y and f| : U →f(U) is a homeomorphism. U Exercise 2.8. (a) Show that every local homeomorphism is an open map. (b) Show that every homeomorphism is a local homeomorphism. (c) Show that a bijective continuous open map is a homeomorphism. (d) Show that a bijective local homeomorphism is a homeomorphism. Closed Sets Becauseoftheimportanceofneighborhoodsinunderstandingatopological space and its continuous maps and convergent sequences, the definition of a topological space takes open sets as the primary objects. There is also a complementary notion that is nearly as important. AsubsetF ofatopologicalspaceX issaidtobeclosedifitscomplement X(cid:3)F isopen.Fromthedefinitionoftopologicalspaces,severalproperties follow immediately: (i) X and ∅ are closed. (ii) Finite unions of closed sets are closed. (iii) Arbitrary intersections of closed sets are closed. AtopologyonasetX canbedefinedbydescribingthecollectionofclosed sets, as long as they satisfy these three properties; the open sets are then just those sets whose complements are closed. Here are some examples of closed subsets of familiar topological spaces.

Topologies 25 Example 2.6 (Closed Sets). (a) Any closed interval [a,b] ⊂ R is a closed set, as are the half-infinite closed intervals [a,∞) and (−∞,b]. (b) Any closed ball in a metric space is a closed set (Exercise A.11(b) in the Appendix). (c) Every subset of a discrete space is closed. It is important to be aware that just as in metric spaces, “closed” is not thesameas“notopen”—setscanbebothopenandclosed,orneitheropen nor closed. For example, in any topological space X, the sets X and ∅ are both open and closed. On the other hand, the half-open interval [0,1) is neither open nor closed in R. Continuity can be detected by closed sets as well as open ones. Lemma 2.7. A map between topological spaces is continuous if and only if the inverse image of every closed set is closed. Exercise 2.9. Prove Lemma 2.7. Given any set A ⊂ X, we define several related sets as follows. The closure of A in X, denoted by A, is the set (cid:4) A= {B ⊂X :B ⊃A and B is closed in X}. The interior of A, written IntA, is (cid:5) IntA= {C ⊂X :C ⊂A and C is open in X}. It is obvious from the properties of open and closed sets that A is closed and IntA is open. In words, A is “the smallest closed set containing A,” and IntA is “the largest open set contained in A.” We also define the exterior of A, written ExtA, as ExtA=X(cid:3)A, and the boundary of A, written ∂A, as ∂A=X(cid:3)(IntA∪ExtA). It follows immediately from the definitions that for any subset A ⊂ X, the whole space X is equal to the disjoint union of IntA, ExtA, and ∂A. The set A always contains all of its interior points and none of its exterior points, and may contain all, some, or none of its boundary points. For many purposes, it is useful to have alternative characterizations of open and closed sets, and of the interior, exterior, closure, and boundary of a given set. The following lemma gives such characterizations. Some of theseareprobablyfamiliartoyoufromyourstudyofEuclideanandmetric spaces. See Figure 2.5 for illustrations of some of these characterizations.

26 2. Topological Spaces ExtA IntA A ∂A FIGURE 2.5. Interior, exterior, and boundary points. Lemma 2.8. Let X be a topological space and A⊂X any subset. (a) A point q is in the interior of A if and only if q has a neighborhood contained in A. (b) A point q is in the exterior of A if and only if q has a neighborhood contained in X(cid:3)A. (c) A point q is in the boundary of A if and only if every neighborhood of q contains both a point of A and a point of X(cid:3)A. (d) IntA and ExtA are open in X, while ∂A is closed in X. (e) A is open if and only if A=IntA. (f) A is closed if and only if it contains all its boundary points, which is true if and only if A=IntA∪∂A. (g) A=A∪∂A=IntA∪∂A. Exercise 2.10. Prove Lemma 2.8. Given a topological space X and a set A ⊂ X, we say that a point q ∈ X is a limit point of A if every neighborhood of q contains a point of A other than q (which might or might not itself be in A). Limit points are also sometimes called accumulation points or cluster points. For example, if X =R and A=(0,1), then every point in [0,1] is a limit point of A. If we let B ={1/n}∞ ⊂R, then 0 is the only limit point of B. n=1 Exercise 2.11. Show that a set A in a topological space is closed if and only if it contains all of its limit points. A subset A of a topological space X is said to be dense in X if A=X.

Bases 27 Exercise 2.12. Show that a subset A ⊂ X is dense if and only if every nonempty open set in X contains a point of A. Exercise 2.13. Show that Rn has a countable dense subset. Analogous to open maps are closed maps: A map f: X → Y is said to be closed if it takes closed sets in X to closed sets in Y. Exercise 2.14. Show that a bijective continuous map is a homeomor- phism if and only if it is open if and only if it is closed. Bases To define a topology on a given set, it is often convenient to single out some “special” open sets and use them to define the rest of the open sets. Forexample,themetrictopologyisdefinedbyfirstdefiningballsandthen declaring a set to be open if it contains a ball around each of its points. This idea can be generalized easily to arbitrary topological spaces, as in the next definition. Suppose X is any set. A basis in X is a collection B of subsets of X satisfying the following conditions: (i) (cid:2)Every element of X is in some element of B; in other words, X = B. B∈B (ii) If B1,B2 ∈B and x∈B1 ∩B2, there exists an element B3 ∈B such that x∈B3 ⊂B1 ∩B2. Proposition 2.9. Let B be a basis in a set X, and let T be the collection of all unions of elements of B. Then T is a topology on X. This topology T is called the topology generated by B. Before we prove the proposition, it will be useful to have an alternative characterization of T, analogous to the definition of open sets in a metric space in terms of balls. Given X and a collection B of subsets of X, we say that a subset U ⊂ X satisfies the basis criterion with respect to B if for every x ∈ U, there exists B ∈B such that x∈B ⊂U. Lemma 2.10. Suppose B is a basis in X. Then the collection T defined in Proposition 2.9 is precisely the set of all subsets of X that satisfy the basis criterion with respect to B. Proof. Let U ⊂ X, and suppose first that U satisfies the basis criterion. Let (cid:5) V = {B ∈B:B ⊂U}.

28 2. Topological Spaces U2 B2 U1 B3 B1 x FIGURE 2.6. Proof that U1 ∩U2 satisfies the basis criterion. Obviously, V ∈T, since V is by definition a union of basis sets. Thus if we can show that U = V, it will follow that U ∈ T. Clearly, V ⊂ U, because V is a union of subsets of U. To show that U ⊂V, let x∈U be arbitrary; sinceU satisfiesthebasiscriterion,theremustexistabasissetB ∈Bsuch that x∈B ⊂U. It follows immediately that x∈V, so we are done. Conversely,supposethatU ∈T.Thism(cid:2)eansthatU isaunionofelements of B, i.e., for some subset A ⊂ B, U = B. In other words, x ∈ U if B∈A and only if x∈B for some B ∈A. In particular, x∈B ⊂U, so U satisfies the basis criterion. Proof of Proposition 2.9. We need to show that T satis(cid:2)fies the three defin- ing conditions for a topology. First, the fact that X = B means that B∈B X ∈T.TheemptysetisalsoinT trivially.(Itisthe“unionofnoelements of B”!) A union of elements of T is a union of unions of elements of B, and thereforeisitselfaunionofelementsofB;thus T isclosedunderarbitrary unions. Finally, to show that T is closed under finite intersections, suppose first that U1,U2 ∈ T. Then, for any x ∈ U1 ∩U2, the basis criterion says that thereexistbasiselementsB1,B2 ∈Bsuchthatx∈B1 ⊂U1 andx∈B2 ⊂ U2 (Figure2.6).Butthenpart(ii)ofthedefinitionofbasisguaranteesthat there exists B3 ∈B such that x∈B3 ⊂B1 ∩B2 ⊂U1 ∩U2. Thus U1 ∩U2 satisfies the basis criterion, so it is again in T. This shows that T is closed under pairwise intersections, and closure under finite intersections follows easily by induction. It often happens that we are given a particular topological space X together with a collection B of open subsets of X, and we would like to know whether B forms a basis for the topology of X. On the face of it, this would require showing first that B satisfies the two conditions in the

Bases 29 definition of a basis, and then that each open subset of X is a union of elements of B (or alternatively satisfies the basis criterion). But as the following lemma shows, once we know that the sets in B are open subsets with respect to some topology, showing that they actually are a basis for that topology is much easier. Lemma 2.11. Suppose X is a topological space, and B is a collection of open subsets of X. If every open subset of X satisfies the basis criterion with respect to B, then B is a basis for the topology of X. Proof. By Lemma 2.10, all we need to show is that B satisfies the two defining conditions for a basis. Forthefirstcondition,sinceX itselfisanopensubset,thebasiscriterion tells us that any point x∈X is in some element B ∈B; thus the union of all the elements of B is X. For the second condition, suppose B1 and B2 are elements of B and x∈B1 ∩B2.ThebasiscriterionappliedtoB1 ∩B2(whichistheintersection of two open subsets of X and therefore open) guarantees the existence of B3 ∈B such that x∈B3 ⊂B1 ∩B2. The next exercise describes bases for the topologies of Exercise 2.1. The precedinglemmamakesthejobofshowingthattheyareindeedbasesquite straightforward. Exercise 2.15. In each of the following cases, prove that the given set B is a basis for the given topology. (a) M is a metric space with the metric topology, and B is the collection of all open balls in M. (b) X is a set with the discrete topology, and B is the collection of all one-point subsets of X. (c) X is a set with the trivial topology, and B={X}. Exercise 2.16. Show that each of the following collections B i is a basis for the Euclidean topology on Rn. (a) B 1 = {Cs(x) : x ∈ Rn and s > 0}, where Cs(x) is the open cube of side s around x: Cs(x)={y=(y1,…,yn):|xi −yi |<s/2, i=1,…,n}. (b) B 2 ={Br(x): r is rational and x has rational coordinates}. When we have a basis for a topology on Y, it is sufficient (and often mucheasier)tocheckcontinuityofmapsintoY usingonlybasisopensets, as the following lemma shows.

30 2. Topological Spaces Lemma 2.12. Let X and Y be topological spaces and let B be a basis for Y. A map f: X → Y is continuous if and only if for every basis open set B ∈B, f−1(B) is open in X. Proof. Clearly, if f is continuous, the inverse image of every basis open set is open. Conversely, suppose f−1(B) is open for every B ∈ B. Let U ⊂ Y be open, and let x∈f−1(U). By the basis criterion, there is a basis set B such that f(x) ∈ B ⊂ U. This implies that x ∈ f−1(B) ⊂ f−1(U), which means that x has a neighborhood contained in f−1(U). Since this is true for every x∈U, U is the union of all these neighborhoods as x ranges over points of U, and therefore is open. Manifolds We are almost ready to give the official definition of manifolds. We need just a few more preliminary definitions. The first one is easy, and captures very precisely the intuitive idea that a manifold should look “locally” like Euclidean space. Let X be a topological space. A topological space M is said to be locally Euclidean of dimension n if every point q ∈ M has a neighborhood that is homeomorphic to an open subset of Rn. Such a neighborhood is called a Euclidean neighborhood of q. Forsomepurposes,itisusefultobemorespecificaboutthekindofopen subset we use to characterize locally Euclidean spaces. The next lemma shows that we could have replaced “open subset” by open ball or by Rn itself. Lemma 2.13. A topological space M is locally Euclidean of dimension n if and only if either of the following properties holds: (a) Every point of M has a neighborhood homeomorphic to an open ball in Rn. (b) Every point of M has a neighborhood homeomorphic to Rn. Proof. It is immediate that any space with property (a) or (b) is locally Euclidean of dimension n. Conversely, suppose M is locally Euclidean of dimension n. Because any open ball in Rn is homeomorphic to Rn itself (Example2.4),properties(a)and(b)areequivalent,soweneedonlyprove (a). Given any point q ∈ M, let U be a neighborhood of q that admits a homeomorphism ϕ: U → V, where V is an open subset of Rn. The fact that V is open means that there is some open ball B around ϕ(q) that is contained in V, and then ϕ−1(B) is a neighborhood of q homeomorphic to an open ball.

Manifolds 31 IfM islocallyEuclideanofdimensionn,ahomeomorphismfromanopen subset U ⊂M to an open subset of Rn is called a coordinate chart (or just a chart) on U. We will call any open subset of M that is homeomorphic to a ball in Rn a Euclidean ball in M. (When M has dimension 2, we will sometimes use the term Euclidean disk.) The preceding lemma shows that everypointinalocallyEuclideanspacehasaEuclideanballneighborhood. The definition of locally Euclidean spaces makes sense even when n=0. Since R0 is by convention a single point, Lemma 2.13(b) implies that a space is locally Euclidean of dimension 0 if and only if each point has a neighborhoodhomeomorphictoaone-pointspace,orinotherwordsifand only if the space is discrete. There are two other properties that we will include in the definition of manifolds, to rule out “pathological” spaces that might otherwise pass as manifolds. Hausdorff Spaces The first property we want to introduce ensures that there are “enough” opensets,sothatneighborhoodsbehavemoreorlessthewayourintuition derived from Euclidean and metric spaces leads us to expect. For example, inametricspace,aone-pointset{q}isalwaysclosed,becausearoundevery pointotherthanqthereisaballthatdoesnotincludeq.Moregenerally,any two points in a metric space always have disjoint neighborhoods. However, these properties do not always hold in topological spaces. Consider the set {1,2,3} with the topology {∅,{1},{2,3},{1,2,3}} (Figure 2.1). In this case 2 and 3 do not have disjoint neighborhoods, since every open set that contains one also contains the other. Moreover, the set {2} is not closed, because its complement is not open. The problem with this example is that there are too few open sets, so neighborhoodsdonothavethesameintuitivemeaningtheyhaveinmetric spaces. In our study of manifolds, we will want to rule out such “patholog- ical” spaces, so we make the following definition. A topological space X is said to be a Hausdorff space if given any pair of distinct points q1,q2 ∈X, there exist neighborhoods U1 of q1 and U2 of q2 with U1 ∩U2 = ∅. This property is often summarized by saying “open sets separate points.” Any metric space is Hausdorff. (If d(q1,q2) = r, then the open balls of radius r/2 around q1 and q2 are disjoint by the triangle inequality.) More generally, any open subset of a Hausdorff space is Hausdorff: If V ⊂ X is open in the Hausdorff space X, and q1,q2 are distinct points in V, then in X thereareopensetsU1,U2 separatingq1 andq2,andthesetsU1 ∩V and U2 ∩V are open in V, disjoint, and contain q1 and q2, respectively. Hausdorff spaces have many of the properties that we expect of metric spaces, such as those expressed in the following lemma. Lemma 2.14. Let X be a Hausdorff space.

32 2. Topological Spaces (a) Every one-point set in X is closed. (b) If a sequence {x } in X converges to a limit x ∈ X, the limit is i unique. Proof. For part (a), choose any q0 ∈ X. Given p (cid:14)= q0, the Hausdorff property says that there exist disjoint neighborhoods U p of q0 and V p of p. This means that the complement of {q0 } is equal to the union of the open sets V p as p ranges over X(cid:3){q0 }, which is open, so {q0 } is closed. To prove that limits are unique, suppose that x and x(cid:5) are two distinct limits of the sequence {x }. By the Hausdorff property, there exist disjoint i neighborhoodsU ofxandU(cid:5) ofx(cid:5).Bydefinitionofconvergence,thereexist N,N(cid:5) suchthati≥N impliesx ∈U andi≥N(cid:5) impliesx ∈U(cid:5).Butsince i i U and U(cid:5) are disjoint, this is a contradiction when i≥max(N,N(cid:5)). Exercise 2.17. Show that the only Hausdorff topology on a finite set is the discrete topology. The non-Hausdorff example above involving {1,2,3} is obviously con- trived, and has little relevance to our study of manifolds. But in Problem 3-8 at the end of the next chapter you will see a space that would be a manifold except for the fact that it fails to be Hausdorff. Second Countability Whereas the Hausdorff property ensures that there are enough open sets, the next property we will introduce ensures that there are not too many. Wesaythatatopologicalspaceissecond countableifitadmitsacountable basis. As the “second” in the name second countable suggests, there is also another weaker notion of countability. It relies on the following definition: If X is a space and q ∈X, a collection B of neighborhoods of q is called a q neighborhood basis at q if every neighborhood of q contains some B ∈ B . q X is said to be first countable if there exists a countable neighborhood basis at each point. Second countability implies first countability: Given a countablebasisforX,thecollectionofbasisopensetscontainingq iseasily seen to be a countable neighborhood basis at q. Oneofthewaysinwhichsecondcountabilityisoftenusedisinreducing thenumberofopensetsoneneedsto“cover”aspace.IfX isanytopological space, a collection U of subsets of X is said to cover X, or to be a cover of X, if every point in X is in one of the sets of U. An open cover of X is a collection of open sets that covers X. Given any cover U, a subcover of U is a subset of U that is still a cover. Lemma 2.15. If X is a second countable space, every open cover of X has a countable subcover.

Manifolds 33 Proof. Let B be a countable basis for X, and let U be an arbitrary open cover of X. Let B(cid:5) denote the subset of B consisting of those basis sets that are entirely contained in some element of U. Because any subset of a countable set is countable, B(cid:5) is a countable set. Now, for each element B ∈ B(cid:5), choose an element U ∈ U such that B B ⊂ U (this is possible by the way we defined B(cid:5)). The collection U(cid:5) = B {U : B ∈ B(cid:5)} is a countable subset of U; the lemma will be proved if we B can show that it still covers X. Ifx∈X isarbitrary,thenx∈U0 forsomeopensetU0 ∈U.Bythebasis criterion for U0, there is some B ∈ B such that x ∈ B ⊂ U0. This means, in particular, that B ∈ B(cid:5), and therefore there is a set U ∈ U(cid:5) such that B x∈B ⊂U . This shows that U(cid:5) is a cover and completes the proof. B Most “reasonable” spaces are second countable. For example, it follows from Exercise 2.16(b) above that Rn is second countable. Moreover, any opensubsetU ofasecondcountablespaceX issecondcountable:Starting with a countable basis for X, just throw away all the elements of the basis that do not lie in U; then it is easy to check that the remaining basis sets form a countable basis for the topology of U. In Problems 3-7 and 4-6 we will see examples of spaces that would be manifolds except for the failure of second countability. Definition of Manifolds We come now to the culmination of this chapter: the official definition of manifolds. An n-dimensional topological manifold is a second countable Hausdorff spacethatislocallyEuclideanofdimension n.Sincetheonlykindofman- ifolds we will be considering in this book are topological manifolds, we will usually simply call them n-dimensional manifolds, or n-manifolds, or evenjustmanifoldsifthedimensionisunderstoodorirrelevant.(Theterm “topological manifold” is usually used only to emphasize that the kind of manifold under consideration is the kind we have defined here, which is a topological space with special properties, rather than other kinds of manifolds that can be defined, such as “smooth manifolds” or “complex manifolds.” We will not treat any of these other kinds of manifolds in this book.) A shorthand notation that is in common use is to write “let Mn be a manifold”tomean“letM beamanifoldofdimensionn.”Thesuperscriptn isnotpartofthenameofthemanifold,andisusuallydroppedafterthefirst time the manifold is introduced. One must be a bit careful to distinguish this notation from the n-fold Cartesian product Mn = M ×···×M, but it is usually clear from the context which is meant. We will not use this shorthand in this book, but you should be aware of it because you will encounter it in your reading.

34 2. Topological Spaces FIGURE 2.7. A manifold with boundary. The most obvious example of an n-manifold is Rn itself. More gener- ally, any open subset of Rn—or in fact of any n-manifold—is again an n-manifold, as the next lemma shows. Lemma 2.16. Any open subset of an n-manifold is an n-manifold. Proof. Let M be an n-manifold, and let V be an open subset of M. Any q ∈V has a neighborhood (in M) that is homeomorphic to an open subset ofRn;theintersectionofthatneighborhoodwithV isstillopen,stillhome- omorphic to an open subset of Rn, and lies in V, so V is locally Euclidean. We remarked above that any open subset of a Hausdorff space is Haus- dorffandanyopensubsetofasecondcountablespaceissecondcountable. Therefore V is an n-manifold. In the next few chapters we will develop many more examples of mani- folds. Manifolds with Boundary Forsomepurposesitisusefulalsotohavethefollowingmoregeneralnotion. An n-dimensional manifold with boundary is a second countable Hausdorff space in which every point has a neighborhood homeomorphic to an open subset of the n-dimensional upper half space Hn = {(x1,…,x n ) ∈ Rn : x ≥0}. Just as in the case of manifolds, we will call any homeomorphism n from an open subset U of M to an open subset of Hn a chart on U. Example 2.17 (Manifolds with Boundary). TheupperhalfspaceHn itself is obviously a manifold with boundary, as is any closed interval in R, any closed disk in R2, or in fact a closed ball in any Euclidean space (see Figure2.7).(Thisisnothardtoseeintuitively.Youcanprobablyconstruct appropriate charts yourself, or you can wait until Chapter 3 and use the ones suggested in Problem 3-4.) The boundary of Hn in Rn is the set of points where x = 0. If M is a n manifold with boundary, a point that is in the inverse image of ∂Hn under some chart is called a boundary point of M, and a point that is in the inverse image of IntHn is called an interior point. The boundary of M (the

Manifolds 35 set of all its boundary points) is denoted by ∂M; similarly, its interior is denoted by IntM. Note that this use of the terms “boundary” and “interior” is distinct from their use earlier in this chapter in reference to subsets of topological spaces:IfM isamanifoldwithboundary,itsboundaryasasubsetofitself is always empty, even though its boundary as a manifold with boundary may not be. Usually the distinction will be clear from the context, but if necessary we can always distinguish the two meanings by referring to the topological boundary or the manifold boundary as appropriate. There is a subtlety about this definition that might not be immediately evident. Although the interior and boundary of a manifold with boundary M are well-defined subsets, and it might seem intuitively rather obvious that they are disjoint sets, we have no way of proving at this stage that a point of M cannot be simultaneously both a boundary point and an interior point. After we have developed some more machinery, you will be asked to prove this fact (for the 1-dimensional case in Problem 4-14, the 2-dimensional case in Problem 8-6, and the general case in Problem 13-9). Nonetheless, we will go ahead and assume it when convenient. Since any open ball in Rn is homeomorphic to an open subset of Hn, an n-manifold is automatically an n-manifold with boundary (with empty boundary). But the converse is not true: For example, an endpoint of a closed interval has no Euclidean neighborhood. Assuming the (as yet un- proved) fact that a boundary point cannot be an interior point, it follows that a manifold with boundary is a manifold if and only if its boundary is empty.

36 2. Topological Spaces Problems 2-1. Let (X1,T 1) and (X2,T 2) be topological spaces and let f: X1 →X2 be a bijective map. Show that f is a homeomorphism if and only if f(T 1) = T 2 in the sense that U ∈ T 1 if and only if f(U) ∈ T 2. [This shows, roughly speaking, that the topology is precisely the in- formation preserved by homeomorphisms, and justifies the definition of topological spaces as the right setting for studying properties pre- served by homeomorphisms.] 2-2. Suppose X is a set, and B is any collection of subsets of X whose union equals X. Let T be the collection of all unions of finite inter- sections of elements of B. (Note that the empty set is the union of the empty collection of sets.) (a) ShowthatT isatopology.(Itiscalledthetopology generated by B, and B is called a subbasis for T.) (b) Show that T is the “smallest” topology for which all the sets in Bareopen;moreprecisely,showthatT istheintersectionofall topologies containing B. 2-3. Let X be an infinite set. Consider the following collections of subsets of X: T 1 ={U ⊂X :X(cid:3)U is finite or is all of X}; T 2 ={U ⊂X :X(cid:3)U is infinite or is empty}; T 3 ={U ⊂X :X(cid:3)U is countable or is all of X}. For each collection, determine whether it is a topology. 2-4. Let X ={1,2,3}. Give a list of topologies on X such that any topol- ogy on X is homeomorphic to exactly one on your list. 2-5. Let X = R2 as a set, but with the topology determined by the fol- lowing basis: B={sets of the form {(c,y):a<y <b}, for fixed a,b,c∈R}. Determine which (if either) of the identity maps X → R2, R2 → X is continuous. 2-6. Let X be a discrete space, Y be a space with the trivial topology, and Z be any topological space. Show that any maps f: X → Z and g: Z →Y are continuous. If Z is Hausdorff, show that the only continuous maps h: Y →Z are constant maps. 2-7. Give examples of maps between subsets of the plane (with the Eu- clidean topology) that are

Problems 37 (a) open but not closed or continuous; (b) closed but not open or continuous; (c) continuous but neither open nor closed; (d) continuous and open but not closed; (e) continuous and closed but not open; (f) open and closed but not continuous. 2-8. Let f: X →Y be a continuous map between topological spaces, and letBbeabasisforthetopologyofX.Letf(B)denotethecollection {f(B) : B ∈ B} of subsets of Y. If f is surjective and open, show that f(B) is a basis for the topology of Y. 2-9. Suppose we are given an indexed collection of nonempty(cid:6)topological spaces {X α } α∈A . Declare a subset of the disjoint union α∈A X α to be open if and only if its intersection with each X is open. α (cid:6) (a) Show that this is a topology on X , called the disjoint α∈A α union topology. (b) Show that a subset of the disjoint union is closed if and only if its intersection with each X is closed. α (c) I(cid:6)f each X α is an n-manifold, show that the disjoint union X is an n-manifold if and only if the index set A is α∈A α countable. 2-10. Suppose X is locally Euclidean of dimension n, and f: X → Y is a surjective local homeomorphism. Show that Y is also locally Eu- clidean of dimension n. 2-11. Show that a topological space is a 0-manifold if and only if it is a countable discrete space. 2-12. LetX beatotallyorderedset(seetheAppendix),andassumethatX has at least two elements. For any a∈X, define sets L(a),R(a)⊂X by L(a)={c∈X :c<a}, R(a)={c∈X :c>a}. GiveX thetopologygeneratedbythesubbasis{L(a),R(a):a∈X}, called the order topology. (a) Show that each set of the form (a,b) is open in X and each set oftheform[a,b]isclosed(where(a,b)and[a,b]aredefinedjust as in R). (b) Show that X is Hausdorff.

38 2. Topological Spaces (c) Showthatforanya,b∈X,(a,b)⊂[a,b].Underwhatconditions does equality hold? 2-13. Let X be a second countable topological space. Show that every col- lection of disjoint open subsets of X is countable. 2-14. ShowthatlocallyEuclideanspacesandmetricspacesarefirstcount- able. 2-15. (a) Show that every second countable space has a countable dense subset. (b) Show that a metric space is second countable if and only if it has a countable dense subset. 2-16. Let X be a first countable space. (a) ForanysetA⊂X andanypointp∈X,showthatp∈Aifand only if there is a sequence {p }∞ in A such that p →p. n n=1 n (b) Show that for any space Y, a map f: X → Y is continuous if and only if f takes convergent sequences in X to convergent sequences in Y. 2-17. Show that any manifold has a basis of Euclidean balls. 2-18. Suppose M is an n-dimensional manifold with boundary. Show that IntM isann-manifoldand∂M isan(n−1)-manifold(withoutbound- ary).

3 New Spaces from Old Inthischapterweintroducethreestandardwaysofconstructingnewtopo- logical spaces from given ones: subspaces, product spaces, and quotient spaces. We will explore how various topological properties are affected by these constructions, and we will show how each topology is characterized by which maps it makes continuous. At the end of the chapter we will explore in some detail the quotient spaces that arise from group actions. Throughout the chapter we will use these tools to build new examples of manifolds. Subspaces We have seen a number of examples of topological spaces that are subsets of Rn, with the topology induced by the Euclidean metric. We have also seen that open subsets of a topological space inherit a topology from the containing space. It turns out that arbitrary subsets of topological spaces can also be viewed as topological spaces in their own right. Let X be a space, and let A ⊂ X be any subset. We define a topology T on A by A T ={U ⊂A:U =A∩V for some open set V ⊂X}. A Inotherwords,theopensetsofT aretheintersectionswithAoftheopen A sets of X (Figure 3.1). Exercise 3.1. Prove that T A is a topology on A.

40 3. New Spaces from Old X U V A FIGURE 3.1. An open set in the subspace A. 0 1 2 3 0…11 1 1 1 54 3 2 B C FIGURE 3.2. Subspaces of R. ThetopologyT iscalledthesubspacetopology(orsometimestherelative A topology) on A. A subset of a topological space X is called a subspace of X if it is endowed with the subspace topology. Henceforth, whenever we mention a subset of a topological space, we will always consider it as a topological space with the subspace topology unless otherwise specified. Exercise 3.2. Let M be a metric space, and let A ⊂ M be any subset. Show that the subspace topology on A is the same as the metric topology obtained by restricting the metric of M to points in A. Example 3.1. Consider the subspaces B = [0,1] ∪ (2,3) and C = {1/n}∞ of R (Figure 3.2). Notice that the set [0,1] is not open in R. n=1 But it is an open subset of B, because [0,1] is the intersection with B of the open interval (−1,2). In C, the one-point sets {1/n} are all open (why?), so the subspace topology on B is discrete. Theseexamplesillustratethatopennessandclosednessarenotproperties of a set by itself, but rather of a set in relation to a particular topological space. An injective continuous map that is a homeomorphism onto its image (in the subspace topology) is called a topological embedding. If f: A → X issuchamap,wecanthinkoftheimagesetf(A)asahomeomorphiccopy of A embedded in X.

Subspaces 41 Example 3.2. Let a: [0,1)→R2 be the map a(s)=(cos2πs,sin2πs). In Exercise 2.7, you showed that a is not a homeomorphism onto its image in the subspace topology (which is the same as the metric topology by Exercise 3.2), so it is an example of a map that is continuous and injective but not an embedding. However, the restriction of a to any interval [0,b) for 0<b<1 is an embedding. The first property we will prove about the subspace topology is so fun- damental that, in a sense we will explain later, it completely characterizes the subspace topology among all the possible topologies on a subset. For anysubsetA⊂X,ι : A(cid:9)→X denotestheinclusionmapofAintoX (see A the Appendix). Theorem 3.3 (Characteristic Property of Subspace Topologies). SupposeA⊂X isasubspace.ForanytopologicalspaceY,amapf: Y →A is continuous if and only if the following composite map from Y to X is continuous: Y −f→A(cid:9)→ιA X. Proof. Directly from the definitions of continuity and the subspace topol- ogy, f: Y →A is continuous ⇐⇒ for all U ⊂ A, f−1(U) ⊂ Y open open ⇐⇒ for all V ⊂ X, f−1(V ∩A) ⊂ Y open open ⇐⇒ for all V ⊂ X, (ι ◦f)−1(V) ⊂ Y A open open ⇐⇒ ι ◦f: Y →X is continuous. A Proposition 3.4 (Other Properties of the Subspace Topology). Suppose A is a subspace of the topological space X. (a) The inclusion map ι : A (cid:9)→ X is continuous, and in fact is a topo- A logical embedding. (b) If f: X →Y is continuous, then its restriction to A is continuous. (c) If f: X →Y is continuous, then f: X →f(X) is continuous. (d) TheclosedsubsetsofAarepreciselytheintersectionsofAwithclosed subsets of X. (e) If B ⊂ A is a subspace of A, then B is a subspace of X; in other words, the subspace topologies that B inherits from A and from X agree.

42 3. New Spaces from Old A V X U B W FIGURE 3.3. A subspace of a subspace. (f) If B ⊂A⊂X, B is open in A, and A is open in X, then B is open in X. (g) If B is a basis for the topology of X, then B ={B∩A:B ∈B} A is a basis for the topology of A. (h) Any subspace of a Hausdorff space is Hausdorff. (i) Any subspace of a second countable space is second countable. Proof. Part (a) follows immediately from the characteristic property, just by taking Y to be equal to A and f to be the identity map. Then (b) followsfrom(a),sincef| =f◦ι .Part(c)followsfromthecharacteristic A A property because the hypothesis says that ι f(X) ◦f: X →Y is continuous, where ι f(X) is the inclusion of f(X) into Y. Forpart(e),letU ⊂B beanysubset.Forthepurposesofthisproof,we sayU is open in B relative to AifU isopeninthesubspacetopologythat B inherits from A; U is open in B relative to X is defined similarly. Then we argue as follows (Figure 3.3):

Subspaces 43 V U X B q A FIGURE 3.4. A basis open set for a subspace. U is open in B relative to A ⇐⇒ U =B∩V for some V ⊂ A open ⇐⇒ U =B∩V, where V =A∩W, W ⊂ X open ⇐⇒ U =B∩A∩W for some W ⊂ X open ⇐⇒ U =B∩W for some W ⊂ X (since B ⊂A) open ⇐⇒ U is open in B relative to X. To prove (g), we have to show that every open subset of A satisfies the basis criterion with respect to B . Let U be an open subset of A, and let A q ∈ U. Then by definition U = A∩V for some open subset V ⊂ X. By the basis criterion for V, there is an element B ∈ B such that q ∈ B ⊂ V (Figure 3.4). It then follows that q ∈B∩A⊂U with B∩A∈B . A Parts (d), (f), (h), and (i) are left as an exercise. Exercise 3.3. Complete the proof of Proposition 3.4. We can now produce many examples of manifolds as subspaces of Eu- clidean spaces. In particular, since the Hausdorff property and second countability are hereditary by parts (h) and (i) of the preceding propo- sition, to show that a subspace of Rn is a manifold we need only verify the locally Euclidean condition. We begin with a very general construction. Example 3.5. IfU ⊂Rn isanopensetandf: U →Rk isanycontinuous map, the graph of f (Figure 3.5) is the set

44 3. New Spaces from Old y1,… ,y k Γ(f) x2,…,x n x1 U FIGURE 3.5. The graph of a continuous function. Γ(f)={(x,y)=(x1,…,x n ,y1,…,y k )∈Rn+k :x∈U and y =f(x)}. To verify that Γ(f) is locally Euclidean, we construct an explicit homeo- morphism between U and Γ(f). Let Φ : U →Γ(f) be the map f Φ (x)=(x,f(x)). f It is continuous because f is, and it is easily seen that its inverse is the restriction to Γ(f) of the map π(x,y) = x (the projection onto the first n coordinates), which is continuous by Proposition 3.4(b). Thus Φ is a f topological embedding. In particular, Γ(f) is (globally) homeomorphic to the open set U ⊂Rn, so it is an n-manifold. Example 3.6. Our next examples are arguably the most important man- ifolds of all, so it is worth taking some time to understand them well. The (unit) n-sphere is the set Sn ={x∈Rn+1 :|x|=1}.

Subspaces 45 Inlowdimensions,spheresareeasytovisualize:S0 isthetwo-pointdiscrete space {±1} ⊂ R; S1 is the unit circle in the plane; and S2 is the familiar sphericalsurfaceofradius1inR3.Inthecaseofthecircle,itisoftenmore convenient to identify the plane R2 with the set C of complex numbers by the correspondence (x,y) ↔ x+iy, and think of the circle as the set of complex numbers with unit modulus: S1 ={z ∈C:|z|=1}. To see that Sn is a manifold, we need to show that each point has a Euclidean neighborhood. The most straightforward way is to show that each point has a neighborhood in which Sn is the graph of a continuous function. For each i = 1,…,n + 1, let U + denote the subset of Rn+1 i where x > 0, and U − the set where x < 0. If x is any point in Sn, some i i i ± ± coordinate x must be nonzero there, so the sets U , … , U cover a i 1 n+1 neighborhood of Sn. On U ± , we can solve the equation |x| = 1 for x and i i find that x∈Sn∩U ± if and only if i (cid:3) x i =± 1−(x1)2−···−(x i−1)2−(x i+1)2−···−(x n+1)2. In other words, the portion of Sn in U ± is the graph of a continuous i function, and is therefore locally Euclidean. This proves that Sn is an n- manifold. Here is another useful way to show that Sn is a manifold. Consider both Rn and Sn as subsets of Rn+1 (identifying Rn with the set of points whose x n+1 coordinateiszero),andletN =(0,…,0,1)denotethe“northpole.” Define stereographic projection σ :Sn(cid:3){N}→Rn to be the map given by σ(x1,…,x n+1)= (x 1 1 − ,.. x . n , + x 1 n ) . Geometrically, it sends a point x ∈ Sn (cid:3){N} to the point u ∈ Rn where the line from N to x intersects Rn (Figure 3.6), as you can easily check. It is a homeomorphism, because it has an inverse given by σ−1(u1,…,u n )= (2u1,… |u ,2 |2 u + n , 1 |u|2−1) . Thus Sn(cid:3){N} is homeomorphic to Rn. In particular, this provides a Eu- clidean neighborhood of every point of Sn except N, and the analogous projection from the south pole works in a neighborhood of N. Example 3.7. Finally,considerthedoughnut surfaceD,whichisthesur- face of revolution in R3 defined by revolving the circle (y−2)2 +z2 = 1 (calledthegeneratingcircle)aroundthez-axis(Figu(cid:3)re3.7).Itischaracter- ized by the equation (r−2)2+z2 =1, where r = x2+y2. This surface canbeparametrizedbytwoanglesθ (measuredaroundthez axisfromthe

46 3. New Spaces from Old N x u=σ(x) FIGURE 3.6. Stereographic projection. xz-plane) and ϕ (measured around the generating circle from the horizon- tallyoutwarddirection).Itismoreconvenientforcalculationstomakethe substitutions ϕ=2πu and θ =2πv, and define a map F: R2 →D by F(u,v)=((2+cos2πu)cos2πv,(2+cos2πu)sin2πv,sin2πu). (3.1) This maps the plane onto D. It is not one-to-one, since F(u+k,v+l) = F(u,v)foranypairofintegers(k,l).However,F isinjectiveifitisrestricted toasmallenoughneighborhoodofanypoint(u0,v0),andastraightforward calculation shows that a local inverse in a neighborhood of (u0,v0) can be constructed from the formulas 1 z 1 y u= tan−1 +k; v = tan−1 +l; 2π r−2 2π x 1 r−2 1 x u= cot−1 +k; v = cot−1 +l 2π z 2π y for suitable choices of k,l. Thus D is a 2-manifold. The next lemma is similar to the local criterion for continuity of Lemma 2.2, in that it asserts the global continuity of a map that is known to be continuous on certain subsets. In this case, however, the subsets must be closed, and there can be only finitely many of them. This lemma will turn out to be extremely useful in our investigations of surfaces. Lemma 3.8 (Gluing Lemma). Let X be a topological space, and sup- pose X = A1 ∪···∪A k , where each A i is closed in X. For each i, let

Subspaces 47 z y ϕ θ x FIGURE 3.7. A doughnut surface of revolution. f i : A i → Y be a continuous map such that f i | Ai ∩Aj = f j | Ai ∩Aj . There exists a unique continuous map f: X →Y such that f| =f for each i. Ai i Exercise 3.4. Prove Lemma 3.8. In choosing a topology for a subset A ⊂ X, there are two competing priorities:Wewouldliketheinclusionmap A(cid:9)→X tobecontinuous(from whichitfollowsbycompositionthattherestrictionto Aofanycontinuous map f: X → Y is continuous); and we would also like continuous maps intoX whoseimageshappentolieinAalsotobecontinuousasmapsinto A. For the first requirement, A needs to have enough open sets, while for the second it should not have too many. The subspace topology is chosen as the optimal compromise between these requirements. As we will see several times in this chapter, natural topologies like the subspacetopologycanusuallybecharacterizedintermsofwhichmapsare continuouswithrespecttothem.Thisiswhythe“characteristicproperty” of the subspace topology (Theorem 3.3) is so named. The next lemma makes this precise. Theorem 3.9 (Uniqueness of the Subspace Topology). Suppose A is a subset of a topological space X. The subspace topology on A is the unique topology for which the characteristic property holds.

48 3. New Spaces from Old Proof. Suppose we are given an arbitrary topology on A that is known to satisfy the characteristic property. For this proof, let A denote A with g the given topology, and let A denote A with the subspace topology. To s show that the given topology is equal to the subspace topology, it suffices to show that the identity map of A is a homeomorphism between A and g A , by Problem 2-1. s First we note that the inclusion map from A into X is continuous, as g follows:Sincetheidentitymapofanyspaceiscontinuous,thecharacteristic property applied to the composition A −I→d A (cid:9)→X g g implies that this composite map is continuous; but this composition is just the inclusion A (cid:9)→ X itself. Of course, the inclusion map A (cid:9)→ X is g s also continuous, because it is the inclusion map of a subspace (Proposition 3.4(a)). Now consider the two composite maps A − Id →sgA (cid:9)→ ιg X, s g A − Id →gsA (cid:9)→ιs X. g s Here both Id and Id represent the identity map of A, and ι and ι gs sg s g represent inclusion of A into X; we decorate them with subscripts only for the purpose of discussing their continuity. Note that ι ◦Id = ι , and ι ◦Id = ι , both of which we have just g sg s s gs g showntobecontinuous.Thus,applyingthecharacteristicpropertytoeach of the compositions above, we conclude that both Id and its inverse Id sg gs are continuous. Therefore, Id is a homeomorphism. sg Product Spaces SupposeX1,…,X n aretopologicalspaces.WedefineabasisintheirCarte- sian product X1 ×···×X n by B={U1 ×···×U n :U i is open in X i , i=1,…,n}. Exercise 3.5. Prove that B is a basis. ThetopologygeneratedbyBiscalledtheproducttopology,andthespace X1 ×···×X n withtheproducttopologyiscalledaproductspace.Thebasis open sets of the form U1 ×···×U n are called product open sets. Forexample,intheplaneR2 =R×R,theproducttopologyisgenerated bysetsoftheform I×J,where I and J areopensetsin R.Atypicalsuch set is an open rectangle.

Product Spaces 49 Exercise 3.6. ShowthattheproducttopologyonRn =R×···×Risthe same as the metric topology induced by the Euclidean distance function. The product topology can also be defined in the more general setting of infinite products, with a slightly more complicated definition. We will not need to use infinite products in this book, but the general definition of the product topology is given in Problem 7-12. For more information, consult [Sie92] or [Mun75]. The characteristic property relates continuity of a map into a product spacetocontinuityofitscomponentfunctions.Inthespecialcaseofamap from Rn to Rm, this reduces to a familiar result from advanced calculus. Theorem 3.10 (Characteristic Property of Product Topologies). Let X1 ×···×X n be a product space. For any topological space B, a map f: B → X1 ×···×X n is continuous if and only if each of its component functions f =π ◦f is continuous: i i X1 ×···×X n (cid:3)(cid:4) (cid:3) f (cid:3) π i (cid:3) (cid:5) (cid:2) B X . f i i Proof. Since it suffices to check continuity on basis open sets, f is continuous ⇐⇒ f−1(U1 ×···×U n ) ⊂ B for all U i ⊂ X i open open ⇐⇒ f 1 −1 (U1)∩···∩f n −1(U n ) op ⊂ en B for all U i op ⊂ en X i . If each f is continuous, the set in the last line above is the intersection of i finitely many open sets and is therefore open in B, which shows that f is continuous. Conversely, if f is continuous, choose j between 1 and n and take U = X for all i except i =j. Then f −1 (U ) = B when i (cid:14)= j, so the i i i i −1 argument above shows that f (U ) is open in B whenever U is open in j j j X , or in other words, f is continuous. j j It follows from the characteristic property that if f1,f2: X → C are complex-valuedcontinuousfunctions,thentheirsum(f1+f2)(x)=f1(x)+ f2(x) is continuous, because it is the composition of the maps f: X →C2 given by f(x)=(f1(x),f2(x)) and s: C2 →C given by s(w,z)=w+z. A similar remark applies to the product (f1f2)(x)=f1(x)f2(x). Just as in the case of the subspace topology, the product topology is uniquely determined by its characteristic property. Theorem 3.11 (Uniqueness of the Product Topology). Let X1, … , X n be topological spaces. The product topology on X1 ×···×X n is the unique topology that satisfies the characteristic property.

50 3. New Spaces from Old Proof. Suppose that X1 ×···×X n is endowed with some topology that satisfiesthecharacteristicproperty.Firstwenotethattheprojectionmaps π arecontinuous(ineithertopology)bythecharacteristicpropertyapplied i to the identity map of X1 ×···×X n . Now inserting X1 ×···×X n with the product topology in place of B shows that the identity map from the product topology to the given topology is continuous, and reversing roles showsthatitsinverseisalsocontinuous.Thusthetwotopologiesareequal. Proposition 3.12 (Other Properties of the Product Topology). Let X1, … , X n be topological spaces. (a) The projection maps π i : X1 ×···×X n →X i are all continuous. (b) The product topology is “associative” in the sense that the three prod- uct topologies X1 ×X2 ×X3, (X1 ×X2)×X3, and X1 ×(X2 ×X3) on the set X1 ×X2 ×X3 are all equal. (c) For any i and any points x j ∈ X j , j (cid:14)= i, the map f i : X i → X1 × ···×X given by n f i (x)=(x1,…,x i−1,x,x i+1,…,x n ) is a topological embedding of X into the product space. i (d) If for each i, B is a basis for the topology of X , then the set i i {B1 ×···×B n :B i ∈B i } is a basis for the product topology on X1 ×···×X n . (e) If A is a subspace of X for i = 1,…,n, the product topology and i i the subspace topology on A1 ×···×A n ⊂X1 ×···×X n are equal. (f) If each X i is Hausdorff, so is X1 ×···×X n . (g) If each X i is second countable, so is X1 ×···×X n . Exercise 3.7. Prove Proposition 3.12. Iff : X →Y aremaps(continuousornot)fori=1,…,k,theirproduct i i i map is f1 ×···×f k : X1 ×···×X k →Y1 ×···×Y k given by f1 ×···×f k (x1,…,x k )=(f1(x1),…,f k (x k )).

Product Spaces 51 Proposition 3.13. A product of continuous maps is continuous, and a product of homeomorphisms is a homeomorphism. Proof. Becauseitsufficestocheckthattheinverseimagesofbasisopensets areopen,thefirstclaimfollowsfromthefactthat(f1 ×···×f k )−1(U1 ×···× U k ) is just the product of the open sets f1(U1), … , f k (U k ). The second claim follows from the first, because the inverse of a bijective product map is itself a product map. Product spaces provide us with another rich source of examples of man- ifolds. The key is the following proposition. Proposition 3.14. IfM1,…,M k aremanifoldsofdimensionsn1,…,n k , respectively, the product space M1 ×···×M k is a manifold of dimension n1+···+n k . Proof. Proposition 3.12 shows that the product is Hausdorff and second countable, so only the locally Euclidean property needs to be checked. Given any point q = (q1,…,q k ) ∈ M1 ×···×M k , for each i there exists a neighborhood U of q and a homeomorphism ϕ from U to an open i i i i subsetofRni.Bytheprecedinglemma,theproductmapϕ1 ×···×ϕ k isa homeomorphismfromaneighborhoodofqtoanopensetinRn1+···+nk. AparticularlyimportantexampleistheproductmanifoldTn =S1×···× S1, which is an n-dimensional manifold called the n-torus. In particular, the 2-torus is usually just called the torus. Because S1 is a subspace of R2, T2 can be considered as a subspace of R4 by Proposition 3.12(e): It is just the set of points (x1,x2,x3,x4) ∈ R4 such that (x1)2 +(x2)2 = 1 and (x3)2+(x4)2 = 1. As the next lemma shows, T2 is homeomorphic to a familiar surface. Lemma 3.15. The torus T2 is homeomorphic to the doughnut surface D of Example 3.7. Proof. The key idea is that both surfaces are parametrized by two angles. For D, the angles are ϕ = 2πu and θ = 2πv as in (3.1); for T2, they are the angles in the two circles. Although one must be careful using angle functions because they cannot be defined continuously on a whole circle, with some care we can eliminate the angles altogether and come up with formulas that are manifestly continuous. Withthisinmind,wewritex1 =cosθ,x2 =sinθ,x3 =cosϕ,x4 =sinϕ. Substituting into (3.1) suggests defining a map G: T2 →D by G(x1,x2,x3,x4)=((2+x3)x1, (2+x3)x2, x4). This is obviously continuous, and a little algebra shows that G maps T2 into D. To see that it is a homeomorphism, just check that its inverse is

52 3. New Spaces from Old given by G−1(x,y,z)=(x/r, y/r, r−2, z), (cid:3) where r = x2+y2 as in Example 3.7. Quotient Spaces Our third technique for constructing new topological spaces from old ones is somewhat more involved than the preceding two. It is a way to identify some points in a given topological space with each other, to obtain a new, smaller space. LetX beatopologicalspace,Y beanyset,andπ: X →Y beasurjective map.DefineatopologyonY bydeclaringasubsetU ⊂Y tobeopenifand only if π−1(U) is open in X. This is called the quotient topology induced by the map π. Exercise 3.8. Show that the quotient topology is indeed a topology. More generally, if X and Y are topological spaces, a map π: X → Y is called a quotient map if it is surjective and continuous and Y has the quotient topology induced by π. If π is known to be surjective, this is the same as saying that U is open in Y if and only if π−1(U) is open in X. An easy example to keep in mind is the map π: Rn+k → Rn given by projection onto the first n coordinates. It is straightforward to check directly from the definition that it is a quotient map. Themostcommonsourceofquotientmapsisthefollowingconstruction. Let ∼ be an equivalence relation on a topological space X (see the Ap- pendix). For each q ∈ X let [q] denote the equivalence class of q, and let X/∼ denote the set of equivalence classes: This is a partition of X, which is a decomposition of X into a collection of disjoint subsets whose union is X. Let π: X →X/∼ be the natural projection sending each element of X to its equivalence class. Then X/∼ together with the quotient topology determined by π is called the quotient space (or sometimes identification space)ofX bythegivenequivalencerelation.Thequotientmapπ iscalled the projection. Alternatively, a quotient space can be defined by explicitly giving a par- titionofX.Whetheragivenquotientspaceisdefinedintermsofanequiv- alence relation or a partition is a matter of convenience. If π: X → Y is a quotient map, a subset U ⊂ X is said to be saturated (with respect to π) if U = π−1(V) for some subset V ⊂ Y. (In fact, you can check that U is saturated if and only if U = π−1(π(U)).) If Y is a quotient space determined by an equivalence relation, the saturated sets are those that are unions of equivalence classes. More generally, for any

Quotient Spaces 53 quotient map π: X → Y, a subset π−1(y) ⊂ X for y ∈ Y is called a fiber of π; a saturated set is one that is a union of fibers. Althoughquotientmapsdonotalwaystakeopensetstoopensets,there is a useful alternative characterization of quotient maps in terms of satu- rated open sets. Lemma 3.16. A continuous surjective map π: X →Y is a quotient map if and only if it takes saturated open sets to open sets, or saturated closed sets to closed sets. Exercise 3.9. Prove Lemma 3.16. Lemma 3.17. Suppose f: X →Y is a quotient map. The restriction of f to any saturated open or closed set is a quotient map. Exercise 3.10. Prove Lemma 3.17. Itisnotalwaysatrivialmattertocheckthatacontinuoussurjectivemap is a quotient map—it may well not be, as the following example shows. Example 3.18. Consider the map a: [0,1)→S1 defined (in complex no- tation) by a(s)=e2πis. It is surjective and continuous, but not a quotient map, because [0,1) is a saturated open subset of [0,1) whose image is not 2 open in S1. The following lemma gives two very useful sufficient conditions for a surjective continuous map to be a quotient map. Lemma 3.19. If π: X →Y is a surjective continuous map that is also an open or closed map, then it is a quotient map. Proof. If π is open, it takes saturated open sets to open sets (because it takes all open sets to open sets). If π is closed, it takes saturated closed sets to closed sets. In either case, it is a quotient map by Lemma 3.16. One simple property of quotient maps is that they behave well with respect to composition. Lemma 3.20 (Composition Property of Quotient Maps). Sup- pose π1: X → Y and π2: Y → Z are quotient maps. Then their composition π2 ◦π1: X →Z is also a quotient map. Proof. Just note that U ⊂ Z is open if and only if π −1 (U) is open in Y, 2 whichistrueifandonlyifπ 1 −1 (π 2 −1 (U))=(π2 ◦π1)−1(U)isopeninX. Asithappens,quotientspacesdonotgenerallybehavewellwithrespect to products, subspaces, bases, or topological properties such as locally Eu- clidean, Hausdorff, or second countable. In particular, quotient spaces of manifoldsaregenerallynotmanifolds.In fact,itisnotdifficulttoconstruct

54 3. New Spaces from Old FIGURE 3.8. A quotient of B2. FIGURE 3.9. A quotient of I×I. aquotientspaceofamanifoldthatsatisfiesallthedefinitionsofamanifold except that it is not Hausdorff (see Problem 3-8). Thus if we wish to prove that a given quotient space is a manifold, we have to prove at least the locally Euclidean and Hausdorff properties directly. The following lemma shows that in many cases this is sufficient. In par- ticular,itshowsthataquotientofamanifoldisagainamanifold,provided that it is locally Euclidean and Hausdorff. Lemma 3.21. Suppose P is a second countable space and π: P →M is a quotient map. If M is locally Euclidean, it is second countable. Proof. Let U be a covering of M by Euclidean balls. The collection {π−1(U) : U ∈ U} is an open cover of P, which has a countable sub- coverbyLemma2.15.IfweletU(cid:5) ⊂UdenoteacountablesubsetofUsuch that {π−1(U) : U ∈ U(cid:5)} covers P, then U(cid:5) is a countable cover of M by Euclidean balls. Each such ball has a countable basis, and the union of all these bases is a countable basis for M. Becausequotientspacesareprobablylessfamiliartoyouthansubspaces or products, we will introduce a number of examples before going any farther. Example 3.22. The map α: [0,1]→S1 given by α(s)=e2πis is a closed map and therefore a quotient map. Example 3.23. Let B2 denote the closed unit disk in R2. Let ∼ be the equivalencerelationonB2 generatedby(x,y)∼(−x,y)forall(x,y)∈∂B2 (Figure 3.8). (You can think of this space as being obtained from B2 by “pasting” the left half of the boundary to the right half.) We will see in Chapter 6 that B2/∼ is homeomorphic to S2. Example 3.24. Let I = [0,1] denote the closed unit interval in the real line; we will generally just call this the unit interval. Define an equivalence relation on the square I ×I by setting (x,0) ∼ (x,1) for all x ∈ I, and (0,y) ∼ (1,y) for all y ∈ I (Figure 3.9). This can be visualized as the space obtained by gluing the top boundary segment of the square to the

Quotient Spaces 55 FIGURE 3.10. Wedge of two lines. FIGURE3.11.Wedgeoftwocircles. bottom to form a cylinder, and then gluing the left-hand boundary circle of the resulting cylinder to the right-hand one. Later we will see that the resulting quotient space is homeomorphic to the torus. Example 3.25. Let X1,…,X k be topological spaces, and let q i ∈ X i . The wedge of X1,…,X k (also called their one-point union), written X1 ∨ ···∨X k , is the quotient space obtained from X1 (cid:20)···(cid:20)X k by identifying q1 ∼ ··· ∼ q k . In other words, we glue the spaces together by identifying all their distinguished points together. For example, the wedge R∨R is homeomorphictotheunionofthex-axisandthey-axisintheplane(Figure 3.10), and the wedge S1∨S1 is homeomorphic to the figure eight space E consisting of the union of the two circles of radius 1 centered at (0,1) and (0,−1) in the plane (Figure 3.11). A wedge of finitely many copies of S1 is sometimes called a bouquet of circles. Example 3.26. Define an equivalence relation on R by declaring x∼y if x and y differ by an integer. We will see below that the resulting quotient space is homeomorphic to the circle. Example 3.27. Consider the map q: Rn+1(cid:3){0}→Sn defined by q(x)= x/|x|. Observe that q is continuous and surjective, and the fibers of q are raysinRn+1(cid:3){0}.Thusthesaturatedsetsaretheunionsofrays,anditis easytocheckthatq takessaturatedopensetstoopensetsandistherefore a quotient map. Example 3.28. Define Pn, the real projective space of dimension n, to be thesetof1-dimensionallinearsubspaces(linesthroughtheorigin)inRn+1. There is a natural map π: Rn+1 (cid:3){0} → Pn defined by sending a point x to its span. We topologize Pn by giving it the quotient topology with respect to this map.

56 3. New Spaces from Old Thisspacecanalsobeviewedinanotherway.Ifwedefineanequivalence relationonRn+1(cid:3){0}bydeclaringtwopointsx,ytobeequivalentifx=λy for some nonzero real number λ, then there is an obvious identification between Pn and the set of equivalence classes. Under this identification, the map π defined above is just the map sending a point to its equivalence class. The Characteristic Property of the Quotient Topology Thecharacteristicpropertyofthequotienttopologyisevenmoreimportant than those of the subspace or product topologies. Theorem 3.29 (Characteristic Property of Quotient Topologies). Let π: X → Y be a quotient map. For any topological space B, a map f: Y → B is continuous if and only if the composite map f ◦ π is continuous: X (cid:6) π (cid:6) f ◦π (cid:6) (cid:5) (cid:6)(cid:7) (cid:2) Y B. f Proof. Iff iscontinuous,f◦π iscontinuousbycomposition.Conversely,if f◦π iscontinuousandU ⊂B isopen,thenπ−1(f−1(U))=(f◦π)−1(U)is openinX,whichimpliesf−1(U)isopeninY bydefinitionofthequotient topology. Thus f is continuous. Thecharacteristicpropertyhasthefollowingextremelyimportantcorol- lary: Corollary 3.30 (Passing to the Quotient). Suppose π: X → Y is a quotient map, B is a topological space, and f: X → B is any continuous map that is constant on the fibers of π (i.e., if π(p) = π(q), then f(p) = f(q)). Then there exists a unique continuous map f (cid:7) : Y → B such that f =f (cid:7)◦π: X (cid:6) π (cid:6) f (cid:6) (cid:5) (cid:6)(cid:7) (cid:2) Y B. (cid:7) f (cid:7) Proof. The existence and uniqueness of f follow from elementary set the- ory: Given q ∈Y, there is some p∈X such that π(p)=q, and we can set (cid:7) (cid:7) f(q)=f(p)foranysuchp.Thehypothesisonf guaranteesthatf isunique

Quotient Spaces 57 (cid:7) andwell-defined.Continuityoff isthenimmediatefromthecharacteristic property. In the situation of the preceding corollary, we say that the map f passes to the quotient or descends to the quotient. In the case of quotient spaces, there are two slightly different ways of phrasing the uniqueness associated with the characteristic property. The first one says that the characteristic property uniquely characterizes quo- tient maps. Theorem 3.31 (Characterization of Quotient Maps). LetX andY be topological spaces, and let π: X → Y be any surjective map. Then π is a quotient map if and only if the characteristic property holds. Proof. Ifπisaquotientmap,thecharacteristicpropertyholdsbyTheorem 3.29. Conversely, suppose π has the characteristic property. Applying the characteristic property to the diagram X (cid:6) π (cid:6) π (cid:6) (cid:5) (cid:6)(cid:7) (cid:2) Y Y Id shows that π is continuous because the identity is. To show that π is a quotientmap,wewillshowthatY withthegiventopologyishomeomorphic to Y with the quotient topology. As before, let Y and Y denote Y with g q the given and quotient topologies, respectively, and let Id , Id , π , and gq qg q π have the obvious meanings. Then the characteristic property applied to g the two diagrams X X (cid:6) (cid:6) π (cid:6) π π (cid:6) π g q q g (cid:6) (cid:6) (cid:5) (cid:6)(cid:7) (cid:5) (cid:6)(cid:7) (cid:2) (cid:2) Y Y Y Y g Id q q Id g gq qg showsthatId andId arebothcontinuous,fromwhichtheresultfollows. qg gq The second uniqueness result says that quotient spaces are uniquely de- terminated up to homeomorphism by the identifications made by their quo- tient maps. Corollary 3.32 (Uniqueness of Quotient Spaces). Suppose π1: X → Y1 and π2: X → Y2 are quotient maps that make the same identifications (i.e., π1(p) = π1(q) if and only if π2(p) = π2(q)). Then there is a unique homeomorphism ϕ: Y1 →Y2 such that ϕ◦π1 =π2.

58 3. New Spaces from Old Proof. By Corollary 3.30, both π1 and π2 pass uniquely to the quotient as in the following diagrams: X X (cid:6) (cid:6) π1 (cid:6) π2 π2 (cid:6) π1 (cid:6) (cid:6) (cid:5) (cid:6)(cid:7) (cid:5) (cid:6)(cid:7) (cid:2) (cid:2) Y1 π(cid:8) 2 Y2 Y2 π(cid:8) 1 Y1. Sincebothdiagramsabovecommute,itfollowsthatπ(cid:8) 1 ◦(π(cid:8) 2 ◦π1)=π(cid:8) 1 ◦π2 = π1. Consider another diagram: X (cid:6) π1 (cid:6) π1 (cid:6) (cid:5) (cid:6)(cid:7) (cid:2) Y1 Y1. If the bottom arrow is interpreted as either π(cid:8) 1 ◦π(cid:8) 2 or the identity map of Y1, this diagram will commute; by the uniqueness part of Corollary 3.30, these maps must be equal. Similarly, π(cid:8) 2 ◦π(cid:8) 1 is the identity on Y2. Thus ϕ=π(cid:8) 2 is the required homeomorphism, and it is the unique such map by the uniqueness statement of Corollary 3.30. Group Actions Ournextconstructionisafar-reachinggeneralizationofExamples3.26and 3.28. A topological group is a group G endowed with a topology such that the maps μ: G×G→G and ι: G→G given by μ(g1,g2)=g1g2, ι(g)=g−1 arecontinuous,wheretheproductandinversearethoseofthegroupstruc- true of G. A discrete group is a topological group that has the discrete topology. Example 3.33. Each of the following is a topological group.

• The real line R with its additive group structure and the Euclidean topology.
• The set R∗ =R(cid:3){0} of nonzero real numbers under multiplication, with the subspace topology.
• The set C∗ = C(cid:3){0} of nonzero complex numbers under complex multiplication, with the subspace topology.

Group Actions 59

• Thegeneral linear groupGL(n,R),whichisthesetofn×ninvertible realmatricesundermatrixmultiplication,withthesubspacetopology inherited from Rn2 .
• Any group with the discrete topology. Exercise 3.11. Verify that each of the above examples is a topological group. Lemma 3.34. Any subgroup of a topological group is a topological group with the subspace topology. Any finite product of topological groups is a topological group with the direct product group structure and the product topology. Exercise 3.12. Prove Lemma 3.34. In view of Lemma 3.34, each of the following is a topological group:
• Euclidean space Rn as a group under vector addition.
• The circle S1 ⊂C∗ under complex multiplication, with the subspace topology.
• The n-torus Tn =S1×···×S1 with the direct product group struc- true.
• The orthogonal group O(n), which is the subgroup of GL(n,R) con- sisting of matrices A such that AAt is the identity. IfGisatopologicalgroupandg ∈G,thelefttranslationmapL : G→G g defined by L (g(cid:5)) = gg(cid:5) is continuous, because it is the restriction of the g multiplication map to {g}×G. Because L ◦L = Id , left translation g g−1 G by any element of g is a homeomorphism of G. Similarly, right translation R (g(cid:5))=g(cid:5)g is also a homeomorphism. g Suppose G is a group and X is a topological space. A left action of G · on X is a map G×X → X, written (g,x) (cid:10)→ g x, with the following properties: · · · (i) For any x∈X and any g1,g2 ∈G, g1 (g2 x)=(g1g2) x. · (ii) For all x∈X, 1 x=x. · Similarly, a right action is a map X×G→X, written (x,g)(cid:10)→x g, with · · thesamepropertiesexceptthatcompositionworksinreverse:(x g1) g2 = · x (g1g2).Anyrightactiondeterminesaleftactioninacanonicalway,and vice versa, by the correspondence g · x=x · g−1.

60 3. New Spaces from Old Thus for many purposes, the choice of left or right action is a matter of taste. We usually choose to focus on left actions because the composi- tion law mimics composition of functions, and unless we specify otherwise, groups will always be understood to act on the left. However, we will see some situations in which an action appears naturally as a right action. If G is a topological group, an action of G on a space X is said to be continuousifthemapG×X →X iscontinuous.Thismeans,inparticular, · thatforeachg ∈Gthemapx(cid:10)→g xiscontinuousfromX toitself,because itistherestrictionoftheactiontothesubspace{g}×X ⊂G×X.In fact, each such map is a homeomorphism, because the definition of a group action guarantees that it has a continuous inverse x (cid:10)→ g−1 · x. When G is discrete, it is easy to check that the action is continuous if and only if · x(cid:10)→g x is continuous for each g ∈G. · · For any x ∈ X, the set G x = {g x : g ∈ G} is called the orbit of x. Theactionissaidtobetransitiveifforeverypairofpoints x,y ∈X,there · is a group element g such that g x=y or equivalently if the only orbit is the entire space X. It is said to be free if the only element of G that has · any fixed points is the identity, i.e., g x=x for some x implies g =1. Example 3.35 (Continuous Group Actions). (a) ThegenerallineargroupGL(n,R)actscontinuouslyontheleftonRn by matrix multiplication, each vector in Rn considered as a column matrix. Because any nonzero vector in Rn can be taken to any other by a linear transformation, there are only two orbits: Rn(cid:3){0} and {0}. (b) The orthogonal group O(n) acts continuously on Rn by matrix mul- tiplication as well; this is just the restriction of the action in part (a)toO(n)×Rn ⊂GL(n,R)×Rn.Sinceorthogonaltransformations preservelengthsofvectors,andanyvectorcanbetakentoanyother of the same length by an orthogonal transformation, the orbits are {0} and the spheres centered at 0. (c) The restriction of the action of O(n) to the unit sphere in Rn yields a transitive action on Sn−1. (d) The group R∗ acts on Rn(cid:3){0} by scalar multiplication. The action isfree,andtheorbitsarethelinesthroughtheorigin(withtheorigin removed). (e) Any topological group G acts freely and transitively on itself on the left by left translation: g · g(cid:5) = L (g(cid:5)) = gg(cid:5). Similarly, G acts on g itself on the right by right translation. Given an action of G on a space X, we define an equivalence relation on · X by setting x1 ∼ x2 if there is an element g ∈ G such that g x1 = x2.

Group Actions 61 The equivalence classes are precisely the orbits of the group action. The resulting quotient space is denoted by X/G, and is called the orbit space of the action. If the action is transitive, the orbit space is a single point, so only nontransitive actions yield interesting examples. Exercise 3.13. Verify that the real projective space Pn of Example 3.28 istheorbitspaceoftheactionofR∗ onRn+1(cid:3){0}byscalarmultiplication. Aparticularlyimportantspecialcaseariseswhenweconsiderasubgroup Γ of a topological group G (with the subspace topology). Group multipli- cation on the left or right defines a left or right action of Γ on G; it is just the restriction of the action of G on itself to Γ×G or G×Γ. This action is continuous and free, but in general not transitive. An orbit of the right action of Γ on G is a set of the form {gγ : γ ∈ Γ}, which is precisely the left coset gΓ. Thus the orbit space of the right action of Γ on G is the set G/Γ of left cosets with the quotient topology. This quotient space is called the(left) coset spaceofGbyΓ.(Itisunfortunatebutunavoidablethatthe right action produces a left coset space and vice versa. If G is abelian, the situation is simpler, because then the left action and right action of Γ are equal to each other.) Example 3.36. As an application, let us consider the coset space R/Z. Because Z (with the discrete topology) is a subgroup of the topological groupR,thereisanaturalfreecontinuousactionofZonRbytranslation: · n x = n+x. (Because R is abelian, we might as well consider it as a left action.) The orbits are exactly the equivalence classes of the relation defined in Example 3.26 above, x ∼ y if and only if x−y ∈ Z. Thus the quotient space of that example is the same as the coset space R/Z. Consider also the map ε: R→S1 defined by ε(s)=e2πis. It is straightforward to check that this is a local homeomorphism and thus an open map, so it is a quotient map. Because it makes the same identifi- cations as the quotient map R → R/Z, the uniqueness of quotient spaces tellsusthatR/ZishomeomorphictoS1.(Wewillbereturningtothismap ε, which we call the exponential quotient map, extensively in this book.) Moregenerally,thediscretesubgroupZnactsfreelyonRnbytranslation. By similar reasoning, the quotient space Rn/Zn is homeomorphic to the n-torus Tn =S1×···×S1. We will see more examples of this technique in the next few chapters.

62 3. New Spaces from Old Problems 3-1. Showthatafiniteproductofopenmapsisopen,andafiniteproduct of closed maps is closed. 3-2. By considering the space X = [0,1] ⊂ R, and the sets A0 = {0}, A = [1/(i+1),1/i] for i = 1,2,…, show that the gluing lemma i (Lemma3.8)isfalseif{A1,…,A k }isreplacedbyaninfinitesequence of closed sets. 3-3. Formulatea“characteristicproperty”forthedisjointuniontopology (Problem 2-9) and prove that the disjoint union topology is uniquely characterized by it. 3-4. Use stereographic projection to show that any closed ball in Rn is an n-dimensional manifold with boundary. 3-5. Let X be a topological space. The diagonal of X ×X is the subset Δ={(x,x):x∈X}⊂X×X.ShowthatX isHausdorffifandonly if Δ is closed in X×X. 3-6. If X1,…,X k are topological spaces, show that the projections π i : X1 ×···×X k →X i are quotient maps. 3-7. Let M =R ×R, where R is the set R with the discrete topology. d d (a) Show that M is homeomorphic to the space X of Problem 2-5. (b) Show that M is locally Euclidean (of what dimension?) and Hausdorff, but not second countable. 3-8. Let X be the subset R×{0}∪R×{1} of R2. Define an equivalence relation on X by declaring (x,0) ∼ (x,1) if x (cid:14)= 0. Show that the quotient space X/∼ is locally Euclidean and second countable, but not Hausdorff. (This space is called the line with two origins.) 3-9. Lemma 3.17 showed that the restriction of a quotient map to a satu- rated open set is still a quotient map. Show that the “saturated” hypothesis is necessary, by giving an example of a quotient map f: X →Y andanopensubsetU ⊂X suchthatf| issurjectivebut U not a quotient map. 3-10. Show that real projective space Pn is an n-manifold. [Hint: Consider the subsets U ⊂Rn+1 where x =1.] i i 3-11. Let CPn denote the set of all 1-dimensional complex subspaces of Cn+1,calledn-dimensional complex projective space.TopologizeCPn as the quotient Cn+1 (cid:3){0}/C∗, where C∗ is the group of nonzero complex numbers acting by scalar multiplication. Show that CPn is a 2n-manifold. [Hint: Mimic what you did in Problem 3-10.]

Problems 63 3-12. Let G be a topological group and let H ⊂ G be a subgroup. Show that H is also a subgroup. 3-13. If G is a group that is also a topological space, show that G is a topologicalgroupifandonlyifthemapG×G→Ggivenby(x,y)(cid:10)→ xy−1 is continuous. 3-14. Let G be a topological group and Γ⊂G be a subgroup. (a) For any g ∈ G, show that left translation L : G → G passes g tothequotientG/Γanddefinesahomeomorphismof G/Γwith itself. (b) AtopologicalspaceX issaidtobehomogeneousifforanyx,y ∈ X, there is a homeomorphism ϕ: X → X taking x to y. Show that every coset space is homogeneous. 3-15. Let G be a topological group acting continuously on a topological space X. (a) Show that the quotient map π: X →X/G is open. (b) Show that X/G is Hausdorff if and only if the orbit relation · {(x1,x2)∈X×X :x2 =g x1 for some g ∈G} is closed in X×X. 3-16. If Γ is a normal subgroup of the topological group G, show that the coset space G/Γ is a topological group. [Hint: It might be helpful to use Problems 3-1 and 3-15(a).]

4 Connectedness and Compactness In this chapter we treat two topological properties that will be of central importance in our study of manifolds: connectedness and compactness. The definition of connectedness is formulated so that connected spaces will behave similarly to intervals in the real line, so, for example, a contin- uous real-valued function on a connected space satisfies the intermediate value theorem. Similarly, compactness is defined so that compact spaces willhavemanyofthesamepropertiesenjoyedbyclosedandboundedsub- sets of Euclidean spaces. In particular, continuous real-valued functions on compact sets always achieve their maxima and minima. Connectedness One of the most important elementary facts about continuous functions is the intermediate value theorem: If f is a continuous real-valued function defined on a closed bounded interval [a,b], then f takes on every value be- tweenf(a)andf(b).Thekeyideahereisthe“connectedness”ofintervals. In this section we generalize this concept to topological spaces. Definitions and Basic Properties IfX isatopologicalspace,aseparationofX isapairofnonempty,disjoint, opensubsetsU,V ⊂X suchthatX =U∪V.WesaythatX isdisconnected if there exists a separation of X, and connected otherwise.

66 4. Connectedness and Compactness FIGURE 4.1. Union of two disks. FIGURE 4.2. The x-axis minus 0. Bythisdefinition,connectednessisapropertyofaspace,notaproperty of subsets like openness or closedness. We can also talk about connected subsets of a topological space, by which we always mean connected in the subspace topology. In this context we can also consider a separation of A to be a pair of open subsets U,V ⊂ X whose intersections with A are nonempty and disjoint, and whose union contains A: This is equivalent to the original definition because the open subsets of A are exactly the open subsets of X intersected with A. Example 4.1. Each of the following subspaces of the plane is discon- nected. (a) X istheunionofthetwodisjointcloseddisksB1(2,0)andB1(−2,0) (Figure 4.1). Each of the disks is open in X, so the pair of disks is a separation of X. (b) Y is the x-axis minus the origin (Figure 4.2). The two sets {(x,0) : x>0} and {(x,0):x<0} separate Y. (c) Z isthesetofpointswithrationalcoordinates.Aseparationisgiven by, say, {(x,y):x<π} and {(x,y):x>π}. On the other hand, it is intuitively clear that the open and closed unit disks, the circle, the whole plane, and the x-axis are all connected, at least in the everyday sense of the word. Proving it, however, is not so easy, because we would have to show that it is impossible to find a separation. We will soon come up with an easy technique for proving connectedness that will work in most practical cases, including that of manifolds. Here is a useful alternative characterization of connectedness. Proposition 4.2. A space X is connected if and only if the only subsets of X that are both open and closed in X are ∅ and X itself. Proof. Suppose first that X is connected, and assume that U ⊂X is open andclosed.ThenV =X(cid:3)U isalsoopenandclosed.IfbothU andV were nonempty, then {U,V} would be a separation of X; therefore, either V is empty, which means that U =X, or U is empty.

Connectedness 67 Conversely, if X is disconnected, we can write X =U ∪V where U and V are nonempty, open, and disjoint. This implies that U is open, closed, not empty, and not equal to X. The most important feature of connectedness is that continuous images of connected sets are connected. Theorem 4.3 (Main Theorem on Connectedness). Let X,Y be topological spaces and let f: X → Y be a continuous map. If X is con- nected, then f(X) is connected. Proof. Suppose f(X) is not connected. Then there exist open sets U,V ⊂ Y whose intersections with f(X) are nonempty and disjoint and whose union contains f(X). It follows immediately that {f−1(U),f−1(V)} is a separation of X, so X is not connected. Proposition 4.4 (Properties of Connected Sets). (a) Suppose X is any space and U,V are disjoint open subsets of X. If A is a connected subset of X contained in U ∪V, then either A⊂U or A⊂V. (b) SupposeX isanyspaceandA⊂X isconnected.ThenAisconnected. (c) Let X be a space, and let {B α } α∈A be a(cid:2)ny collection of connected subspaces with a point in common. Then B is connected. α∈A α (d) Any product of finitely many connected spaces is connected. (e) Any quotient space of a connected space is connected. Proof. For part (a), if A contained points in both U and V, then {A∩ U,A∩V} would be a separation of A. To prove part (b), suppose U and V are disjoint open subsets of X that separate A. By (a), A is contained in one of the sets, say A ⊂ U. Each point of V has a neighborhood (namely V) disjoint from A, so every point of V is exterior to A. Therefore, A ⊂ U, which means that A∩V = ∅, a contradiction. For part (c), let(cid:2)q be a point contained in each B α , and suppose {U,V} is a separation of B . Suppose without loss of generality that q lies α∈A α in U. By part (a), each B must be entirely contained in U, and thus so is α their union. Forpart(d),sinceX1 ×···×X k =(X1 ×···×X k−1)×X k ,byinductionit sufficestoconsideraproductoftwospaces.ThusletX andY beconnected, andsuppose{U,V}isaseparationofX×Y.Chooseanypoint(x0,y0)∈U. The set {x0 }×Y is connected because it is homeomorphic to Y; since it contains the point (x0,y0)∈U, it must be entirely contained in U by part (a). For each y ∈ Y, the set X ×{y} is also connected and has a point

68 4. Connectedness and Compactness U V a c b FIGURE 4.3. Proof that an interval is connected. (x0,y) ∈ U, so it must also be contained in U. Since the sets X ×{y} exhaust X×Y, the result follows. Finally, (e) follows from Theorem 4.3 and the fact that quotient maps are surjective. Although this proposition gives us a number of ways of building new connected spaces out of given ones, so far we have no examples of spaces to start with that are known to be connected (except a one-point space, which does not carry us very far). The one example of a space that can be shown to be connected by “brute force” is the one that enters into the proof of the intermediate value theorem: an interval in the real line (see the Appendix). Proposition 4.5. A nonempty subset of R is connected if and only if it is an interval. Proof. First assume that J ⊂R is an interval. If it is not connected, there are open subsets U,V ⊂R that separate J. Choose a∈U ∩J, b∈V ∩J, and assume (interchanging U and V if necessary) that a<b (Figure 4.3). Then [a,b] ⊂ J because J is an interval. Since U and V are both open in R, there exists ε>0 such that [a,a+ε)⊂U ∩J and (b−ε,b]⊂V ∩J. Letc=sup(U∩[a,b]).Byourchoiceofε,a+ε≤c≤b−ε.Inparticular, c is between a and b, so c∈J ⊂U ∪V. But if c were in U, it would have a neighborhood (c−δ,c+δ)⊂U, which would contradict the definition of c. Similarly, c∈V leads to a contradiction. Therefore, J is connected. Conversely,assumethatJ isnotaninterval.Thismeansthatthereexist a < c < b with a,b ∈ J but c (cid:14)∈ J. Then the sets (−∞,c) and (c,∞) separate J, so J is not connected. Animmediateconsequenceofthispropositionisthefollowinggeneralized intermediate value theorem. Theorem 4.6 (Intermediate Value Theorem). Suppose X is a con- nected topological space, and f is a continuous real-valued function on X. If p,q ∈X, then f takes on all values between f(p) and f(q). Proof. The image set f(X) is connected, so it must be an interval.

Connectedness 69 Path Connectedness Now we can give a simple but powerful sufficient condition for connected- ness, based on the following definitions. Let X be a topological space and p,q ∈X. A path in X from p to q is a continuous map f: [0,1]→X such that f(0) = p and f(1) = q. We say that X is path connected if for every p,q ∈X, there is a path in X from p to q. Theorem 4.7. Path connectedness implies connectedness. Proof. SupposethatX ispathconnectedbutnotconnected,andlet{U,V} be a separation of X. We can choose p∈U and q ∈V (since neither set is empty), and find a path f from p to q in X. Then f−1(U) and f−1(V) are disjoint open subsets of [0,1] that cover [0,1]; moreover, 0 ∈ f−1(U) and 1∈f−1(V),soneithersetisempty.Thisimpliesthat[0,1]isdisconnected, which is a contradiction. Example 4.8. The following spaces are all easily shown to be path con- nected, and therefore they are connected. (a) Rn. (b) Any subset B ⊂Rn that is convex, which means that for any x,x(cid:5) ∈ B, the line segment from x to x(cid:5) lies entirely in B. (c) Rn(cid:3){0} for n≥2. Example 4.9. The following spaces are also connected. (a) Sn forn≥1,becauseitisaquotientspaceofRn+1(cid:3){0}byExample 3.27. (b) The n-torus Tn, because it is a product of connected spaces. On the other hand, path connectedness is stronger in general than con- nectedness. Here is an example of a space that is connected but not path connected. Example 4.10. Define subsets of the plane by A={(x,y):x=0 and y ∈[−1,1]}; B ={(x,y):y =sin(1/x) and x∈(0,1]}. Let X = A∪B (Figure 4.4). X is called the topologist’s sine curve. In Problem 4-5 you will show that it is connected but not path connected.

70 4. Connectedness and Compactness B A FIGURE 4.4. The topologists’s sine curve. Components and Path Components Look back at Example 4.1. Our first example of a disconnected set, the union of two disjoint closed disks, could be separated in only one way, becauseanyotherseparationwouldinduceaseparationofoneoftheclosed disks, which is path connected. The same reasoning applies to the second example, the x-axis minus the origin. The set of rational points in the plane,however,admitsinfinitelymanypossibleseparations.Identifyingthe possible separations of a space amounts to finding “maximal” connected subsets, a concept we now explore more fully. Let X be a topological space. Define a relation on X, called the connec- tivity relation, by saying that p∼q if there exists a connected subset of X containing both p and q. Lemma 4.11. The connectivity relation is an equivalence relation. Proof. It is reflexive because {q} is a connected subset containing q, and symmetric because p ∼ q and q ∼ p both mean that there is a connected subset containing p and q. To prove transitivity, suppose p∼q and q ∼r, which means that there are connected subsets A containing {p,q} and B containing {q,r}. Since A and B have the point q in common, A∪B is connected by Proposition 4.4(c). Thus A∪B is a connected set containing {p,r}, so p∼r. The equivalence classes in X under the connectivity relation are called the components of X.

Connectedness 71 Lemma 4.12. The components of X are exactly the maximal connected subsets of X, that is, connected sets that are not contained in any larger connected set. Proof. Given q ∈X, let A be the component of X containing q, and let B be the union of all connected sets containing q. Then B itself is connected by Proposition 4.4(c), and is thus a maximal connected subset. If p ∈ B, thenp,qlieintheconnectedsubsetB,sop∼qandthusp∈A.Conversely, if p∈A, then p∼q, so p lies in some connected subset containing q. Since B is the union of all such subsets, p∈B. Example 4.13. Consider the disconnected subsets of Example 4.1. (a) ThecomponentsofX (theunionoftwodisjointcloseddisks)arethe two disks themselves. (b) The components of Y (the x-axis minus the origin) are the positive x-axis and the negative x-axis. (c) InthesetZ ofpointswithrationalcoordinates,ifpandq aredistinct points of Z, they must differ in one of their coordinates, say their x-coordinates. Choosing an irrational number α between the two x- coordinates, the sets where x < α and x > α give a separation of Z in which p and q lie in different subsets. Therefore, p and q cannot both be contained in any connected subset, so p is not equivalent to q. Thus the components of Z are the one-point subsets. Proposition 4.14 (Properties of Components). Let X be any space. (a) Each component of X is closed in X. (b) Any connected subset of X is contained in a single component. Proof. If B is any component of X, it follows from Proposition 4.4(b) that B is a connected set containing B. Since components are maximal connected sets, B =B, so B is closed. Suppose A ⊂ X is connected. Since the components cover X, A has a point in common with some component B. By Proposition 4.4(c) A∪B is connected, so by maximality of B, it must be equal to B. This means that A⊂B. Althoughcomponentsarealwaysclosed,theymaynotbeopeningeneral, sotheydonotnecessarilyseparatethespace.ConsiderthesetZ ofrational points in the plane, for example: Its components are single points, which are not open sets. Wecanalsoapplytheconstructionusedtodefinecomponentswithpath connectedness in place of connectedness. Define the path connectivity rela- tion for points p,q in a space X by saying p ∼ q if there is a path in X p from p to q.

72 4. Connectedness and Compactness Exercise 4.1. Show that ∼ is an equivalence relation. p The equivalence classes under ∼ are called the path components of X. p Proposition 4.15 (Properties of Path Components). Let X be any space. (a) Each path component is contained in a single component, and each component is a disjoint union of path components. (b) If A ⊂ X is path connected, then A is contained in a single path component. Exercise 4.2. Prove Proposition 4.15. WesaythataspaceX islocallyconnectedifitadmitsabasisofconnected open sets, and locally path connected if it admits a basis of path connected open sets. To put it more concretely, for any p∈X and any neighborhood U of p, p has a (path) connected neighborhood contained in U. Clearly, any locally path connected space is locally connected. A space can be connected but not locally connected, as is, for example, thetopologist’ssinecurve(seeProblem4-5);anditcanbelocallyconnected but not connected, as is the disjoint union of two closed disks. Lemma 4.16. Let X be any space. (a) If X is locally connected, then each component of X is open. (b) If X is locally path connected, then each path component of X is open, its path components are the same as its components, and it is connected if and only if it is path connected. Proof. FirstassumethatX islocallyconnected,andletAbeacomponent of X. If p ∈ A, then p has a connected neighborhood U by local connect- edness, and this neighborhood must lie entirely in A by Lemma 4.14(b). Thus every point of A has a neighborhood in A; in other words, A is open. NowassumethatX islocallypathconnected.Thesameargument,with “connected” replaced by “path connected,” shows that each path compo- nent is open. Let q ∈ X, and let A be the component containing q, and B the path component. By Proposition 4.15(a), we know that B ⊂A and A can be written as a disjoint union of path components, each of which is open in X and thus in A. If B is not the only path component in A, then the pair {B,A(cid:3)B} is a separation of A, which is a contradiction because Aisconnected.ThisprovesthatA=B.Finally,X beingconnectedmeans it has only one component, which by the above argument is the same as havingonlyonepathcomponent,whichinturnisequivalenttobeingpath connected.

Compactness 73 Proposition 4.17. Every manifold is locally path connected. Exercise 4.3. Prove Proposition 4.17. This proposition means that in our work with manifolds we can use connectedness and path connectedness interchangeably. This will simplify many arguments because path connectedness is so much easier to check. Compactness Another fundamental fact about continuous functions is the extreme value theorem (Theorem A.10 in the Appendix): A continuous real-valued func- tiononaclosed,boundedsubsetofRn attainsitsmaximumandminimum values. Thistheorem,ofcourse,failsingeneralformetricspaces,and“bounded” doesnotevenmakesenseintopologicalspaces.Buttheessentialpropertyof closedandboundedsubsetsofRn thatmakestheproofwork,compactness, makessenseinarbitrarytopologicalspaces.Thispropertyisthesubjectof the rest of this chapter. Definitions and Basic Properties Recall that an open cover of a space X is a collection U of open subsets of X whose union is X, and a subcover of U is a subcollection of U that still covers X. A topological space X is said to be compact if every open cover of X has a finite subcover; or in other words, if given any open cover U of X,therearefinitelymanysetsU1,…,U k ∈UsuchthatX =U1 ∪···∪U k . As in the case of connectedness, when we say that a subset A of a topo- logical space X is compact, we always mean with respect to the subspace topology unless otherwise specified. A subspace A ⊂ X is compact if and only if given any collection of open subsets of X whose union contains A (which we also call an open cover of A), there is a finite subcover. Themostimportantfactaboutcompactspacesisthatcontinuousimages of compact spaces are compact. Theorem 4.18 (Main Theorem on Compactness). Let X,Y be topological spaces and let f: X →Y be a continuous map. If X is compact, then f(X) is compact. Proof. Let U be an open cover of f(X). (As noted in the remark above, we can take the elements of U either to be open subsets of f(X) in the subspacetopology,ortobeopensubsetsofY whoseunioncontainsf(X).) ForeachU ∈U,f−1(U)isanopensubsetofX.SinceUcoversf(X),every point of X is in some set f−1(U), so the collection {f−1(U):U ∈U} is an open cover of X. By compactness of X, some finite number of these, say

74 4. Connectedness and Compactness U p1 A p1 V p1 U q V p2 V p2 U p2 FIGURE 4.5. The case B ={q}. {f−1(U1),…,f−1(U k )}, cover X. Then it follows that {U1,…,U k } cover f(X). Proposition 4.19 (Properties of Compact Spaces). (a) Every closed subset of a compact space is compact. (b) In a Hausdorff space X, compact sets can be separated by open sets. That is, if A,B ⊂X are disjoint compact subsets, there exist disjoint open sets U,V ⊂X such that A⊂U and B ⊂V. (c) Every compact subset of a Hausdorff space is closed. (d) Every finite product of compact spaces is compact. (e) Every quotient of a compact space is compact. Proof. For part (a), let U be a cover of A by open subsets of X. Then U ∪ {X (cid:3) A} is an open cover of X, which has a finite subcover {U1,…,U k ,X(cid:3)A}. Therefore, A must be covered by the finite collection {U1,…,U k }. To prove (b), first consider the case in which B = {q} is a one-point set. For each p ∈ A, there exist disjoint open sets U containing p and V p p containing q by the Hausdorff property. The collection {U : p ∈ A} is an p open cover of A, so it has a finite subcover: Call it {U ,…,U } (Figure p1 pk 4.5). Let U = U ∪···∪U and V = V ∩···∩V . Then U and V are p1 pk p1 pk disjoint open sets with A⊂U and {q}⊂V, so this case is proved. Next consider the case of a general compact subset B. The argument above shows that for each q ∈B there exist disjoint open subsets U ,V ⊂ q q X such that A ⊂ U and q ∈ V . By compactness of B, finitely many of q q these, say {V ,…,V }, cover B. Then setting U =U ∩···∩U and q1 qm q1 qm V =V ∪···∪V proves the result. q1 qm

Compactness 75 Y U1 U2 X x U3 FIGURE 4.6. Finding a finite cover of the strip Zx ×Y. For (c), suppose X is Hausdorff and A ⊂ X is compact. For any point q ∈ X (cid:3)A, by part (b) there exist disjoint open sets U containing A and V containing q. In particular, V is a neighborhood of q disjoint from A, so every such q is exterior to A. This means that A is closed. Toprove(d),itsufficesbyinductiontoconsideraproductX×Y oftwo compact spaces. Let U be an open cover of X ×Y. Choose any x ∈ X. The “slice” {x}×Y is homeomorphic to Y, so finitely many of the sets of U cover it, say U1,…,U k (Figure 4.6). Because product open sets are a basis for the product topology, for each y ∈ Y there is a product open set V ×W ⊂ X ×Y such that (x,y) ∈ V ×W ⊂ U1 ∪···∪U k . Finitely many of these product sets cover {x}×Y, say V1 ×W1,…,V m ×W m . If we set Z x =V1 ∩···∩V m , then it is evident that the whole “strip” Z x ×Y is actually contained in U1 ∪···∪U k . Thus we have shown the following: For each x∈X, there exists an open subsetZ ⊂X suchthatZ ×Y iscoveredbyfinitelymanyofthesetsinU. x x The collection {Z :x∈X} is an open cover of X, which by compactness x hasafinitesubcover,say{Z ,…,Z }.SincefinitelymanysetsofUcover x1 xk eachstripZ ×Y,andfinitelymanysuchstripscoverX×Y,wearedone. xi Finally, part (e) is immediate from Theorem 4.18, since a quotient of a compact space is the image of a compact space by a continuous map. Part (d) is actually true in the more general context of infinite products (see [Sie92] or [Mun75]); in its general form, it is known as Tychonoff’s theorem.

76 4. Connectedness and Compactness Exercise 4.4. LetXbeacompactspace,andsuppose{Fn }isacountable collectionofnonemptyclosedsubse(cid:9)tsofXthatarenested,whichmeansthat Fn ⊃Fn+1 for each n. Show that n Fn is nonempty. One of the main applications of compactness is the following generaliza- tion of the extreme value theorem of elementary calculus. Theorem 4.20 (Extreme Value Theorem). If X is a compact space and f: X →R is continuous, then f is bounded and attains its maximum and minimum values on X. Proof. By the main theorem on compactness, f(X) is a compact subset of R,sobyPropositionA.6itisclosedandbounded.Inparticular,itcontains its supremum and infimum. The next lemma expresses an important property of compact metric spaces, which we will use frequently later in the book. Recall that the di- ameterofasetS inametricspaceisdefinedtobediam(S)=sup{d(x,y): x,y ∈S}. If U is an open cover of a metric space, a number δ >0 is called a Lebesgue number for the cover if any set whose diameter is less than δ is contained in one of the sets U ∈U. Lemma 4.21 (Lebesgue Number Lemma). Any open cover of a compact metric space has a Lebesgue number. Proof. LetUbeanopencoverofthecompactmetricspace M.Eachpoint x∈M is in some set U ∈U. Since U is open, there is some r(x)>0 such that B2r(x)(x) ⊂ U. The balls {B r(x)(x) : x ∈ M} form an open cover of M, so finitely many of them, say B r(x1)(x1),…,B r(xn )(x n ), cover M. Wewillshowthatδ =min{r(x1),…,r(x n )}isaLebesguenumberforU. To see why, suppose S ⊂M is a nonempty set whose diameter is less than δ. Let y be any point of S; then there is some x i such that y ∈ B r(xi )(x i ) (Figure 4.7). It suffices to show that any other point of S is in B2r(xi )(x i ), since the latter set is by construction contained in some U ∈ U. If z ∈ S, the triangle inequality gives d(z,x )≤d(z,y)+d(y,x )<δ+r(x )≤2r(x ), i i i i which proves the claim. Sequential and Limit Point Compactness The definition of compactness in terms of open covers lends itself to sim- ple proofs of some rather powerful theorems, but it does not convey much intuitive content. There are two other properties that are equivalent to compactness for manifolds and metric spaces (though not for arbitrary topological spaces), and that give a more vivid picture of what compact- ness really means. A space X is said to be limit point compact if every

Compactness 77 U 2r(x i ) x r(x ) i i S y z FIGURE 4.7. Proof of the Lebesgue number lemma. infinite subset of X has a limit point in X, and sequentially compact if every sequence of points in X has a subsequence that converges to a point in X. Proposition 4.22. Compactness implies limit point compactness. Proof. Suppose X is compact, and let S ⊂ X be an infinite subset. If S hasnolimitpoint,theneverypointx∈X hasaneighborhoodU suchthat U ∩S is either empty or {x}. Finitely many of these neighborhoods cover X.Butsinceeachsuchneighborhoodcontainsatmostonepointof S,this implies that S is finite, which is a contradiction. Problem 4-7 shows that the converse of this proposition is not true in general. Lemma 4.23. For first countable Hausdorff spaces, limit point compact- ness implies sequential compactness. Proof. Suppose X is first countable, Hausdorff, and limit point compact, and let {p } be any sequence of points in X. If the sequence takes on only n finitelymanyvalues,thenithasaconstantsubsequence,whichiscertainly convergent. So we may suppose it takes on infinitely many values. By hypothesis the set of values {p } has a limit point q ∈ X. If q is n actuallyequaltop forinfinitelymanyvaluesofn,againthereisaconstant n subsequence and we are done; so by discarding finitely many terms at the beginningofthesequenceifnecessarywemayassumep (cid:14)=qforalln.First n countability of X means that there is a countable neighborhood basis atq, say {B n : n = 1,2,…}. By replacing B n with B1 ∩···∩B n if necessary, we may assume that the neighborhood basis is nested: B1 ⊃B2 ⊃···. For

78 4. Connectedness and Compactness such a neighborhood basis, it is easy to see that any subsequence {p } ni such that p ∈B converges to q. ni i Since q is a limit point, we can choose n1 such that p n1 ∈ B1. Suppose by induction that we have chosen n1 < n2 < ··· < n k with p ni ∈ B i . By theHausdorffproperty,q hasaneighborhoodU disjointfromthefiniteset {p n : 1 ≤ n ≤ n k }, and by definition of limit point there is some n k+1 (necessarily greater than n k ) such that p nk+1 ∈U ∩B k+1. This completes the induction, and proves that there is a subsequence {p } converging to ni q. Thenextresultshowsthatformanifoldsandmostoftheotherspaceswe willbeconsideringinthisbook,wecanuseallthreenotionsofcompactness interchangeably. Proposition 4.24. For metric spaces and second countable Hausdorff spaces, compactness, limit point compactness, and sequential compactness are all equivalent. Proof. We have shown that compactness implies limit point compactness for all spaces, and limit point compactness implies sequential compactness for first countable Hausdorff spaces, which include both metric spaces and second countable Hausdorff spaces. So it remains to show that a metric space or second countable Hausdorff space that is sequentially compact is actually compact. Suppose first that X is second countable and sequentially compact. (For this part we do not need the Hausdorff property.) Any open cover U of X has a countable subcover {U : n = 1,2,…} by Lemma 2.15. Suppose n that no finite subcollection of U ’s covers X. This means that for each n n there exists q n ∈ X such that q n (cid:14)∈ U1 ∪···∪U n . By hypothesis, the sequence {q } has a convergent subsequence q → q. Now, q ∈ U for n nk m somembecausetheU ’scoverX,andthenconvergenceofthesubsequence n means that there exists some N such that q ∈ U whenever k ≥ N. nk m But by construction, q nk (cid:14)∈ U1 ∪ ··· ∪ U m as soon as n k ≥ m, which is a contradiction. This proves that finitely many of the U ’s cover X. n Therefore, second countable sequentially compact spaces are compact. Finally, let M be a sequentially compact metric space. We will show that M is second countable, which by the above argument implies that M iscompact.FromProblem2-15,itsufficestoshowthatM hasacountable dense subset. The key idea is to show first that sequential compactness implies the following weak form of compactness for metric spaces: For each ε>0, the opencoverofM consistingofallε-ballshasafinitesubcover.Supposethis is not true for some ε. Construct a sequence as follows. Let q1 ∈ M be arbitrary. Since B ε (q1) (cid:14)= M, there is a point q2 (cid:14)∈ B ε (q1). Similarly, since B ε (q1)∪B ε (q2) (cid:14)= M, there is a point q3 in neither of the two preceding ε-balls. Proceeding by induction, we construct a sequence {q } such that n

Compactness 79 for each n, q n+1 (cid:14)∈B ε (q1)∪···∪B ε (q n ). (4.1) Replacing this sequence by a convergent subsequence (which still satisfies (4.1)),wecanassumeq →q ∈M.SinceconvergentsequencesareCauchy, n as soon as n is large enough we have d(q n+1,q n ) < ε, which contradicts (4.1). Now, for each n let q (n) ,…,q (n) be finitely many points such that the 1 kn balls of radius 1/n around these points cover M. The collection of points {q (n)} is countable, and is easily seen to be dense. This shows that M is i second countable and completes the proof. Exercise 4.5. Show that every compact metric space is complete. The Closed Map Lemma The next lemma, though simple, is among the most useful results in this entire chapter. Lemma 4.25 (Closed Map Lemma). Suppose F is a continuous map from a compact space to a Hausdorff space. (a) F is a closed map. (b) If F is surjective, it is a quotient map. (c) If F is injective, it is a topological embedding. (d) If F is bijective, it is a homeomorphism. Proof. Let F: X → Y be such a map. If A ⊂ X is closed, it is com- pact, since any closed subset of a compact space is compact (Proposition 4.19(a)). Therefore, F(A) is compact by the main theorem on compact- ness, and closed in Y because compact subsets of Hausdorff spaces are closed(Proposition4.19(c)).ThisshowsthatF isaclosedmap.Ifinaddi- tion F is surjective, it is a quotient map by Lemma 3.19. If it is bijective, the fact that it is closed implies that its inverse is continuous, so it is a homeomorphism (Exercise 2.14). Finally, if F is injective, it is bijective onto its image, so the fact that it is an embedding follows from (d). Here are some immediate applications of the closed map lemma. In Example 3.24 we constructed a quotient space of the square I ×I by gluing the side boundary segments together and the top and bottom boundarysegmentstogether,andweclaimedthatitwashomeomorphicto thetorus.Hereisaproof.Constructanothermapq: I×I →T2 bysetting q(u,v)=(cos2πu,sin2πu,cos2πv,sin2πv).Bytheclosedmaplemma,this is a quotient map. Since it makes the same identifications as the quotient

80 4. Connectedness and Compactness mapwestartedwith,theoriginalquotientofI×I mustbehomeomorphic to the torus by the uniqueness of quotient spaces. InLemma3.15weshowedthatthedoughnutsurfaceishomeomorphicto the torus by a rather laborious explicit computation. Now that lemma can be proved much more simply, as follows. Consider the map F: R2 → R3 defined in Example 3.7. The restriction of this map to I ×I is a quotient map by the closed map lemma. Since it makes the same identifications as the map q in the preceding paragraph, the two quotient spaces D and S1 ×S1 are homeomorphic. (The homeomorphism is the map that sends q(u,v) to F(u,v).) Another application of the closed map lemma is the following useful result. You should notice how the closed map lemma, invoked twice in the proof, allows us to avoid ever having to prove continuity directly by ε-δ estimates. Proposition 4.26. Let K be a compact convex subset of Rn with nonempty interior. Then K is homeomorphic to the closed unit ball Bn, by a homeomorphism that sends Sn−1 to ∂K. Proof. Let q be an interior point of K. By replacing K with its image under the translation x (cid:10)→ x−q (which is a homeomorphism of Rn with itself), we can assume 0 ∈ IntK. Then there is some ε > 0 such that the ball B (0) is contained in K; using the dilation x (cid:10)→ x/ε, we can assume ε Bn =B1(0)⊂K. Thecoreoftheproofisthefollowingclaim:Eachraystartingattheorigin intersects∂K inexactlyonepoint.SinceK iscompact,itsintersectionwith each closed ray is compact; thus there is a point x0 in this intersection at which the distance to the origin assumes its maximum. This point is easily seen to lie in the boundary of K. To see that there can be only one such point, we will show that the line segment from 0 to x0 consists entirely of interior points of K, except for x0 itself. Since B1(0) ⊂ K, every line segment from x0 to a point y ∈B1(0) is contained in K. As y ranges over B1(0),theselinesegmentssweepoutasetC shapedlikeanicecreamcone (Figure 4.8). Around each point λx0 for 0≤λ<1, it is easy to check that there is a ball of radius 1−λ contained in C and hence in K. Thus x0 is the only boundary point of K on the ray. Now we define a map f: ∂K →Sn−1 by x f(x)= . |x| In words, f(x) is the point where the line segment from the origin to x intersects the sphere. Since f is the restriction of a continuous map, it is continuous, and the discussion in the preceding paragraph shows that it is bijective. Since ∂K is compact, f is a homeomorphism by the closed map lemma.

Locally Compact Hausdorff Spaces 81 K x0 C λx0 0 FIGURE 4.8. Proof that there is only one boundary point on a ray. Finally, define F: Bn →K by (cid:10) (cid:11) x F(x)=|x|f−1 . |x| Then F is continuous because f−1 is. Geometrically, F takes each radial line segment from 0 to a point ω ∈ Sn linearly onto the radial segment from 0 to the point f−1(ω) ∈ ∂K. By convexity, F takes its values in K. The map F is injective, since points on distinct rays are mapped to distinctrays,andeachradialsegmentismappedlinearlytoitsimage.Itis surjective because each point y ∈ K is on some ray from 0. By the closed map lemma, F is a homeomorphism. Locally Compact Hausdorff Spaces Compact Hausdorff spaces have many of the familiar properties of sub- setsofEuclideanspaces.However,whileallmanifoldsareHausdorff,many interestingmanifoldsarenotcompact.Nonetheless,manyoftheniceprop- erties of compact Hausdorff spaces carry over to a more general class of spaces, which we now define. AtopologicalspaceX issaidtobelocallycompactifforeveryq ∈X there isacompactsubsetofX containinganeighborhoodofq.Inthisgenerality, thedefinitionisnotparticularlyuseful,anddoesnotseemparalleltoother

82 4. Connectedness and Compactness definitions of what it means for a topological space to possess a property “locally,” which usually entails the existence of a basis of open sets with a particularproperty.ButwhencombinedwiththeHausdorffproperty,local compactness is much more useful. A subset A of a topological space X is said to be precompact or relatively compact if A is compact. Proposition 4.27. Let X be a Hausdorff space. The following are equiv- alent. (a) X is locally compact. (b) Each point of X has a precompact neighborhood. (c) X has a basis of precompact open sets. Proof. Clearly, (c) =⇒ (b) =⇒ (a), so all we have to prove is (a) =⇒ (c). It suffices to show that if X is locally compact Hausdorff, then each pointx∈X hasaneighborhoodbasisofprecompactopensets.LetK ⊂X beacompactsetcontaininganeighborhoodU ofx.ThecollectionVofall neighborhoods of x contained in U is clearly a neighborhood basis at x. BecauseX isHausdorff,K isclosedinX.IfV ∈V,thenV ⊂K (because V ⊂U ⊂K andK isclosed),andthereforeV iscompact(becauseaclosed subset of a compact set is compact). Thus V is the required neighborhood basis. Lemma 4.28 (Shrinking Lemma). Let X be a locally compact Haus- dorff space. If x ∈ X and U is any neighborhood of x, there exists a pre- compact neighborhood V of x such that V ⊂U. Proof. Suppose x∈X and U is a neighborhood of x. If W is any precom- pactneighborhoodofx,thenW(cid:3)U isclosedinW andthereforecompact. Because open sets separate compact sets in a Hausdorff space, there are disjoint open sets Y containing x and Y(cid:5) containing W (cid:3)U (Figure 4.9). Let V = Y ∩W. Because V ⊂ W, V is compact. Because V ⊂ Y, which is disjoint from Y(cid:5), we have V ⊂W (cid:3)Y(cid:5). Now the fact that W (cid:3)U ⊂Y(cid:5) means that W (cid:3)Y(cid:5) ⊂U, so V ⊂U. Lemma 4.29. Any open or closed subset of a locally compact Hausdorff space is locally compact Hausdorff. Proof. LetX bealocallycompactHausdorffspace.Notethatanysubspace ofX isHausdorff,soonlylocalcompactnessneedstobechecked.IfY ⊂X is open, the shrinking lemma says that any point in Y has a neighborhood whose closure is compact and contained in Y, so Y is locally compact. Suppose Z ⊂ X is closed. Any x ∈ Z has a precompact neighborhood U in X. Since U ∩Z = U ∩Z is a closed subset of the compact set U, it is compact, so U ∩Z is a precompact neighborhood of x in Z. Thus Z is locally compact.

Locally Compact Hausdorff Spaces 83 Y(cid:5) x Y W U FIGURE 4.9. Proof of the shrinking lemma. Exercise 4.6. Show that any finite product of locally compact Hausdorff spaces is locally compact Hausdorff. Example 4.30 (Locally Compact Hausdorff Spaces). (a) Euclidean space Rn is locally compact Hausdorff, because any closed ball B (x) is a precompact neighborhood of x. Thus every open or ε closed subset of Rn is locally compact Hausdorff. (b) Let M be a manifold, and let U be a cover of M by Euclidean balls. Each U ∈ U has a basis of open sets that are precompact in U and thusalsoinM,andtheunionofallsuchbasesisabasisforM.Thus any manifold is locally compact Hausdorff. The last example shows that every manifold has a basis of precompact open sets. For later use, we will need the following refinement of that fact. LetM beann-manifold.AEuclideanballB ⊂M iscalledregularifithas the following properties: (i) There is a Euclidean ball B(cid:5) ⊂M containing B. (ii) For some r >0, there is a chart ϕ: B(cid:5) →B2r (0)⊂Rn that sends B onto B (0). r Lemma 4.31. Every manifold has a countable basis of regular Euclidean balls. Proof. Let M be an n-manifold. Every point of M is contained in a Eu- clidean neighborhood, and since M is second countable, a countable col- lection U={U :i∈N} of such neighborhoods covers M by Lemma 2.15. i For each of these open sets U , choose a homeomorphism ϕ from U to an i i i open set U (cid:7) ⊂Rn. i Now let B be the collection of all open subsets of M of the form ϕ −1 (B (x)), where x ∈ U (cid:7) is a point with rational coordinates and r is i r i

84 4. Connectedness and Compactness (cid:7) U i M ϕ i 3r ϕ (q) U i 2r i x r q V FIGURE 4.10. Showing that the collection B is a basis. any positive rational number such that B2r (x) ⊂ U (cid:7) i . Since there are only countably many such balls for each U , the collection B is countable. For i any such set B =ϕ − i 1 (B r (x))∈B, let B(cid:5) =ϕ − i 1 (B2r (x)). To show that B isaregularball,weneedtoshowthatB ⊂B(cid:5) andϕ (B)=B (x).Wewill i r −1 −1 show,equivalently,thatϕ (B (x))=B.Now,ϕ (B (x))iscompactand i r i r thereforeclosedinM (becauseM isHausdorff),soB ⊂ϕ −1 (B (x))⊂B(cid:5). i r This means that the closure of B in M is equal to its closure in B(cid:5), which −1 is ϕ (B (x)), since ϕ is a homeomorphism. i r i To show that the collection B is a basis, it suffices to show that each opensubsetofM satisfiesthebasiscriterionwithrespecttoit.LetV ⊂M beanyopensubsetandq ∈V.Thenq ∈U forsomei,andϕ (V ∩U )isan i i i (cid:7) opensubsetofU containingϕ (q)(Figure4.10).Choosearationalnumber i i r >0 small enough that B3r (ϕ i (q))⊂ϕ i (V ∩U i ), and then choose a point x ∈ ϕ (V ∩U ) with rational coordinates such that |x−ϕ (q)| < r. Then i i i ϕ i (q) ∈ B r (x), and it follows from the triangle inequality that B2r (x) ⊂ ϕ (V ∩U ).Therefore,B =ϕ −1 (B (x))isinB,containsq,andiscontained i i i r in V, thus completing the proof. The closed map lemma is powerful, but it applies only when the domain is compact. The following proposition provides a useful generalization of the closed map lemma to noncompact spaces. A continuous map is said to be proper if the inverse image of each compact subset of Y is compact. Proposition 4.32. Suppose f: X → Y is a continuous map between lo- cally compact Hausdorff spaces. If f is proper, it is a closed map.

Locally Compact Hausdorff Spaces 85 Proof. Let K ⊂X be a closed set. We will show that f(K) contains all of its boundary points, which means that it is closed in Y. If y ∈ Y is a boundary point of f(K), let U be a precompact neigh- borhood of y. An easy verification shows that y is also a boundary point of f(K)∩U. Because f is proper, f−1(U) is compact, which implies that K∩f−1(U)iscompact.Bycontinuity, f(K∩f−1(U))=f(K)∩U iscom- pact and therefore closed in Y. In particular, y ∈ f(K)∩U ⊂ f(K), so f(K) is closed. Theorem 4.33 (Baire Category Theorem). In a locally compact Hausdorff space or a complete metric space, any countable collection of dense open subsets has dense intersection. Proof. Suppose {V n } n∈N is a countable collection of dense open subsets of suchaspaceX.Weneedto(cid:9)showthatifU ⊂X isanonemptyopensubset, the intersection of U with V is nonempty. n n FirstconsiderthecaseinwhichX islocallycompactHausdorff.SinceV1 isdense,U∩V1isnonempty,sobytheshrinkinglemmathereisanonempty precompact open set W1 such that W1 ⊂ U ∩ V1. Similarly, there is a nonemptyprecompactopensetW2 suchthatW2 ⊂W1 ∩V2 ⊂U∩V1 ∩V2. Continuingbyinduction,weobtainasequenceofnestednonemptycompact sets W1 ⊃ W2 ⊃ ··· ⊃ W n ⊃ ··· (cid:9)such that W n ⊂ U ∩V1 ∩···∩ (cid:9)V n . By Exercise 4.4, there is a point x∈ W , which is clearly in U ∩ V as n n n n well. In the case that X is a complete metric space, we modify the above proof as follows. At the inductive step, since W n−1 ∩ V n is open and nonempty,thereissomeballB (x )containedintheintersection.Choos- εn n ing r < min(ε ,1/n), we obtain a sequence of nested closed balls such n n that B rn (x n )⊂U ∩V1 ∩···∩V n . Because r n →0, the c(cid:9)enters {x n } form a Cauchy sequence, which converges to a point x∈U ∩ V . n n The Baire category theorem has a useful complementary reformulation. A subset F of a topological space X is said to be nowhere dense if its closure contains no nonempty open set. Corollary 4.34. In a locally compact Hausdorff space or a complete met- ricspace,anycountablecollectionofnowheredensesetshasemptyinterior. Proof. Let X be such a space, and let {F } be a countable collection of n nowhere dense subsets of X. Replacing each F by its closure, we may n assume that the sets are closed. Then their(cid:9)complements U n are open and d(cid:2)ense, so by t(cid:9)he Baire category theorem n U n is dense. It follows that F =X(cid:3) U cannot contain any nonempty open set. n n n n For example, it is easy to show that the solution set to any polynomial equation in two variables is nowhere dense in R2. Since there are only

86 4. Connectedness and Compactness countably many polynomials with rational coefficients, this corollary im- plies that there are points in the plane (a dense set of them, in fact) that satisfy no rational polynomial equation. The name of the theorem derives from the (astonishingly unedifying) terminology used by Baire: He defined a set of the first category to be a countable union of nowhere dense sets, and a set of the second category to be any set that is not of the first category. The theorem proved by Baire wasthatforspacessatisfyingthehypothesis,everyopensetisofthesecond category. Although the category terminology is mostly ignored nowadays, the name of the theorem has stuck. As we mentioned in Chapter 3, quotient maps do not generally behave well with respect to products. In particular, it is not always true that the product of two quotient maps is again a quotient map. However, it turns out that the product of a quotient map with the identity map of a locally compact Hausdorff space is indeed a quotient map, as the next lemma shows.ThiswillbeusedinChapter7;theproofisrathertechnicalandcan safely be skipped on first reading. Lemma 4.35. Suppose π: X → Y is a quotient map and K is a locally compact Hausdorff space. The map π×Id: X×K →Y ×K is a quotient map. Proof. We need to show that π × Id takes saturated open sets in X × K to open sets in Y × K. Let U ⊂ X × K be a saturated open set. Given (x0,k0) ∈ U, we will show that (x0,k0) has a saturated product neighborhoodW×J containedinU.Itthenfollowsthatπ(W)×J contains (π(x0),k0),iscontainedinπ×Id(U),andisopen(sinceπ(W)istheimage ofasaturatedopensetunderthequotientmapπ).Thusπ×Id(U)isopen in Y ×K. Now we proceed to prove the existence of the desired saturated prod- uct neighborhood. For any subset W ⊂ X, we define its saturation to be Sat(W)=π−1(π(W)); it is the smallest saturated subset containing W. By definition of the product topology, (x0,k0) has a product neighbor- hoodW0 ×J0 ⊂U.Bytheshrinkinglemma,thereisaprecompactneighbor- hood J of k0 such that J ⊂J0, and thus (x0,k0)∈W0 ×J ⊂W0 ×J0 ⊂U (Figure 4.11). Because U is saturated, it follows that Sat(W0)×J ⊂ U. Now,Sat(W0)×J isasaturatedsubsetofX×K,butnotnecessarilyopen (since π may not be an open map). We will show that there exists an open set W1 ⊂X containing Sat(W0) such that W1 × J ⊂ U. To prove this, fix some x ∈ Sat(W0). For any k ∈ J, (x,k) has a product neighborhood in U. Finitely many of these cover the compact set {x}×J; call them V1 ×J1,…,V m ×J m . If we set V x =V1 ∩···∩V m ,thenV x isaneighborhoodof{x}suchthatV x ×J ⊂U. Taking W1 to be the union of all such sets V x for x ∈ Sat(W0) proves the claim.

Locally Compact Hausdorff Spaces 87 K U J (x0,k0) W0 x X Sat(W0) W1 FIGURE 4.11. Finding a saturated product neighborhood. Repeating this construction, we obtain a sequence of open sets W ⊂X i such that W0 ⊂Sat(W0)⊂W1 ⊂Sat(W1)⊂··· and W ×J ⊂ U. Let W be the union of all the W ’s. Then W is open i i because it is a union of open sets, and W ×J ⊂ U. Moreover, W ×J is saturated: If (x,k) ∈ W ×J, then x is in some W ; and if (x(cid:5),k) is any i point in the same fiber, then x(cid:5) ∈ W i+1, so (x(cid:5),k) ∈ W ×J as well. Thus W ×J is the required saturated product neighborhood of (x0,k0).

88 4. Connectedness and Compactness Problems 4-1. (a) If U is any open subset of R and x ∈ U, show that U (cid:3){x} is disconnected. (b) Show that a topological space cannot be both a 1-manifold and an n-manifold for any n>1. 4-2. Show that the union of the x-axis and the y-axis in R2 is not a manifold in the subspace topology. 4-3. Showthatany n-manifoldisadisjointunionofcountablymanycon- nected n-manifolds. 4-4. Suppose f: X → Y is a surjective local homeomorphism. If X is locally connected, locally path connected, or locally compact, show that Y has the same property. 4-5. Let X be the topologist’s sine curve (Example 4.10). (a) Show that X is connected but not path connected or locally connected. (b) Determine the components and the path components of X. 4-6. Like Problem 3-7, this problem constructs a space that is locally Eu- clideanandHausdorffbutnotsecondcountable.Unlikethatexample, however, this one is connected. (a) Recallthatatotallyorderedsetissaidtobewell-orderedifevery nonempty subset has a smallest element (see the Appendix). Showthatthewell-orderingtheorem(TheoremA.2)impliesthat there exists an uncountable well-ordered set Y such that for every p∈Y, there are only countably many q <p. [Hint: Let X be any uncountable well-ordered set. If X does not satisfy the desired condition, let Y be an appropriate subset of X.] (b) Now let (cid:12) (cid:13) L= Y ×[0,1) (cid:3){(a0,0)}, where a0 is the smallest element of Y. We give L the dictionary order: This means that (p,q) < (r,s) if either p < r, or p = r and q < s. With the order topology, L is called the long line. Show that L is locally Euclidean and Hausdorff but not second countable. (c) Show that L is path connected. 4-7. Define a topology on Z by declaring a set A to be open if and only if n ∈ A implies −n ∈ A. Show that Z with this topology is second countable and limit point compact but not compact.

Problems 89 4-8. Let V be a finite-dimensional real vector space. A norm on V is a real-valued function on V, written v (cid:10)→|v|, satisfying

• Positivity: |v|≥0, and |v|=0 if and only if v =0.
• Homogeneity: |cv|=|c||v| for any c∈R and v ∈V.
• Triangle inequality: |v+w|≤|v|+|w|. Anormdeterminesametricbyd(v,w)=|v−w|.Showthatallnorms determine the same topology on V. [Hint: Consider the restriction of the norm to the unit sphere.] 4-9. SupposeK andLarecompactconvexsetsinRn,bothwithnonempty interior. Show that any continuous map f: ∂K →∂L has a continu- ous extension to a map F: K → L. If f is a homeomorphism, show that F can be chosen to be a homeomorphism also. 4-10. Let X be a noncompact, locally compact Hausdorff space. The one- point compactification of X is the topological space X∗ defined as follows.Let∞besomeobjectnotinX,andletX∗ =X(cid:20){∞}with the following topology: T ={open subsets of X} ∪{U ⊂X∗ :X∗(cid:3)U is a compact subset of X}. (a) Show that T is a topology. (b) Show that X∗ is a compact Hausdorff space. (c) Show that X is open and dense in X∗ and has the subspace topology. 4-11. If X and Y are noncompact, locally compact Hausdorff spaces, show that a continuous map f: X → Y extends to a continuous map f∗: X∗ →Y∗ if and only if it is proper. 4-12. Let σ : Sn (cid:3){N} → Rn be stereographic projection, as defined in Example 3.6. Show that σ extends to a homeomorphism of Sn with the one-point compactification of Rn. 4-13. If M is a noncompact n-manifold, show that its one-point compact- ification is an n-manifold if and only if there exists a precompact open subset U ⊂M such that M (cid:3)U is homeomorphic to Rn(cid:3)Bn. [Hint: You may find the inversion map I:Rn(cid:3)Bn →Bn defined by I(x)=x/|x|2 useful.] 4-14. Suppose M is a 1-dimensional manifold with boundary. Show that the interior and boundary of M are disjoint. Use this to conclude that M is a manifold if and only if ∂M =∅.

5 Simplicial Complexes In this chapter we give a brief introduction to simplicial complexes. These are spaces constructed from building blocks called simplices, which are points, line segments, filled-in triangles, solid tetrahedra, and their higher- dimensional analogues. They provide a highly useful way of constructing topological spaces, and play a fundamental role in geometry and algebraic topology. As we did with manifolds, we will define simplicial complexes in two stages, starting with a very concrete version and proceeding to the most general definition. Concretely, we think of a simplicial complex as a col- lection of simplices in some Euclidean space that overlap “nicely.” More abstractly, a simplicial complex is an abstract “vertex scheme,” specify- ing which sets of vertices are supposed to span simplices. We will see that any abstract simplicial complex determines a topological space, called a polyhedron, in a natural way. Then we apply these ideas to manifolds by asking which manifolds are homeomorphic to polyhedra. Any such homeomorphism is called a trian- gulation of the manifold, and any manifold that admits such a homeomor- phism is said to be triangulable. We will give a complete proof that every 1-manifold is triangulable, and will give a brief sketch of the proof for 2- manifolds.Theseresultswillbeusedinthenextchapterassteppingstones toward classifying curves and surfaces up to homeomorphism. Attheendofthechapterweexploretwocombinatorialpropertiesofsim- plicial complexes that are important in the study of manifolds. The first is the concept of an orientation of a complex, which generalizes and sys- tematizes the intuitive notions of “direction” in 1-dimensional complexes,

92 5. Simplicial Complexes FIGURE 5.1. Simplices. “clockwise” and “counterclockwise” in 2-dimensional ones, and “handed- ness” in 3-dimensional ones. The second is the Euler characteristic, which is the alternating sum of the numbers of simplices in different dimensions, and generalizes Euler’s classical formula for compact convex polyhedra in R3. Euclidean Simplicial Complexes We begin with a little linear algebra. An affine map between vector spaces is a map f: V → W of the form f(x) = a(x)+b, where a is a linear map andb∈W.Anaffinesubspaceofavectorspaceisthezerosetofsomeaffine map:{x:a(x)+b=0}.Itsdimensionisthedimensionofthekernelofthe linear part of the affine map. The special case of an affine subspace of V whosedimensionisonelessthanthatofV iscalledanaffine hyperplanein V.Elementarylinearalgebrashowsthatifn≥k,anyk+1pointsv0,…,v k in Rn are contained in some k-dimensional affine subspace (just choose a linearmapa: Rn →Rn−k whosekernelcontains{v1 −v0,…,v k −v0 }and letb=−a(v0)).Wesaythatk+1pointsareingeneral positioniftheyare not contained in any (k−1)-dimensional affine subspace, or equivalently if {v1 −v0,…,v k −v0 } are linearly independent. Given points v0,…,v k in general position in Rn, the simplex (plural: simplices) spanned by them is the set of all points in Rn of the form (cid:14)k (cid:14)k t v , where 0≤t ≤1 and t =1, (5.1) i i i i i=0 i=0 with the subspace topology. Each of the points v is called a vertex of the i simplex. We will sometimes use the notation (cid:22)v0,…,v k (cid:23) to denote the simplex spanned by v0,…,v k . The integer k (one less than the number of vertices)iscalleditsdimension,andak-dimensionalsimplexisoftencalled a k-simplex. A 0-simplex is a single point, a 1-simplex is a line segment, a 2-simplex is a (filled-in) triangle, and a 3-simplex is a solid tetrahedron (Figure 5.1). For any subset A ⊂ Rn, the convex hull of A is defined to be the inter- sectionofallconvexsetscontainingA.Itisimmediatethattheconvexhull is itself a convex set, in fact, the smallest convex set containing A.

Euclidean Simplicial Complexes 93 Lemma 5.1. A simplex is the convex hull of its vertices. Exercise 5.1. Prove Lemma 5.1. Let σ be a simplex. Each simplex spanned by a nonempty subset of the vertices is called a face of σ. The faces that are not equal to σ itself are called its proper faces. The 0-dimensional faces of σ are just its vertices, and the 1-dimensional faces are called its edges. The (k−1)-dimensional faces of a k-simplex are sometimes called its boundary faces. A map f: σ →τ between simplices is called a simplicial map if it is the restriction of an affine map that takes vertices of σ to vertices of τ. As the next exercise shows, simplicial maps between a given pair of simplices are in one-to-one correspondence with maps between their vertices. Exercise 5.2. (a) Show that given any map f0 from the set of vertices of σ to the set of vertices of τ, there is a unique simplicial map f: σ → τ whose restriction to the vertices of σ is f0. (b) Showthatanytwok-simplicesarehomeomorphicbyasimplicialhome- omorphism. (c) Show that every k-simplex is homeomorphic to Bk. [Hint: Work with a particular simplex in Rk and use Proposition 4.26.] It follows from part (c) of the preceding exercise that a k-simplex is a k-dimensional manifold with boundary. Thus we define the boundary of a simplex to be the union of its boundary faces (which is the same as the union of all of its proper faces), and its interior to be the simplex minus its boundary. The interior of a k-simplex(cid:15)is sometimes called an open k- simplex;itisthesetofpointsoftheform t i v i where{v0,…,v k }arethe vertices of σ and none of the coefficients t are zero. For example, if σ is a i 0-simplex, Intσ = σ, and if σ is a 1-simplex, Intσ is σ minus its vertices. Note that an open simplex is generally not an open subset of Rn, and the interiorandboundaryofσ asasimplexmaynotbeequaltoitstopological interior and boundary as a subset of Rn. A Euclidean simplicial complex is a collection K of simplices in some Euclidean space Rn satisfying the following conditions: (i) If σ ∈K, then every face of σ is in K. (ii) The intersection of any two simplices in K is either empty or a face of each. (iii) Local Finiteness: Every point in a simplex of K has a neighbor- hood that intersects at most finitely many simplices of K.

94 5. Simplicial Complexes FIGURE 5.2. A complex in R2. FIGURE 5.3. Not a complex. ThedimensionofK isdefinedtobethemaximumdimensionofanysimplex in K (which is well-defined, since simplices in Rn have dimension at most n). Figure 5.2 shows an example of a 2-dimensional simplicial complex in R2. The set of simplices shown in Figure 5.3 is not a simplicial complex, because the intersection condition is violated. Given a Euclidean complex K, the union of all the simplices in K, with thesubspacetopologyinheritedfromRn,isatopologicalspacedenotedby |K| and called the (Euclidean) polyhedron of K. Many of the spaces we have seen so far are homeomorphic to Euclidean polyhedra. Here are some simple examples. Example 5.2 (Euclidean Polyhedra). (a) Any n-simplex together with its faces is a simplicial complex whose polyhedron is homeomorphic to Bn. (b) The proper faces of an n-simplex constitute an (n−1)-dimensional complex whose polyhedron is homeomorphic to Sn−1. (c) The set of all unit-length intervals [n,n+1]⊂R for n∈Z, together with their endpoints, is a simplicial complex whose polyhedron is R. (d) For any integer m ≥ 3, let P be a regular m-sided polygon in the m plane. The set of edges and vertices of P is a simplicial complex m whose polyhedron is homeomorphic to S1. Example 5.3. Thesetofclosedlinesegmentsintheplanefromtheorigin to the points (1,1/n) for n ∈ N, together with their vertices (Figure 5.3), is not a simplicial complex, because the local finiteness condition fails at the origin. Exercise 5.3. Prove the claims made in the two preceding examples. You might wonder why we should focus on building spaces out of sim- plices, and not out of cubes or some other sort of geometric object. The simple answer is that simplicial complexes are the most general: It is not hardtoshowthatalocallyfinite“polyhedralcomplex”(underanyreason- able definition) can be subdivided to form a simplicial complex.

Euclidean Simplicial Complexes 95 FIGURE 5.4. Failure of local finiteness. Let K be a Euclidean simplicial complex. Any subset K(cid:5) ⊂ K that is itself a simplicial complex is called a subcomplex of K. It is clear that the only condition that needs to be checked is that the faces of each simplex in K(cid:5) are in K(cid:5). In particular, for any nonnegative integer k, the subset K(k) ⊂K consisting of all simplices of dimension less than or equal to k is a subcomplex, called the k-skeleton of K. Let K and L be two Euclidean simplicial complexes. A continuous map f: |K| → |L| whose restriction to each simplex of K is a simplicial map to a simplex of L is called a simplicial map, and is denoted by f: K →L. The restriction of f to K(0) is called the vertex map of f. Exercise 5.4. Let K and L be Euclidean simplicial complexes. (a) Let f0: K(0) → L(0) be any map with the property that whenever {v0,…,vk }aretheverticesofasimplexofK,{f0(v0),…,f0(vk)}are the vertices of a simplex of L (possibly with repetitions). Show that there is a unique simplicial map f: K →L whose vertex map is f0. (b) Now let f0 be as in (a), and assume in addition that f0 is bijective and {v0,…,vk } are the vertices of a simplex of K if and only if {f0(v0),…,f0(vk)} are the vertices of a simplex of L. Show that |K| and |L| are homeomorphic by a simplicial map. All the considerations of this section carry over without change if we replace Rn by an arbitrary finite-dimensional vector space V. We give V the metric topology induced by any norm; Problem 4-8 shows that the resulting topology is independent of the norm. The only properties of Rn that we use are its vector space structure and its topology, and since any choice of basis gives a linear homeomorphism of V with Rn, all the results of this section are true with Rn replaced by V. We will use this slightly more general setting in the next section.

96 5. Simplicial Complexes Abstract Simplicial Complexes Just as it is too restrictive to define manifolds to be subsets of Euclidean spaces,Euclideansimplicialcomplexesarenotsufficientlygeneralformany important applications. In this section we will define a more general kind of simplicial complex. The key idea is already implicit in Exercise 5.4(b), whichsaysthatasimplicialcomplexiscompletelydetermined,uptosimpli- cialhomeomorphism,byknowledgeofwhichsetsofverticesspansimplices. Motivated by this observation, we define an abstract simplicial complex to be a collection K of nonempty finite sets called (abstract) simplices, subject only to one condition: If σ ∈ K, then every nonempty subset of σ is in K. Any element of a simplex σ ∈ K is called a vertex of σ, and any nonemptysubsetofσiscalledafaceofσ.(Wemakenodistinctionbetween avertexv andthecorrespondingface{v}.)Todistinguishthesimpliceswe defined earlier (as convex subsets of some Euclidean space) from abstract simplices in this sense, we will sometimes refer to the former as Euclidean simplices. Thedimensionofanabstractsimplexconsistingofk+1verticesisdefined to be k. The dimension of K is the maximum dimension of any simplex in K, if it exists; if there are simplices of arbitrarily high dimensions, K is said to be infinite-dimensional. We say that K is a finite complex if K is a finite set, and locally finite if every vertex belongs to only finitely many simplices. AsubsetofKthatisitselfasimplicialcomplex(i.e.,thatcontainsallthe faces of each of its simplices) is called a subcomplex of K. The set K(k) of all simplices of dimension at most k is a k-dimensional subcomplex called the k-skeleton of K. Given two abstract complexes K,L, a map f: K→L is called a simpli- cial map if it is of the form f({v0,…,v k })={f0(v0),…,f0(v k )} for some map f0: K(0) → L(0), called the vertex map of f (which must have the property that {f(v0),…,f(v k )} ∈ L whenever {v0,…,v k } ∈ K). A sim- plicial map f is called an isomorphism if f0 is a bijection and {v0,…,v k } is a simplex of K if and only if {f0(v0),…,f0(v k )} is a simplex of L. One way of constructing an abstract simplicial complex, as you have probably already guessed, is the following. Given a Euclidean simplicial complex K, let K denote the set of all those finite subsets {v0,…,v k } ⊂ K(0) that consist of the vertices of some simplex of K. It is immediate that K is an abstract simplicial complex, called the vertex scheme of K. It is an immediate consequence of Exercise 5.4(b) that two Euclidean com- plexes are simplicially homeomorphic if and only if their vertex schemes are isomorphic. Exercise 5.5. Show that every finite abstract complex is the vertex scheme of a Euclidean simplicial complex. [Hint: Use basis vectors ei = (0,…,1,…,0) as vertices.]

Abstract Simplicial Complexes 97 Not all abstract simplicial complexes are vertex schemes of Euclidean complexes, however. Such an abstract complex must obviously be finite- dimensionalandlocallyfinite.Moreover,sincethelocalfinitenesscondition forces the vertex set of a Euclidean complex to be a discrete subset of Rn, its vertex scheme can have only countably many simplices. Problem 5-5 shows that these conditions are also sufficient. Thetheoryofquotientspacesgivesausefulwayofconstructingtopolog- ical spaces out of abstract complexes without these restrictions. The first step is to construct a canonical Euclidean k-simplex for each abstract k- simplex. Using equation (5.1) as(cid:15)a guide, we wish to think of our simplex as “a set of points of the form t v .” The trouble is that the vertices i i v of an abstract simplex are just abstract objects and not points in some i Euclideanspace,sothisexpressionnolongermakesliteralsenseasavector sum. Instead, we consider such a sum as a “formal linear combination” of theverticesv .(Theword“formal”isusedheretoindicatethattheexpres- i sion has the form of a linear combination, but may not actually represent addition of vectors in a vector space.) To make this precise, we introduce a bit of algebraic terminology. Given a set S, we wish to define a vector space whose elements we can think of as “formal linear combinations” of the elements of S. The main property of such a linear combination is that it is completely determined bythecoefficienttattachedtoeachv ∈S.Thusweareledtothefollowing definition:AformallinearcombinationofelementsofSisafunctiont: S → R such that t(v)=0 for all but finitely many v ∈S. Under the operations of pointwise addition and multiplication by constants, the set of all such functionsisavectorspace,denotedbyR(cid:22)S(cid:23)andcalledthefreevectorspace on S. Any element t∈R(cid:22)S(cid:23) can be represented symbolically as (cid:14)k t= t v , (5.2) i i i=0 where v are the (finitely many) elements of S for which t(v) (cid:14)= 0, and i t = t(v ). To be a bit more precise, each v ∈ S determines in a natural i i way a function from S to R, also denoted by v for simplicity, given by (cid:16) 1, w =v, v(w)= 0, w (cid:14)=v. Itiseasytocheckthateacht∈R(cid:22)S(cid:23)hasauniqueexpressionasafinitelin- earcombinationofthesefunctions;thisistheappropriatewaytointerpret (5.2). Now consider any abstract simplex {v0,…,v k }. We define its geometric realization to be the k-simplex (cid:22)v0,…,v k (cid:23) in the finite-dimensional vector space R(cid:22)v0,…,v k (cid:23). With the topology on R(cid:22)v0,…,v k (cid:23) induced by any norm,thisgeometricrealizationishomeomorphictoaEuclideank-simplex.

98 5. Simplicial Complexes π FIGURE 5.5. The quotient map defining the topology of |K|. Since each abstract simplex determines its geometric realization and vice versa, we will sometimes use the term “simplex” and the notation (cid:22)v0,…,v k (cid:23) interchangeably to refer either to an abstract simplex or to its geometric realization. When we need to distinguish between the two, we will use the notation |σ| for the geometric realization of σ. As in the Euclidean case, the open simplex Int|σ| is the subset of |σ| consisting of points all of whose coefficients t are nonzero. i ForanyabstractsimplicialcomplexK,let|K|denotethesetofallformal (cid:15) linear combinations of the form k i=0 t i v i with (cid:22)v (cid:15)0,…,v k (cid:23) a simplex of K and with coefficients satisfying 0 ≤ t ≤ 1 and k t = 1. This can be i i=0 i thought of abstractly as a subset of the free vector space R(cid:22)K(0)(cid:23) on the vertex set of K. More concretely, |K| is just the union of all the geometric realizations of the simplices of K, with points in two simplices identified whenevertheyhavethesameexpressionasli(cid:6)nearcombinationsofvertices. Wetopologize|K|inthefollowingway.Let |σ|bethedisjointunion σ∈K ofthegeometricrealizationsofallthesimplicesofK,withthedisjo(cid:6)intunion topology as in Problem 2-9 (this just means that a set is open in |σ| σ∈K if and only if its intersection with each |σ| is open in |σ|), and let (cid:17) π: |σ|→|K| σ∈K be the natural map that sends each simplex |σ| to itself (Figure 5.5). We give|K|thequotienttopologywithrespecttoπ.Unwindingthedefinitions,

Abstract Simplicial Complexes 99 thisisthesameassayingthatasubsetof|K|isopen(orclosed)ifandonly ifitsintersectionwitheachsimplexisopen(orclosed).Withthistopology, |K| is called the geometric realization of K. Thiswayofcharacterizingatopologyturnsouttobeofgreatimportance, soithasaname.Givenanycollection{S α } α∈A ofsubspacesofatopological space X whose union is X, the topology of X is said to be coherent with the subspaces S if a set is open in X if and only if its intersection with α each S is open in S . It is easy to check that this is equivalent to saying α α thatasetisclosedinX ifandonlyifitsintersectionwitheachS isclosed α in S . α Lemma 5.4. LetKbeanabstractsimplicialcomplexand|K|itsgeometric realization. (a) Each simplex |σ| is a closed, compact subset of |K|. (b) If dimK = n, then each open n-simplex Int|σ| is an open subset of |K|. (c) The topology of |K| is the unique topology coherent with the collection of subspaces {|σ|:σ ∈K}. (d) A map F: |K|→|L| is continuous if and only if its restriction to |σ| is continuous for each σ ∈K. Exercise 5.6. Prove Lemma 5.4. Any simplicial map f: K → L between abstract complexes induces in an obvious way a map |f|: |K|→|L|. (On each simplex |σ|, |f| is just the Euclidean simplicial map determined by the vertex map of f.) Since the restriction of |f| to each simplex is continuous, |f| is a continuous map. Lemma 5.5. Let K, L, and M be simplicial complexes. (a) If Id: K→KdenotestheidentitymapofK,then|Id|istheidentity map of |K|. (b) Iff: K→Landg: L→Maresimplicialmaps,then|g◦f|=|g|◦|f|. (c) Isomorphic complexes have homeomorphic geometric realizations. Exercise 5.7. Prove Lemma 5.5. Lemma 5.6. If K is the vertex scheme of a Euclidean simplicial complex K, then the geometric realization of K is homeomorphic to |K|. Proof. Recall that an abstract simplex σ ∈ K is just the set of vertices of some Euclidean simplex σ(cid:7) ∈K. Let π(cid:5) denote the natural map (cid:17) π(cid:5): |σ|→|K|, σ∈K

100 5. Simplicial Complexes which, restricted to each simplex |σ|, is the obvious simplicial homeomor- phism |σ|→σ(cid:7). This map makes the same identifications as the map π we used above to define the topology of |K|. If we can show that π(cid:5) is a quo- tient map, the lemma will follow from the uniqueness of quotient spaces. To show that it is a quotient map is the same as showing that |K| has the topology coherent with its simplices. To verify this, let G ⊂ |K| be an arbitrary subset. If G is closed, then clearly its intersection with any simplex is closed, because it is an inter- section of closed sets. Conversely, suppose the intersection of G with each simplex is closed. If x ∈ |K| is any limit point of G, by local finiteness x has a neighborhood U that intersects only finitely many simplices. Thus G∩U is the union of finitely many closed subsets of U and hence closed in U.Thisimpliesx∈G,soGisclosedin|K|.(Thisisthereasonweinsisted on local finiteness in the definition of Euclidean simplicial complexes.) Any topological space that is homeomorphic to the geometric realiza- tion of some simplicial complex is called a polyhedron. A particular such homeomorphismiscalledatriangulationofX.Anyspacethatadmitsatri- angulation (i.e., any polyhedron) is said to be triangulable. Sometimes one canobtainabetterunderstandingofthetopologyofanunknownspaceby firstshowingthatitistriangulable;thiswillbeourapproach,forexample, to the classification of 1-dimensional and 2-dimensional manifolds. Example 5.7. The following abstract complexes are isomorphic to the vertex schemes of the Euclidean complexes of Example 5.2, and therefore yield triangulations of the indicated spaces. (a) Thesetofallnonemptysubsetsof{0,1,2,…,n}isanabstractcom- plex whose geometric realization is homeomorphic to Bn. (b) Thesetofallpropernonemptysubsetsof{0,1,2,…,n}isanabstract complex whose geometric realization is homeomorphic to Sn−1. (c) Let K ∞ be the abstract complex consisting of 0-simplices {{n}:n∈ Z} and 1-simplices {{n,n+1} : n ∈ Z}. Its geometric realization is homeomorphic to R. (d) For any integer m ≥ 3, let K be the abstract complex whose m 0-simplices are {{1},{2},…,{m}}, and whose 1-simplices are {{1,2},{2,3},…,{m − 1,m},{m,1}}. Its geometric realization is homeomorphic to S1. Example 5.8 (Graphs). We define a graph to be a 1-dimensional poly- hedron with a given triangulation. (For some applications, it is useful to have a more general definition, allowing two edges to share more than one vertex,oroneedgetobeginandendatthesamevertex;butthiswillsuffice for our purposes. The kind of graph we have defined is sometimes called a

Abstract Simplicial Complexes 101 simplegraphtodistinguishitfromothermoregeneraltypes.Notethatthis use of the word graph has no relation to the graph of a function as defined in Chapter 3.) A subgraph of a graph is the polyhedron of a 1-dimensional subcomplex.Agraphissaidtobefiniteifitsassociatedsimplicialcomplex is finite. We will illustrate the utility of simplicial complexes by showing how a topological property of polyhedra—connectedness—can be detected by purely combinatorial means. Let K be a simplicial complex. An edge path inKisafiniteorinfinitesequenceofverticessuchthatanytwoconsecutive vertices span an edge. An edge path is said to bereduced if in addition any three consecutive vertices are all distinct. (The idea is that a reduced edge pathcontainsno“dead-endexcursions”likev,w,v.)WesaythatKisedge path connected if any two vertices can be joined by a finite edge path. Proposition 5.9. Let K be a simplicial complex. Then |K| is connected if and only if K is edge path connected, in which case any two vertices can be joined by a reduced edge path. Proof. SupposeKisedgepathconnected.Becausesimplicesareconnected, any two vertices that span an edge lie in the same component of |K|. It follows easily by induction that any two vertices joined by a finite edge path lie in the same component. Thus if K is edge path connected, all the verticeslieinthesamecomponentV0 of|K|.Anypointx∈|K|liesinsome simplex |σ|, and since |σ| contains at least one vertex in common with V0, it must be contained in V0. This shows that V0 =|K|, so |K| is connected. Conversely, suppose |K| is connected. Choose a vertex v ∈ K, and let C denote the subcomplex of K consisting of all vertices that are traversed in edge paths starting from v, together with all the simplices of K they span. If σ is a simplex of K that has a vertex w ∈ C, then every vertex w(cid:5) of σ must lie in C, because we can form an edge path from v to w(cid:5) by starting with an edge path to w and then appending w(cid:5) (since (cid:22)w,w(cid:5)(cid:23) is an edge of σ). Thus σ ∈ C as well. It follows that |C| is both open and closed in |K|,becauseitsintersectionwitheachsimplexiseitheremptyortheentire simplex. Thus C=K, which shows that K is edge path connected. Now suppose K is edge path connected. Given any two vertices v,v(cid:5) ∈ K, there is an edge path (v,…,v(cid:5)) connecting them. If this edge path is not reduced, it must have three consecutive vertices of the form w,w(cid:5),w for some pair of vertices w,w(cid:5) that span an edge. It is easy to see that the sequence obtained by replacing these three vertices with the single vertexw isstillanedgepathconnectingthesametwovertices.Repeatedly shorteningtheedgepathinthiswayuntilitisimpossibletoshortenitany more, we obtain a reduced edge path joining the same two vertices.

102 5. Simplicial Complexes Triangulation Theorems In the next chapter we will begin to study the problem of classifying man- ifolds up to homeomorphism. Our approach to classifying 1-manifolds and 2-manifolds will be to start with a triangulated manifold and study the combinatorialpropertiesofthetriangulation.Forthiswewillneedtoknow that all manifolds of dimensions 1 and 2 are triangulable. Theorem 5.10 (Triangulation Theorem for 1-Manifolds). Every1- manifold can be triangulated by a 1-dimensional simplicial complex. Proof. We begin by showing that there exists a sequence of compact sub- spaces G ⊂ M, n = 1,2,…, whose union is M, satisfying the following n conditions: (i) Each G is a finite graph. n (ii) For each n, G n is a subgraph of G n+1. (iii) For each n there exists m>n such that G ⊂IntG . n m By Lemma 4.31, M admits a countable cover {B } by regular Euclidean i balls. From the definition of regular balls, it is evident that the closure of eachregularballishomeomorphictoa1-simplexwhosetopologicalinterior in M is equal to its interior as a simplex. Begin by letting G1 be the graph consisting of the single 1-simplex B1 and its vertices. Now let n > 1, and assume by induction that we have found finite graphs G1 ⊂ G2 ⊂ ··· ⊂ G n satisfying (i) and (ii) with G n

B1 ∪···∪ B n . Consider the next 1-simplex B n+1. Some of the vertices of G n may lie in B n+1 (the interior of B n+1); the ones that do define a subdivisionofB n+1 intoafinitegraphS,withthepropertythatnovertex of G lies in the interior of any edge of S. n For each of the edges e ⊂ S, we will prove the following claim: Either e intersectseachoftheedgesofG onlyatvertices,oreisentirelycontained n in one of the edges of G . To prove this, suppose e has an interior point n that lies in some edge e(cid:5) ⊂G (Figure 5.6). By the remark above, it must n be an interior point of e(cid:5) as well. NotethatInte∩Inte(cid:5) isopeninInte.Ontheotherhand,e(cid:5) isacompact subset of the Hausdorff space M, so it is closed in M, and therefore Inte∩ Inte(cid:5) =Inte∩e(cid:5) isclosedinInte.Byconnectedness,therefore,Inte∩Inte(cid:5) isallofInte.Inotherwords,Inte⊂Inte(cid:5),whichimpliese⊂e(cid:5) andproves the claim. Now simply throw away those edges of S that are contained in G , and n redefine S to be the graph consisting of the remaining edges and their vertices. Let G n+1 =G n ∪S. We wish to show that G n+1 is a finite graph containing G as a subgraph. The edges of S intersect each of the edges n of G only at vertices; but it may happen that both vertices of some edge n

Triangulation Theorems 103 G n e(cid:5) e S FIGURE 5.6. Proof that e⊂e(cid:3). e in S intersect a single edge e(cid:5) in G . If so, simply subdivide e into two n edges by adding a new vertex in its interior. With this modification, each of the edges of S intersects each edge of G at most in a single vertex. n Since G n+1 has only finitely many simplices, its topology is coherent with the simplices by Problem 5-1. Thus the foregoing argument proves that G n+1 is a finite graph whose polyhedron is B1 ∪···∪B n+1 and that containsG asasubgraph.Continuingbyinduction,weobtainanincreas- n ing sequence {G : n = 1,2,…} of graphs such that every point of M is n contained in G for some n. Since the interiors of the Euclidean balls B n i cover M, for each n there is some m>n such that the compact set G is n covered by B1 ∪···∪B m , and therefore G n ⊂IntG m . This completes the induction. Let K be the abstract simplicial complex whose 0-skeleton is the union of the 0-skeletons of G for all n, and whose 1-simplices are the pairs n {v,v(cid:5)} that span an edge in some G . There is an obvious bijective map n |K| → M, defined by choosing a homeomorphism from each 1-simplex of |K| onto the corresponding edge in M. To see that this map is a homeo- morphism, we need only show that M has the topology coherent with the simplices. Clearly, any set that is closed in M has closed intersection with eachsimplex,becausethesimpliceshavethesubspacetopology.Conversely, suppose K ⊂M is a subset whose intersection with each simplex is closed. If x ∈ M is a limit point of K, choose n large enough that x ∈ IntG . n SincetheintersectionofK witheachofthe(finitelymany)simplicesofG n isclosedinG ,itfollowsthatK∩IntG isclosedinIntG .Inparticular, n n n x∈K, which proves that K is closed in M and thus the topology of M is coherent with the simplices. For use in the next chapter, we will need the following property of tri- angulated 1-manifolds. Proposition 5.11. If K is a simplicial complex whose geometric realiza- tion is a 1-manifold, each vertex of K lies on exactly two edges.

104 5. Simplicial Complexes Proof. Let v be any vertex, and let V be the union of {v} together with the interiors of all the edges that have v as a vertex. Since the intersection of V with each simplex is open in the simplex, V is open in |K|. Because|K|isa1-manifold,v hasaneighborhoodU ⊂V homeomorphic to an open interval. It follows that U (cid:3){v} has exactly two components. Foreachedgeecontainingv,Inte∩(U(cid:3){v})isanopensubsetofU(cid:3){v}. These sets are disjoint (because all the edges have disjoint interiors), and nonempty(because∩U isnonemptyandopenineandthusmustcontain some interior points of e). Therefore, if v lies on more than two edges, we have a separation of U (cid:3){v} into more than two nonempty disjoint open subsets,contradictingthefactthatithasonlytwocomponents.Thisshows that each vertex lies on at most two edges. On the other hand, if some vertex v lies on only one edge e, the con- struction above shows that v has a Euclidean neighborhood U contained entirely in e. This means that v has a neighborhood Y ⊂ U such that Y (cid:3){v} is connected—just take Y to be the image of [0,ε) under some homeomorphism ϕ: [0,1] → e taking 0 to v. But any Euclidean neighbor- hood minus a point is disconnected by Problem 4-1(a), so there can be no such vertex. We turn our attention next to 2-manifolds. The following theorem was proved by Tibor Rad´o [Rad25] in 1925. Theorem 5.12 (Triangulation Theorem for Surfaces). Every 2- manifold admits a triangulation by a 2-dimensional simplicial complex, in which each edge lies on exactly two 2-simplices. Sketch of proof. The basic approach is analogous to the proof of triangu- lability of 1-manifolds: Cover the manifold with countably many regular disks, and inductively show that each successive disk can be triangulated in a way that is compatible with the triangulations that have already been defined,sothatthemanifoldisultimatelywrittenasanincreasingunionof polyhedra. In the case of surfaces, however, finding a triangulation of each successivediskthatiscompatiblewiththepreviousonesismuchmoredif- ficult, primarily because the boundary of the new disk might intersect the boundaries of the already-defined simplices infinitely many times. Even if thereareonlyfinitelymanyintersections,showingthattheregionsdefined by the intersecting curves are homeomorphic to closed disks, and therefore triangulable, requires a delicate topological result known as the Scho¨nflies theorem,whichassertsthatanytopologicalembeddingofthecircleintoR2 extends to an embedding of the closed disk. The details of the proof are long and intricate and would take us too far from our main goals, so we leave it to the reader to look it up. A readable presentation can be found in [Moi77].

Orientations 105 v1 v2 v3 v4 v4 v5 v6 v1 FIGURE 5.7. The Mo¨bius band. Finally,althoughwewillnotuseit,wementionthefollowingmorerecent result, proved by Edwin Moise in 1977 [Moi77]. Theorem 5.13 (Triangulation Theorem for 3-Manifolds). Every 3-manifold is triangulable. Beyonddimension3,mattersarenotnearlysonice.Ithasrecentlybeen shown that there are manifolds of dimension 4 that admit no triangula- tions; and it is still not known whether all manifolds of dimension greater than 4 can be triangulated. See [Ran96] for a history of the subject of triangulations and a summary of the current state of the art. Orientations The Mo¨bius band is the famous topological space obtained by identifying two edges of the square I ×I according to the relation (0,t) ∼ (1,1−t) (Figure 5.7). It is a manifold with boundary (though not a manifold), and it is triangulable (one triangulation is shown in Figure 5.7). If you have ever made a paper model (it is best to start with a long, narrow rectangle instead of a square), you have undoubtedly noticed that it has the curious property that it is impossible to consistently pick out which is the “front” side and which is the “back”—you cannot continuously color one side gray and the other side white. By using simplicial theory, we can make this notion precise and extend it to complexes of other dimensions as well. Instead of choosing which side of each triangle to call the front, we will, in effect, choose which direction of travel around the vertices to consider “counterclockwise.” Let σ be an abstract k-simplex. Given any two orderings (v ,…,v ) i0 ik and (v ,…,v ) of the vertices of σ, there is a permutation s of the set j0 jk {0,…,k} such that s(i )=j for p=0,…,k. Define an equivalence rela- p p tion on the set of all orderings by saying that two orderings are equivalent if they differ by an even permutation (see the Appendix). A choice of an equivalence class of vertex orderings is called an orientation of σ. For ex- ample, an orientation of a 1-simplex is just a choice of initial and terminal vertices, which can be indicated schematically by drawing an arrow along

106 5. Simplicial Complexes v3 v2 v1 v0 v0 v0 v1 v1 v2 FIGURE 5.8. Orientations of simplices. the simplex (see Figure 5.8). An orientation of a 2-simplex is a choice of a preferred direction of rotation, which can be indicated by a circular arrow; andanorientationofa3-simplexisachoiceof“handedness”:Thepreferred hand is the one whose fingers curl around the first three vertices in order while the thumb points toward the fourth. Since there is only one way to order a single vertex, by convention an orientation for a 0-simplex is just a choice of a plus or minus sign. An oriented simplex is a simplex together with a choice of orientation. We will write [v0,…,v k ] for the k-simplex (cid:22)v0,…,v k (cid:23) oriented by the vertex ordering (v0,…,v k ), and we will let −[v0,…,v k ] denote the same simplex with the opposite orientation. Thus, for example, for 1-simplices and 2-simplices we have [v,w]=−[w,v], [v,w,x]=[w,x,v]=[x,v,w]=−[v,x,w]=−[x,w,v]=−[w,v,x]. Any n-simplex in Rn automatically gets an orientation, which we call the natural orientation, by declaring [v0,…,v n ] to be oriented if and only ifdet(v1 −v0,v2 −v0,…,v n −v0)>0.Toseethatthisiswell-defined,first note that the n vectors {v1 −v0,…,v n −v0 } are independent precisely when the vertices {v0,…,v n } are in general position. Interchanging two vertices other than v0 has the same effect as interchanging two rows of the determinant, which changes its sign. If v0 is interchanged with another vertex v i , the determinant becomes det(v1 −v i ,…,v0 −v i ,…,v n −v i ); multiplyingtheithrowby−1(whichchangesthesignofthedeterminant) and then adding the ith row to each other row (which leaves the deter- minant unchanged) transforms the new determinant back to the original one. Thus a transposition of two vertices always changes the sign of the determinant, so an arbitrary permutation of the vertices changes the sign ofthedeterminantifandonlyifitiseven,whichshowsthatthisrulegives a well-defined orientation. Geometrically, for a 1-simplex in R the natural orientationisfromthesmallertothelargervertex;fora2-simplexin R2 it

Orientations 107 − + FIGURE 5.9. Induced orientations of boundary faces. is the counterclockwise direction of rotation; and for a 3-simplex in R3 it is the right-handed orientation. (These statements can be taken as mathe- matical definitions of the terms “counterclockwise” and “right-handed.”) If σ = [v0,…,v k ] is an oriented k-simplex, the orientation of σ deter- mines an orientation on each of its boundary faces (i.e., faces of dimension k−1), called the induced orientation, by the following rule: The induced orientation on the face τ i = (cid:22)v0,…,v(cid:18) i ,…,v k (cid:23) (where the hat indicates that v i is omitted) is defined to be (−1)i[v0,…,v(cid:18) i ,…,v k ]. To check that thisiswell-defined,weneedtoshowthattheinducedorientationofτ isun- i changed if the vertices of σ are subjected to an even permutation. Because everypermutationcanbewrittenasacompositionoftranspositionsofad- jacentvertices(seeExerciseA.19intheAppendix),itsufficestoshowthat the induced orientation is reversed if two adjacent vertices of σ are trans- posed. This is clear if neither of the vertices is v . If v is transposed with i i v i±1, the induced orientation becomes (−1)i±1[v0,…,v i−1,v i+1,…,v k ], which is the opposite of what it was originally. For an oriented 1-simplex [v0,v1], the induced orientation gives a minus signtotheinitialvertexv0 andaplussigntotheterminalvertexv1 (Figure 5.9). For an oriented 2-simplex [v0,v1,v2], the induced orientations on the edges are [v1,v2], −[v0,v2] = [v2,v0], and [v0,v1]. Thus the arrow on each edge points in the preferred direction of rotation. Now suppose K is an n-dimensional simplicial complex in which every (n−1)-simplexisafaceofnomorethantwon-simplices.(Itcanbeshown, though we will not do so, that any triangulated manifold has this form.) If σ and σ(cid:5) are two n-simplices that share a boundary face τ, we say that orientations of σ and σ(cid:5) are consistent if they induce opposite orientations on τ. An orientation of K is a choice of orientation of each n-simplex in such a way that any two simplices that intersect in an (n−1)-face are consistentlyoriented.Figure5.10givesschematicindicationsoforientations of 1-dimensional and 2-dimensional complexes. If a complex K admits an orientation, it is said to be orientable. Example 5.14. ThetriangulationoftheMo¨biusbandshowninFigure5.7 isnotorientable.Toseewhy,supposethereexistsanorientation.Reversing the orientations of all the 2-simplices if necessary, we may assume that the leftmost triangle is oriented as [v1,v4,v5] (i.e., in the counterclockwise

108 5. Simplicial Complexes FIGURE 5.10. Orientations of simplicial complexes. direction). Then the consistency condition implies that the next simplex is oriented as [v1,v5,v2]. Similarly, each of the succeeding simplices must be oriented in the counterclockwise direction. But then both the leftmost and the rightmost 2-simplices induce the same orientation [v1,v4] on their commonedge,whichcontradictstheconsistencycondition.Therefore,there exists no orientation. Example 5.15. Letσ =(cid:22)v0,…,v n+1 (cid:23)bean(n+1)-simplexinRn+1,and let K be the set of proper faces of σ, which is a triangulation of Sn. Give σ the natural orientation inherited from Rn+1, and give the n-simplices of K the induced orientation. Each (n−1)-simplex of K is of the form τ = (cid:22)v0,…,v(cid:18) i ,…,v(cid:18) j ,…,v n+1 (cid:23), and belongs to two n-simplices: the one oppositev andtheoneoppositev .Itiseasytocheckthattheorientations i j induced on τ by these two faces are opposite, because in one case v is i removed first and then v , while in the other case the order is reversed. j Thus we have produced an orientation of K. Proposition 5.16. LetK beanyn-dimensionalEuclideancomplexinRn. The natural orientation of each n-simplex determines an orientation of K. Proof. First we show that no more than two n-simplices in K can have an (n−1)-face in common. Let τ be an (n−1)-simplex in K. The vertices of τ determine a unique affine hyperplane (i.e., (n−1)-dimensional affine subspace) in Rn, whose complement has exactly two components, which we call the sides of τ. Because the vertices of σ are in general position, the additionalvertexofσ thatisnotinτ mustlieononesideofτ ortheother, and therefore all of σ(cid:3)τ must lie on the same side (since it is connected). For any x ∈ Intτ and any sufficiently small ε > 0, B (x) (cid:3) τ has two ε components, one lying on each side of τ. Any n-simplex that contains τ must contain exactly one of these components for ε small enough. Thus if more than two n-simplices contain τ, two of them must contain the same

Combinatorial Invariants 109 component of B (x) (cid:3) τ and therefore have interior points in common, ε which contradicts the definition of a Euclidean complex. Now we must show that the natural orientations of any two sim- plices σ,σ(cid:5) ∈ K that share an (n − 1)-face τ are consistent. Write τ =(cid:22)v0,…,v n−1 (cid:23), σ =(cid:22)v0,…,v n−1,v n (cid:23), and σ(cid:5) =(cid:22)v0,…,v n−1,v n (cid:5)(cid:23). The function f(v) = det(v1 −v0,…,v −v0) is an affine function of v that is zero precisely when v lies in the affine hyperplane determined by τ, so it must be positive on one side of τ and negative on the other side. Since the argument above implies that v and v(cid:5) lie on opposite sides n n of τ, it follows that det(v1 −v0,…,v n (cid:5) −v0) has the opposite sign from det(v1 − v0,…,v n − v0). If the vertices have been ordered so that the natural orientation of σ is [v0,…,v n ], then the natural orientation of σ(cid:5) is −[v0,…,v n (cid:5)], and it is immediate that they induce opposite orientations on τ. Combinatorial Invariants Simplicial complexes were invented in the hope that they would enable topologicalquestionsaboutmanifoldstobereducedtocombinatorialques- tions about simplicial complexes. To make sense of this, we need a notion ofequivalenceofcomplexesthatisweakerthansimplicialisomorphismbut strong enough to imply that they have homeomorphic geometric realiza- tions, and that can be detected purely from the combinatorial structure of the abstract complexes. The most natural way to modify a simplicial complex to obtain another one with a homeomorphic geometric realization is to “subdivide” the sim- plices of the original complex into smaller ones. We can then consider two complexestobeequivalentiftheybothhaveacommonsubdivision.Inthis section we make this notion precise, and study one important property of complexes that is preserved by this kind of equivalence. For our purposes, it is sufficient and simpler to restrict our attention to finite complexes, although many of the definitions can be extended to the general case. Let K be a finite Euclidean simplicial complex. A subdivision of K is a simplicial complex K(cid:5) with the following properties:

• |K(cid:5)|=|K|.
• Each simplex of K(cid:5) is contained in a simplex of K.
• Each simplex of K is a finite union of simplices of K(cid:5). Some examples of subdivisions are shown in Figure 5.11. Example 5.17 (Barycentric Subdivision). Aparticularlyusefulkind of subdivision is obtained in the following way. Let σ = (cid:22)v0,…,v k (cid:23) be a

110 5. Simplicial Complexes K K(cid:5) M M(cid:5) L L(cid:5) N N(cid:5) FIGURE 5.11. Subdivisions. Euclidean k-simplex in Rn. If v is a point not in the k-dimensional affine subspace determined by σ, we define v∗σ =(cid:22)v,v0,…,v k (cid:23). This is a (k+1)-simplex, called the cone on σ from v. Now let K be a finite Euclidean complex. For each k-simplex σ = (cid:22)v0,…,v k (cid:23)∈K, define the barycenter of σ to be the point (cid:14)k 1 b = v ∈Intσ. σ k+1 i i=0 It is the “center of gravity” of the vertices of σ. (The name comes from Greekbarys,meaning“heavy.”)Forexample,thebarycenterofa1-simplex is just its midpoint; the barycenter of a vertex v is v itself. We will define a complex SK, called the barycentric subdivision of K, whose vertices are the barycenters of all the simplices in K. It is easiest to define by induction on the dimension of K. If dimK = 0, then we set SK =K (you cannot subdivide a point!). Assuming that we have defined SK for all finite complexes of dimension less than n, we define SK for a complex of dimension n as the union of S(K(n−1)) with the set of all simplices of the form b ∗τ where σ is an n-simplex of K and τ is any σ simplex of S(K(n−1)) contained in a face of σ. It is straightforward to check that SK is indeed a subdivision of K (see Problem 5-7). Examples of barycentric subdivisions are pictured in Figure 5.12. The key fact about barycentric subdivision is that it reduces the sizes of all the simplices by a uniform ratio, as the following lemma shows. Lemma 5.18. If σ is a Euclidean k-simplex in Rm, the diameter of each simplex in the barycentric subdivision of σ is at most k/(k+1) times that of σ.

Combinatorial Invariants 111 K SK S(SK) FIGURE 5.12. Barycentric subdivisions. Proof. Note first that for any point x ∈ σ, the maximum of the function |x − y| for y ∈ σ is achieved when y is a vertex. To see why, let R be the maximum distance from x to any vertex of σ; since σ is the convex hull of its vertices and the closed ball B (x) is a convex set containing the R vertices, σ ⊂ B (x), which proves the claim. It follows immediately that R the diameter of σ is the maximum of the distances between its vertices. The following computation shows that the distance from the barycenter b τ ofaq-simplexτ =(cid:22)v0,…,v q (cid:23)toanyofitsverticesv j isatmostq/(q+1) times the diameter of τ: (cid:19) (cid:19) (cid:19)(cid:14)q (cid:19) (cid:19) 1 (cid:19) |b −v |=(cid:19) v −v (cid:19) τ j (cid:19) q+1 i j(cid:19) (cid:19)i=0 (cid:19) (cid:19)(cid:14)q (cid:14)q (cid:19) (cid:19) 1 1 (cid:19) =(cid:19) v − v (cid:19) (cid:19) q+1 i q+1 j(cid:19) i=0 i=0 (cid:14)q 1 ≤ |v −v | q+1 i j i=0 q ≤ diamτ. q+1 Now if σ(cid:5) is any face of the barycentric subdivision of σ and w1, w2 are any two vertices of σ(cid:5), by Problem 5-7 each w is the barycenter of a k - j j dimensional face τ j of σ, and we may assume that τ1 is a face of τ2. By the computation above, the distance from w2 to any point of τ2 is at most k2/(k2+1) times the diameter of τ2. Since w1, in particular, is a point in τ2, we have |w1 −w2 |≤ k2 diamτ2 ≤ k diamσ. k2+1 k+1

112 5. Simplicial Complexes It follows from the remark at the beginning of the proof that diamσ(cid:5) sat- isfies the same inequality. To express subdivisions in terms of the combinatorics of abstract com- plexes, we will show how to decompose an arbitrary subdivision into a sequence of subdivisions that are combinatorially simpler. Suppose K(cid:5) is a subdivision of K. We say that it is an elementary subdivision if K(cid:5) con- tains precisely one more vertex than K. (For example, the 3-dimensional complex M(cid:5) in Figure 5.11 is an elementary subdivision ofM, obtained by adding one vertex in the bottom face of M.) Suppose we start with a finite Euclidean complex K and choose a k- simplex σ = (cid:22)v0,…,v k (cid:23) ∈ K, choose a point v ∈ Intσ, and replace each simplex (cid:22)v0,…,v k ,w1,…,w m (cid:23) that has σ as a face (including σ itself) by the set of all simplices of the form (cid:22)v,v i1 ,…,v ij ,w1,…,w m (cid:23) as { ch v e i1 c , k .. th ., a v t ij K } (cid:5) r i a s n a g n es el o e v m e e r n p t r a o r p y e s r ub su d b iv se is t i s on of o { f v K 0, , . a . n . d ,v t k h } a . t T e h v e e n ry it el i e s m e e a n s t y ar t y o subdivision is of this form. Moreover, if K(cid:5)(cid:5) is any subdivision of K, there is a finite sequence K = K0,K1,…,K m = K(cid:5)(cid:5) of complexes such that K i+1 is an elementary subdivision of K i . One advantage of working with elementary subdivisions is that the effect of an elementary subdivision on the vertex scheme of K is clearly determined solely by the choice of σ, and soelementarysubdivisionscanbedefinedbytherecipeaboveforarbitrary abstract complexes as well. Twofinitesimplicialcomplexesaresaidtobe combinatorially equivalent if they become isomorphic after finitely many elementary subdivisions. It was conjectured by Ernst Steinitz and Heinrich Tietze in 1908 that if two finite simplicial complexes have homeomorphic polyhedra, they are com- binatorially equivalent; this conjecture became known as the Hauptvermu- tung (main conjecture) of combinatorial topology. It is now known to be true for all finite complexes of dimension 2 and for triangulated compact manifolds of dimension 3, but false in all higher dimensions. (See [Ran96] for a nice discussion of the history of this problem.) Thus the hope of re- ducing topological questions about manifolds to combinatorial ones about simplicial complexes has not been realized. Nonetheless, simplicial theory hasprovideduswithanumberofextremelyusefulcombinatorialinvariants thatturnouttohaveimportanttopologicalramifications.Weconcludethis chapterwithanintroductiontooneofthem,calledtheEulercharacteristic. The Euler Characteristic One of the oldest results in global surface theory is Euler’s formula: If P ⊂ R3 is a compact polyhedral surface that is the boundary of a convex openset,andP hasF faces,E edges,andV vertices,thenV −E+F =2. This quantity has a natural generalization to arbitrary finite simplicial complexes: If K is a finite simplicial complex of dimension n, we define the

Combinatorial Invariants 113 Euler characteristic of K, denoted by χ(K), by (cid:14)n χ(K)= (−1)kn , k k=0 where n is the number of k-dimensional simplices in K. Although we are k not yet in a position to prove Euler’s formula in full generality, we can at least show that the Euler characteristic of a simplicial complex is a combinatorial invariant. Theorem 5.19. If K and L are combinatorially equivalent finite simpli- cial complexes, then χ(K)=χ(L). Proof. ItclearlysufficestoprovethattheEulercharacteristicisunchanged by an elementary subdivision. Let K(cid:5) be an elementary subdivision of K obtained by adding a vertex v in the k-simplex σ = (cid:22)v0,…,v k (cid:23), and let Δχ=χ(K(cid:5))−χ(K). We must show that Δχ=0. For each simplex τ = (cid:22)v0,…,v k ,w1,…,w m (cid:23) of K that has σ as a face, K(cid:5) has one less (k + m)-simplex. In its place, for each j-element proper subset {v i1 ,…,v ij } ⊂ {v0,…, (cid:12) v k }, (cid:13) K(cid:5) has a new (j+m)-simplex (cid:22)v,v i1 ,…,v ij ,w1,…,w m (cid:23). There are k+ j 1 = j!( ( k k + + 1 1 − )! j)! such subsets, so each such τ makes a contribution to Δχ of (cid:20) (cid:21) (cid:20) (cid:21) (cid:14)k k(cid:14)+1 k+1 k+1 −(−1)k+m+ (−1)j+m = (−1)j+m. j j j=0 j=0 By the binomial theorem, this last sum is the expansion of the polynomial (−1)m(x+1)k+1 evaluated at x=−1, and therefore is equal to zero. Note that we are not claiming yet that the Euler characteristic is a topologicalinvariant,becausetwotriangulationsofthesamecompactspace are not necessarily combinatorially equivalent. In fact, it is a topological invariant. For compact surfaces, this will follow from the classification of surfaces,whichwewillcompleteinChapter10.Formoregeneralsimplicial complexes, the proof will require techniques of homology theory, which we will develop in Chapter 13.

114 5. Simplicial Complexes Problems 5-1. Suppose X is a topological space, and G1,…,G k are finitely many closed subspaces of X whose union is X. Show that the topology of X iscoherentwiththesesubspaces.Explainwhatthishastodowith the gluing lemma (Lemma 3.8). 5-2. Let v be a vertex of the simplicial complex K, and let Stv (the open starofv)betheunionoftheopensimplicesIntσ asσ rangesoverall simplices that have v as a vertex. Show that Stv is a neighborhood of v in |K|, and the collection of open stars of all the vertices is an open cover of |K|. 5-3. ShowthateverypolyhedronisHausdorffandlocallypathconnected. 5-4. LetKbeanabstractsimplicialcomplex.Showthat|K|iscompactif and only if K is finite, and locally compact if and only if K is locally finite. 5-5. Show that an abstract simplicial complex is the vertex scheme of a Euclidean complex if and only if it is finite-dimensional, locally finite, and countable. [Hint: If the complex has dimension n, let the vertices be the points v = (k,k2,k3,…,k2n+1) ∈ R2n+1. Use the k fundamentaltheoremofalgebratoshowthatno2n+2verticesliein aproperaffinesubspace,soany2n+2orfewerverticesareingeneral position. If two simplices σ, τ with vertices in this set intersect, let σ0, τ0 be the smallest face of each containing an intersection point, and consider the set consisting of all the vertices of σ0 and τ0. (This proof is from [Sti93].)] 5-6. DefineanabstractsimplicialcomplexKtobethefollowingcollection of abstract 2-simplices together with all of their faces: {{a,b,e},{b,e,f},{b,c,f},{c,f,g},{a,c,g},{a,e,g}, {e,f,h},{f,h,j},{f,g,j},{g,j,k},{e,g,k},{e,h,k}, {a,h,j},{a,b,j},{b,j,k},{b,c,k},{c,h,k},{a,c,h}}. Show that the geometric realization of K is homeomorphic to the torus. [Hint: Look at Figure 5.13.] 5-7. IfK isafiniteEuclideansimplicialcomplex,showthatitsbarycentric subdivisionSK isinfactasubdivisionofK,andthatthesimplicesof SK are those of the form (cid:22)b ,…,b (cid:23) in which each σ is a simplex σ0 σk j in K and σ j is a face of σ j+1 for j =0,…,k−1. 5-8. LetKbeafinitecomplex.Giveanexplicitalgorithmforobtainingthe barycentricsubdivisionofKasasequenceofelementarysubdivisions.

Problems 115 a b c a e f g e h j k h a b c a FIGURE 5.13. Triangulation of the torus. 5-9. Let K be the 1-dimensional abstract complex whose vertices are the nonnegative integers and whose 1-simplices are {(cid:22)0,n(cid:23):n∈N}, and let S be the subspace of R2 obtained by taking the union of all the line segments in Example 5.3. Define a map F: |K| → S by sending 0 to the origin, sending each n>0 to the point (1,1/n), and sending each 1-simplex (cid:22)0,n(cid:23) linearly onto the corresponding line segment. Show that F is continuous and bijective but not a homeomorphism. 5-10. SupposeKisanysimplicialcomplexwhosegeometricrealizationisa 1-manifold. Show that K is 1-dimensional. 5-11. Show that every triangulated 1-manifold is orientable. 5-12. Showthatorientabilityisacombinatorialinvariantoffinitesimplicial complexes. [Hint: It suffices to prove that if K is a finite Euclidean complex and K(cid:5) is a subdivision of K, then an orientation of either K or K(cid:5) determines an orientation of the other. Show that if σ is an oriented Euclidean n-simplex and σ(cid:5) is an n-simplex in some subdi- vision of σ, there is a unique orientation of σ(cid:5) such that any affine embedding σ → Rn that determines the given orientation of σ also determines the given orientation of σ(cid:5).]

6 Curves and Surfaces In this chapter we undertake a detailed study of curves (1-manifolds) and surfaces(2-manifolds).Thesearethemanifoldsthataremostfamiliarfrom our everyday experience, and about which the most is known mathemat- ically. They are thus excellent prototypes for the study of manifolds in higher dimensions. We begin by proving the classification theorem for 1-manifolds, which says that every connected 1-manifold is homeomorphic to S1 or R. Using the triangulation theorem for 1-manifolds proved in Chapter 5, this is a simple exercise in the combinatorics of graphs. We then proceed to a general discussion of 2-manifolds and a detailed examination of the basic examples of compact surfaces: the sphere, the torus, and the projective plane. Next we show how to form other compact surfaces by the technique of connected sums, a way of patching together simpler surfaces to form more complicated ones. To unify these results, we introducethenotionofpolygonalpresentationsofsurfaces,whichgeneralize simplicial complexes by representing surfaces as a collection of polygons (not necessarily triangles) with edges identified in pairs. Thecentralpartofthechapterpresentsthemainpartoftheclassification theorem for compact surfaces, which says that every compact, connected surface is homeomorphic to a sphere, a connected sum of tori, or a con- nected sum of projective planes. Again, the triangulation theorem reduces the problem to one of showing that every polygonal presentation can be reduced to a standard presentation of one of the model surfaces.

118 6. Curves and Surfaces In the last section we revisit orientations and the Euler characteristic, introduced in the last chapter for simplicial complexes, and reinterpret them in the context of polygonal surface presentations. Classification of Curves Our first goal in this chapter is to prove that up to homeomorphism, the onlyconnected1-manifoldsarethelineandthecircle.(Ofcourse,thisclas- sifies the disconnected ones too, because it implies that each component of adisconnected1-manifoldisalineoracircle,soevery1-manifoldishome- omorphic to a countable disjoint union of lines and/or circles.) You can think of this classification theorem as a warm-up for the more complicated classification of surfaces to follow; but it is also quite important in its own right. Theorem 6.1 (Classification of 1-Manifolds). A connected 1- manifold is homeomorphic to S1 if it is compact and to R if it is not. Proof. Bythetriangulationtheorem(Theorem5.10)andProposition5.11, M is homeomorphic to a graph in which every vertex lies on exactly two edges. Let K denote the abstract simplicial complex associated with this graph. We will show that K is isomorphic either to one of the complexes K of Example 5.7(d) (the vertex scheme of a regular m-gon, whose poly- m hedron is homeomorphic to S1) or to the complex K ∞ of Example 5.7(c) (whose polyhedron is homeomorphic to R). Since isomorphic complexes have homeomorphic polyhedra, this suffices to prove the theorem. ThefirststepistoshowthateveryvertexinM iscontainedinareduced edge path {v : n ∈ Z} that extends indefinitely in both directions. Start n withanyvertexv0.Byassumptionv0 liesontwoedges,sowecanlabelthe other vertices of those edges arbitrarily as v1 and v−1. Now by induction define v n+1 for each n ≥ 1 to be the unique vertex other than v n−1 such that v n and v n+1 span an edge. Similarly, we define v−n by induction on n. Let U be the union of all the edges (cid:22)v n ,v n+1 (cid:23) ⊂ M for v n in the edge path. The set U is closed in M (because its intersection with each sim- plex is closed and M has the coherent topology), and open (because the edges minus their vertices are open, and each vertex has a neighborhood intersecting only two edges, both of which must be in U). Thus U =M. Now we distinguish two cases. Case I: v (cid:14)=v for any n(cid:14)=k. In this case, the correspondence sending n k n(cid:10)→v n is easily seen to give an isomorphism between K ∞ and K, so M is homeomorphic to R.

Surfaces 119 v i+m+1 v i+1 v v i+m i v i+m−1 FIGURE 6.1. Periodic edge path. Case II: v n = v n+m for some n ∈ Z and some m > 0. We may assume that m and n have been chosen so that m is the least positive integer with this property. (Note that m ≥ 3 because the edge path is reduced.) We will show by induction that the edge path is periodic, in the sense that v i = v i+m for every i. By hypothesis this is true for i = n. If it is true for some i≥n, we argue as follows. The two vertices v i+m−1 and v i+m+1 are the only vertices that are connected to v i+m by edges (Figure 6.1). Since v i =v i+m ,thevertexv i+1 alsoisconnectedbyanedgetov i+m ,soitmust beequaltoeitherv i+m−1 orv i+m+1.Byminimalityofmitcannotbeequal tov i+m−1,sov i+1 =v i+m+1,completingtheinductionfori≥n.Asimilar induction takes care of i≤n. Now let K be the complex of Example 5.7(d), and define a map m f: K(0) → K(0) by f(n) = v for n = 1,…,m. Again, it is straightfor- m n ward to check that {f(i),f(j)} are the vertices of a simplex of K if and only if {i,j} are the vertices of a simplex of K . Thus f extends to a m simplicial isomorphism, so M is homeomorphic to S1. Surfaces The rest of this chapter is devoted to the study of compact surfaces. We have already seen several important examples: the sphere S2, the torus T2, and the projective plane P2 (i.e., the projective space of dimension 2). As we will soon see, these examples are fundamental because every compact surface can be built up from these three. In order to systematize our knowledge of surfaces, it will be useful to haveauniformwaytorepresentthemthatissomewhatmoregeneralthan simplicial complexes. The prototype is the representation of the torus as a

120 6. Curves and Surfaces π FIGURE 6.2. The sphere as a quotient of the disk. quotient of the square by identifying the edges in pairs (Example 3.24). It turns out that every compact surface can be represented as a quotient of a polygonal region in the plane by an equivalence relation that identifies its edges in pairs. Let us begin by seeing how our three basic compact examples can be so represented. We have already shown how to represent the torus as a quotient of a square (Example 3.24), so we focus on the sphere and the projective plane. Proposition 6.2. The sphere S2 is homeomorphic to the following quo- tient spaces. (a) The closed disk B2 ⊂R2 modulo the equivalence relation generated by (x,y)∼(−x,y) for x∈∂B2 (Figure 6.2). (b) The square I×I modulo the equivalence relation generated by (0,t)∼ (t,0) and (t,1)∼(1,t) for 0≤t≤1 (Figure 6.3). Proof. To see that each of these spaces is homeomorphic to the sphere, all we need to do is exhibit a quotient map from the given space to the sphere that makes the same identifications, and then appeal to uniqueness of quotient spaces (Corollary 3.32). For (a), define a map from the disk to the sphere by wrapping each horizontal line segment around a “latitude circle” (Figure 6.2). Formally, π: B2 →S2 is given by π(x,y) ⎧(cid:26) (cid:27) ⎪⎪⎪⎨ − (cid:3) 1−y2cos(cid:3) πx ,− (cid:3) 1−y2sin(cid:3) πx ,y , y (cid:14)=±1; 1−y2 1−y2

⎪⎪⎪⎩ (0,0,y), y =±1.

Surfaces 121 α β γ π FIGURE 6.3. The sphere as a quotient of the square. This is a quotient map by the closed map lemma; it is straightforward to checkthatitmakesexactlythesameidentificationsasthegivenequivalence relation. Toprove(b),wewillconstructaquotientmapfromI×I tothesphereas a composition of several simpler maps (Figure 6.3). First let S denote the square {(x,y) : |x|,|y| ≤ 1}, and define a homeomorphism α: I ×I → S by first scaling both coordinates by a factor of 2 and then translating the new center (1,1) back to the origin: α(x,y) = (2x−1,2y−1). Then let β: S → B2 be the homeomorphism whose existence is guaranteed by Proposition 4.26; it sends each radial line segment between the origin and the boundary of S linearly onto the parallel segment between the center of the disk and its boundary. Next let γ: B2 → B2 be the counterclockwise rotation through π/4 radians (obviously a homeomorphism), and consider the composite map ϕ=π◦γ◦β◦α: I×I −α→S −β→B2 −γ→B2 −π→S2, where π is the quotient map of the preceding paragraph. Since this is a compositionofquotientmaps,itisaquotientmap.Threadingthroughthe definitions (with help from the pictures!), you will see that it makes the same identifications as the quotient map defined in (b), thus completing the proof. Proposition 6.3. The projective plane P2 is homeomorphic to each of the following quotient spaces (Figure 6.4).

122 6. Curves and Surfaces x −x (a) (b) (c) FIGURE 6.4. Representations of P2 as a quotient space. (a) The sphere S2 modulo the equivalence relation x ∼ −x for each x ∈ S2. (b) The closed disk B2 modulo the relation (x,y) ∼ (−x,−y) for each (x,y)∈∂B2. (c) ThesquareI×I modulotherelation(t,0)∼(1−t,1),(0,1−t)∼(1,t) for 0≤t≤1. Proof. Let S2/∼ denote the quotient space of S2 obtained by identifying eachpointxwithitsantipodalpoint−x,andletp: S2 →S2/∼denotethe quotient map. Consider also the composite map S2 (cid:9)→ι R3(cid:3){0}−π→P2, where ι is inclusion and π is the quotient map defining P2. Note that π◦ι is a quotient map by the closed map lemma. It makes exactly the same identificationsasp,sobyuniquenessofquotientspacesP2ishomeomorphic to S2/∼. If F: B2 → S2 is(cid:3)the map sending the disk onto the upper hemisphere by F(x,y) = (x,y, 1−x2−y2), then p◦F: B2 → S2/∼ is easily seen to be surjective, and is thus a quotient map by the closed map lemma. It identifies only (x,y) ∈ ∂B2 with (−x,−y) ∈ ∂B2, so P2 is homeomorphic to the resulting quotient space. Part (c) is left as an exercise. Exercise 6.1. Prove Proposition 6.3(c). When doing geometric “cutting and pasting” constructions like the ones in the last two propositions, it is often safe to rely on pictures and a few wordstodescribethemapsandidentificationsbeingconstructed.Sofar,we havebeencarefultogiveexplicitdefinitions(oftenwithformulas)ofallour maps, together with rigorous proofs that they do in fact give the results

Surfaces 123 v q FIGURE 6.5. An edge point. FIGURE 6.6. A vertex. we claim; but as your sophistication increases and you become adept at carrying out such explicit constructions yourself, you can leave out many of the details. The main thing is that before you skip any such details, youshouldbeabsolutelysurethatyoucouldquicklywritethemdownand checkyourclaimsrigorously;thisistheonlywaytobesurethatyouarenot hidingrealdifficultiesbehind“hand-waving.”Inthisbookwewillbeginto leave out some such details in our proofs; for a while, you should fill them in for yourself to be sure that you know how to turn an argument based on pictures into a complete proof. Now we describe a general method for building surfaces by identifying edgesofgeometricfigures.LetussaythatasubsetP oftheplaneisapolyg- onalregionifitisacompactsubsetwhoseboundaryisafinite1-dimensional Euclidean simplicial complex, satisfying the following conditions: (i) Eachpointq ofanedgeotherthanavertexhasaneighborhoodU in R2 suchthatP∩U isequaltotheintersectionwithU ofsomeclosed half-plane {(x,y):ax+by+c≥0} (Figure 6.5). (ii) Each vertex v has a neighborhood V in R2 such that P ∩V is equal totheintersectionofV withtwoclosedhalf-planeswhoseboundaries intersect only at v (Figure 6.6). Any finite collection of disjoint 2-simplices in the plane is easily seen to beapolygonalregion,asisafilled-insquare,oranycompactconvexregion boundedbyfinitelymany1-simplices.Below,wewillseesomemoreexam- ples of manifolds obtained as quotients of polygonal regions by identifying the edges in pairs. It is a general fact that such a quotient space is always a surface. Proposition 6.4. Let P be a polygonal region in the plane with an even number of edges, and suppose we are given an equivalence relation that

124 6. Curves and Surfaces α1 q1 U1 q2 U2 α2 FIGURE 6.7. Euclidean neighborhood of an edge point. identifies each edge with exactly one other edge by means of a simplicial homeomorphism. The resulting quotient space is a compact 2-manifold. Proof. LetM bethequotientspace,andletπ: P →M denotethequotient map. Clearly, M is compact, because it is the continuous image of the compact space P. Since the equivalence relation identifies only edges with edges and ver- tices with vertices, the points of M fall into three disjoint sets: (a) face points, whose inverse images in P are in the interior of P; (b) edge points, whose inverse images are on edges but not vertices; and (c) vertex points, whose inverse images are vertices. To prove that M is locally Euclidean, we consider the three types separately. Because π is injective on IntP, IntP is a saturated open subset of P, so the restriction of π to IntP is a one-to-one quotient map and therefore a homeomorphism. Thus π(IntP) is a Euclidean neighborhood of each face point. An edge point q(cid:7) has exactly two inverse images q1 and q2, each on a different edge. Using condition (i) in the definition of polygonal region, there exist disjoint neighborhoods U1 of q1 and U2 of q2 such that P ∩U i is a closed half-disk (Figure 6.7). Let V = P ∩U . It is straightforward i i to construct affine homeomorphisms α1 taking V1 to a half-disk in the upper half-plane and α2 taking V2 to a half-disk in the lower half-plane, in such a way that q1 and q2 both go to the origin and the boundary identifications are respected. Define α: V1 ∪V2 →R2 by letting α=α1 on V1 and α=α2 on V2. Shrinking V1 and V2 if necessary, we can ensure that V1 ∪V2 is a saturated open set in P (this just means that for each point in V1 ∩∂P, the corresponding boundary point is in V2, and conversely). Then the restriction of π to V1 ∪V2 is a quotient map, so α descends to a mapα(cid:7): V (cid:7) →R2,whereV (cid:7) =π(V1 ∪V2).Itsimagecontainsaneighborhood

Surfaces 125 v2 v3 v1 v4 FIGURE 6.8. Euclidean neighborhood of a vertex point. (cid:7) of the origin by construction, and its domain V is the image under π of a saturated open set and therefore open. This shows that q(cid:7)has a locally Euclidean neighborhood. Similarly, a vertex point v(cid:7) has as its inverse image a finite set of ver- tices {v1,…,v k } ⊂ P. For each i, choose a homeomorphism from a neighborhood of v in P to an open subset in a closed “wedge” of an- i gle 2π/k in the plane, which is a set described in polar coordinates by {(r,θ) : θ0 ≤ θ ≤ θ0+2π/k}. (If we place v i at the origin, such a homeo- morphismisgiveninpolarcoordinatesbyafan transformationoftheform (r,θ)(cid:10)→(r,θ0+cθ) for suitable constants θ0,c.) Because each edge is paired with exactly one other, the k wedges can be mapped onto a set containing a neighborhood of the origin by rotating and piecing them together (Figure 6.8). However, this may not respect the edge identifications. To correct this, we can subject each wedge to a preliminary transformation that rescales its edges independently. First, by a rotation followed by a fan transformation, take the wedge to the first quadrantsothatoneedgeliesalongthepositivex-axisandtheotheralong the positive y-axis. Then rescale the two axes by a linear transformation (x,y) (cid:10)→ (ax,by). Finally, use another fan transformation to insert the wedge into its place. (The case k = 1 deserves special comment. This case can occur only if the two edges adjacent to the single vertex v1 are identified with each other; then you can check that our construction maps a neighborhood of v1 onto a neighborhood of the origin, with both edges going to the same ray.) In each case, we end up with a map defined on a saturated open set in P, which descends to a homeomorphism from a neighborhood of v(cid:7)to a neighborhood of the origin. The quotient is second countable by Lemma 3.21. To prove that it is Hausdorff is the same as showing that the fibers of π can be separated by

126 6. Curves and Surfaces FIGURE 6.9. The Klein bottle. saturated open sets. It is straightforward to check on a case-by-case basis that the inverse images of sufficiently small Euclidean balls will do. Hereisanotherexampleofamanifoldformedasaquotientofapolygonal region. Example 6.5. The Klein bottle is the 2-manifold K obtained by iden- tifying the edges of the square I × I according to (0,t) ∼ (1,t) and (t,0)∼(1−t,1) for 0≤t≤1. To visualize K, think of gluing the left and right edges together to form a cylinder, and then passing the upper end of the cylinder through the cylinder wall near the lower end, in order to glue the upper circle to the lower one “from the inside” (Figure 6.9). Of course, this cannot be done with a physical model; in fact, it can be shown that the Klein bottle is not homeomorphic to any subspace of R3. Nonetheless, the preceding proposition shows that it is a 2-manifold. Connected Sums To construct other examples of surfaces, we now introduce an important way of producing manifolds by gluing together simpler ones, called “con- nected sum.” Although we will use this primarily for surfaces, it works for manifolds of any dimension, so we will define it in arbitrary dimensions. Let M1 and M2 beconnected n-manifolds.Geometrically,theconnected sum is obtained by cutting out a small open ball from each of M1 and M2 andgluingtheresultingspacestogetheralongtheirboundaryspheres.More precisely,letB ⊂M beregularEuclideanballs.Chooseahomeomorphism i i σ: ∂B1 →∂B2 (suchahomeomorphismexistsbecausebothboundariesare homeomorphic to Sn−1). Let M(cid:5) = M (cid:3)B , and define a quotient space i i i of M 1 (cid:5) (cid:20)M 2 (cid:5) by identifying each q ∈ ∂B1 with σ(q) ∈ ∂B2 (Figure 6.10). The resulting quotient space is called a connected sum of M1 and M2 and is denoted by M1#M2. Proposition 6.6. IfM1 andM2 areconnectedn-manifolds,anyconnected sum M1#M2 is a connected n-manifold.

Connected Sums 127 M1 M2 B1 B2 π M1#M2 FIGURE 6.10. A connected sum. Proof. FirstweshowthatM1#M2 islocallyEuclidean.Letπ: M 1 (cid:5)(cid:20)M 2 (cid:5) → M1#M2 denote the quotient map, and let S = π(∂B1 ∪∂B2). Since π is one-to-one and thus a homeomorphism away from S, M1 #M2 (cid:3)S is a manifoldandthereforelocallyEuclidean.Thusweneedonlyconsiderpoints in S. For any nonempty interval J ⊂ [0,∞), let us use the notation A to J denotetheannulus{x∈Rn :|x|∈J}.Thus,forexample,A[1,2) isequalto B2(0)(cid:3)B1(0). Our regular balls B i for i = 1,2 come with neighborhoods U i containingB i andhomeomorphismsϕ i : U i →B2(0)takingU i (cid:3)B i onto A[1,2) (Figure 6.11). (If ϕ i takes U i onto a ball of radius different from 2, wecanadjustitbyadilationtomakesurethatitsimageis B2(0).)Notice that ϕ sends ∂B to the unit sphere. i i Thefirstthingweneedtodoistoadjustoneofthesemapstocompensate for the fact that ϕ − 2 1◦ϕ1 does not make the same identification between ∂B1 and ∂B2 as σ does. To correct this, observe that the composite map β =ϕ2 ◦σ◦ϕ − 1 1 from Sn−1 ⊂Rn to itself is a homeomorphism: Sn−1 −ϕ− 1→ 1 ∂B1 −σ→∂B2 −ϕ→2 Sn−1. (cid:7) Defineahomeomorphismβ fromB2(0)toitselfbysendingtheraythrough each point x0 ∈ Sn−1 linearly onto the ray through β(x0); formally, since

128 6. Curves and Surfaces ϕ1 β (cid:7) U1 (cid:3)B1 A[1,2) A[1,2) I ϕ(cid:7) 1 ϕ2 U2 (cid:3)B2 A(1/2,2) FIGURE 6.11. Euclidean neighborhood of points in S. x/|x| is the point where the ray through x intersects Sn−1, (cid:20) (cid:21) β (cid:7) (x)=|x|β x . |x| Let ϕ(cid:7) 1 =β (cid:7)◦ϕ1. It is immediate from the definitions that ϕ(cid:7) 1 =ϕ2 ◦σ on ∂B1. (6.1) TheinversionmapI(x)=x/|x|2 mapstheannulusA[1,2) homeomorphi- cally onto the annulus A(1/2,1], and restricts to the identity map on Sn−1. We define a map Φ from the open set (U1 (cid:3)B1)∪(U2 (cid:3)B2) ⊂ M 1 (cid:5) (cid:20)M 2 (cid:5) to Rn by (cid:16) I◦ϕ(cid:7) 1(q), q ∈U1 (cid:3)B1, Φ(q)= ϕ2(q), q ∈U2 (cid:3)B2. Let us check that Φ respects the identifications made by π. If q ∈ ∂B1 and p=σ(q)∈∂B2, then I◦ϕ(cid:7) 1(q)=ϕ(cid:7) 1(q) (because I is the identity on Sn−1) =ϕ2 ◦σ(q) (by (6.1)) =ϕ2(p). ThusΦpassestothequotientanddefinesahomeomorphismfromaneigh- borhoodofS totheopenannulusA(1/2,2),showingthatM1#M2 islocally

Polygonal Presentations of Surfaces 129

=

FIGURE 6.12. Connected sum with a sphere. Euclidean. The proofs that it is second countable and Hausdorff are anal- ogous to those in Proposition 6.4, and are left as an exercise. To show that M1#M2 is connected, just note M1#M2 is the union of the two sets π(M(cid:5)) and π(M(cid:5)), which are both connected and have points 1 2 of S in common. Exercise 6.2. Prove that M1#M2 is Hausdorff and second countable. Our definition of M1#M2 depends on several choices—the sets B i and the homeomorphism σ. When M1 and M2 are surfaces, it can be shown that different choices yield homeomorphic connected sums. We will not prove this in full generality, but in the case of compact surfaces it will turn out to be a consequence of the classification theorem (see Problem 10-4). Nevertheless, we will sometimes use the imprecise terminology “the connected sum M1#M2” to refer to any connected sum between M1 and M2. Example 6.7. IfM isanyn-manifold,aconnectedsumM#Sn ishomeo- morphictoM,atleastifwechoosethesettocutoutofSncarefully(Figure 6.12). Let B2 ⊂ Sn be the open lower hemisphere, so (Sn)(cid:5) = Sn (cid:3)B2 is theclosedupperhemisphere,whichishomeomorphictoaclosedball.Then M#Sn isobtainedfromM bycuttingouttheopenballB1 andgluingback a closed ball along the boundary sphere, so we have not changed anything. Example 6.8. The n-fold connected sum T2 # T2 # ··· # T2 (Figure 6.13), is called the n-holed torus, or the sphere with n handles. The lat- terterminologyreferstothefactthatthissurfaceisalsohomeomorphicto S2#T2#···#T2, and each added torus looks like a “handle” attached to the sphere (Figure 6.14). Polygonal Presentations of Surfaces As we mentioned earlier in this chapter, for the classification theorem we needauniformwaytodescribesurfaces.Wewillrepresentallofoursurfaces

130 6. Curves and Surfaces FIGURE 6.13. A connected sum of tori. FIGURE 6.14. A sphere with handles. asquotientsof2n-sidedpolygonalregions.Informally,wecandescribeany edge equivalence relation by labeling the edges with letters a1,…,a n , and giving each edge an arrow pointing toward one of its vertices, in such a way that edges with the same label are to be identified, with the arrows indicating which way the vertices match up. With each such labeling of a polygon we associate a sequence of symbols, obtained by reading off the boundarylabelscounterclockwisefromthetop,andforeachboundarylabel −1 a , placing a in the sequence if the arrow points counterclockwise and a i i i if it points clockwise. For example, the equivalence relation on I ×I of Example3.24thatyieldsthetorusmightresultinthesequenceofsymbols aba−1b−1. Formally,givenasetS,wedefineawordinS tobeanorderedk-tupleof symbols, each of the form a or a−1 for some a∈S. (To be more precise, if youlike,youcandefineawordtobeafinitesequenceoforderedpairsofthe form(a,1)or(a,−1)fora∈S,andthendefineaanda−1 asabbreviations for (a,1) and (a,−1), respectively.) A polygonal presentation, written P=(cid:22)S |W1,…,W k (cid:23), isafinitesetS togetherwithfinitelymanywordsW1,…,W k inS oflength 3 or more, such that every symbol in S appears in at least one word. As a

Polygonal Presentations of Surfaces 131 a a a a S2 P2 FIGURE 6.15. Presentations of S2 and P2. matter of notation, when the set S is described by listing its elements, we willleaveoutthebracessurroundingtheelementsofS,andwilldenotethe wordsW byjuxtaposition,soforexamplethepresentationwithS ={a,b} i and the single word W = (a,b,a−1,b−1) will be written (cid:22)a,b | aba−1b−1(cid:23). WealsoallowasaspecialcaseanypresentationinwhichS hasoneelement and there is a single word of length 2. There are only four such: (cid:22)a | aa(cid:23), (cid:22)a|a−1a−1(cid:23), (cid:22)a|aa−1(cid:23), and (cid:22)a|a−1a(cid:23). Any polygonal presentation P determines a topological space |P|, called the geometric realization of P, by the following recipe:

1. For each word W , let P denote the convex k-sided polygonal region i i in the plane that has its center at the origin, sides of length 1, and one vertex on the positive y-axis. (Here k is the length of the word W .) i
2. Define a one-to-one correspondence between the symbols of W and i the edges of P in counterclockwise order, starting at the vertex on i the y-axis. (cid:6)
3. Let |P| denote the quotient space of P determined by identifying i i each pair of edges that have the same edge symbol according to the simplicialhomeomorphismthatmatchesupthefirstverticesincoun- terclockwise order if both edges have the same label a or a−1, and matches the first vertex of one with the second vertex of the other if the edges are labeled a and a−1. IfPisoneofthespecialpresentationswithawordoflength2,motivated by Propositions 6.2 and 6.3 we define |P| to be the sphere if the word is aa−1 or a−1a, and the projective plane if it is aa or a−1a−1 (Figure 6.15). The interiors, edges, and vertices of the polygonal regions P are called i thefaces,edges,andverticesofthepresentation.Thenumberoffacesisthe sameasthenumberofwords,whilethenumberofedgesisthetotalnumber of symbols that occur in all the words. For an edge labeled a, the initial vertex is the first one in counterclockwise order, and the terminal vertex

132 6. Curves and Surfaces F p1 p2 P1 P2 FIGURE 6.16. Simplicial homeomorphism between polygons. is the other one; for an edge labeled a−1, these definitions are reversed. In terms of our informal description above, if we label each edge with an arrow pointing counterclockwise when the symbol is a and clockwise when it is a−1, the arrow points from the initial vertex to the terminal vertex. The geometric realization of a presentation with only one face is con- nected, because it is a quotient of a single connected polygon; with more thanoneface,itmightormightnotbeconnected.Althoughfordefiniteness we have defined the geometric realization as a quotient of a disjoint union of a specific collection of polygonal regions, the following lemma shows that we could have used arbitrary disjoint convex polygonal regions in the planewiththesamenumbersofedges,becausebetweenanytwopolygonal regions with the same sequence of edge labels there is a homeomorphism that respects the edge labels. Because of this, in the arguments that fol- low we will illustrate our presentations with any convex polygons that are convenient. Lemma 6.9. LetP1,P2 beconvexpolygonswiththesamenumberofedges, and let f: ∂P1 → ∂P2 be a simplicial homeomorphism. Then f extends to a homeomorphism F: P1 →P2. Proof. Choose any points p ∈ IntP , i = 1,2. By convexity, the line seg- i i mentfromp toeachvertexofP liesentirelyinP .Theedgesandvertices i i i of P together with p , these new line segments, and the triangles they i i bound form a Euclidean simplicial complex whose polyhedron is P (Fig- i ure 6.16). The required map F: P1 → P2 is the simplicial map whose restriction to ∂P1 is f and that takes p1 to p2. A polygonal presentation is called a surface presentation if each symbol a ∈ S occurs exactly twice in W1,…,W k (counting either a or a−1 as one occurrence). By Proposition 6.4, the geometric realization of a surface presentation is a compact surface.

Polygonal Presentations of Surfaces 133 a a a a b a b b b b b b b a a a S2 T2 P2 K FIGURE 6.17. Standard presentations. Example 6.10. Thefollowingsurfacesaredeterminedbythegivenpolyg- onalpresentations,whichwecalltheirstandardpresentations(Figures6.15 and 6.17). (a) The sphere: (cid:22)a|aa−1(cid:23) or (cid:22)a,b|abb−1a−1(cid:23) (Proposition 6.2). (b) The torus: (cid:22)a,b|aba−1b−1(cid:23) (Example 3.24). (c) The projective plane: (cid:22)a|aa(cid:23) or (cid:22)a,b|abab(cid:23) (Proposition 6.3). (d) The Klein bottle: (cid:22)a,b|abab−1(cid:23) (Example 6.5). IfKisa2-dimensionalsimplicialcomplexinwhicheverysimplexiscon- tainedinsome2-simplex,thenKdeterminesinanobviouswayapolygonal presentation in which each 2-simplex corresponds to a word of length 3. For later use in proving the classification theorem, we will develop some generalrulesfortransformingpolygonalpresentations.Iftwopresentations P 1 andP 2 havehomeomorphicgeometricrealizations,wewillsaythatthey are topologically equivalent and write P 1 ≈P 2. The following operations are called elementary transformations of a polygonal presentation. As a matter of notation, in what follows S will denote any sequence of symbols; a,b,c,a1,a2,… will denote any symbols fromS ortheirinverses;ewilldenoteanysymbolnotinS;andW1,W2,… will denote any words made from the symbols in S. Given two words W1,W2, the notation W1W2 denotes the word formed by concatenating W1 and W2 together. We adopt the convention that (a−1)−1 =a.

• Relabeling:Changingalloccurrencesofasymbolatoanewsymbol not already in the presentation, interchanging all occurrences of two symbols a and b, or interchanging all occurrences of a and a−1 for some a∈S.
• Subdividing: Replacing every occurrence of a by ae and every oc- currence of a−1 by e−1a−1, where e is a new symbol not already in the presentation.

134 6. Curves and Surfaces a1 a5 a5 a1 a1 a5 a2 a1 a2 a2 a2 a3 a3 a4 a4 a3 a3 a4 a4 a5 FIGURE 6.18. Reflecting. FIGURE 6.19. Rotating. a e a c e e W1 W2 W1 e W2 b e c b FIGURE 6.20. Cutting/pasting. FIGURE 6.21. Folding/unfolding.

• Consolidating:Ifaandbalwaysoccuradjacenttoeachothereither as ab or b−1a−1, replacing every occurrence of ab by a and every occurrence of b−1a−1 by a−1, provided that the result is one or more words of length at least 3 or a single word of length 2.
• Reflecting (Figure 6.18): (cid:22)S |a1…a m , W2,…,W k (cid:23)(cid:10)→(cid:22)S |a− m 1…a 1 −1 , W2,…,W k (cid:23).
• Rotating (Figure 6.19): (cid:22)S |a1a2…a m , W2,…,W k (cid:23)(cid:10)→(cid:22)S |a2…a m a1, W2,…,W k (cid:23).
• Cutting (Figure 6.20): If W1 and W2 both have length at least 2, (cid:22)S |W1W2, W3,…,W k (cid:23)(cid:10)→(cid:22)S,e|W1e, e−1W2, W3,…,W k (cid:23).
• Pasting (Figure 6.20): (cid:22)S,e|W1e, e−1W2, W3,…,W k (cid:23)(cid:10)→(cid:22)S |W1W2, W3,…,W k (cid:23).
• Folding (Figure 6.21): If W1 has length at least 3, (cid:22)S,e|W1ee−1, W2,…,W k (cid:23)(cid:10)→(cid:22)S |W1, W2,…,W k (cid:23). We also allow W1 to have length 2, provided that the presentation has only one word.

Polygonal Presentations of Surfaces 135

• Unfolding (Figure 6.21): (cid:22)S |W1, W2,…,W k (cid:23)(cid:10)→(cid:22)S,e|W1ee−1, W2,…,W k (cid:23). Proposition 6.11. Each elementary transformation of a polygonal pre- sentation produces a topologically equivalent presentation. Proof. Clearly, subdividing and consolidating are inverses of each other, as are cutting/pasting and folding/unfolding, so by symmetry only one of each pair needs to be proved. We demonstrate the techniques by proving the proposition for cutting and folding, and leave the rest as exercises. To prove that cutting produces a homeomorphic geometric realization, let P1 and P2 be convex polygons labeled W1e and e−1W2, respectively, and let P(cid:5) be a convex polygon labeled W1W2. For the moment, let us assume that these are the only words in their respective presentations. Let π: P1 (cid:20)P2 → S and π(cid:5): P(cid:5) → S(cid:5) denote the respective quotient maps. The line segment going from the terminal vertex of W1 in P(cid:5) to its initial vertex lies in P(cid:5) by convexity; label this segment e. By Lemma 6.9 there is a continuous map f: P1 (cid:20)P2 →P(cid:5) that takes each edge of P1 or P2 to the edge in P(cid:5) with the corresponding label, and whose restriction to each P i isahomeomorphism.Bytheclosedmaplemma,f isaquotientmap.Since f identifies the two edges labeled e and e−1 but nothing else, the quotient mapsπ(cid:5)◦f andπ makepreciselythesameidentifications,sotheirquotient spaces are homeomorphic. If there are other words W3,…,W k , we just extendf bydeclaringittobetheidentityontheirrespectivepolygonsand proceed as above. For folding, as before we can ignore the additional words W2,…,W k . If W1 has length 2, we can subdivide to lengthen it, then perform the folding operation, and then consolidate, so we assume that W1 has length at least 3.AssumefirstthatW1 =abchaslengthexactly3.LetP andP(cid:5) beconvex polygonswithedgelabelsabcee−1 andabc,respectively,andletπ: P →S, π(cid:5): P(cid:5) → S(cid:5) be the quotient maps. Adding edges as shown in Figure 6.21 turnsP andP(cid:5) intopolyhedraofEuclideansimplicialcomplexes,andthere isauniquesimplicialmapf: P →P(cid:5) thattakeseachedgeofP totheedge of P(cid:5) with the same label. As before, π(cid:5)◦f and π are quotient maps that make the same identifications, so their quotient spaces are homeomorphic. IfW1 haslength4ormore,wecanwriteW1 =XbcforsomeX oflength at least 2. Then we cut along a to obtain (cid:22)S,b,c,e|Xbcee−1(cid:23)≈(cid:22)S,a,b,c,e|Xa−1,abcee−1(cid:23), and proceed as before. Exercise 6.3. ProvetherestofProposition6.11.Notethatyouwillhave to consider a word of length 2 as a special case when treating subdividing or consolidating.

136 6. Curves and Surfaces v a W1 v W1 f b a a c b v c c v b P 1 (cid:5) Q P1 π(cid:5) π (cid:7) f M1 B1 M1 FIGURE 6.22. The presentation (cid:8)S1,a,b,c|W1c−1b−1a−1,abc(cid:9). Next we need to find standard polygonal presentations for connected sums. The key is the following proposition. Proposition 6.12. Let M1 and M2 be surfaces determined by presenta- tions (cid:22)S1 |W1 (cid:23) and (cid:22)S2 |W2 (cid:23), respectively, in which S1 and S2 are disjoint sets and each presentation has a single face. Then (cid:22)S1,S2 | W1W2 (cid:23) is a presentation of a connected sum M1#M2. (Here W1W2 denotes the word formed by concatenating W1 and W2 together.) Proof. Consider the presentation (cid:22)S1,a,b,c|W1c−1b−1a−1,abc(cid:23) (pictured in the left half of Figure 6.22). Pasting along a and folding twice, we see that this presentation is equivalent to (cid:22)S1 | W1 (cid:23) and therefore is a pre- sentation of M1. Let B1 denote the image in M1 of the interior of the polygonal region bounded by triangle abc. We will show below that B1 is a regular Euclidean disk in M1. Assuming this, it follows immediately that the geometric realization of (cid:22)S1,a,b,c | W1c−1b−1a−1(cid:23) is homeomor- phic to M1 (cid:3)B1 (which we denote by M 1 (cid:5)), and ∂B1 is the image of the edges c−1b−1a−1. A similar argument shows that (cid:22)S2,a,b,c | abcW2 (cid:23) is a presentationofM2 withaEuclideandiskremoved(denotedbyM 2 (cid:5)).There- fore, (cid:22)S1,S2 |W1c−1b−1a−1,abcW2 (cid:23) is a presentation of M 1 (cid:5)(cid:20)M 2 (cid:5) with the boundaries of the respective disks identified, which is M1 #M2. Pasting along a and folding twice, we arrive at the presentation (cid:22)S1,S2 |W1W2 (cid:23). ItremainsonlytoshowthatB1isaregulardiskinM1,i.e.,thatithasan open disk neighborhood in which B1 corresponds to a smaller closed disk. LetP1,P 1 (cid:5),andQbeconvexpolygonswithedgeslabeledbythewordsW1, W1c−1b−1a−1, and abc, respectively. Triangulating the polygons as shown in Figure 6.22, we obtain a simplicial map f: P 1 (cid:5)(cid:20)Q→P1 such that π◦f

Classification of Surface Presentations 137 v FIGURE 6.23. Showing that B1 is a regular disk. makesthesameidentificationsasπ(cid:5),andsodescendstoahomeomorphism f (cid:7) : M1 →M1. The inverse image of f (cid:7) (B1) under π is a small triangle in P1 sharing one vertex v in common with P1. Now look back at the proof in Proposition 6.4 that the quotient space of a surface presentation is a manifold. In constructing a Euclidean neigh- borhood of a vertex point, we assembled “wedges” at the various vertices intoaEuclideandisk.Applyingthatconstructiontothevertexv,thesmall triangleistakentoasetthatishomeomorphictoacloseddiskintheplane (Figure 6.23), and it is an easy matter to extend that homeomorphism to a slightly larger open disk. Example 6.13. Usingtheprecedingproposition,wecanaugmentourlist of presentations of known surfaces as follows:

• Connected sum of n tori: (cid:22)a1,b1,…,a n ,b n |a1b1a − 1 1 b − 1 1 …a n b n a− n 1b− n 1(cid:23).
• Connected sum of n projective planes: (cid:22)a1,…,a n |a1a1…a n a n (cid:23). We call these the standard presentations of these surfaces. Classification of Surface Presentations We are now ready to state the main result in the classification of surfaces. This theorem was first proved in 1907 by Max Dehn and Poul Heegaard [DH07] under the assumption that the surface had some polygonal presen- tation. Using the triangulation theorem of Chapter 5, we now know that every compact surface has a triangulation, which determines a polygonal presentation.

138 6. Curves and Surfaces a a b c b a a b a b b b c b b c a a c c c FIGURE 6.24. Transforming the Klein bottle to P2#P2. Theorem 6.14 (Classification of Surface Presentations). Any sur- face presentation is equivalent by a sequence of elementary transformations to a presentation of one of the following: (a) the sphere S2; (b) a connected sum T2#···#T2; or (c) a connected sum P2#···#P2. Therefore, every compact surface is homeomorphic to one of the surfaces in this list. Before we prove the theorem, we need to make one important observa- tion. You might have noticed that some surfaces appear to be absent from the list—the Klein bottle, for example, and T2 #P2, and, for that mat- ter, every connected sum involving both tori and projective planes. These apparent deficiencies are explained by the following two lemmas. Lemma 6.15. The Klein bottle is homeomorphic to P2#P2. Proof. Byasequenceofelementarytransformations,wefindthattheKlein bottle has the following presentations (see Figure 6.24): (cid:22)a,b|abab−1(cid:23) ≈(cid:22)a,b,c|abc, c−1ab−1(cid:23) (cut along c) ≈(cid:22)a,b,c|bca,a−1cb(cid:23) (rotate and reflect) ≈(cid:22)b,c|bbcc(cid:23). (paste along a and rotate) This last is a presentation of the connected sum of two projective planes. Lemma 6.16. TheconnectedsumT2#P2 ishomeomorphictoP2#P2#P2. Proof. Start with (cid:22)a,b,c|abab−1cc(cid:23) (Figure 6.25), which is a presentation ofK#P2,andthereforebytheprecedinglemmaofP2#P2#P2.Following Figure 6.25, we cut along d, paste along c, cut along e, and paste along b, rotating and reflecting as necessary, to obtain (cid:22)a,d,e|a−1d−1adee(cid:23), which is a presentation of T2#P2.

Classification of Surface Presentations 139 a a d a d b c d b c c a b a a c e b d b d b a a d d d d b b e a e a e e FIGURE 6.25. Transforming P2#P2#P2 to T2#P2. Proof of the classification theorem. By the triangulation theorem we can assumethatM comeswithagivenpresentation.Wewillprovethetheorem by transforming this presentation to one of our standard presentations in several steps. Let us say that a pair of edges that are to be identified are complementary if they appear in the presentation as both a and a−1, and twisted if they appear as a,…,a or as a−1,…,a−1. (The terminology reflects the fact that if a polygonal region is cut from a piece of paper, you have to twist the paper to paste together a twisted edge pair, but not for a complementary pair.) Step 1: M admits a presentation with only one face. Since M is con- nected, if there are two or more faces, some edge in one face must be identified with an edge in a different face; otherwise, M would be the dis- jointunionofthequotientsofitsfaces,andsinceeachsuchquotientisopen andclosed,theywouldprovideaseparationofM.Thusbyperformingsuc- cessive pasting transformations (together with rotations and reflections as necessary), we can reduce the number of faces in the presentation to one. Step2:EitherM ishomeomorphictothesphere,orM admitsapresen- tation in which there are no adjacent complementary pairs. Each adjacent complementary pair can be eliminated by folding, unless it is the only pair of edges in the presentation; in this case the presentation is equivalent to (cid:22)a|aa−1(cid:23) and M is homeomorphic to the sphere. From now on, we assume that the presentation is not the standard pre- sentation of the sphere. Step 3:M admits a presentation in which all twisted pairs are adjacent. If a twisted pair is not adjacent, the presentation is described by a word of the form VaWa, where neither V nor W is empty. Figure 6.26 shows how to transform the word VaWa into VW−1bb by cutting along b, reflecting, and pasting along a. (Here W−1 denotes the word obtained from W by reflecting).Inthislastpresentation,thetwistedpaira,ahasbeenreplaced

140 6. Curves and Surfaces V V VW−1 a a b b W−1 a b a W b b FIGURE 6.26. Making a twisted pair adjacent. byanothertwistedpairb,b,whichisnowadjacent.Moreover,nootherad- jacent pairs have been separated. We may have created some new twisted pairs when we reflected W, but we decreased the total number of nonadja- cent pairs by at least one. Thus, after finitely many such operations, there willbenomorenonadjacenttwistedpairs.Wemayalsohavecreatedsome new adjacent complementary pairs. These can be eliminated by repeating Step 2, which does not increase the number of nonadjacent pairs. Step 4: M admits a presentation in which all vertices are identified to a single point. Choose some equivalence class of vertices, and call it v. If there are vertices that are not identified with v, there must be some edge that goes from a v vertex to a vertex in some other equivalence class; label the edge a and the other vertex class w (Figure 6.27). The other edge that touches a at its v vertex cannot be identified with a: If it were complementarytoa,wewouldhaveeliminatedbothedgesinStep3,while ifitformedatwistedpairwitha,thenthequotientmapwouldidentifythe initial and terminal vertices of a with each other, which we are assuming is not the case. So label this other edge b, and label its other vertex x (this one may be identified with v, w, or neither one). Somewhereinthepolygonisanotheredgelabeledborb−1.Letusassume for definiteness that it is b−1; the argument for b is similar except for an extrareflection.Thuswecanwritetheworddescribingthepresentationin the form baXb−1Y, where X and Y are unknown words, not both empty. NowcutalongcandpastealongbasinFigure6.27.Inthenewpresentation, the number of vertices labeled v has decreased, and the number labeled w has increased. We may have introduced a new adjacent complementary pair, so perform Step 2 again to remove it. This may again decrease the number of vertices labeled v (for example, if a v vertex lies between edges labeled aa−1 that are eliminated by folding), but it cannot increase their number. So repeating this sequence a finite number of times—decrease the v vertices by one, then eliminate adjacent complementary edges—we eventually eliminate the vertex class v from the presentation altogether. Iterate this procedure for each vertex class until there is only one left. Step 5: If the presentation has any complementary pair a,a−1, then it has another b,b−1 that occurs intertwined with the first, as in a,…,b,…,a−1,…,b−1. If this is not the case, then the presentation is of theformaXa−1Y,whereX containsonlymatchedcomplementarypairsor

Classification of Surface Presentations 141 v w c x a b w x X Y c b X Y v x a c v b x w FIGURE 6.27. Reducing the number of vertices equivalent to v. c c W b a b b b c Z d X c Y Z d c Y X Z b W a b a b b a Y c Y X W X d Z c W FIGURE 6.28. Bringing intertwined complementary pairs together. adjacenttwistedpairs.ThuseachedgeinX isidentifiedonlywithanother edge in X, and the same is true of Y. But this means that the terminal verticesoftheaanda−1 edges,bothofwhichtouchonlyX,canbeidenti- fiedonlywithverticesinX,whiletheinitialverticescanbeidentifiedonly with vertices in Y. This is a contradiction, since all vertices are identified together by Step 4. Step6:M admitsapresentationinwhichallintertwinedcomplementary pairs occur together with no other edges intervening: aba−1b−1. If the pre- sentation is given by the word WaXbYa−1Zb−1, perform the elementary transformations indicated in Figure 6.28 (cut along c, paste along a, cut alongd,andpastealongb)toobtainthenewwordcdc−1d−1WZYX.This replaces the old intertwined set of pairs with a new adjacent set cdc−1d−1, withoutseparatinganyedgesthatwerepreviouslyadjacent.Repeatthisfor each set of intertwined pairs. (Note that this step requires no reflections.) Step 7: M is homeomorphic to either a connected sum of tori or a con- nected sum of projective planes.Fromwhatwehavedonesofar,alltwisted pairs occur adjacent to each other, and all complementary pairs occur in intertwinedgroupslikeaba−1b−1.Thisisapresentationofaconnectedsum

142 6. Curves and Surfaces of tori (presented by aba−1b−1) and projective planes (presented by cc). If there are only tori or only projective planes, we are done. The only remaining case is that in which the presentation contains both twistedandcomplementarypairs.Inthatcase,sometwistedpairmustoc- cur next to a complementary one; thus the presentation is described either byawordoftheformaba−1b−1ccX orbyoneoftheformccaba−1b−1X.In eithercase,thisisaconnectedsumofatorus,aprojectiveplane,andwhat- ever surface is described by the word X. But Lemma 6.16 shows that the standardpresentationofT2#P2 canbetransformedtothatofP2#P2#P2. Making this transformation, we eliminate one of the occurrences of T2 in the connected sum. Iterating this procedure, we eliminate them all, thus completing the proof. Combinatorial Invariants Two of the properties of simplicial complexes introduced in Chapter 5— orientations and the Euler characteristic—generalize easily to polygonal surface presentations, and give us interesting information about surfaces. Letusextendthenotionofcombinatorialinvariancebysayingthataprop- erty of a polygonal presentation is a combinatorial invariant if it is un- changed by elementary transformations. Of course, any topological invari- ant of the geometric realization (such as connectedness) is automatically a combinatorial invariant, because elementary transformations yield homeo- morphic surfaces. It would not be difficult to show that every polygonal presentationcanbetriangulatedandthatthetwonotionsofcombinatorial invariants coincide; however, in this case it is easier just to work directly with the polygonal presentations. The Euler Characteristic The Euler characteristic can be generalized to polygonal presentations in the following form. If P is a polygonal presentation, define the Euler char- acteristic of P to be V −E+F, where V is the number of vertices (after identifications), E is the number of edge symbols (which is equal to the number of edges after identifications), and F is the number of faces. If P is the presentation determined by a 2-dimensional simplicial complex, it is easytocheckthatthisdefinitionagreeswiththedefinitiongiveninChapter 5. Theorem 6.17. The Euler characteristic of a polygonal presentation is unchanged by elementary transformations. Proof. It is obvious that relabeling, rotating, and reflecting do not change the Euler characteristic of a presentation, because they leave the numbers

Combinatorial Invariants 143 of vertices, edges, and faces individually unchanged. For the other trans- formations, we need only check that the changes to these three numbers cancelout.Subdividingincreasesboththenumberofedgesandthenumber ofverticesbyone,leavingthenumberoffacesunchanged.Cuttingincreases both the number of edges and the number of faces by one, and leaves the numberofverticesunchanged.Unfoldingincreasesthenumberofedgesand the number of vertices by one, and leaves the number of faces unchanged. Finally, consolidating, pasting, and folding leave the Euler characteristic unchanged, since they are the inverses of subdividing, cutting, and unfold- ing, respectively. Proposition 6.18 (Euler Characteristics of Compact Surfaces). The Euler characteristic of a standard surface presentation is equal to (a) 2 for the sphere; (b) 2−2n for the connected sum of n tori; (c) 2−n for the connected sum of n projective planes. Proof. Just compute. These results allows us to conclude a great deal about a surface from a given presentation, without actually carrying out the reduction to a stan- dardpresentation.Forexample,anypresentationwithEulercharacteristic 2 gives the sphere, and a presentation with Euler characteristic 0 gives either the torus or the Klein bottle ≈P2#P2. At the moment, we do not know that the Euler characteristic is a topo- logical invariant, for the simple reason that we still do not know that the standard surfaces on our list are not homeomorphic to each other. (If you donotbelievethis,justtrytoprove,forexample,thattheprojectiveplane is not homeomorphic to the torus using the techniques we have developed so far!) The problem is that we cannot yet rule out the possibility that P2, say, could have a presentation that is so exotic that it is not related to the standardonebyaseriesofelementarytransformations,butsomehowman- ages to reduce to a presentation of the torus after following the algorithm oftheclassificationtheorem.WewillremedythisdeficiencyinChapter10, when we show that all of our standard compact surfaces are topologically distinct;onlythenwillwebeabletocompletetheclassificationofcompact surfaces. TheEulercharacteristiccanbeusedbyitselftodistinguishpresentations that reduce to connected sums of different numbers of tori or connected sums of different numbers of projective planes. However, to distinguish a presentationoftheconnectedsumof ntorifromoneoftheconnectedsum of 2n projective planes (for example, the torus from the Klein bottle), we will need one more property: orientability.

144 6. Curves and Surfaces Orientability In Chapter 5 we introduced the notion of orientation of a simplicial com- plex. In this section we show how to extend it to polygonal presentations. Note that a permutation of a set with three elements is even if and only if it is a cyclic permutation (i.e., of the form i (cid:10)→ j (cid:10)→ k (cid:10)→ i). Thus an orientation of a 2-simplex is just an equivalence class of vertex orderings modulo cyclic permutations. Suppose P is a surface presentation arising from a simplicial complex, so that every word has length 3. Each word determines an orientation of the associated 2-simplex, by ordering the ver- tices in counterclockwise order. It is easy to check that when two simplices are glued together via an edge pairing, their orientations are consistent if and only if the edge pair is complementary. Motivated by this, we make the following definition. A surface presentation P is said to be oriented if it has no twisted edge pairs. Intuitively, this means that you can decide whichisthe“front”side(or“outside”)of|P|bycoloringthetopsurfaceof each polygon white and the bottom side gray; the condition on edge pairs ensures that the colors will match up when edges are pasted together. A compact surface is said to be orientable if it admits an oriented pre- sentation. By looking a little more closely at the proof of the classification theorem, we can identify exactly which compact surfaces are orientable. Proposition 6.19. A compact surface is orientable if and only if it is homeomorphic to the sphere or a connected sum of one or more tori. Proof. The standard presentations of the sphere and the connected sums oftoriareoriented,sothesesurfacesarecertainlyorientable.Toshowthat an orientable surface is homeomorphic to one of these, let M be any sur- face that admits at least one orientable presentation. Starting with that presentation, follow the algorithm described in the proof of the classifi- cation theorem to transform it to one of the standard presentations. The only elementary transformation that can introduce a twisted pair into an oriented presentation is reflection. The only steps in which reflections are used are Steps 3, 4, and 7, and you can check that none of those steps require any reflections if there were no twisted pairs to begin with. Thus the classification theorem tells us that the presentation can be reduced to one of the standard ones with no twisted pairs, which means that M is homeomorphic to a sphere or a connected sum of tori. Because of this result, the connected sum of n tori is also known as the orientablesurfaceofgenus n,andtheconnectedsumofnprojectiveplanes iscalledthenonorientable surface of genus n.Byconvention,thesphereis the (unique, orientable) surface of genus 0. Technically, this terminology is premature,becausewestilldonotknowthataconnectedsumofprojective planes is not homeomorphic to an oriented surface. But for now we will go ahead and use this standard terminology with the caveat that all we

Combinatorial Invariants 145 know about the “nonorientable surface of genus n” is that its standard presentation is not oriented. Before moving away from classification theorems, it is worth remarking onthesituationwithhigher-dimensionalmanifolds.Becauseofthetriangu- lation theorem for 3-manifolds stated in Chapter 5, one might hope that a similarapproachtoclassifying3-manifoldsmightbearfruit.Unfortunately, the combinatorial problem of reducing any given 3-manifold triangulation to some standard form is, so far, unsolved. And this approach cannot get us very far in dimensions higher than 3, because we do not have trian- gulation theorems. Thus, in order to make any progress in understanding higher-dimensional manifolds, as well as to resolve the question of whether the standard surfaces are distinct, we will need to develop more powerful tools. This we will do in the remainder of the book.

146 6. Curves and Surfaces Problems 6-1. For each of the following surface presentations, compute the Euler characteristic, and then apply the algorithm of the classification the- orem to determine which of our standard surfaces it is. (a) (cid:22)a,b,c|abacb−1c−1(cid:23). (b) (cid:22)a,b,c|abca−1b−1c−1(cid:23). (c) (cid:22)a,b,c,d,e,f |abc, bde, c−1df, e−1fa(cid:23). (d) (cid:22)a,b,c,d,e,f,g,h,i,j,k,l,m,n,o| abc, bde, dfg, fhi, haj, c−1kl, e−1mn, g−1ok−1, i−1l−1m−1, j−1n−1o−1(cid:23). 6-2. Showthataconnectedsumofoneormoreprojectiveplanescontains a subspace that is homeomorphic to the M¨obius band. 6-3. Show that the projective plane is homeomorphic to the quotient ob- tainedbygluingaM¨obiusbandandadisktogetheralongtheircom- mon boundary. 6-4. ShowthattheKleinbottleishomeomorphictothequotientobtained by gluing two M¨obius bands together along their common boundary.

7 Homotopy and the Fundamental Group The results of the preceding chapter left a serious gap in our attempt to classify compact 2-manifolds up to homeomorphism: Although we have exhibitedalistofsurfacesandshownthateverycompact,connectedsurface is homeomorphic to one on the list, we still have no way of knowing when two surfaces are not homeomorphic. For all we know, all the surfaces on our list might be homeomorphic to the sphere! (Think, for example, of the unexpected homeomorphism between P2#P2#P2 and T2#P2.) To distinguish nonhomeomorphic surfaces, we need topological invari- ants. For some surfaces, the properties we already know suffice. For exam- ple,the2-sphereisnothomeomorphictotheplanebecauseoneiscompact, while the other is not. The plane, the disjoint union of two planes, and the disjoint union of three planes are all topologically distinct, because they have different numbers of components. It follows from Problem 4-1 that thelineisnothomeomorphictotheplane;theproofinvolvedarathersub- tle use of connectedness. But to decide whether, for example, the sphere is homeomorphictothetorus,ortheplaneishomeomorphictothepunctured plane R2(cid:3){0}, we need to introduce some new invariants. Inthischapterwebeginourstudyofthefundamentalgroup,analgebraic groupthatmeasuresthenumberof“holes”inaspace,inacertainsense.To set the stage, let us think about the difference between the plane and the punctured plane. Both are connected, noncompact 2-manifolds, so they cannot be distinguished by any of the basic topological properties that we have discussed so far. Yet intuition suggests that they should not be homeomorphic to each other because the punctured plane has a “hole,” while the full plane does not.

148 7. Homotopy and the Fundamental Group Toseehowthisdistinctionmightbedetectedtopologically,observethat every closed curve in R2 can be continuously shrunk to a point (you will prove this rigorously in Exercise 7.3 below); by contrast, it is intuitively clear that a circle drawn around the hole in the space R2(cid:3){0} can never becontinuouslyshrunktoapointwhileremaininginthespace,andinfact cannotbedeformedintoanyclosedpaththatdoesnotgoaroundthehole. Wewilldefineanequivalencerelationonclosedpathswithafixedstart- ing and ending point: Two paths are equivalent if one can be continuously deformed into the other while keeping the starting and ending point fixed. Thesetofequivalenceclassesiscalledthefundamentalgroupofthespace; the product of two elements of the group is obtained by first following one path and then the other. After making the basic definitions, we will prove that homeomorphic spaces have isomorphic fundamental groups. Then we willprovethatthefundamentalgroupsatisfiesanevenstrongerinvariance property, that of homotopy invariance. As a consequence, we will be able toreducethecomputationsoffundamentalgroupsofmanyspacestothose of simpler ones. Provingthatthefundamentalgroupofaspaceisnottrivialturnsoutto be somewhat harder, and we will not do so until the next chapter. Homotopy Let X and Y be topological spaces, and let f,g: X → Y be continuous maps. A homotopy from f to g is a continuous map H: X×I →Y (where I =[0,1] is the unit interval) such that for all x∈X, H(x,0)=f(x); H(x,1)=g(x). (7.1) If there exists a homotopy from f to g, we say that f and g are homo- topic, and write f (cid:25) g (or H: f (cid:25) g if we want to emphasize the specific homotopy). Ahomotopydefinesaone-parameterfamilyofmapsH (x)=H(x,t)for t 0≤t≤1 (Figure 7.1), and condition (7.1) says that H0 =f and H1 =g. We usually think of the parameter t as time, and think of H as giving a “deformation” of f into g as t goes from 0 to 1. The continuity of H guarantees that this deformation proceeds continuously without breaks or jumps. Lemma 7.1. For any topological spaces X and Y, homotopy is an equiv- alence relation on the set of all continuous maps from X to Y. Proof. Anymapf ishomotopictoitselfviathetrivialhomotopyH(x,t)= f(x), so homotopy is reflexive. Similarly, if H: f (cid:25) g, then a homotopy from g to f is given by H (cid:7) (x,t) = H(x,1−t), so homotopy is symmetric. Finally, if F: f (cid:25)g and G: g (cid:25)h, define H: X×I →Y by following F at

Homotopy 149 f1 Y X×I H f0 FIGURE 7.1. A homotopy between f0 and f1. double speed for 0 ≤ t ≤ 1, and then following G at double speed for the 2 remainder of the unit interval. Formally, (cid:16) F(x,2t), 0≤t≤ 1; H(x,t)= 2 G(x,2t−1), 1 ≤t≤1. 2 Since F(x,1) = g(x) = G(x,0), the two definitions of H agree at t = 1, where they overlap. Thus H is continuous by the gluing lemma, and 2 is therefore a homotopy between f and h. This shows that homotopy is transitive. Lemma 7.2. The homotopy relation is preserved by composition: If f0,f1: X →Y and g0,g1: Y →Z are continuous maps with f0 (cid:25)f1 and g0 (cid:25)g1, then g0 ◦f0 (cid:25)g1 ◦f1. Proof. SupposeF: f0 (cid:25)f1 andG: g0 (cid:25)g1 arehomotopies.DefineH: X× I → Z by H(x,t) = G(F(x,t),t). At t = 0, H(x,0) = G(f0(x),0) = g0(f0(x)), and at t = 1, H(x,1) = G(f1(x),1) = g1(f1(x)). Thus H is a homotopy from g0 ◦f0 to g1 ◦f1. Example 7.3. Define maps f: R→R2 by f(x)=(x,x2); g(x)=(x,x). Then the map H(x,t)=(x,x2−tx2+tx) is a homotopy from f to g. Example 7.4. Let B ⊂ Rn and let X be any topological space. Suppose f,g: X → B are any two continuous maps with the property that for all

150 7. Homotopy and the Fundamental Group Rn X×I B FIGURE 7.2. A straight-line homotopy. x∈X, the line segment from f(x) to g(x) lies in B. This will be the case, forexample,ifB isconvex.DefineahomotopyH: f (cid:25)g bylettingH(x,t) trace out the line segment from f(x) to g(x) at constant speed as t goes from 0 to 1 (Figure 7.2): H(x,t)=(1−t)f(x)+tg(x). This is called the straight-line homotopy between f and g. It shows, in particular,thatallmapsfromagivenspaceintoaconvexsetarehomotopic. The Fundamental Group Recall that a path in a topological space X is a continuous map f: I →X. The points p=f(0) and q =f(1) are called the initial point and terminal point of f, respectively, and we say that f is a path “from p to q.” We will use paths to detect “holes” in a space. Example 7.5. Consider the path α: I → R2 (cid:3){0} defined (in complex notation) by α(s)=e2πis and the map H: I×I →R2(cid:3){0} by H(s,t)=e2πist. Ateachtimet,H isapaththatfollowsthecircleonlyasfarasangle2πt, t so H0 is the constant path c1(s) ≡ 1 and H1 = α. Thus H is a homotopy from the constant path to α.

The Fundamental Group 151 This last example shows that the circular path around the origin is ho- motopic in R2 (cid:3){0} to a constant path, so that simply asking whether a closed path is homotopic to a constant is not sufficient to detect holes. To remedy this, we will need to consider homotopies of paths throughout whichtheendpointsstayfixed.Moregenerally,itwillbeusefultoconsider homotopies that fix an arbitrary subset of the domain. Let X and Y be topological spaces, and A ⊂ X an arbitrary subspace. A homotopy H between maps f,g: X → Y is called a homotopy relative to A if H(x,t)=f(x) for all x∈A, t∈I. In other words, for each t, the map H agrees with f on A. If there exists t such a homotopy, we say that f and g are homotopic relative to A and write f (cid:25) A g. Notice that this implies g| A =H1 | A =f| A , so for two maps to be homotopic relative to A they must first of all agree on A. Nowsupposef andg aretwopathsinX.Apathhomotopyfromf tog is a homotopy relative to {0,1}, that is, a homotopy that fixes the endpoints for all time. If there exists a path homotopy between f and g, we say they are path homotopic, and write f ∼g. By the remark above, this is possible only if f and g both have the same initial point and the same terminal point. Lemma 7.6. For any points p,q ∈ X, path homotopy is an equivalence relation on the set of all paths from p to q. Exercise 7.1. Prove Lemma 7.6. For any path f in X, we denote the path homotopy equivalence class of f by [f], and call it the path class of f. For our purposes, we will be most interested in paths that start and end at the same point. Such a path is calledaloop.Iff isaloopwhoseinitialandterminalpointisq ∈X,wesay that f is based at q, and we call q the base point of f. The set of all loops in X based at q is denoted by Ω(X,q). The constant loop c ∈ Ω(X,q) is q the map c (s) ≡ q. If a loop is path homotopic to the constant loop, we q say that it is null homotopic. One (not very interesting, but sometimes useful) way to get homotopic pathsisbythefollowingconstruction.Areparametrizationofapathf: I → X is a path of the form f ◦ϕ for some homeomorphism ϕ: I →I fixing 0 and 1. Lemma 7.7. Any reparametrization of a path f is path homotopic to f. Proof. Suppose f ◦ϕ is a reparametrization of f, and let H: I ×I → I denote the straight-line homotopy from the identity map to ϕ. Then f◦H is a homotopy from f to f ◦ϕ.

152 7. Homotopy and the Fundamental Group t f1 g1 f1 g1 H F G f0 g0 s f0 g0 FIGURE 7.3. Homotopy invariance of path multiplication. Lemma 7.6 says that path homotopy is an equivalence relation on Ω(X,q). We define the fundamental group of X based at q, denoted by π1(X,q), to be the set of path classes of loops based at q. Tomakeπ1(X,q)intoagroup,wemustdefineamultiplicationoperation. This is done first on the level of paths: The product of two paths f and g is the path obtained by first following f and then following g, both at double speed. For future use, we will define products of paths in a more general setting—instead of requiring that both paths start and end at the samepoint,wewillrequiresimplythatthesecondonestartwherethefirst ends. Thus let f,g: I →X be paths with f(1)=g(0). We define their product · f g: I →X by (cid:16) f · g(s)= f(2s); 0≤s≤ 1 2 ; g(2s−1); 1 ≤s≤1. 2 (Here and throughout the book we will consistently use s as the “space variable” parametrizing individual paths, and reserve t for the “time vari- · able” in homotopies.) The condition f(1) = g(0) guarantees that f g is continuous by the gluing lemma. Lemma 7.8 (Homotopy Invariance of Path Multiplication). Path multiplicationiswell-definedonpathclasses.Moreprecisely,iff0 ∼f1 and · · g0 ∼ g1, and if f0 g0 is defined (that is, if f0(1) = g0(0)), then f1 g1 is · · also defined and f0 g0 ∼f1 g1. Proof. Let F: f0 ∼ f1 and G: g0 ∼ g1 be homotopies (Figure 7.3). The · · required homotopy H: f0 g0 ∼f1 g1 is given by (cid:16) F(2s,t); 0≤s≤ 1, 0≤t≤1; H(s,t)= 2 G(2s−1,t); 1 ≤s≤1, 0≤t≤1. 2

The Fundamental Group 153 t t c p f −1 f f p s s c f p FIGURE 7.4. f ∼cp · f. FIGURE 7.5. cp ∼f · f−1. Again, this is continuous by the gluing lemma. Withthisresult,itmakessensetodefinemultiplicationofpathclassesby · · · setting [f] [g]=[f g] whenever f g is defined. In particular, it is always defined for [f],[g] ∈ π1(X,q). We wish to show that π1(X,q) is a group under this multiplication, which amounts to proving associativity of path class multiplication and the existence of an identity and inverses. Again, it will be useful to prove these properties in a slightly more general setting, forpathsthatdonotnecessarilyhavethesameinitialandterminalpoints. For any path f, we define the reverse path f−1 by f−1(s)=f(1−s); this just retraces f from its terminal point to its initial point. Recall that c q denotes the constant loop at q. Theorem 7.9 (Properties of Path Multiplication). Let f be any path from p to q in a space X, and let g and h be any paths in X. Path multiplication satisfies the following properties: · · (a) [c ] [f]=[f] [c ]=[f]. p q (b) [f] · [f−1]=[c ]; [f−1] · [f]=[c ]. p q · · · · (c) [f] ([g] [h])=([f] [g]) [h] whenever either side is defined. · Proof. For(a),letusshowthatc f ∼f;theproducttheotherwayworks p similarly. Define H: I×I →X (Figure 7.4) by ⎧ ⎨p,(cid:20) (cid:21) t≥2s; H(s,t)= 2s−t ⎩f , t≤2s. 2−t Geometrically, this maps the portion of the square on the left of the line t = 2s to the point p, while it maps the portion on the right along the path f at increasing speeds as t goes from 0 to 1. (The slanted lines in the picture are the level sets of H, i.e., the lines along which H takes the same

154 7. Homotopy and the Fundamental Group value.) This map is continuous by the gluing lemma, and you can check · · that H(s,0)=f(s) and H(s,1)=c f(s). Thus H: f ∼c f. p p For(b),wewilljustshowthatf · f−1 ∼c .Since(f−1)−1 =f,theother p relationfollowsbyinterchangingtherolesoff andf−1.Defineahomotopy H: c ∼ f · f−1 by the following recipe (Figure 7.5): At any time t, the p path H follows f as far as f(t) at double speed while the parameter s is t in the interval [0,t/2]; then for s ∈ [t/2,1−t/2] it stays at f(t); then it retraces f at double speed back to q. Formally, ⎧ ⎪⎨f(2s), 0≤s≤t/2; H(s,t)= f(t), t/2≤s≤1−t/2; ⎪⎩ f(2−2s) 1−t/2≤s≤1. It is easy to check that H is a homotopy from c to f · f−1. p · · · · Finally,toproveassociativity,weneedtoshowthat(f g) h∼f (g h). The first path follows f and then g at quadruple speed for s ∈ [0,1], and 2 then follows h at double speed for s ∈ [1,1], while the second follows f 2 at double speed and then g and h at quadruple speed. The two paths are therefore reparametrizations of each other and thus homotopic. Corollary 7.10. For any space X and any point q ∈ X, π1(X,q) is a group. Note that path multiplication is not associative on the level of paths, onlyonthelevelofpathhomotopyclasses.Fordefiniteness,letusagreeto interpret products of more than two paths as being grouped from left to · · · · right if no parentheses are present, so that f g h means (f g) h. The next question we need to address is how the fundamental group depends on the choice of base point. The first thing to notice is that if X is not path connected, we cannot expect the fundamental groups based at points in different path components to have any relationship to each other; π1(X,q) can give us information only about the path component containing q. Therefore, the fundamental group is usually used only to study path connected spaces. When X is path connected, it turns out that the fundamental groups at different points are all isomorphic; the next theorem gives an explicit isomorphism between them. Theorem 7.11 (Change of Base Point). Suppose X is path con- nected, p,q ∈ X, and g is any path from p to q. The map Φ g : π1(X,p) → π1(X,q) defined by Φ [f]=[g−1] · [f] · [g] g is an isomorphism. Proof. Beforewebegin,weshouldverifythatΦ makessense(Figure7.6): g

The Fundamental Group 155 f g q p FIGURE 7.6. Change of base point. Since g goes from p to q and f goes from p to p, paths in the class [g−1] · [f] · [g] go from q to p (by g−1), then from p to p (by f), and then from p back to q (by g), so Φ g (f) does indeed define an element of π1(X,q). To check that Φ is a group homomorphism, use Theorem 7.9: g · Φ g [f1] Φ g [f2] =[g−1] · [f1] · [g] · [g−1] · [f2] · [g] =[g−1] · [f1] · [c p ] · [f2] · [g] =[g−1] · [f1] · [f2] · [g] · =Φ g ([f1] [f2]). (This is one reason why we needed to prove the properties of Theorem 7.9 for paths that start and end at different points.) Finally, the fact that Φ is an isomorphism follows easily from the fact g that it has an inverse, given by Φ g−1 : π1(X,q)→π1(X,p). Because of this theorem, when X is path connected we sometimes use the imprecise notation π1(X) to refer to the fundamental group of X with respect to an unspecified base point, if the base point is irrelevant. For ex- ample, we might say “π1(X) is trivial” if π1(X,q)={[c q ]} for any q ∈X; or we might say “π1(X) ∼ =Z” if there exists an isomorphism π1(X,q)→Z for some (hence any) q. However, we cannot dispense with the base point altogether: Since different paths from p to q may give rise to different iso- morphisms,whenweneedtorefertoaspecificelementofthefundamental group, or to a specific homomorphism between fundamental groups, we must be careful to specify all base points.

156 7. Homotopy and the Fundamental Group X k g I×I F h f FIGURE 7.7. The map of Lemma 7.12. If X is path connected and π1(X) is trivial, we say that X is simply connected. This means that every loop in X can be continuously shrunk to a constant loop while its endpoints are kept fixed. Exercise 7.2. Let X be a topological space. (a) Letf,g: I →X betwopathsfromptoq.Showthatf ∼gifandonly · if f g−1 ∼cp. (b) Show that X is simply connected if and only if any two paths in X with the same initial and terminal points are path homotopic. Exercise 7.3. Show that any convex subset of Rn is simply connected. In particular, Exercise 7.3 shows that the plane is simply connected. We will see later that the punctured plane is not, thus proving that the two spacesarenothomeomorphic.In fact,wewillshowthatbothR2(cid:3){0}and S1 have infinite cyclic fundamental groups, generated by the path class of a loop that winds once around the origin. The proof will occupy most of our efforts for the remainder of this chapter and the next. Lemma 7.12. Let F: I ×I → X be a continuous map, and let f, g, h, and k be the paths in X defined by f(s)=F(s,0); g(s)=F(1,s); h(s)=F(0,s); k(s)=F(s,1). · · (See Figure 7.7.) Then f g ∼h k. Exercise 7.4. Prove Lemma 7.12. Considernowtheloopα: I →S1 givenbyα(s)=e2πis.Thismapwraps the interval once around the circle counterclockwise, and maps 0 and 1 to

The Fundamental Group 157 f 0 1 X α q 1 (cid:7) f FIGURE 7.8. The circle representative of a loop. the base point 1 ∈ S1. By the closed map lemma, it is a quotient map. If f: I → X is any loop in a space X, it passes to the quotient to give a unique map f (cid:7) : S1 → X such that f (cid:7)◦α = f (Figure 7.8), which we call (cid:7) the circle representative of f. Conversely, any continuous map f from the circle to X is the circle representative of the map f =f (cid:7)◦α. The next proposition gives a convenient criterion for detecting null ho- motopic loops in terms of their circle representatives. Proposition 7.13. A loop in a space X is null homotopic if and only if its circle representative extends to a map from the closed disk into X. Proof. Let f: I → X be a loop based at q ∈ X, and f (cid:7) : S1 → X its circle representative. Suppose first that f (cid:7) extends to a map F: B2 → X. Since B2 isconvex,theloopαisnullhomotopicwhenthoughtofasaloopinB2. Let A: I ×I → B2 be a path homotopy from α to the constant loop c1. BecauseA(s,t)∈S1 whent=0or1,thecompositemapF◦A: I×I →X satisfies F ◦A(s,0)=f (cid:7)◦A(s,0)=f (cid:7)◦α(s)=f(s), F ◦A(s,1)=f (cid:7)◦A(s,1)=f (cid:7)◦c1(s)=f(1)=q, and is therefore a homotopy from f to c . q Conversely, suppose f is null homotopic, and let H: I ×I → X be a homotopy from c to f. Observe that H sends the sides and bottom of the q square to the point q. We will show below that there is a quotient map π: I × I → B2 that sends these three sides to 1 ∈ S1, makes no other identifications, and restricts to α on the top side I ×{1}. Granting this for the moment, the homotopy H passes to the quotient to give a map

158 7. Homotopy and the Fundamental Group b a b c A B C h h d e a d e c FIGURE7.9.Aquotientmapsendingthreesidesofthesquaretoapoint. H (cid:7) : B2 → X satisfying H (cid:7) ◦π = H. The restriction of H (cid:7) to the circle is (cid:7) clearly the circle representative of f, so H is the desired extension. We will construct the claimed quotient map π: I ×I → B2 in several steps.LetT beanequilateraltriangleofaltitude2intheplane.Triangulate I×I and T as shown in Figure 7.9, and let A: I×I →T be the simplicial homeomorphismdeterminedbytheindicatedvertexmap.ThenletB: T → B2 be the (nonsimplicial) map that sends each horizontal line segment in T linearly onto the horizontal segment at the same height in the disk. The resultingcompositemapB◦Aisaquotientmapbytheclosedmaplemma, and it identifies the sides and bottom of the square to a point but makes no other identifications. It is not quite the map we are seeking, because it takes the sides and bottom to −i instead of 1, and maps the top interval around the circle in the wrong direction. A suitable rotation and reflection of the disk, which we denote by C, corrects these problems. Let β: I →S1 denotetherestrictionofC◦B◦AtothetopsegmentI×{1}ofthesquare (identified with I). Although β is still not equal to α, the two maps differ only by a homeomorphism of S1 (since both α and β are quotient maps that make the same identifications). This homeomorphism extends to a homeomorphism of the closed disk by Problem 4-9, and composing with the inverse of this homeomorphism yields the desired map. Homomorphisms Induced by Continuous Maps Inthissectionweexploretheeffectofacontinuousmaponthefundamental groups of its domain and range. The first thing we need to know is that continuous maps preserve the path homotopy relation. Lemma 7.14. The path homotopy relation is preserved by composition with continuous maps. That is, if f0,f1: I → X are path homotopic and ϕ: X →Y is continuous, then ϕ◦f0 and ϕ◦f1 are path homotopic. Exercise 7.5. Prove Lemma 7.14

Homomorphisms Induced by Continuous Maps 159 An immediate consequence of this lemma is that any continuous map ϕ: X → Y induces a well-defined map ϕ∗: π1(X,q) → π1(Y,ϕ(q)) simply by setting ϕ∗[f]=[ϕ◦f]. Lemma 7.15. For any continuous map ϕ, ϕ∗ is a group homomorphism. Proof. Just note that · · · ϕ∗([f] [g])=ϕ∗[f g]=[ϕ◦(f g)]. · · Thusitsufficestoshowthatϕ◦(f g)=(ϕ◦f) (ϕ◦g).Thisisimmediate, because expanding both sides using the definition of path multiplication results in identical formulas. The homomorphism ϕ∗: π1(X,q) → π1(Y,ϕ(q)) is called the homomor- phism induced by ϕ. It has the following properties: Proposition 7.16 (Properties of the Induced Homomorphism). (a) Let ϕ: X →Y and ψ: Y →Z be continuous maps. Then (ψ◦ϕ)∗ = ψ∗ ◦ϕ∗. (b) If Id : X →X denotes the identity map of X, then for any q ∈X, X (Id X )∗ is the identity map of π1(X,q). Proof. Compute: ψ∗(ϕ∗[f])=ψ∗[ϕ◦f]=[ψ◦ϕ◦f]=(ψ◦ϕ)∗[f]; (Id X )∗[f]=[Id X ◦f]=[f]. Corollary 7.17 (Topological Invariance of π 1). Homeomorphic spaces have isomorphic fundamental groups. Specifically, if ϕ: X → Y is a homeomorphism, then ϕ∗: π1(X,q)→π1(Y,ϕ(q)) is an isomorphism. Proof. Ifϕisahomeomorphism,then(ϕ−1)∗ ◦ϕ∗ =(ϕ−1◦ϕ)∗ =(Id X )∗ = Id π1(X,q), and similarly ϕ∗ ◦(ϕ−1)∗ is the identity on π1(Y,ϕ(q)). Be warned that injectivity or surjectivity of a continuous map does not necessarily imply that the induced homomorphism has the same property. Forexample,acceptingforthemomentthefactthatπ1(S1)isinfinitecyclic (we will prove it in the next chapter), the inclusion map S1 (cid:9)→R2 is injec- tive, but its induced homomorphism is not, while the map ϕ: [0,1) → S1 of Exercise 2.7 that wraps the interval once around the circle is surjective (infactbijective),butitsinducedhomeomorphismisthetrivialhomomor- phism because [0,1) is convex and therefore simply connected.

160 7. Homotopy and the Fundamental Group B E A FIGURE 7.10. S1 is a retract of T2. FIGURE 7.11. Figure eight space. There is, however, one case in which the homomorphism induced by inclusion can be easily shown to be injective. Let X be a space and A ⊂ X a subspace. A continuous map r: X → A is called a retraction if the restrictionofrtoAistheidentitymapofA,orequivalentlyifr◦ι =Id , A A where ι : A(cid:9)→X is the inclusion map. If there exists a retraction from X A to A, we say that A is a retract of X. Proposition 7.18. Suppose A is a retract of X. If r: X → A is any retraction, then for any q ∈A, (ι A )∗: π1(A,q)→π1(X,q) is injective and r∗: π1(X,q)→π1(A,q) is surjective. Proof. Sincer◦ι A =Id A , r∗ ◦(ι A )∗ istheidentityon π1(A,q),fromwhich it follows that (ι A )∗ is injective and r∗ is surjective. Here are some examples of retractions. For these examples we will use without proof the fact that the fundamental group of the circle is infinite cyclic. Example 7.19. It is easy to check that the map r: Rn+1 (cid:3) {0} → Sn given by r(x) = x/|x| is a retraction. Thus the homomorphism π1(Sn) → π1(Rn+1(cid:3){0}) induced by inclusion is injective. In particular, in the case n=1, this means that π1(R2(cid:3){0}) has an infinite cyclic subgroup. Example 7.20. The torus T2 = S1 ×S1 has a subspace A = S1 ×{1} homeomorphic to S1 (Figure 7.10), and the map r: T2 → A given by r(p,q) = (p,1) is easily seen to be a retraction. Thus the image of the map (ι A )∗: π1(S1)→π1(T2) is an infinite cyclic subgroup of π1(T2). Example 7.21. Consider the figure eight space E ⊂ R2 (Figure 7.11), which is the union of the circles of radius 1 around (0,1) and (0,−1). Let B denote the upper circle. There are at least two different retractions of E onto B—one that maps the entire lower circle to the origin and is the identityonB,andanotherthat“folds”thelowercircleontotheupperone (formally, (x,y)(cid:10)→(x,|y|)). Thus π1(E) has an infinite cyclic subgroup.

Homotopy Equivalence 161 Homotopy Equivalence Although retractions are sometimes useful tools for showing that a certain fundamental group is not trivial, it is much more useful to have a criterion under which a continuous map induces an isomorphism of fundamental groups. In this section we explore a very general such criterion. Let ϕ: X → Y be a continuous map. We say that another continuous mapψ: Y →X isahomotopy inverseforϕifψ◦ϕ(cid:25)Id andϕ◦ψ (cid:25)Id . X Y Ifthereexistsahomotopyinverseforϕ,ϕiscalledahomotopyequivalence. Inthiscase,wesaythatX ishomotopy equivalenttoY,orX hasthesame homotopy type as Y, and we write X (cid:25)Y. Lemma 7.22. Homotopy equivalence is an equivalence relation. Exercise 7.6. Prove Lemma 7.22. One kind of homotopy equivalence is relatively easy to visualize. A sub- space A ⊂ X is said to be a deformation retract of X if there exists a retraction r: X → A such that the identity of X is homotopic to ι ◦r. A The homotopy H: Id (cid:25) ι ◦r is called a deformation retraction. Intu- X A itively, this means that X can be continuously deformed into A in such a way that points in A end up where they started. We say that A is a strong deformation retract of X if in addition Id (cid:25) (ι ◦r), which means that X A A points in A remain fixed throughout the deformation. In this case, the homotopy H can be called a strong deformation retraction. Example 7.23. InExample7.19weshowedthatSn isaretractofRn+1(cid:3) {0}. In fact, it is a strong deformation retract: The deformation retraction is given by H: (Rn+1(cid:3){0})×I →Rn+1(cid:3){0}, where x H(x,t)=(1−t)x+t . |x| Thisisjustthestraight-linehomotopyfromtheidentitymaptotheretrac- tion onto the sphere (Figure 7.12). If A is a deformation retract of X, since ι ◦r (cid:25)Id and r◦ι =Id , A X A A the inclusion A(cid:9)→X is a homotopy equivalence. Our main goal in this section is the following theorem, which is a much strongerinvariancepropertythanhomeomorphisminvariance,andwillen- able us to compute the fundamental groups of many more spaces. Theorem 7.24 (Homotopy Invariance of π 1). If ϕ: X →Y is a ho- motopy equivalence, then for any point q ∈X, ϕ∗: π1(X,q)→π1(Y,ϕ(q)) is an isomorphism. Before proving the theorem, let us look at several important examples. Example 7.25. Let X be any space. If the identity map of X is homo- topic to a constant map, we say that X is contractible. Other equivalent

162 7. Homotopy and the Fundamental Group FIGURE 7.12. Strong deformation retraction of R2(cid:3){0} onto S1. definitions are that any point of X is a deformation retract of X, or X is homotopy equivalent to a one-point space (Exercise 7.7). Concretely, con- tractibility means that there exists a continuous map H: X×I →X such that H(x,0)=x for all x∈X; H(x,1)=q for all x∈X. In other words, the whole space X can be continuously shrunk to a point. Some obvious examples of contractible spaces are convex subsets of Rn, and, more generally, any subset B ⊂ Rn that is star-shaped, which means thatthereissomepointq0 ∈B suchthatforeveryq ∈B,thelinesegment fromq0 toq iscontainedinB.Sinceaone-pointspaceissimplyconnected, it follows that every contractible space is simply connected. Exercise 7.7. Show that the following are equivalent: (a) X is contractible. (b) X is homotopy equivalent to a one-point space. (c) Any point of X is a deformation retract of X. Example 7.26. Example 7.23 showed that the circle is a strong deforma- tion retract of R2 (cid:3){0}. Therefore, inclusion S1 (cid:9)→ R2 (cid:3){0} induces an isomorphism of fundamental groups. Once we show that π1(S1) is infinite cyclic, this will characterize π1(R2(cid:3){0}) as well. Example 7.27. The figure eight space E of Example 7.21 and the theta space, defined by Θ={(x,y)∈R2 :x2+y2 =4, or y =0 and −2≤x≤2},

Homotopy Equivalence 163 FIGURE 7.13. Deformation retractions onto E and Θ. FIGURE 7.14. A tree. FIGURE 7.15. Not a tree. are both strong deformation retracts of R2 with the two points (0,1) and (0,−1) removed. The deformation retractions, indicated schematically in Figure 7.13, are defined by carving the space up into regions in which straight-line homotopies are easily defined; the resulting maps are con- tinuous by the gluing lemma. Therefore, since homotopy equivalence is transitive, E and Θ are homotopy equivalent to each other. Example 7.28 (Finite Trees). LetΓbeagraph.AcycleinΓisaclosed finiteedgepath,thatis,anedgepath(v0,…,v n )suchthatv0 =v n .Atree is a connected graph that contains no reduced cycles (see Figures 7.14 and 7.15). We will show that any finite tree T is contractible and thus simply connected. TheproofisbyinductiononthenumberofedgesinT.Ifthereisonlyone edge,thenT ishomeomorphictoaclosedintervalinR,whichisconvexand therefore contractible. So suppose every tree with n edges is contractible, and let T be a tree with n+1 edges.

164 7. Homotopy and the Fundamental Group v1 v0 T0 FIGURE 7.16. Proof that a tree is contractible. IfeveryvertexinT liesonatleasttwoedges,thenarguingasintheproof of the classification theorem for 1-manifolds (Theorem 6.1), there must be an infinite reduced edge path {v : i ∈ Z} in T. Because T is finite, there i mustbesomeintegersnandn+k >nsuchthatv n =v n+k .Ifnandk are chosen so that k is the minimum positive integer with this property, this meansthat(v n ,…,v n+k )isareducedcycle,contradictingtheassumption thatT isatree.Thustheremustbeatleastonevertexv0 thatliesononly one edge. Since T is connected, v0 lies on exactly one edge, say (cid:22)v0,v1 (cid:23). LetT0 denotethesubgraphofT withthevertexv0 andtheedge(cid:22)v0,v1 (cid:23) deleted (Figure 7.16). The straight-line homotopy defines a strong defor- mation retraction of (cid:22)v0,v1 (cid:23) onto v1; extending this to be the identity on T0 yieldsastrongdeformationretractionofT ontoT0,whichiscontinuous becauseitsrestrictiontoeachsimplexiscontinuous.Therefore,T ishomo- topy equivalent to T0, which is contractible by the induction hypothesis. Now we turn to the proof of Theorem 7.24. Roughly speaking, we would like to prove the theorem by showing that if ψ is a homotopy inverse for ϕ, then ψ◦ϕ(cid:25)Id X implies that ψ∗ ◦ϕ∗ is the identity map, and similarly forϕ∗ ◦ψ∗.Thiswouldrequireustoshowthathomotopicmapsinducethe samefundamentalgrouphomomorphisms.However,thereisanimmediate problemwiththisapproach:IftwomapsF0 andF1 arehomotopic,wehave no guarantee that both maps take the base point q ∈X to the same point in Y, so their induced homomorphisms do not even map into the same group! The following rather complicated-looking lemma is a substitute for the claimthathomotopicmapsinducethesamefundamentalgrouphomomor- phism. It says, in effect, that homotopic maps induce the same homomor- phism up to a canonical change of base point. Lemma 7.29. Suppose ϕ,ψ: X → Y are continuous, and H: ϕ (cid:25) ψ is a homotopy. For any q ∈X, let h be the path in Y from ϕ(q) to ψ(q) defined by h(t) = H(q,t). Let Φ h : π1(Y,ϕ(q)) → π1(Y,ψ(q)) be the isomorphism

Homotopy Equivalence 165 ψ◦f ψ ψ(q) Y h H ◦f H h(t) t I ϕ(q) q f ϕ◦f X ϕ FIGURE 7.17. Induced homomorphisms of homotopic maps. defined in Theorem 7.11. Then the following diagram commutes: π1(Y,ϕ(q)) (cid:3)(cid:4) ϕ∗ (cid:3) (cid:3) π1(X,q) Φ h (cid:6) ψ∗ (cid:6) (cid:6)(cid:7) (cid:5) π1(Y,ψ(q)). Proof. Let f be any loop in X based at q. What we need to show is ψ∗[f]=Φ h (ϕ∗[f]) ⇐⇒ ϕ∗[f]=Φ h−1 (ψ∗[f]) ⇐⇒ [ϕ◦f]=[h] · [ψ◦f] · [h−1] ⇐⇒ ϕ◦f ∼h · (ψ◦f) · h−1. Atanytimet,themapH ◦f isaloopbasedath(t)(Figure7.17).Thus t we can define a homotopy G from ϕ◦f to h · (ψ◦f) · h−1 by letting G t be the “lasso-shaped” loop that first follows h as far as h(t), then follows H ◦f at triple speed, then follows h back to ϕ(q). Formally, t ⎧ ⎪⎨h(3ts), 0≤s≤ 1, 3 G(s,t)= H(f(3s−1),t), 1 ≤s≤ 2, ⎪⎩ 3 3 h(3t(1−s)), 2 ≤s≤1. 3

166 7. Homotopy and the Fundamental Group It is straightforward to check that G is continuous by the gluing lemma. · · T ho h m e o p t a o t p h ic G t 0 o i ϕ s ◦ a f re ; p a a n r d am G e 1 tr is iz a at r io ep n a o r f am c ϕ e ( t q r ) iza ( t ϕ io ◦ n f o ) f h c ·ϕ( ( q ψ ), ◦ w f h ) ic · h h− is 1. path Proof of Theorem 7.24. Suppose ϕ: X → Y is a homotopy equivalence, and let ψ: Y →X be a homotopy inverse for it. Consider the sequence of maps π1(X,q)−ϕ→∗ π1(Y,ϕ(q))−ψ→∗ π1(X,ψ(ϕ(q)))−ϕ→∗ π1(Y,ϕ(ψ(ϕ(q)))). (7.2) We need to prove that the first ϕ∗ above is bijective. As we mentioned earlier, ψ∗ is not an inverse for it, because it does not map into the right space. Since ψ◦ϕ(cid:25)Id , Lemma 7.29 shows that there is a path h in X such X that the following diagram commutes: π1(X,q) (cid:3)(cid:4) Id (cid:3) (cid:3) π1(X,q) Φ h (cid:6) ψ∗ ◦ϕ∗ (cid:6) (cid:6)(cid:7) (cid:5) π1(X,ψ(ϕ(q))). (7.3) Thusψ∗ ◦ϕ∗ =Φ h ,whichisanisomorphism.Inparticular,thismeansthat the first ϕ∗ in (7.2) is injective and ψ∗ is surjective. Similarly, the homotopy ϕ◦ψ (cid:25)Id leads to the diagram Y π1(Y,ϕ(q)) (cid:3)(cid:4) Id (cid:3) (cid:3) π1(Y,ϕ(q)) Φ k (cid:6) ϕ∗ ◦ψ∗ (cid:6) (cid:6)(cid:7) (cid:5) π1(Y,ϕ(ψ(ϕ(q)))), from which it follows that ϕ∗ ◦ψ∗: π1(Y,ϕ(q)) → π1(Y,ϕ(ψ(ϕ(q)))) is an isomorphism.Thismeansinparticularthatψ∗ isinjective;sincewealready showed that it is surjective, it is an isomorphism. Therefore, going back to (7.3), we conclude that ϕ∗ = (ψ∗)−1 ◦Φ h : π1(X,q) → π1(Y,ϕ(q)) is also an isomorphism. Homotopy Equivalence and Deformation Retraction In Example 7.27 we showed that the figure eight and theta spaces are homotopy equivalent by showing that they are both deformation retracts

Homotopy Equivalence 167 (cid:7) X X×I π f (cid:7) Y Y Z f FIGURE 7.18. The mapping cylinder. ofasinglelargerspace.Thisexampleisnotasspecialasitmightseem.As thenextpropositionshows,twospacesarehomotopyequivalentifandonly if both are homeomorphic to deformation retracts of a single larger space. This gives a rather concrete way to think about homotopy equivalence. Let X and Y be topological spaces, and let f: X → Y be a continuous map. Define the mapping cylinder Z of f to be the quotient space of f (X ×I)(cid:20)Y by the equivalence relation generated by (x,0)∼f(x) for all x∈X. Let π denote the quotient map. The space Z can be visualized as f a “top hat” (Figure 7.18) formed by gluing the “cylinder” X×I to Y (the “brim”) by attaching each point (x,0) on the bottom of the cylinder to its image f(x) in Y. ThesubspaceX×{1}⊂(X×I)(cid:20)Y isasaturatedclosedsubsethomeo- morphic to X. The restriction of π to this subset is thus a one-to-one quo- (cid:7) (cid:7) tientmap,soitsimageX isalsohomeomorphictoX.Similarly,Y =π(Y) is homeomorphic to Y. Proposition 7.30. With notation as above, if f is a homotopy equiva- (cid:7) (cid:7) lence, then Y and X are deformation retracts of Z . Thus two spaces are f homotopy equivalent if and only if they are both homeomorphic to defor- mation retracts of a single space. Proof. For any (x,s) ∈ X ×I, let [x,s] = π(x,s) denote its equivalence class in Z ; similarly, [y]=π(y) is the equivalence class of y ∈Y. f (cid:7) FirstweshowthatY isastrongdeformationretractofZ ,assumingonly f that f is continuous. We define a retraction A: Z → Z , which collapses f f

168 7. Homotopy and the Fundamental Group f G F g H1 H2 H3 FIGURE 7.19. Homotopies of the mapping cylinder. (cid:7) Z down onto Y, by f A[x,s]=[x,0]; A[y]=[y]. To be a bit more precise, we should define a map A (cid:7) : (X×I)(cid:20)Y →Z by f (cid:7) (cid:7) A(x,s)=[x,0]andA(y)=[y].Thismapisevidentlycontinuousbecauseits restrictions to X×I and Y are compositions of continuous maps. Because (cid:7) (cid:7) (cid:7) A(x,0)=[f(x)]=A(f(x)), A respects the identifications made by π, so it passes to the quotient to yield the continuous map A defined above. This kind of standard argument will be used repeatedly to show that a map from Z is continuous; we will generally abbreviate it by saying something f like “A is well-defined and continuous because A[x,0]=[f(x)]=A[f(x)].” Define a homotopy H1: Z f ×I →Z f (Figure 7.19) by H1([x,s],t)=[x,s(1−t)]; H1([y],t)=[y]. Because H1([x,0],t) = [x,0] = [f(x)] = H1([f(x)],t), H1 is well-defined. To check that it is continuous, we need only observe that it respects the identificationsmadebythemapπ×Id: ((X×I)(cid:20)Y)×I →Z ×I,which f is a quotient map by Lemma 4.35. Since H1(ζ,0)=ζ and H1(ζ,1)=A(ζ) for any ζ ∈ Z f , H1 is a homotopy between the identity map of Z f and A. Since, moreover, H1([y],t) = [y] for all y ∈ Y, it is in fact a strong deformation retraction. Now suppose f is a homotopy equivalence, and let g: Y → X be a homotopy inverse for it. Thus there exist homotopies F: Y ×I → Y and G: X ×I → X such that F: f ◦g (cid:25) Id and G: g◦f (cid:25) Id . Define two Y X

Higher Homotopy Groups 169 more homotopies H2 and H3 by H2([x,s],t)=[F(f(x),1−t)]; H2([y],t)=[F(y,1−t)]; H3([x,s],t)=[G(x,st),t]; H3([y],t)=[g(y),t]. The straightforward verification that H2 and H3 are well-defined and con- tinuous is left to the reader. Geometrically, H2 deforms all of Z f into the (cid:7) (cid:7) image of f in Y along the homotopy F, and then H3 collapses Z f onto X by deforming each point along the homotopy G (Figure 7.19). Inserting t=0 and t=1 into the definitions of H2 and H3, we find that H2: A(cid:25)B and H3: B (cid:25)C, where B[x,s]=[g(f(x)),0]; B[y]=[g(y),0]; C[x,s]=[G(x,s),1]; C[y]=[g(y),1]. Because homotopy is transitive, the three homotopies H1,H2,H3 yield Id (cid:25) A (cid:25) B (cid:25) C. Since G(x,1) = x, we find that C[x,1] = [x,1], so Zf (cid:7) (cid:7) C is a retraction onto X, which shows that X is a deformation retract of Z . f Higher Homotopy Groups You might have wondered what the subscript 1 stands for in π1(X). As the notation suggests, the fundamental group is just one in a series of groups associated with a topological space, all of which measure “holes” of various dimensions. In this section we introduce the basic definitions withoutmuchdetail,justsothatyouwillrecognizethisconstructionwhen you see it again. We will not use this material anywhere else in the book. The definition of the higher homotopy groups is motivated by the fact that by identifying loops with their circle representatives as described ear- lier in this chapter, we can regard the fundamental group π1(X,q) as the setofequivalenceclassesofmapsfromS1 intoX taking1toq,moduloho- motopy relative to the base point 1. Generalizing this, for any nonnegative integer n, we define π (X,q) to be the set of equivalence classes of maps n

170 7. Homotopy and the Fundamental Group from Sn into X taking (1,0,…,0) to q, modulo homotopy relative to the base point. Just as in the case of the fundamental group, it can be shown that π (X,q) is a topological invariant. n Thesimplestcaseisn=0.SinceS0 ={±1},amapfromS0 toX sending the base point 1 to q is determined by where it sends −1. Two such maps are homotopic if and only if the two images of −1 lie in the same path component of X. Therefore, π0(X,q) can be identified with the set of path components of X. There is no canonical group structure on π0(X,q); it is merelyasetwithadistinguishedelement(thecomponentcontainingq).It is conventional to define π0(X) to be the set of path components without any distinguished element. For n > 1, π (X,q) has a multiplication operator (which we will not n describe here) under which it turns out to be an abelian group, called the nthhomotopygroupofX basedatq.Thesegroupsmeasuretheinequivalent waysofmappingSn intoX,andtellus,inasense,aboutthen-dimensional “holes”inX.Forexample,wewillseelaterthatπ1(R3(cid:3){0})istrivial;but it can be shown that the inclusion S2 (cid:9)→ R3(cid:3){0} represents a nontrivial element of π2(R3(cid:3){0}). The higher homotopy groups are notoriously hard to compute. In fact, only a limited amount is known about π (Sn) for k much larger than n. k Thestructureandcomputationofthesegroupsformtheembarkationpoint for a vast branch of topology known as homotopy theory. See [Whi78] for an excellent introduction to the subject. Categories and Functors In this section we digress a bit to give a brief introduction to category theory, a powerful idea that unifies many of the concepts we have seen so far, and indeed much of mathematics. We will only touch on these ideas fromtimetotimeinthisbook,butyouwillusethemextensivelyifyoudo more advanced work in algebraic topology, so it is important to familiarize yourself with the basic concepts. A category C consists of the following:

• a class (not necessarily a set) of objects;
• for each pair of objects X,Y a set HomC(X,Y) of morphisms; and
• for each triple X,Y,Z of objects a function called composition: HomC(X,Y)×HomC(Y,Z)→HomC(X,Z), written (f,g)(cid:10)→g◦f; such that the following axioms are satisfied: (i) Composition is associative: (f ◦g)◦h=f ◦(g◦h).

Categories and Functors 171 (ii) For each object X there exists an identity morphism Id ∈ X HomC(X,X) such that for any morphism f ∈ HomC(X,Y) we have Id ◦f =f =f ◦Id . Y X Therearemanyalternativenotationsinuse.ThesetHomC(X,Y)isalso sometimes denoted by C(X,Y) or even just Hom(X,Y) if the category in question is understood. An element f ∈ HomC(X,Y) is often written f: X →Y.Forthemostpartyoucanthinkoftheobjectsinacategoryas sets with some special structure and the morphisms as maps that preserve the structure, although the definitions do not require this, and we will see below that there are natural examples that are not of this type. Here are some familiar examples of categories, which we describe by specifying their objects and morphisms; the composition laws and identity morphisms are the obvious ones.

• SET: Sets and functions.
• GROUP: Groups and group homomorphisms.
• AB: Abelian groups and group homomorphisms.
• RING: Rings and ring homomorphisms.
• CRING: Commutative rings and ring homomorphisms.
• VECT R: Real vector spaces and R-linear maps.
• VECT C: Complex vector spaces and C-linear maps.
• TOP: Topological spaces and continuous maps.
• TOP ∗:Pointed topological spaces—topologicalspacestogetherwitha choice of base point in each—and base-point-preserving continuous maps.
• SIMP: Abstract simplicial complexes and simplicial maps. Ineachcase,theverificationoftheaxiomsofacategoryisstraightforward. Themainpointistoshowthatacompositionoftheappropriatestructure- preserving maps again preserves the structure. Associativity is automatic because it holds for composition of maps. InanycategoryC,amorphismf ∈HomC(X,Y)iscalledanisomorphism if there exists a morphism g ∈ HomC(Y,X) such that f ◦g = Id Y and g◦f =Id .Forexample,inSET,theisomorphismsarejustthebijections; X in GROUP they are the group isomorphisms; and in TOP they are the homeomorphisms. AsubcategoryofCisacategoryDwhoseobjectsare(someofthe)objects of C and whose sets of morphisms are subsets of the morphisms in C, with the composition law and identities inherited from C. A full subcategory is

172 7. Homotopy and the Fundamental Group one in which HomD(X,Y)=HomC(X,Y) whenever X,Y are objects of D. For example, AB is a full subcategory of GROUP. The real power of category theory becomes apparent when we consider relations between categories. Suppose C and D are categories. A covariant functorF fromCtoDisarulethatassignstoeachobjectX inCanobject F(X) in D, and to each morphism f ∈HomC(X,Y) an induced morphism F(f)∈HomD(F(X),F(Y)), in such a way that composition and identities are preserved: F(g◦h)=F(g)◦F(h); F(Id X )=IdF(X). In many cases, if the functor is understood, it is traditional to write the induced morphism F(g) as g∗. It is also frequently useful to consider contravariant functors, which are defined in exactly the same way as covariant functors, except that the induced morphisms go in the reverse direction: If g: X → Y, then F(g): F(Y)→F(X); and the composition law becomes F(g◦h)=F(h)◦F(g). It is common for the morphism F(f) induced by a contravariant functor F to be written f∗ if the functor is understood. (Note the upper star: The use of a lower star to denote a covariant induced morphism and an upper star to denote a contravariant one is universal.) Here are some important examples of functors. Example 7.31 (Covariant Functors).

• The fundamental group functor π1: TOP ∗ →GROUP assigns to each pointed topological space (X,q) its fundamental group based at q, and to each base-point-preserving continuous map its induced homo- morphism. The fact that it is a covariant functor is the content of Proposition 7.16.
• Thefunctorπ0: TOP→SETassignstoeachtopologicalspaceitsset ofpathcomponents.Foranycontinuousmapf: X →Y,theinduced mapf∗: π0(X)→π0(Y)justtakesapathcomponentX0 ofX to the path component of Y containing f(X0).
• Theforgetful functor F: TOP→SETjustassignstoeachtopological space its underlying set, and to each continuous map its underlying set map. In fact, such a functor exists for any category whose ob- jects are sets with some extra structure and whose morphisms are structure-preserving maps.
• The geometric realization functor from SIMP to TOP takes an ab- stract complex K to its geometric realization, and a simplicial map

Categories and Functors 173 f: K → L to the continuous map |f|: |K| → |L|. It is a functor because of Lemma 5.5. Example 7.32 (Contravariant Functors).

• The dual space functor from VECT R to itself assigns to each vector space V its dual space V∗ (the vector space of linear maps V → R), and to each linear map ϕ: V → W the dual map or transpose ϕ∗: W∗ → V∗ defined by ϕ∗(f)x = f(ϕx). The verification of the functorial properties can be found in most linear algebra texts.
• ThefunctorC: TOP→CRINGassignstoeachtopologicalspaceX its commutative ring C(X) of continuous real-valued functions f: X → R.Foranycontinuousmapϕ: X →Y,theinducedmapϕ∗: C(Y)→ C(X) is given by ϕ∗(f)=f ◦ϕ.
• If X and Z are abelian groups, the set Hom(X,Z) of group homo- morphismsisalsoanabeliangroupunderpointwiseaddition.Weget a contravariant functor from AB to itself by sending each group X to the group Hom(X,Z), and each homomorphism f: X → Y to the dual homomorphism f∗: Hom(Y,Z) → Hom(X,Z) defined by f∗(g)=g◦f. An immediate consequence of the definitions is that any (covariant or contravariant) functor from C to D takes isomorphisms in C to isomor- phismsinD:Theproofisexactlythesameastheproofforthefundamental group functor (Corollary 7.17). The examples considered so far are all categories whose objects are sets with some structure and whose morphisms are structure-preserving maps. Here are some examples that are not of this type. Example 7.33 (Homotopy Categories).
• ThehomotopycategoryHTOPisthecategorywhoseobjectsaretopo- logicalspacesasinTOP,butwhosemorphismsarehomotopy classes of continuous maps. Since composition preserves the homotopy rela- tion,thisisindeedacategory.Theisomorphismsinthiscategoryare the (homotopy classes of) homotopy equivalences.
• A closely related category is the pointed homotopy category HTOP ∗, which has the same objects as TOP ∗ but whose morphisms are the equivalence classes of continuous maps modulo homotopy relative to the base point. One consequence of the homotopy invariance of the fundamental group is that π1 defines a functor from HTOP ∗ to GROUP.

174 7. Homotopy and the Fundamental Group Example 7.34 (Groups as Categories). Suppose C is a category with oneobject,inwhicheverymorphismisanisomorphism.Ifwecalltheobject X,theentirestructureofthecategoryiscontainedinthesetHomC(X,X) of morphisms and its composition law. The axioms for a category say that any two morphisms can be composed to obtain a third morphism, that composition is associative, and that there is an identity morphism. The additional assumption that every morphism is an isomorphism means that each morphism has an inverse. In other words, HomC(X,X) is a group! Functors between such categories are just group homomorphisms. In fact, every group can be identified with such a category. One way to see this is to identify a group G with the subcategory of SET consisting of the one object G and the maps L : G→G given by left translation. g Another ubiquitous and useful technique in category theory goes by the nameof“universalmappingproperties.”Thesegiveaunifiedwaytodefine common constructions that arise in many categories, such as products and sums. Let{X :α∈A}beanyindexedcollectionofobjectsinacategoryC.An α object P together with a set of morphisms π : P →X called projections α α is said to be a product of the objects {X } if given any object W in C and α morphisms f : W →X , there exists a unique morphism f: W →P such α α that each of the following diagrams commutes: P (cid:3)(cid:4) f (cid:3) π α (cid:3) (cid:5) (cid:2) W X . f α α Lemma 7.35. If a product exists in any category, it is unique up to an isomorphism that respects the projections. Proof. If (P,{π }) and (P(cid:5),{π(cid:5) }) are both products of objects {X }, α α α the defining property guarantees the existence of maps f: P → P(cid:5) and f(cid:5): P(cid:5) →P satisfying π(cid:5) ◦f =π and π ◦f(cid:5) =π(cid:5) . Arguing exactly as in α α α α theproofthattheproducttopologyonX1 ×···×X n istheuniqueonesat- isfying the characteristic property (Theorem 3.10), we conclude that f◦f(cid:5) and f(cid:5)◦f are both identity maps. Inanyparticularcategory,productsmayormaynotexist.Herearesome examples of familiar categories in which products always exist. Example 7.36 (Categorical Products). (a) The product of a collection of sets in the category SET is just their Cartesian product.

Categories and Functors 175 (b) In the category TOP of topological spaces and continuous maps, the productoffinitelymanyspacesX1,…,X n isthespaceX1 ×···×X n with the product topology. The characteristic property of the prod- uct topology guarantees that the product space satisfies the defining conditionforaproduct.(Thecategoricaldefinitionofproduct,bythe way, determines the correct definition of the product topology on a product of infinitely many spaces; see Problem 7-12.) (c) The produ(cid:28)ct of groups {G α :α∈A} in GROUP is their direct prod- uct group G , with the group structure obtained by multiplying α α elements componentwise. Exercise 7.8. Prove that each of the above constructions satisfies the defining property of a product in its category. If we reverse all the morphisms in the definition of a product, we get a dual concept called a sum. A sum of objects {X } in a category C is α an object S together with morphisms ι : X → S called injections such α α that given any object W in C and morphisms f : X → W, there exists α α a unique morphism f: S → W such that each of the following diagrams commutes: S (cid:8)(cid:6) ι α (cid:6)f (cid:6)(cid:7) (cid:2) X W. α f α Lemma 7.37. If a sum exists in a category, it is unique up to an isomor- phism that respects the injections. Exercise 7.9. Prove Lemma 7.37. Some examples of categorical sums are given in the problems.

176 7. Homotopy and the Fundamental Group Problems 7-1. Let B ⊂ Rn be any convex set, X any topological space, and A any subset of X. Show that any two continuous maps f,g: X → B that agree on A are homotopic relative to A. 7-2. Prove the following facts about retracts. (a) A retract of a Hausdorff space is closed. (b) A retract of a connected space is connected. (c) A retract of a compact space is compact. (d) A retract of a simply connected space is simply connected. (e) A retract of a retract is a retract, i.e., if A ⊂ B ⊂ X, A is a retract of B, and B is a retract of X, then A is a retract of X. 7-3. Show that the M¨obius band (see Chapter 5) is homotopy equivalent to S1. 7-4. Let X be a path connected space and p,q ∈ X. Determine an alge- braic necessary and sufficient condition on the fundamental group of X underwhichallpathclassesfromptoqgivethesameisomorphism of π1(X,p) with π1(X,q). 7-5. For any compact surface S, show that S with one point removed is homotopy equivalent to a bouquet of finitely many circles. 7-6. Let K be a simplicial complex and σ ∈K any simplex. Show that |σ| has a neighborhood U ⊂ |K| such th(cid:2)at |σ| is a strong deformation retract of U. [Hint: Let U = |K| (cid:3) {|τ| : τ ∩ σ = ∅}. For any k-simplex τ = (cid:22)v0,…,v k (cid:23) whose intersection with σ is nonempty, define π: τ ∩U →τ ∩U by (cid:20) (cid:21) (cid:20) (cid:21) (cid:20) (cid:21) (cid:14)k (cid:14)l −1 (cid:14)l π t v = t t v , i i i i i i=0 i=0 i=0 wherev0,…,v l aretheverticesofτ∩σ,andletH: (τ∩U)×I →τ∩U bethestraight-linehomotopybetweenIdandπ.ShowthatH defines a strong deformation retraction of U onto |σ|.] 7-7. GiveanotherproofofLemma7.29byconsideringthemapF: I×I → Y defined by F(s,t)=H(f(s),t) and applying Lemma 7.12. 7-8. Show that any two vertices in a tree are joined by a unique reduced edge path.

Problems 177 7-9. Given any collection(cid:6) {X α : α ∈ A} of topological spaces, show that theirdisjointunion α X α ,togetherwith(cid:6)thedisjointuniontopology and the natural inclusions ι : X (cid:9)→ X , is their sum in the α α α α category TOP. (cid:29) 7-10. Given any collection of abelian groups(cid:28) {G α : α ∈ A}, let α G α be the subgroup of the direct product G consisting of all those α α elements {g α } α∈A such that g α is the identity element(cid:29)t in G α for all but finitely many α; and for each α let ι : G (cid:9)→ G be the α α α α obvious injection. Show that this group, called the direct sum of the groups G , is the sum of the G ’s in the category AB. α α 7-11. Show that the construction described in Problem 7-10 does not yield the sum in the category GROUP as follows: Take G1 = G2 = Z, andfindhomomorphismsf1 andf2 fromZtosome(necessarilynon- abelian) group H such that no homomorphism f: Z⊕Z→H makes the following diagram commute for i=1,2: Z⊕Z (cid:8)(cid:6) ι i (cid:6)f (cid:6)(cid:7) (cid:2) Z H. f i (We will see how to construct the sum in GROUP in Chapter 9.) 7-12. Given any collection {X α : α ∈ (cid:28)A} of topological spaces, define a basis in th(cid:28)e Cartesian product α X α consisting of product sets of the form α U α , where U α is open in X α and U α =(cid:28)X α for all but finitely many α. Show that this is a basis, and that X with this α α topology is the product of the spaces X in the category TOP. α 7-13. Showthatanyvertexinaconnectedfinitetreeisastrongdeformation retract of the tree.

8 Circles and Spheres So far, we have not actually computed any nontrivial fundamental groups. The main goal of this chapter is to remedy this by computing the funda- mental group of the circle. We will show, as promised, that π1(S1,1) is an infinite cyclic group generated by the path α that goes once around the circle counterclockwise at constant speed. Thus each element of π1(S1,1) is uniquely determined by an integer, called its “winding number,” which counts the net number of times and in which direction the path winds around the circle. Here is the essence of the plan. We would like to show that any loop in the circle is in the path class [α]n for a unique integer n. The idea is to represent a path by giving its angle θ(s) as a function of the parameter, and then the winding number should be essentially 1/(2π) times the total change in angle, θ(1)−θ(0). Sincetheangleθisnotawell-definedcontinuousfunctiononthecircle,in ordertomakerigoroussenseofthis,weneedtoundertakeadetailedstudy oftheexponentialquotientmapε: R→S1 definedattheendofChapter3. Ananglefunctionforaloopf isjust(uptoaconstantmultiple)a“lift”of f to a path in R. Because R is simply connected, we can always construct a homotopy between two lifts that have the same total change in angle. Therearethreekeyliftinglemmasthatmakethisallwork.Inthebegin- ning of the chapter we state those lifting lemmas, and then we use them to prove that the fundamental group of the circle is infinite cyclic. In the next section we prove the lifting lemmas. These three properties will make another very important appearance later in the book, when we discuss covering spaces.

180 8. Circles and Spheres At the end of the chapter we compute the fundamental groups of the higher-dimensional spheres and product spaces, and show that the funda- mental group of any manifold is countable. The Fundamental Group of the Circle Throughout this chapter we will think of the circle as lying in the complex plane, and we will always use the base point 1 ∈ C, which corresponds to (1,0)∈R2. Let α: I → S1 denote the loop α(s) = e2πis. The complete structure of the fundamental group of the circle is described by the following theorem. Theorem 8.1. The group π1(S1,1) is infinite cyclic, with generator [α]. To prove this theorem we will use a concrete representative of the path class[α]n,definedasfollows.Foranyinteger n,letα : I →S1 betheloop n α n (s)=e2πins. It is easy to see that α n is a reparametrization of α n−1 · α, so by induction [α ]=[α]n. n As mentioned in the introduction to this chapter, the proof of the theo- remisbasedonacloseexaminationofthequotientmapε: R→S1 defined by ε(x) = e2πix. If ϕ: B → S1 is any continuous map, a lift of ϕ is a continuous map ϕ(cid:7): B →R such that the following diagram commutes: R (cid:3)(cid:4) (cid:3) ϕ(cid:7) ε (cid:3) (cid:3) (cid:5) (cid:2) B ϕ S1. For example, if f: I → S1 is a path in S1, then a lift f (cid:7) of f can be (cid:7) interpreted geometrically by observing that θ(s) = 2πf(s) is a continuous choice of angle function such that f(s)=eiθ(s). It is important to be aware that some maps may have no lifts at all. For example, suppose σ: S1 →R were a lift of the identity map of S1: R (cid:3)(cid:4) (cid:3) σ ε (cid:3) (cid:3) (cid:5) (cid:2) S1 S1. IdS1 Then the equation ε ◦ σ = IdS1 means that 2πσ is a continuous choice of angle function on the circle. It is intuitively evident that this cannot exist, because any choice of angle function would have to change by 2π as one goes once around the circle, and thus could not be continuous on

The Fundamental Group of the Circle 181 the whole circle. Assuming for the moment the result of Theorem 8.1, we can prove rigorously that σ cannot exist just by noting that if there were such a map, the induced homomorphism ε∗ ◦σ∗ would be the identity on π1(S1,1), which would mean in particular that ε∗: π1(R,0) → π1(S1,1) was surjective. Since π1(R,0) is the trivial group and π1(S1,1) is not, this is impossible. The first important fact about lifts is the following uniqueness lemma. This is not actually used directly in the computation of π1(S1), but it is necessary for proving the other two lifting properties. Lemma 8.2 (Unique Lifting Property of the Circle). SupposeB is connected, ϕ: B →S1 is continuous, and ϕ(cid:7) 1,ϕ(cid:7) 2: B →R are lifts of ϕ that agree at some point of B. Then ϕ(cid:7) 1 ≡ϕ(cid:7) 2. The next lifting lemma shows that paths in the circle, at least, always have lifts. Lemma 8.3 (Path Lifting Property of the Circle). Suppose f: I → S1 is any path, and r0 ∈ R is any point in the fiber of ε over f(0). Then there exists a unique lift f (cid:7) : I → R of f such that (cid:7) f(0)=r0. Our third lifting lemma concerns lifts of homotopies: It says that lifts of path homotopic paths are path homotopic, as long as they both start at the same point. Lemma 8.4 (Homotopy Lifting Property of the Circle). Suppose f0,f1: I →S1 are path homotopic, and f (cid:7) 0,f (cid:7) 1: I →R are lifts of f0 and f1 with the same initial points. Then f (cid:7) 0 ∼f (cid:7) 1. Assuming these lifting lemmas, let us now carry out the proof of our theorem about the fundamental group of the circle. Proof of Theorem 8.1. Define a map j: Z → π1(S1,1) by j(n) = [α]n. It suffices to show that j is an isomorphism (considering Z as an additive group). Because [α]n+m = [α]n[α]m, j is a homomorphism. We will show that it is injective and surjective. To prove surjectivity, let [f] be any element of π1(S1,1). By the path lifting property of the circle, f has a lift f (cid:7) : I → R such that f (cid:7) (0) = 0 (Figure 8.1). Now, e2πif (cid:7)(1) = ε◦f (cid:7) (1) = f(1) = 1, so f (cid:7) (1) is an integer n. We will show that [f]=[α ]=j(n). n If we let b : I → R be the path b (s) = ns, the two paths f (cid:7) and b n n n bothstartat0andendatn.BecauseRissimplyconnected,f (cid:7)∼b .Since n continuous maps preserve path homotopy, this implies f =ε◦f (cid:7)∼ε◦b = n α , thus proving that j is surjective. n To prove injectivity, suppose some n∈Z is mapped by j to the identity element [c1] ∈ π1(S1,1), or in other words, [α]n = [c1]. Representing the

182 8. Circles and Spheres n (cid:7) f b n 0 ε f 1 FIGURE 8.1. Proof that every path in S1 is homotopic to αn. path class [α]n by α n as above, the assumption is α n ∼ c1. If α(cid:7) n and (cid:7)c1 are lifts of α n and c1 starting at 0, the homotopy lifting property of the circleguaranteesthatα(cid:7) n ∼(cid:7)c1 inR.Inparticular,theybothhavethesame terminal point. Now the lift of α starting at 0∈R is easily seen to be b , n n and the lift of c1 is the constant loop c0. Thus n=b n (1)=c0(1)=0, so j is injective. Acloseexaminationofthisproofshowsthatwehaveactuallyconstructed an explicit inverse for j, which is of interest in its own right. Define a map N: π1(S1,1)→Zasfollows:Forany[f]∈π1(S1,1),letN[f]=f (cid:7) (1),where f (cid:7) is the lift of f starting at 0 ∈ R. Such a lift exists by the path lifting (cid:7) property,andf(1)isindependentofthechoiceoff bythehomotopylifting property. The proof above shows that N =j−1. (cid:7) If we think of 2πf as a continuous choice of angle function for f, then 2πN[f] represents, intuitively, the total change in the angle of f(s) as s goes from 0 to 1, and N[f] represents the number of times f winds around the circle. For this reason, N[f] is called the winding number of the path f.

Proofs of the Lifting Lemmas 183 (cid:7) U n R ε S1 q U FIGURE 8.2. Evenly covered neighborhood in S1. Proofs of the Lifting Lemmas Before proving the three lifting lemmas, we need some preliminary results. The first is a precise description of the behavior of the quotient map ε. Lemma 8.5. Each point q ∈ S1 has a neighborhood U such that ε−1(U) (cid:7) is a disjoint union of countably many open intervals U , on each of which n (cid:7) the restriction of ε is a homeomorphism from U onto U (Figure 8.2). n Proof. This is just a straightforward computation from the definition of ε. We can take, for example, U = S1(cid:3){−q}; then the sets U (cid:7) are the open n intervals of the form (r +n,r +n+1), where r is a fixed number such that ε(r)=−q and n ranges over the integers. For each n, ε: U (cid:7) →U is a n bijective open map and therefore a homeomorphism.

184 8. Circles and Spheres An open set U ⊂ S1 that has the properties described in this lemma is saidtobeevenlycovered.Themostimportantpropertyofanevenlycovered open set is that it admits local right inverses for ε, as we now describe. First, a bit of terminology. If p: X →Y is any surjective continuous map, a section of p is a continuous map σ: Y →X such that p◦σ =Id (i.e., a Y right inverse for p): X p σ Y. If U ⊂ Y is an open set, a local section of p over U is a continuous map σ: U →X such that p◦σ =Id . U Lemma 8.6 (Local Section Property of the Circle). Let U ⊂S1 be an evenly covered open set. For any q ∈U and any r in the fiber of ε over q, there is a local section σ of ε over U such that σ(q)=r. Proof. By definition of an evenly covered open set, r is contained in some open set U (cid:7) ⊂ R such that ε: U (cid:7) → U is a homeomorphism. Thus σ = n n (ε| (cid:7) )−1 is the desired local section. Un Proof of the Unique Lifting Property. LetS={b∈B :ϕ(cid:7) 1(b)=ϕ(cid:7) 2(b)}.By hypothesis S is not empty. Since B is connected, if we can show that S is open and closed in B, it must be all of B. To show that S is open, let b ∈ S. Write r = ϕ(cid:7) 1(b) = ϕ(cid:7) 2(b) and q = ε(r) = ϕ(b). Let U ⊂ S1 be an evenly covered neighborhood of q, and let U (cid:7) be the component of ε−1(U) containing r (Figure 8.3). If we set V = ϕ(cid:7)− 1 1 (U (cid:7) )∩ϕ(cid:7)− 2 1 (U (cid:7) ), then V is a neighborhood of b on which both ϕ(cid:7) 1 and ϕ(cid:7) 2 take their values in U (cid:7) . Now, the fact that ϕ(cid:7) 1 and ϕ(cid:7) 2 are lifts of ϕ translates to ϕ = ε◦ϕ(cid:7) 1 = ε◦ϕ(cid:7) 2. Since ε is injective on U (cid:7) , we conclude that ϕ(cid:7) 1 and ϕ(cid:7) 2 agree on V, which is to say that V ⊂S, so S is open. To show that it is closed, we will show that its complement is open. Let b (cid:14)∈ S, and set r1 = ϕ(cid:7) 1(b) and r2 = ϕ(cid:7) 2(b), so that r1 (cid:14)= r2. As above, let q = ε(r1) = ε(r2) = ϕ(b), and let U be an evenly covered neighborhood (cid:7) (cid:7) of q. Then there are disjoint neighborhoods U1 of r1 and U2 of r2 such (cid:7) (cid:7) that ε is a homeomorphism from U1 to U and from U2 to U. Letting V = ϕ(cid:7) 1 −1 (U (cid:7) 1)∩ϕ(cid:7)− 2 1 (U (cid:7) 2), we conclude that ϕ(cid:7) 1(V) ⊂ U (cid:7) 1 and ϕ(cid:7) 2(V) ⊂ U (cid:7) 2, soϕ(cid:7) 1 (cid:14)=ϕ(cid:7) 2 onV,whichistosaythat V ∩S=∅.ThusSisclosed,and the proof is complete. Proof of the path lifting property. Let f: I → S1 be a path, and r0 ∈ ε−1(f(0)) as in the statement of the lemma. If U is an open cover of the circle by evenly covered open sets, the collection {f−1(U) : U ∈ U} is an open cover of I. Let δ be a Lebesgue number for this cover. Choosing an

Proofs of the Lifting Lemmas 185 ϕ(cid:7) 1 r (cid:7) U ϕ(cid:7)−1 (U (cid:7) ) b 1 ϕ(cid:7) 2 V ϕ(cid:7)−1 (U (cid:7) ) ε 2 ϕ B q U FIGURE 8.3. Proof of the unique lifting property. integer n large enough that 1/n < δ, the Lebesgue number lemma says that each interval [k/n,(k+1)/n] of length 1/n is contained in one of the sets f−1(U), which is to say that it is mapped by f into an evenly covered open set. Wedefinetheliftf (cid:7) : I →Rinductivelyasfollows.Firstchooseanevenly covered open set U0 such that f[0,1/n] ⊂ U0. Letting σ0: U0 → R denote the local section such that σ0(f(0))=r0, we set f (cid:7) =σ0 ◦f on [0,1/n]. (cid:7) (cid:7) It follows immediately that f is continuous and satisfies f(0) = r0 and ε◦f (cid:7) =f. Proceeding by induction, suppose we have defined a continuous lift of f onanintervaloftheform[0,k/n]forsomeintegerk.Asbefore,f[k/n,(k+ 1)/n] is contained in an evenly covered set U . Letting σ : U →R be the k k k (cid:7) local section such that σ (f(k/n))=f(k/n), we set k f (cid:7) =σ ◦f on [k/n,(k+1)/n]. k

186 8. Circles and Spheres The resulting map is continuous by the gluing lemma. By induction we (cid:7) obtainaliftf definedonallofI.Itisuniquebytheuniqueliftingproperty. Proof of the homotopy lifting property. Nowsupposef0,f1 arepathhomo- (cid:7) (cid:7) topic paths in the circle, and f0,f1 are any lifts of them starting at the same point r0 ∈ R. Let H: f0 ∼ f1 be a path homotopy. This means that H: I×I →S1 satisfies H(s,0)=f0(s); H(s,1)=f1(s); H(0,t)=f0(0)=f1(0); H(1,t)=f0(1)=f1(1). We will show below that there exists a lift of H to a map H (cid:7) : I×I →R (cid:7) suchthatH(0,0)=r0.Assumingthis,weargueasfollows.First,notethat ε◦H (cid:7) (0,t)=H(0,t)=f0(0).Therefore,t(cid:10)→H (cid:7) (0,t)isaliftoftheconstant loopatf0(0)toapathstartingatr0;bytheuniqueliftingproperty,itmust (cid:7) be the constant loop at r0, so H(0,t) = r0 for all t. The same argument (cid:7) (cid:7) showsthatH(1,t)isconstantforallt,soH isapathhomotopy.Moreover, (cid:7) (cid:7) (cid:7) H0(s) = H(s,0) is a lift of f0 starting at r0, and similarly, H1 is a lift of f1 starting at r0. By the unique lifting property, these must be equal to (cid:7) (cid:7) (cid:7) the given lifts f0 and f1, respectively, and H provides a path homotopy between them. (cid:7) All that remains is to prove the existence of the lift H. As in the proof ofthepathliftingproperty,thereexistsδ >0suchthatanysubsetofI×I whosediameterislessthanδismappedbyH intoanevenlycoveredsubset ofS1.Choosenlar√geenoughthateachsquareofside1/nhasdiameterless than δ. (Any n> 2/δ will do.) For any integers i,j such that 0≤i,j ≤n−1, let S denote the square ij [i/n,(i+1)/n]×[j/n,(j+1)/n] (Figure 8.4). For any point x ∈ R in the (cid:7) fiber of ε over H(i/n,j/n), there exists a unique lift H of H over S ij ij satisfying H (cid:7) (i/n,j/n) = x, given by H (cid:7) = σ◦H , where σ is the local ij ij ij section of ε such that σ(H(i/n,j/n))=x. WedefineH (cid:7) inductivelyonI×I asfollows.OnS00,letH (cid:7) 00 betheliftof (cid:7) (cid:7) H such that H(0,0)=r0. On the next square to the right, S10, let H10 be (cid:7) (cid:7) theliftsuchthatH10(1/n,0)=H00(1/n,0).Wenowhavetwoliftsdefined onthelinesegment{(1/n,t):0≤t≤1/n}wherethetwosquaresoverlap. But on this line segment, the paths t (cid:10)→ H (cid:7) 00(1/n,t) and t (cid:10)→ H (cid:7) 10(1/n,t) arebothliftsofthepatht(cid:10)→H(1/n,t)startingatthesamepoint;thusby the unique lifting property they are equal. Continuing in this way, we define lifts on each of the squares S i0, i = 0,…,n−1,andthenonthesquaresofthesecondrow,andsoon.Suppose by induction that we have defined lifts H (cid:7) i(cid:3)j(cid:3) on all squares S i(cid:3)j(cid:3) for j(cid:5) <j,

Fundamental Groups of Spheres 187 (cid:7) U (cid:7) H(i/n,j/n) (cid:7) H ε S H ij j/n H(i/n,j/n) i/n U FIGURE 8.4. Proof of the homotopy lifting property (inductive step). andforj(cid:5) =j andi(cid:5) <i,andallsuchliftsagreewheretheyoverlap.Welet (cid:7) H betheuniqueliftofH onS thatagreeswithany(henceall)previous ij ij liftsatthelowerleftcorner(i/n,j/n).Atatypicalsuchsquare,wehaveto checkthatthenewliftagreeswithtwodifferentoldones:onecomingfrom thesquareS i−1,j totheleft,andonecomingfromthesquareS i,j−1 below. Just as in the preceding paragraph, the unique lifting property guarantees that the old and new lifts agree on both of these line segments. (cid:7) (cid:7) (cid:7) In the end we obtain the desired lift H by letting H = H on S ; it is ij ij continuous by the gluing lemma. Fundamental Groups of Spheres The situation is much simpler for the higher-dimensional spheres. The sphere minus the north pole is homeomorphic to Rn by stereographic pro- jection(seeExample3.6).In fact,composingstereographicprojectionwith a suitable rotation of the sphere, it is easy to see that the sphere minus any point is homeomorphic to Rn. Therefore, if we knew that any loop in

188 8. Circles and Spheres Sn omitted at least one point in the sphere, we could consider it as a loop in Rn; since it is null homotopic there, it is null homotopic in Sn. Unfortunately,anarbitraryloopmightnotomitanypoints.Forexample, thereisacontinuoussurjectivemapf: I →I×I (a“space-fillingcurve”— see, e.g., [Rud76]). Composing this with a surjective map I×I →S2 such as the one constructed in Proposition 6.2(b) yields a path whose image is all of S2. But as the proof of the next proposition shows, we can modify any curve by a homotopy so that it does miss a point. Theorem 8.7. For n≥2, Sn is simply connected. Proof. Let N =(0,…,0,1) denote the north pole, and S =−N the south pole.BoththeopensetsU =Sn(cid:3){N}andV =Sn(cid:3){S}arehomeomorphic toRn.Iff: I →Sn isanypath,bytheLebesguenumberlemmathereisan integer n such that on each subinterval [k/n,(k+1)/n], f takes its values either in U or in V. Now, V (cid:3){N} is homeomorphic to Rn(cid:3){0}, which is connected. (Here is where the dimensional restriction comes in—when n=1, Rn(cid:3){0} is disconnected.) Thus, for each such segment that lies in V,thereisanotherpathinV withthesameendpointsthatmissesN;since V is simply connected, these two paths are path homotopic in V and thus inSn.Ofcourse,eachsegmentthatliesin U alreadymissesN.Thisshows thatf ishomotopictoapathinSn(cid:3){N}≈Rn,sof isnullhomotopic. Corollary 8.8. For n≥3, Rn(cid:3){0} is simply connected. Proof. The map F: Rn (cid:3){0} → Sn−1 given by F(x) = x/|x| is a strong deformation retraction (by a straight-line homotopy). Fundamental Groups of Product Spaces In this section we will show how to compute the fundamental group of an arbitrary(finite)productoftopologicalspacesintermsofthefundamental groups of the factors. As an application, we will compute the fundamental groups of tori. Let X1,…,X n be topological spaces, and let p i : X1 ×···×X n → X i denoteprojectionontheithfactor.(Weareavoidingourusualnotationπ i for the projections here so as not to create confusion with the notation π1 for the fundamental group.) Choosing base points q ∈X , we get maps i i p i∗ : π1(X1 ×···×X n ,(q1,…,q n ))→π1(X i ,q i ). Putting these together, we define a map p(cid:7): π1(X1 ×···×X n ,(q1,…,q n ))→π1(X1,q1)×···×π1(X n ,q n )

Fundamental Groups of Manifolds 189 by p(cid:7)[f]=(p1∗ [f],…,p n∗ [f]). (8.1) Proposition 8.9 (Fundamental Group of a Product). If X1, … , X n are any topological spaces, the map p(cid:7): π1(X1 ×···×X n ,(q1,…,q n ))→ π1(X1,q1)×···×π1(X n ,q n ) defined by (8.1) is an isomorphism. Proof. First we will show that p(cid:7) is surjective. Let [f i ] ∈ π1(X i ,q i ) be arbitrary for i = 1,…,n. Define a loop f in the product space by f(s) = (f1(s),…,f n (s)). Since the component functions of f satisfy f i =p i ◦f,wecomputep(cid:7)[f]=(p1∗ [f],…,p n∗ [f])=([p1 ◦f],…,[p n ◦f])= ([f1],…,[f n ]). To show injectivity, suppose f is a loop in the product space, and p(cid:7)[f] is the identity element of π1(X1,q1)×···×π1(X n ,q n ). Writing f in terms of itscomponentfunctionsasf(s)=(f1(s),…,f n (s)),thehypothesismeans that [c ] = p [f] = [p ◦f] = [f ] for each i. If we choose homotopies qi i∗ i i H i : f i ∼ c qi , it follows easily that the map H: X1 × ··· × X n × I → X1 ×···×X n given by H(x1,…,x n ,t)=(H1(x1,t),…,H n (x n ,t)) is a homotopy from f to the constant loop c(q1,…,qn ). Corollary 8.10 (Fundamental Groups of Tori). Let Tn =S1×···× S1 be the n-dimensional torus, and let α denote the standard loop in the i ith copy of S1: α (s)=(1,…,1,e2πis,1,…,1). i Using q = (1,…,1) as base point, the map ϕ: Zn → π1(Tn,q) given by ϕ(k1,…,k n )=[α1]k1···[α n ]kn is an isomorphism. Fundamental Groups of Manifolds We conclude this chapter by proving an important theorem about funda- mentalgroupsofmanifolds.Thisdoesnothavetodowithcirclesorspheres per se, but it does use techniques similar to those used in the other proofs in this chapter, so this is a convenient time to insert it. Theorem 8.11. The fundamental group of a manifold is countable. Proof. Let M be a manifold, and let U be a countable cover of M by Euclidean balls. For each U,U(cid:5) ∈ U the intersection U ∩U(cid:5) has at most countably many components; choose a point in each such component and

190 8. Circles and Spheres x k−1 pU xk k −1,xk x k g k−1 f k g k (cid:12) (cid:13) (cid:12) (cid:13) f k f k−1 n n U k−1 f U k U k+1 FIGURE 8.5. Proof that a manifold has countable fundamental group. let X denote the (countable) set consisting of all the chosen points as U,U(cid:5) range over all the sets in U. For each U ∈ U and x,x(cid:5) ∈ X such that x,x(cid:5) ∈U, choose a definite path pU from x to x(cid:5) in U. x,x(cid:3) Now choose any point q ∈X as base point. Let us say that a loop based at q is special if it is a finite product of paths of the form pU . Because x,x(cid:3) both U and X are countable sets, there are only countably many special loops.Eachspecialloopdeterminesanelementofπ1(M,q).Ifwecanshow that every element of π1(M,q) is obtained in this way, we will be done, because we will have exhibited a surjective map from a countable set onto π1(M,q). So suppose f is any loop based at q. By the Lebesgue number lemma there is an integer n such that f maps each subinterval [(k −1)/n,k/n] into one of the balls in U; call this ball U k . Let f · k

· f| [(k−1)/n,k/n], reparametrized on the unit interval, so that [f]=[f1] ··· [f n ]. Sinceforeachk =1,…,n−1,f(k/n)∈U k ∩U k+1,thereissomex k ∈X that lies in the same component of U k ∩U k+1 as f(k/n). Choose a path g k in U k ∩U k+1 from x k to f(k/n) (Figure 8.5), and set f (cid:7) k =g k−1 · f k · g k −1 (taking x = q and g to be the constant path c when k = 0 or n). It is k k q immediatethat[f]=[f (cid:7) 1] · ··· · [f (cid:7) n ],becausealltheg k ’scancelout.Butfor (cid:7) eachk,f k isapathinU k fromx k−1tox k ,andsinceU k issimplyconnected, (cid:7) f k is path homotopic to pU xk k −1xk . This shows that f is path homotopic to a special loop and completes the proof.

Problems 191 Problems 8-1. Prove that the circle is not a retract of the closed disk. 8-2. Identifying the circle with the subspace S1 ×{1} ⊂ T2, prove that the circle is not a deformation retract of the torus. 8-3. Provethatthefundamentalgroupofanytopologicalgroupisabelian. [Hint: If f and g are loops based at 1∈G, consider the map F from I×I into G given by F(s,t)=f(s)g(t) and use Lemma 7.12.] 8-4. Suppose U ⊂ R2 is an open set and x ∈ U. Show that U (cid:3){x} is not simply connected. [Hint: Let S be a small circle around x, and consider the sequence of inclusions S (cid:9)→U (cid:3){x}(cid:9)→R2(cid:3){x}.] 8-5. Showthatatopologicalspacecannotbesimultaneouslya2-manifold andann-manifoldforsomen>2.[Hint:Ifn>2,anyn-manifoldhas a basis of open sets in which the complement of any point is simply connected.] 8-6. Let M be a 2-dimensional manifold with boundary. Show that the setofboundarypointsofM isdisjointfromthesetofinteriorpoints. Conclude that a 2-manifold with boundary is a manifold if and only if its boundary is empty. 8-7. Let ϕ: S1 → S1 be a continuous map such that ϕ(1) = 1. Because π1(S1,1) is infinite cyclic, there is an integer n, called the degree of ϕ and denoted by degϕ, such that ϕ(γ) = γn for all γ ∈ π1(S1,1). If ϕ: S1 → S1 is an arbitrary continuous map, we define the degree of ϕ to be the degree of ρ ◦ ϕ, where ρ: S1 → S1 is the rotation ρ(z)=z/ϕ(1) (in complex notation), which takes ϕ(1) to 1. (a) Show that two maps ϕ,ψ: S1 →S1 are homotopic if and only if they have the same degree. (b) Showthatdeg(ϕ◦ψ)=degϕdegψforanytwocontinuousmaps ϕ,ψ: S1 →S1. (c) For each n ∈ Z, compute the degrees of the nth power map p (z)=zn and its conjugate p (z)=zn. n n (d) Show that ϕ: S1 → S1 has an extension to a continuous map Φ: B2 →S1 if and only if it has degree zero. 8-8. Provethefundamentaltheoremofalgebra:Everycomplexpolynomial ofpositivedegreehasazero.[Hint:Ifp(z)=zn+a n−1zn−1+···+a0, write p (z) = εnp(z/ε) and show that there exists ε > 0 such that ε |p ε (z)−zn| < 1 when z ∈ S1. If p has no zeros, prove that p ε | S1 is homotopictop (z)=zn,andusetheresultsofProblem8-7toderive n a contradiction.]

192 8. Circles and Spheres 8-9. The Brouwer fixed point theorem says that any continuous map f: Bn → Bn has a fixed point (i.e., a point x such that f(x) = x). Prove this in the case n = 2 as follows: If f: B2 → B2 has no fixed point, define ϕ: B2 → S1 by ϕ(x) = (f(x)−x)/|f(x)−x|. Derive a contradiction by showing that the restriction of ϕ to S1 is homotopic to the identity. [If you crumple a map of the country that you are in and drop it on the ground, this theorem guarantees that some point on the map will lie exactly over the point it represents.] 8-10. Let V be a vector field on R2, i.e., a continuous map V : R2 →R2. A point q ∈R2 is called a singular point of V if V(q)=0, and a regular pointifV(q)(cid:14)=0.Asingularpointisisolatedifithasaneighborhood containing no other singular points. Let R ⊂ R2 denote the set of V regular points of V. For any loop f: I → R , define the index of V V withrespecttof,denotedbyInd(V,f),tobethewindingnumberof the loop f (cid:7) : I →S1, given by (cid:7) V(f(s)) f(s)= . |V(f(s))| (a) Show that Ind(V,f) depends only on the path class of f. (b) If q is an isolated singular point of V, show that Ind(V,f ) is ε independentofεforεsufficientlysmall,wheref (s)=q+εα(s), ε and α is the standard counterclockwise loop around the unit circle. This integer is called the index of V at q, and is denoted by Ind(V,q). (c) If V has finitely many singular points in the closed unit disk, all in the interior, show that the index of V with respect to the loop α around the unit circle is equal to the sum of the indices of V at the interior singular points. (d) Compute the index of each of the following vector fields at the origin: V1(x,y)=(x,y); V2(x,y)=(−x,−y); V3(x,y)=(x+y,x−y).

9 Some Group Theory Inthischapterwedepartfromtopologyforawhiletodiscussgrouptheory. Ourgoal,ofcourse,istousethegrouptheorytosolvetopologicalproblems, and in the next chapter we will compute the fundamental groups of all compact surfaces, and use them to show, among other things, that the different surfaces listed in the classification theorem of Chapter 6 are not homeomorphic to each other. Beforewedoso,however,weneedtodevelopsometoolsforconstructing and describing groups. We will discuss four such tools in this chapter: free products of groups, free groups, presentations of groups by generators and relations, and free abelian groups. These will all play central roles in our computations of fundamental groups in the next chapter, and the material on free abelian groups will also be used in the discussion of homology in Chapter 13. This chapter assumes that you are familiar with the basic facts of group theory as summarized in the Appendix. If your group theory is rusty, this wouldbeagoodtimetopulloutanalgebratextandrefreshyourmemory. Free Products Thereisafamiliarwaytocreateagroupasaproductoftwoormoreother groups: The direct product of groupsG1,…,G n (see the Appendix) is the Cartesian product set G1 ×···×G n with the group structure obtained by multiplying the entries in two n-tuples component by component.

194 9. Some Group Theory For each i, the direct product G1 ×···×G n has a subgroup {1}×···× {1}×G ×{1}×···×{1} isomorphic to G , and it is easy to verify that i i elements of two distinct such subgroups commute with each other. As we mentionedinChapter7,thisconstructionyieldstheproductinthecategory of groups. In our study of fundamental groups, we will need to build another kind of product, in which the elements of different groups are not assumed to commute.Thissituationarises,forexample,incomputingthefundamental group of the wedge X ∨Y of two spaces X and Y, defined in Example 3.25. As we will see in the next chapter, the fundamental group of X ∨Y containssubgroupsisomorphictoπ1(X)andπ1(Y),andanyloopinX∨Y is equivalent to a product of loops lying in one space or the other. But in general, path classes of loops in X do not commute with those in Y. In this section we will introduce a more complicated product of groups G1,…,G n that includes each G i as a subgroup, but in which elements of the different subgroups do not commute with each other. It is called the “free product,” and roughly speaking, it is just the set of expressions you can get by formally multiplying together elements of the different groups, with no relations assumed other than those that come from the multipli- cation in each group G . It turns out (despite its name) to be the sum in i the category of groups. Because terms such as “expressions you can get” and “multiplying ele- mentsofdifferentgroups”aretoovaguetouseinmathematicalarguments, theactualconstructionofthefreeproductisratherinvolved.Webeginwith some preliminary terminology. Let {G α } α∈A be an indexed collection of groups. The index set A can be finite or infinite; for our applications we will need only the finite case, so you are free to think of finite collections throughout this chapter. We will usually omit mention of A and denote the collection simply by {G }, α withGreeklettersunderstoodtorangeoverallelementsofsomeimplicitly understood index set. A word in {G(cid:6)α } is a finite sequence of length m ≥ 0 of elements of the disjoint union G . In other words, a word is an ordered m-tuple of the α α form (g1,…,g m ), where each g i is an element of some G α . (Recall that formally, an element of the disjoint union is a pair (g,α), where α is a “tag” to distinguish which group g came from. We will suppress the tag in our notation, but remember that elements of groups corresponding to different indices have to be considered distinct, even if the groups are the same.) The sequence of length zero, called the empty word, is denoted by ( ). Let W denote the set of all words in {G }. We denote the identity α element of G by 1 . α α Define a multiplication operation in W by concatenation: (g1,…,g m )(h1,…,h l )=(g1,…,g m ,h1,…,h l ).

Free Products 195 Clearly,thismultiplicationisassociative,andhastheemptywordasatwo- sidedidentityelement.However,therearetwoproblemswiththisstructure asitstands:First,Wisnotagroupunderthisoperationbecausethereare noinverses;andsecond,thegroupstructuresofthevariousgroupsG have α not played a role in the definition so far. Tosolvebothoftheseproblems,wedefineanequivalencerelationonthe set of words as follows. An elementary reduction is an operation of one of the following forms: (g1,…,g i ,g i+1,…,g m )(cid:10)→(g1,…,g i g i+1,…,g m ) if g i ,g i+1 ∈ some G α ; (g1,…,g i−1,1 α ,g i+1,…,g m )(cid:10)→(g1,…,g i−1,g i+1,…,g m ). The first operation just multiplies together two consecutive entries, pro- vided that they are elements of the same group, and the second deletes any identity element that appears in a word. We say that two words are equivalent, written W ∼ W(cid:5), if one can be transformed into the other by a finite sequence of elementary reductions or their inverses; this is obvi- ously an equivalence relation. The set of equivalen∗ce classes is called the free product of the groups {G }, and is denoted by G . In the case of a α α α finite set of groups, we just write G1 ∗···∗G n . Lemma 9.1. Given any collection of groups {G }, their free product is a α groupunderthemultiplicationoperationinducedbymultiplicationofwords. Proof. First we need to check that multiplication of words respects the equivalence relation. If V(cid:5) is obtained from V by an elementary reduction, then it is easy to see that V(cid:5)W is similarly obtained from VW, as is WV(cid:5) from WV. If V ∼V(cid:5) and W ∼W(cid:5), it follows by induction on the number of elementary reductions that VW ∼ V(cid:5)W(cid:5). Thus multiplication is well- defined on equivalence classes. The equivalence class of the empty word ( ) is obviously an identity element, and multiplication is associative on equivalence classes because it already is on words. Finally, for any word (g1,…,g m ), it is easy to check that (g1,…,g m )(g m −1,…,g 1 −1 )∼( )∼(g m −1,…,g 1 −1 )(g1,…,g m ), so the equivalence class of (g−1,…,g −1 ) is an inverse for that of m 1 (g1,…,g m ). Henceforth, we will denote the identity element of the free product (the equivalence class of the empty word) by 1. Formanypurposesitisimportanttohaveauniquerepresentativeofeach equivalence class in the free product. We say that a word (g1,…,g m ) is reduced if it cannot be shortened by an elementary reduction. Specifically, this means that no element g is the identity of its group, and no two i

196 9. Some Group Theory consecutive elements g i ,g i+1 come from the same group. It is easy to see that any word is equivalent to a reduced word: Just perform elementary reductions until it is impossible to perform any more. What is not so easy to see is that the reduced word representing any given equivalence class is unique. ∗ Proposition 9.2. Every element of G is represented by a unique re- α α duced word. Proof. We showed above that every equivalence class contains a reduced word,soweneedonlycheckthattworeducedwordsrepresentingthesame equivalenceclassmustbeequal.Thisamountstoconstructinga“canonical reduction algorithm.” Let R denote the set of reduced words. We begin by constructing a map W×R→R, which sends (g1,…,g m )∈W and (h1,…,h l )∈R to a reduced word that · we denote by (g1,…,g m ) (h1,…,h l ) ∈ R. We will define the map by induction on m, the length of the word in W. For m=0, define · ( ) (h1,…,h l )=(h1,…,h l ). For m=1 and g ∈G , set α ⎧ ⎪⎪⎪⎨ (h2,…,h l ), h1 ∈G α and gh1 =1 α ; (g) · (h1,…,h l )= ⎪⎪⎪⎩ ( ( g h h 1, 1 . , . . . . , . h ,h l ) l , ), h h 1 1 ∈ (cid:14)∈ G G α α a a n n d d g g h

1 (cid:14)= 1 α 1 ; α ; (g,h1,…,h l ), h1 (cid:14)∈G α and g (cid:14)=1 α . (Theideaisjusttomultiplythetwowordsandreducethemintheobvious way; what is important about this definition is that there are no arbitrary choices involved.) For m>1, define the map recursively: (cid:12) (cid:13) · · · (g1,…,g m ) (h1,…,h l )=(g1) (g2,…,g m ) (h1,…,h l ) · · · · =(g1) (g2) ··· (g m ) (h1,…,h l ), where we understand the dot operation to be performed from right to left: · · · · U V W =U (V W). The key feature of this operation is that it takes equivalent words to the samereducedword:IfW ∼W(cid:5),thenW · V =W(cid:5)· V forallreducedwords V. To prove this, it suffices to assume that W(cid:5) is obtained from W by an elementary reduction. There are two cases, corresponding to the two types of elementary reduction. Suppose first that W =(g1,…,g i ,g i+1,…,g m ),

Free Products 197 and W(cid:5) =(g1,…,g i g i+1,…,g m ) is obtained by multiplying together two consecutive elements g i ,g i+1 from the same group G α . Then · · · · · · · · · W V =(g1) ··· (g i−1) (g i ) (g i+1) (g i+2) ··· (g m ) V. · · · Writing (g i+2) ··· (g m ) V =(h1,…,h l ), it suffices to show that · · · (g i ) (g i+1) (h1,…,h l )=(g i g i+1) (h1,…,h l ). Applyingthedefinitionofthedotoperatortwiceandkeepingcarefultrack of the various cases, you can compute · · (g i ) (g i+1) (h1,…,h l ) ⎧ ⎪⎪⎪⎨ (h2,…,h l ), h1 ∈G α , g i g i+1h1 =1 α ;

(g i g i+1h1,…,h l ), h1 ∈G α , g i g i+1h1 (cid:14)=1 α ; ⎪⎪⎪⎩ (h1,…,h l ), h1 (cid:14)∈G α , g i g i+1 =1 α ; (g i g i+1,h1,…,h l ), h1 (cid:14)∈G α , g i g i+1 (cid:14)=1 α . · On the other hand, (g i g i+1) (h1,…,h l ) is equal to the same value by definition. The second case, in which W contains an identity element that isdeletedtoobtainW(cid:5),followsinasimilarwayfromthefactthat(1 ) · V = α V. Now we define our canonical reduction operator r: W → R by r(W) = · W ().Clearly,ifW isalreadyreduced,thenr(W)=W.Moreover,bythe argument above, if W ∼ W(cid:5), then r(W) = r(W(cid:5)). Thus, if W ∼ W(cid:5) and both are reduced words, we have W = r(W) = r(W(cid:5)) = W(cid:5). This proves the proposition. ∗ ForeachgroupG ,thereisacanonicalmapι : G → G ,definedby α α α α α sendingg ∈G totheequivalenceclassoftheword(g).Eachofthesemaps α is a homomorphism, since (g1g2) ∼ (g1)(g2) for g1,g2 ∈ G α . Each map is also injective: If g (cid:14)= 1 , both the words ι (g) = (g) and ι (1 ) = ( ) are α α α α reduced,andthereforecannotrepresentthesameequivalenceclassbecause of the preceding proposition. We usually identify G with its image under α the injection ι , and write the equivalence class of the word (g) simply as α g.Therefore,theequivalenceclassofaword(g1,g2,…,g m )canbewritten g1g2 ···g m ;byaslightabuseofterminology,wewillalsocallsuchaproduct a word, and say that it is reduced if the word (g1,g2,…,g m ) is reduced. Multiplication in the free product is indicated by juxtaposition of such words. Thus we have finally succeeded in making mathematical sense of products of elements in different groups. Example 9.3. Let Z/(cid:22)2(cid:23) denote the group of integers modulo 2. The free product Z/(cid:22)2(cid:23)∗Z/(cid:22)2(cid:23) can be described as follows. If we letβ and γ denote

198 9. Some Group Theory thenontrivialelementsofthefirstandsecondcopiesofZ/(cid:22)2(cid:23),respectively, each element of Z/(cid:22)2(cid:23)∗Z/(cid:22)2(cid:23) other than the identity has a unique repre- sentationasastringofalternatingβ’sandγ’s.Multiplicationisperformed byconcatenatingthestringsanddeletingallconsecutivepairsofβ’sorγ’s. For example, (βγβγβ)(γβγβ)=βγβγβγβγβ; (γβγβ)(βγβγβ)=β. This group is not abelian. Example 9.4. Later we will need to consider the free product π1(S1,1)∗ π1(S1,1).Lettingα(s)=e2πis asintheprecedingchapter,andlettingβ,γ denote the path classes of α in the first and second copies of π1(S1,1), respectively, each element of π1(S1,1)∗π1(S1,1) other than the identity has a unique expression of the form βi1γj1···βimγjm, where i1 or j m may be zero, but none of the other exponents is zero. The free product of groups has an important characteristic property. Theorem 9.5 (Characteristic Property of the Free Product). For any group H and any collection of∗homomorphisms ϕ α : G α → H, there exists a unique homomorphism Φ: G →H such that for each α α α the following diagram commutes: ∗ G α α (cid:8)(cid:6) ι α (cid:6)Φ (cid:6)(cid:7) G (cid:2) H. (9.1) α ϕ α Proof. Supposewearegivenacollectionofhomomorphismsϕ : G →H. α α The requirement that Φ◦ι =ϕ implies that the desired homomorphism α α Φ must satisfy Φ(g)=ϕ (g) if g ∈G , (9.2) α α where,asusual,weidentifyG withitsimageunderι .SinceΦissupposed α α to be a homomorphism, it must satisfy Φ(g1 ···g m )=Φ(g1)···Φ(g m ). (9.3) (cid:7) Therefore, if Φ and Φ both satisfy the conclusion, they must be equal because both must satisfy (9.2) and (9.3). This proves that Φ is unique if it exists. ToproveexistenceofΦ,weuse(9.2)and(9.3)todefineit.Thisisclearly a homomorphism that satisfies the required properties, provided that it is

Free Groups 199 well-defined.Toverifythatitiswell-defined,weneedtocheckthatitgives the same result when applied to equivalent words. As usual, we need only check elementary reductions. If g i ,g i+1 ∈G α , we have Φ(g i g i+1)=ϕ α (g i g i+1)=ϕ α (g i )ϕ α (g i+1)=Φ(g i )Φ(g i+1), from which it follows that the definition of Φ is unchanged by multiply- ing together successive elements of the same group. Similarly, Φ(1 ) = α ϕ (1 )=1∈H, which shows that Φ is unchanged by deleting an identity α α element. This completes the proof. Corollary 9.6. The free product is the sum in the category of groups. Proof. The characteristic property is exactly the defining property of the sum in a category. Corollary 9.7. The free product is the unique group (up to isomorphism) satisfying the characteristic property. Proof. Lemma 7.37 shows that sums in any category are unique up to isomorphism. In some texts, a free product is defined as any group satisfying the char- acteristicproperty,orasthesuminthecategoryofgroups.Onemustthen prove the existence of such a group by some construction such as the one wehavegivenbeforeoneisentitledtotalkabout“the”freeproduct.Once existenceisproved,uniquenessfollowsautomaticallyfromcategorytheory. Thenicethingaboutthisuniquenessresultisthatnomatterwhatspecific construction is used to define the free product (and you will find many in the literature), they are all the same up to isomorphism. Free Groups In this section we will use the free product construction to create a new class of groups called “free groups,” consisting of all possible products of a set of “generators,” with no relations imposed at all. We begin with a few more definitions. LetGbeagroup.AsubsetS ⊂GissaidtogenerateG,andtheelements of S are called generators for G, if every element of G can be written as a product of elements of S. Of course, any group has a set of generators, since we can take S to be the whole group G. But it is more interesting to find a small set of generators when possible. For example, a cyclic group is a group with one generator (see the Ap- pendix). The cyclic groups are all isomorphic either to Z or to Z/(cid:22)n(cid:23) for some n (Exercise A.26).

200 9. Some Group Theory In this section we will be concerned mostly with infinite cyclic groups. Given any object α, we can form an infinite cyclic group generated by α, denoted by (cid:22)α(cid:23), as follows: (cid:22)α(cid:23) is the set {α}×Z with multiplication (α,m)(α,k) = (α,k+m). We identify α with the element (α,1); thus we can abbreviate (α,m) by αm, and think of (cid:22)α(cid:23) as the group of all integral powers of α with the obvious multiplication. Now suppose we are given any set S. We define the free group on S, denoted by (cid:22)S(cid:23), to be the free product of all the infinite cyclic groups generated by elements of S: ∗ (cid:22)S(cid:23)= (cid:22)α(cid:23). α∈S There is a natural injection ι: S (cid:9)→(cid:22)S(cid:23), defined by sending each α∈S to the word α ∈ (cid:22)S(cid:23). Thus we can consider S as a subset of (cid:22)S(cid:23), and each elementof(cid:22)S(cid:23)canbeexpressedasawordα 1 n1α 2 n2···α m nm,whereeachα i is someelementofS andeachn isaninteger.Multiplicationisperformedby i juxtapositionandcombiningconsecutivepowersofthesameα bytherule i α i nα i k = α i n+k. In case S = {α1,…,α n } is a finite set, we denote the free group on S by (cid:22)α1,…,α n (cid:23) instead of the more accurate but cumbersome notation (cid:22){α1,…,α n }(cid:23). (We will rely on the context and typographical differences to make clear the distinction between the free group (cid:22)S(cid:23) on the elements of the set S and the infinite cyclic group (cid:22)α(cid:23), which is also equal to the free group on the singleton {α}.) Example 9.8. The free group on the empty set, denoted by (cid:22) (cid:23), is by convention just the trivial group {1}. The free group on a singleton {α} is the infinite cyclic group (cid:22)α(cid:23). The free group on the two-element set {β,γ} is (cid:22)β,γ(cid:23)=(cid:22)β(cid:23)∗(cid:22)γ(cid:23), which is the same as the group described in Example 9.4. Theorem 9.9 (Characteristic Property of the Free Group). Let S be a set. For any group H and any map ϕ: S →H, there exists a unique homomorphism ϕ(cid:7): (cid:22)S(cid:23)→H extending ϕ: (cid:22)S(cid:23) (cid:8)(cid:6) ι (cid:6)ϕ(cid:7) (cid:6)(cid:7) (cid:2) (9.4) S ϕ H. Proof. This can be proved directly as in the proof of Theorem 9.5. Alter- natively, recalling that the free group is defined as a free product, we can proceed as follows. There is a one-to-one correspondence between set func- tions ϕ: S → H and collections of homomorphisms ϕ : (cid:22)α(cid:23) → H for all α

Presentations of Groups 201 α∈S, by the equation ϕ (αn)=ϕ(α)n. α Translating the characteristic property of the free product to this special case and using this correspondence yields the result. The details are left as an exercise. Exercise 9.1. Carry out the details of the proof of Theorem 9.9. Exercise 9.2. Prove that the free group on S is the unique group (up to isomorphism) satisfying the characteristic property. Presentations of Groups It is often convenient to describe a group by giving a set of generators for it,andlistingafewrules,or“relations,”thatdescribehowtomultiplythe generators together. For example, the cyclic group of order n generated by γ might be described as the group generated by γ with the single relation γn = 1; all other relations in the group, such as γ3n = 1 or γk−n = γk, follow from this one. The direct product group Z×Z might be described as the group with two generators β,γ satisfying the relation βγ =γβ. The free group (cid:22)β,γ(cid:23) can be described as the group generated by β,γ with no relations. Sofar,thisismathematicallyveryvague.Whatdoesitmeantosaythat “all other relations follow from a given one”? In this section we develop a way to make these notions precise. We define a group presentation to be an ordered pair, denoted by (cid:22)S|R(cid:23), whereS isanarbitrarysetandRisasetofelementsofthefreegroup(cid:22)S(cid:23). TheelementsofSandRarecalledthegeneratorsandrelators,respectively, of the presentation. A group presentation defines a group, also denoted by (cid:22)S|R(cid:23), as the following quotient: (cid:22)S|R(cid:23)=(cid:22)S(cid:23)/R, where R is the normal closure of R in (cid:22)S(cid:23), which is the intersection of all normal subgroups of (cid:22)S(cid:23) containing R; thus R is the “smallest” normal subgroup containing R. Since the quotient of a group by a normal subgroup is again a group (see the Appendix), (cid:22)S|R(cid:23) is indeed a group. Each of the generators s∈S determines an element in (cid:22)S|R(cid:23) (its coset in the quotient group), which we usually write also as s. Each of the relators r ∈ R represents a particular product of generators and their inverses that is equal to 1 in the quotient.

202 9. Some Group Theory Hereistheintuitionbehindthisconstruction.IfGisanygroupgenerated by S, there is a surjective homomorphism Φ: (cid:22)S(cid:23)→G, whose existence is guaranteedbythecharacteristicpropertyofthefreegroup.Ifallthewords of R are to be equal to the identity in G, then the kernel of Φ must at least contain R, and since it is normal, it must contain R; thus by the first isomorphism theorem (Theorem A.13 in the Appendix), G is isomorphic to a quotient of (cid:22)S(cid:23) by a normal subgroup containing R. By dividing out exactly R, we ensure that the only relations that hold in (cid:22)S|R(cid:23) are those that are forced by the relators in R. Thus, in a certain sense, (cid:22)S|R(cid:23) is the “largest” group generated by S in which all the products represented by elements of R are equal to 1. If G is a group and there exists an isomorphism (cid:22)S|R(cid:23)∼ =G, we say that (cid:22)S|R(cid:23) is a presentation of G. At this point, the question naturally arises whether every group has a presentation. In fact, the answer is yes, but the result is not as satisfying as we might have hoped. Given a group G, the set of all elements of G certainly generates G. By the characteristic property of the free group, the identity map of G to itself has a unique extension to a homomorphism Φ: (cid:22)G(cid:23)→G. If we set R=KerΦ, then the first isomorphism theorem says that G ∼ = (cid:22)G(cid:23)/R. Since R is normal, it is equal to its normal closure, and therefore G has the presentation (cid:22)G|R(cid:23). This is highly inefficient, of course, since both (cid:22)G(cid:23) and R are vastly larger than G itself. If G admits a presentation (cid:22)S|R(cid:23) in which both S and R are finite sets, wesaythatGisfinitelypresented.Inthiscase,weusuallywritethepresen- tationas(cid:22)α1,…,α n |r1,…,r m (cid:23).Sincether i actuallyallbecomeequalto the identity in the group defined by the presentation, it is also often con- venient to replace the relators by the equations obtained by setting them equal to the identity, called relations of the presentation, as in (cid:22)α1,…,α n |r1 =1,…,r m =1(cid:23) or even (cid:22)α1,…,α n |r1 =q1,…,r m =q m (cid:23). We take this to be an alternative notation for (cid:22)α1,…,α n | r1q 1 −1 , … , r q−1(cid:23). m m We conclude this section by describing one important example in detail. Proposition 9.10. ThegroupZ×Zhasthepresentation(cid:22)β,γ |βγ =γβ(cid:23). Proof. For brevity, write G = (cid:22)β,γ | βγ = γβ(cid:23) = (cid:22)β,γ | βγβ−1γ−1(cid:23). As usual, we will use the symbols β and γ to denote either the generators of the free group (cid:22)β,γ(cid:23) or their images in the quotient group G. We begin by noting that G is abelian: The equation βγβ−1γ−1 = 1, which holds in G by definition, immediately implies βγ = γβ, and then a simple induction

Free Abelian Groups 203 shows that any products of powers of β and γ commute with each other. Since β and γ generate G, this suffices. WewillprovethepropositionbydefininghomomorphismsΦ: G→Z×Z and Ψ: Z×Z → G and showing that they are inverses of each other. To define Φ, we first define Φ (cid:7) : (cid:22)β,γ(cid:23) → Z×Z by setting Φ (cid:7) (β) = (1,0) and (cid:7) (cid:7) Φ(γ)=(0,1); this uniquely determines Φ by the characteristic property of (cid:7) the free group. Explicitly, Φ is given by Φ (cid:7) (βi1γj1···βimγjm)=(i1+···+i m ,j1+···+j m ). (9.5) Because βγβ−1γ−1 ∈ KerΦ (cid:7) by direct computation, Φ (cid:7) descends to a map Φ: G→Z×Z still given by (9.5). In the other direction, we define Ψ: Z×Z→G by Ψ(m,n)=βmγn. It follows from the fact that G is abelian that Ψ is a homomorphism. A simple computation shows that Ψ ◦ Φ(β) = β, Ψ ◦ Φ(γ) = γ, and Φ ◦ Ψ(m,n)=(m,n). Thus Φ and Ψ are inverses, so G ∼ =Z×Z. In some ways, a presentation gives a very simple and concrete way to understandthepropertiesofagroup,andwewilldescribethefundamental groups of surfaces in the next chapter by giving presentations. However, you should be aware that even with a finite presentation in hand, some very basic questions about a group may still be difficult or impossible to answer. For example, two of the most basic problems concerning group presentations were first posed around 1910 by topologists Heinrich Tietze and Max Dehn, shortly after the invention of the fundamental group: The isomorphismproblemforgroupsistodecide,giventwofinitepresentations, whether the resulting groups are isomorphic; and the word problem is to decide, given a finite presentation (cid:22)S|R(cid:23) and a specific word formed from elements of S, whether that word represents the identity element of the group (cid:22)S|R(cid:23). It was shown in the 1950s that there is no algorithm for solving either of these problems that is guaranteed to yield an answer for everypresentationinafiniteamountoftime!(See[Sti82]forreferencesand historical background.) These ideas form the basis for the subject called combinatorial group theory, which is a lively research field at the intersec- tion of algebra, topology, and geometry. Free Abelian Groups There is an analogue of free groups in the category of abelian groups. In this section, since all our groups will be abelian, we will always write the group operation additively, and denote the identity element by 0 and the

204 9. Some Group Theory inverseofxby−x.IfGisanabeliangroup,g ∈G,andn∈Z,thenotation ng means the n-fold sum g+···+g, and nG is the subgroup {ng :g ∈G}. GivenanonemptysetS,letZ(cid:22)S(cid:23)denotethesetofallfunctionsk: S →Z such that k(s)=0 for all but finitely many s∈S. This is easily seen to be an abelian group under addition, called the free abelian group on S. Just as we did for the free vector space defined in Chapter 5, we can identify each s ∈ S with the element of Z(cid:22)S(cid:23) that takes the value 1 on s and zero oneveryotherelementofS,soweconsiderS asasubsetofZ(cid:22)S(cid:23),andeach element of Z(cid:22)S(cid:23) can be written uniquely as a finite sum of the form (cid:14)n k s i i i=1 where s i are elements of S and k i are integers. When S ={s1,…,s n } is a finiteset,wewillusuallywritethefreeabeliangrouponS asZ(cid:22)s1,…,s n (cid:23). By convention, the free abelian group on the empty set is the trivial group {0} (we consider a “linear combination of no elements” to sum to 0). Lemma 9.11 (Properties of Free Abelian Groups). Let S be a nonempty set. (a) Characteristic Property: Given any abelian group H and any set map f: S →H, there exists a unique homomorphism f (cid:7) : Z(cid:22)S(cid:23)→ H extending f. (cid:29) (b) Z(cid:22)S(cid:23)isisomorphictothedirectsum (cid:22)s(cid:23)ofalltheinfinitecyclic s∈S groups generated by elements of S. (c) If S = {s1,…,s n } is finite, then Z(cid:22)s1,…,s n (cid:23) is isomorphic to Zn via the map (k1,…,k n )(cid:10)→k1s1+···+k n s n . Exercise 9.3. Prove Lemma 9.11. Let G be an abelian group. By analogy with vector spaces, a finite sum of elements of G with integer coefficients is called a linear combination of elementsofG.AnonemptysubsetS ⊂Gissaidtobelinearly independent if the only linear combination of elements of S that equals zero is the one forwhichallthecoefficientsarezero.AbasisforGisalinearlyindependent subset that generates G. Just as in the case of vector spaces, if S is a basis for G, every element of G can be written uniquely as a linear combination of elements of S. For example, S is a basis for the free abelian group Z(cid:22)S(cid:23). The set of elements e = (0,…,1,…,0) (with a 1 in the ith place) for i i=1,…,n is a basis for Zn, which we call the standard basis. If a group G is isomorphic to Z(cid:22)S(cid:23) for some set S, G is also said to be free abelian.

Free Abelian Groups 205 Exercise 9.4. (a) Showthatanabeliangroupisfreeabelianifandonlyifithasabasis. (b) Show that any two free abelian groups whose bases have the same cardinality are isomorphic. Lemma 9.12. If G has a finite basis, then every finite basis has the same number of elements. Proof. Suppose G has a basis with n elements. Then G ∼ = Zn by Lemma 9.11(c), and the quotient group G/2G is easily seen to be isomorphic to (Z/(cid:22)2(cid:23))n, which has exactly 2n elements. Since the order of G/2G is inde- pendent of the choice of basis, every finite basis must have n elements. In view of this lemma, if G is a free abelian group with a finite basis, we define the rank of G to be the number of elements in any finite basis. (In fact, in that case every basis is finite—see Problem 9-6.) If G has no finite basis, we say it has infinite rank. Proposition 9.13. Suppose G is a free abelian group of finite rank. Any subgroup of G is free abelian of rank less than or equal to that of G. Proof. We may assume without loss of generality that G = Zn. We will provethepropositionbyinductiononn.Forn=1,itfollowsfromthefact that any subgroup of a cyclic group is cyclic. Suppose the result is true for subgroups of Zn−1, and let H be any subgroup of Zn. Identifying Zn−1 with the subgroup {(k1,…,k n−1,0)} of Zn, the inductive hypothesis guarantees that H ∩Zn−1 is free abelian of rank m−1 ≤ n−1, so has a basis {h1,…,h m−1 }. If H ⊂ Zn−1, we are done. Otherwise, the image of H under the projection π : Zn → Z onto n thenthfactorisanontrivialcyclicsubgroupofZ.Letc∈Zbeagenerator of this subgroup, and let h be an element of H such that π (h ) = c. m n m The proof will be complete once we show that {h1,…,h m } is a basis for H. Suppose a1h1 +···+a m h m = 0. Applying π n to this equation yields a m c = 0, so a m = 0. Then a1 = ··· = a m−1 = 0 because of the inde- pendence of {h1,…,h m−1 }, so {h1,…,h m } is linearly independent. Now suppose h ∈ H is arbitrary. Then π (h) = ac for some integer a, so n h−ah ∈H∩Zn−1. This element can be written as a linear combination m of {h1,…,h m−1 }, which shows that H is generated by {h1,…,h m }. We will need to extend the notion of rank to finitely generated abelian groups that are not necessarily free abelian. To that end, we say that an element g of an abelian group G is a torsion element if ng = 0 for some nonzero n ∈ Z. If ng = n(cid:5)g(cid:5) = 0, then nn(cid:5)(g +g(cid:5)) = 0, so the set of all torsion elements is a subgroup Gtor of G, called the torsion subgroup. We saythatGistorsion freeiftheonlytorsionelementis0.Itiseasytocheck that the quotient group G/Gtor is torsion free.

206 9. Some Group Theory Proposition 9.14. Anyabeliangroupthatisfinitelygeneratedandtorsion free is free abelian of finite rank. Proof. SupposeGissuchagroup.ForanylinearlyindependentsubsetS ⊂ G, we will extend our notation slightly and let Z(cid:22)S(cid:23) denote the subgroup of G generated by S. It is easily seen to be free abelian with S as a basis, so this is consistent with our earlier notation. The crux of the proof is the following claim: There exists a nonzero integer n and a finite linearly independent set S ⊂G such that nG⊂Z(cid:22)S(cid:23). Assuming this, the rest of the proof goes as follows. Let ϕ: G → G be the homomorphism ϕ(g) = ng. It is injective because G is torsion free, and the claim implies that ϕ(G) ⊂ Z(cid:22)S(cid:23). Thus G is isomorphic to the subgroup ϕ(G) of the free abelian group Z(cid:22)S(cid:23), so by Proposition 9.13, G is free abelian of finite rank. We will prove the claim by induction on the number of elements in a generating set for G. If G is generated by one element g, the claim is true with n = 1, because the fact that G is torsion free implies that {g} is a linearly independent set. Now assume that the claim is true for any torsion-free group generated by m−1 elements, and suppose G is generated by a set T ={g1,…,g m } with m elements. If T is linearly independent, we just take S = T. If not, there is a relation of the form a1g1+···+a m g m = 0 with at least one of the coefficients, say a , not equal to zero. Letting G(cid:5) denote the subgroup m of G generated by {g1,…,g m−1 }, this means that a m g m ∈G(cid:5). Since G(cid:5) is generated by m−1 elements, by induction there exist a nonzero integer n(cid:5) and a finite linearly independent set S ⊂ G(cid:5) such that n(cid:5)G(cid:5) ⊂ Z(cid:22)S(cid:23). Let n=a n(cid:5). Since G is generated by T, for any g ∈G we have m ng =a m n(cid:5)(b1g1+···+b m g m ) =n(cid:5)(a m b1g1+···+a m b m−1g m−1)+n(cid:5)b m (a m g m ). Both terms above are in n(cid:5)G(cid:5) ⊂ Z(cid:22)S(cid:23). It follows that nG ⊂ Z(cid:22)S(cid:23), which completes the proof. Now let G be any finitely generated abelian group. Because G/Gtor is finitely generated and torsion free, the preceding proposition implies that it is free abelian of finite rank. Thus we can define the rank of G to be the rank of G/Gtor. Example 9.15. The rank of Zn is n, and the rank of any finite group is 0 (since every element is a torsion element). The rank of a product group of the form G = Zn ×Z/(cid:22)k1 (cid:23)×···×Z/(cid:22)k m (cid:23) is n, because Gtor = Z/(cid:22)k1 (cid:23)×···×Z/(cid:22)k m (cid:23). Proposition 9.16. If G is a free abelian group of finite rank and f: G→ H is a surjective homomorphism, then rankG=rankH +rank(Kerf).

Free Abelian Groups 207 Proof. Write K = Kerf. By Proposition 9.13, K is a finitely generated free abelian group, so we can choose elements k1,…,k p ∈ G that form a basis for K. Sincef issurjective,ittakesasetofgeneratorsforGtoasetofgenerators for H; thus H is finitely generated and its rank is the rank of H/Htor. Chooseabasis (cid:7) h1,…, (cid:7) h q forH/Htor,andliftthemtoelementsh1,…,h q ∈ H. By surjectivity, there are elements g ∈G such that f(g )=h . j j j Th(cid:15)e set {k i ,g(cid:15)j } is linearly independent, because a relation of the form g = m k + n g =0 implies i i i j j j (cid:14) (cid:14) 0=f(g)= n f(g )= n h . j j j j j j Projecting (cid:15) this to the quotient group H/Htor, we obtain a relation of (cid:7) the (cid:15)form j n j h j = 0, which implies n j = 0 for each j. Therefore, g = m k =0, so m =0. i i i i Let Z(cid:22)k ,g (cid:23) denote the subgroup of G generated by {k ,g }. It is free i j i j abelian of rank p + q = rankK + rankH, so from Proposition 9.13 we conclude that rankK+rankH ≤rankG. The proof will be complete once we show the reverse inequality. Because Htor is a finitely generated torsion group, there is an integer N such that Nt = 0 for every t ∈ Htor. Let g ∈ G be arbitrary, and let [f(g)] den (cid:15) ote the equivalence class (cid:15) of f(g) in the quotient (cid:15) H/Htor. Writing [f(g)]= j n j (cid:7) h j , w(cid:15)e have [f(g− j n j g j )]=0, so f(g− j n j g j )∈Htor. This implies N(g− n g )∈K, so we can write j j j (cid:14) (cid:14) Ng = Nn g + m k . j j i i j i Letting ϕ: G → G be the homomorphism ϕ(g) = Ng as in the proof of Proposition 9.14, we have shown that ϕ(G) ⊂ Z(cid:22)k ,g (cid:23). Moreover, ϕ is i j injectivebecauseGistorsionfree,sobyProposition9.13againweconclude that rankG≤p+q =rankK+rankH.

208 9. Some Group Theory Problems 9-1. Showthatanyfreeproductoftwoormorenontrivialgroupsisinfinite and nonabelian. 9-2. Show that the following groups have the given presentations. (a) Z/(cid:22)n(cid:23)∼ =(cid:22)β |βn =1(cid:23). (b) Z/(cid:22)m(cid:23)×Z/(cid:22)n(cid:23)∼ =(cid:22)β,γ |βm =1, γn =1, βγ =γβ(cid:23). 9-3. The center of a group G is the set Z of elements of G that commute with every element of G: Z ={g ∈G:gh=hg for all h∈G}. Show that a free group on two or more generators has center consisting of the identity alone. 9-4. Let G1,G2,H1,H2 be groups, and let f i : G i → H i be group homo- morphisms for i=1,2. (a) Using the characteristic property of the free product, show that there is a unique homomorphism f1 ∗f2: G1 ∗G2 → H1 ∗H2 such that the following diagram commutes for i=1,2: G1 ∗G2 f1 ∗f2(cid:2) H1 ∗H2 (cid:8) (cid:8) ι i ι(cid:5) i (cid:2) G H , i f i i where ι i : G i →G1 ∗G2 and ι(cid:5) i : H i →H1 ∗H2 are the canonical injections. (b) Show that Kerf1 ∗f2 = Imj1 ∗j2, where j i : Kerf i (cid:9)→ G i is inclusion: Kerf1 ∗Kerf2 j−1∗ →j2 G1 ∗G2 f−1∗ →f2 H1 ∗H2. 9-5. Show that the free abelian group on a set S is uniquely determined up to isomorphism by the characteristic property (Lemma 9.11(a)). 9-6. Suppose G is a free abelian group of finite rank. Show that every basis of G is finite.

10 The Seifert–Van Kampen Theorem In this section we will develop the techniques needed to compute the fun- damental groups of all compact surfaces, and a good many other spaces as well. The basic tool is the Seifert–Van Kampen theorem, which gives a formula for the fundamental group of a space that can be decomposed as the union of two open, path connected subsets whose intersection is also path connected. Inthefirstsectionwestatearathergeneralversionofthetheorem.Then we examine two special cases in which the formula simplifies considerably. The first special case is that in which the intersection of the two subsets is simply connected: Then the theorem says that the fundamental group of thebigspaceisthefreeproductofthefundamentalgroupsofitssubspaces. Asanapplication,wecomputethefundamentalgroupsofawedgeofspaces and of a graph (a one-dimensional simplicial complex). The second special caseisthatinwhichoneofthetwosubsetsisitselfsimplyconnected:Then the fundamental group of the big space is the quotient of the fundamental group of the non–simply connected piece by the path classes in the inter- section. We use this formula to compute the fundamental groups of the compact surfaces. Aftertheseapplications,wegiveadetailedproofoftheSeifert–VanKam- pen theorem. Then in the last section of the chapter we revisit the classi- fication of compact surfaces, and prove finally that the different surfaces on our list are all topologically distinct by showing that their fundamental groups are not isomorphic.

210 10. The Seifert–Van Kampen Theorem Statement of the Theorem Here is the situation in which we will be able to compute fundamental groups. Suppose we are given a space X that is the union of two open subsets U,V ⊂ X, and suppose we can compute the fundamental groups of U, V, and U∩V, each of which is path connected. As we will see below, any loop in X can be written as a product of loops, each of which lies in either U or V; such a loop can be thought of as representing an element of thefreeproductπ1(U)∗π1(V).ButanyloopinU∩V willrepresentonlya single element of π1(X), even though it represents two distinct elements of the free product (one in π1(U) and one in π1(V)). Thus the fundamental group of X can be thought of as the quotient of this free product modulo some relations coming from π1(U ∩V) that express this redundancy. Let us set the stage for the precise statement of the theorem. Let X be any topological space, let U,V ⊂X be open subsets whose union is X and whose intersection is nonempty, and choose any base point q ∈U∩V. The four inclusion maps U (cid:3)(cid:4) (cid:6) i (cid:3) (cid:6)k (cid:3) (cid:6)(cid:7) U ∩V X (cid:6) (cid:3)(cid:4) j (cid:6) (cid:3)l (cid:6)(cid:7) (cid:3) V induce fundamental group homomorphisms π1(U,q) (cid:3)(cid:4) (cid:6) i∗ (cid:3) (cid:6)k∗ (cid:3) (cid:6)(cid:7) π1(U ∩V,q) π1(X,q). (cid:6) (cid:3)(cid:4) j∗ (cid:6) (cid:3)l∗ (cid:6)(cid:7) (cid:3) π1(V,q) Now insert the free product group π1(U,q)∗π1(V,q) into the middle of the picture, and let ι U : π1(U,q) (cid:9)→ π1(U,q)∗π1(V,q) and ι V : π1(V,q) (cid:9)→ π1(U,q)∗π1(V,q) be the canonical injections. By the characteristic prop- erty of the free product, k∗ and l∗ induce a homomorphism Φ: π1(U,q)∗ π1(V,q)→π1(X,q) such that the right half of the following diagram com-

Statement of the Theorem 211 mutes: π1(U,q) (cid:3)(cid:4) (cid:6) (cid:3) (cid:6) (cid:3) (cid:6) (cid:3) i∗(cid:3) ι U (cid:6) k∗ (cid:6) (cid:3) (cid:6) (cid:3) (cid:6) (cid:3) (cid:5) (cid:6)(cid:7) π1(U ∩V,q) F(cid:2) π1(U,q)∗π1(V,q) Φ(cid:2) π1(X,q). (10.1) (cid:6) (cid:8) (cid:3)(cid:4) (cid:3) (cid:6) (cid:3) (cid:6) (cid:3) j∗ (cid:6) ι V (cid:3) l∗ (cid:6) (cid:3) (cid:6) (cid:3) (cid:6) (cid:6)(cid:7) (cid:3) π1(V,q) Finally, we define a map F: π1(U ∩V,q) → π1(U,q)∗π1(V,q) by setting F(γ) = (i∗γ)−1(j∗γ). (F is not a homomorphism.) Let F(π1(U ∩V,q)) denote the normal closure of the image of F in π1(U,q)∗π1(V,q). Theorem 10.1 (Seifert–Van Kampen). Let X be a topological space. Suppose U,V ⊂X are open subsets whose union is X, and suppose U, V, and U ∩V are path connected. Then, for any q ∈ U ∩V, the homomor- phism Φ defined by (10.1) is surjective, and its kernel is F(π1(U ∩V,q)). Therefore, (cid:30) π1(X,q) ∼ =π1(U,q)∗π1(V,q) F(π1(U ∩V,q)). (10.2) When the fundamental groups in question are finitely presented, the theorem has a useful reformulation in terms of generators and relations. Corollary 10.2. In addition to the hypotheses of the Seifert–Van Kampen theorem, assume that the fundamental groups of U, V, and U∩V have the following finite presentations: π1(U,q) ∼ =(cid:22)α1,…,α m |ρ1,…,ρ r (cid:23); π1(V,q) ∼ =(cid:22)β1,…,β n |σ1,…,σ s (cid:23); π1(U ∩V,q) ∼ =(cid:22)γ1,…,γ p |τ1,…,τ t (cid:23). Then π1(X,q) has the presentation π1(X,q) ∼ =(cid:22)α1,…,α m ,β1,…,β n |ρ1,…,ρ r ,σ1,…,σ s ,u1 =v1,…,u p =v p (cid:23),

212 10. The Seifert–Van Kampen Theorem whereforeacha=1,…,p,u a isanexpressionfori∗γ a ∈π1(U,q)interms of the generators {α1,…,α m }, and v a similarly expresses j∗γ a ∈ π1(V,q) in terms of {β1,…,β n }. Theproofsofthesetwotheoremsarerathertechnical,sowewillpostpone themuntillaterinthechapter.Beforeprovingthem,wewillillustratetheir use by computing a number of fundamental groups. It is worth remarking here that the Seifert–Van Kampen theorem can be generalized to a covering of X by any number, finite or infinite, of path connected open sets containing the base point. This generalization can be found in [Sie92] or [Mas89]. Applications AllofourapplicationsoftheSeifert–VanKampentheoremwillbeinspecial cases in which one of the sets U, V, or U ∩V is simply connected. First Special Case: Simply Connected Intersection The first special case of the Seifert–Van Kampen theorem we will con- sider is that in which U ∩V is simply connected. In that case, the group F(π1(U ∩V,q)) is trivial, so the following corollary is immediate. Corollary 10.3. Assume the hypotheses of the Seifert–Van Kampen the- orem, and suppose in addition that U ∩V is simply connected. Then Φ is an isomorphism between π1(U,q)∗π1(V,q) and π1(X,q). As our first application, we will compute the fundamental group of a wedge of spaces. Recall from Example 3.25 that the wedge of spaces X1,…,X n(cid:6)withbasepointsq j ∈X j isthespaceX1 ∨···∨X n definedasthe quotient of j X j by the equivalence relation generated by q1 ∼···∼q n . A point q in a topological space X is said to be a nondegenerate base point if q has a neighborhood that admits a strong deformation retraction ontoq and{q}isclosedinX.Forexample,anybasepointinamanifoldis nondegenerate, because any Euclidean ball neighborhood admits a strong deformation retraction onto any point. (In more advanced treatments of homotopy theory a slightly more restrictive definition of nondegenerate base point is used, but this one will(cid:6)suffice for our purposes.) Observe that inclusion of X into X followed by projection onto the j j j quotient induces continuous injective maps ι j : X j (cid:9)→ X1 ∨ ··· ∨ X n . If each base point q is nondegenerate, then these are closed maps and thus j embeddings. Identifying each X with its image under ι , we will consider j j X j asasubspaceofX1 ∨···∨X n .Welet∗denotethepointinX1 ∨···∨X n that is the equivalence class of the base points q1,…,q n .

Applications 213 Lemma 10.4. Suppose q ∈ X is a nondegenerate base point for i = i i 1,…,n. Then ∗ is a nondegenerate base point in X1 ∨···∨X n . Proof. For each i, choose a neighborhood U i of q i that admits a(cid:6)strong defor(cid:6)mationretractionH i : U i ×I →U i onto{q i }.DefineamapH: i U i × I → i U i bylettingH =H i(cid:6)onU i ×I.Therestrictionofthequotientmap π to the saturated open set U is a quotient map to a neighborhood U i i of ∗. Since π◦H respects the identifications made by π, it descends to the quotient and yields a strong deformation retraction of U onto {∗}. To see that {∗} is closed in X1 ∨···∨X n , we need only observe that its inverse imageunderπ is{q1,…,q n },whichisclosedinthedisjointunionbecause its intersection {q } with X is closed in X by hypothesis. Thus {∗} is i i i closed. Proposition 10.5. Let X1,…,X n be spaces with nondegenerate base points q ∈X . The map j j Φ: π1(X1,q1)∗···∗π1(X n ,q n )→π1(X1 ∨···∨X n ,∗) induced by ι j∗ : π1(X j ,q j )→π1(X1 ∨···∨X n ,∗) is an isomorphism. Proof. FirstconsiderthewedgeoftwospacesX1 ∨X2.Wewouldliketouse Corollary10.3totheSeifert–VanKampentheoremwithU =X1,V =X2, andU∩V ={q}.ThetroubleisthatthesespacesarenotopeninX1 ∨X2, so the corollary does not apply directly. To remedy this, we replace them by slightly “thicker” spaces using the nondegenerate base point condition. Choose neighborhoods W in which q is a strong deformation retract, i i andletU =π(X1 (cid:20)W2),V =π(X2 (cid:20)W1),whereπ: X1 (cid:20)X2 →X1 ∨X2 is the quotient map (Figure 10.1). Since X1 (cid:20)W2 and X2 (cid:20)W1 are saturated opensetsinX1 (cid:20)X2,therestrictionofπ toeachofthemisaquotientmap and U and V are open in the wedge. The key fact is that the three inclusion maps {∗}(cid:9)→U ∩V, X1 (cid:9)→U, X2 (cid:9)→V areallhomotopyequivalences,becauseeachsubspaceontheleft-handside aboveisastrongdeformationretractofthecorrespondingright-handside. For U ∩ V, this follows immediately from the preceding lemma. For U, choose a strong deformation retraction H2: W2 ×I → W2 of W2 onto q2, and define G1: X1 (cid:20)W2 ×I →X1 (cid:20)W2 to be the identity on X1 ×I and H2 onW2 ×I;itdescendstoastrongdeformationretractionofU ontoX1. A similar construction shows V (cid:25)X2.

214 10. The Seifert–Van Kampen Theorem X1 W1 q1 q2 W2 X2 π V ∗ X1 ∨X2 U FIGURE 10.1. Computing the fundamental group of a wedge. Because U ∩V is contractible, Corollary 10.3 implies that the inclusion maps U (cid:9)→X1 ∨X2 and V (cid:9)→X1 ∨X2 induce an isomorphism π1(U,∗)∗π1(V,∗)→π1(X1 ∨X2,∗). Moreover,theinjectionsι1: X1 (cid:9)→U andι2: X2 (cid:9)→V,whicharehomotopy equivalences,induceisomorphismsπ1(X1,q1)→π1(U,∗)andπ1(X2,q2)→ π1(V,∗).Composingtheseisomorphismsprovesthepropositioninthecase n=2. The case of n>2 spaces follows by induction, because Lemma 10.4 guarantees that the hypotheses of the proposition are satisfied by X1 and X2 ∨···∨X n . Example 10.6. The preceding proposition shows that the bouquet S1 ∨ ···∨S1 ofncircleshasfundamentalgroupisomorphictoZ∗···∗Z,whichis afreegrouponngenerators.In fact,itshowsmore:Sincetheisomorphism is induced by inclusion of each copy of S1 into the bouquet, we can write explicitgeneratorsofthisfreegroup.Ifα denotesthestandardloopinthe i ith copy of S1, then the fundamental group of the bouquet is just the free group (cid:22)[α1],…,α n . As a second application, we will compute the fundamental group of a finite graph. Let Γ be a finite connected graph, and choose a vertex v as base point. A maximal tree in Γ is a subgraph that is a tree and is not contained in any larger tree. The vertex v is contained in a (nonunique) maximal tree: Just start with a single edge containing v and keep adding edges until it is impossible to add another edge and still obtain a tree. Let T ⊂Γ be such a maximal tree.

Applications 215 w1 w(cid:5) 1 w3 w(cid:5) 3 v w2 w(cid:5) 2 FIGURE 10.2. Generators for the fundamental group of a graph. We will construct a set of generators for the fundamental group of Γ as follows. Let (cid:22)w1,w 1 (cid:5)(cid:23),…,(cid:22)w n ,w n (cid:5)(cid:23) be the edges of Γ that are not in T (Figure 10.2). For each i, by maximality of T there is a reduced cycle in T ∪(cid:22)w ,w(cid:5)(cid:23) that does not lie in T. This cycle must therefore traverse the i i edge(cid:22)w ,w(cid:5)(cid:23),whichimpliesthatbothoftheverticesw andw(cid:5) alsobelong i i i i to edges in T. Thus we can choose paths g and h in T from v to w and i i i w(cid:5), respectively. Let f denote the loop in Γ obtained by first following g i i i fromv tow ,thentraversing(cid:22)w ,w(cid:5)(cid:23),andthenfollowingh −1 fromw(cid:5) back i i i i i tov.Notethatthepathclass[f ]isindependentofthechoicesofg andh , i i i because any two paths in T with the same endpoints are path homotopic. Theorem 10.7 (Fundamental Group of a Finite Graph). Thefun- damental group of a finite connected graph Γ based at a vertex v is the free group on the path classes [f1],…,[f n ] constructed above. Proof. We prove the theorem by induction on n. If n=0, then Γ is a tree and hence simply connected, so there is nothing to prove. For n = 1, we must show that Γ is the infinite cyclic group generated by [f1]. By assumption, there is a reduced cycle (v0,…,v m ) in Γ (Figure 10.3(a)). If v i =v j for some i(cid:14)=j other than v0 =v m , we can replace the cycle by the shorter one (v ,…,v ). So we may assume that no vertex is i j repeated in this cycle except v0 = v m . This cycle must traverse the edge (cid:22)w1,w 1 (cid:5)(cid:23),becauseotherwiseitwouldbeareducedcycleinT.Thesubgraph C ⊂ Γ consisting of the vertices {v0,…,v m } and the intervening edges is homeomorphic to S1 because it is isomorphic as a simplicial complex to the complex K of Example 5.7(d). We will show that inclusion C ⊂Γ is m a homotopy equivalence. LetK betheunionofalltheedgesinΓ(cid:3)C togetherwiththeirvertices. Each component K of K is a connected subgraph of Γ contained in T, i and is therefore a tree (since a reduced cycle in K would also be one i in T). Moreover, each such component shares exactly one vertex y with i

216 10. The Seifert–Van Kampen Theorem K1 w(cid:5) 1 x1 x2 y1 x n C y2 y3 w1 x3 K2 K3 (a) n=1. (b) n>1. FIGURE 10.3. Proof that the fundamental group of a graph is free. C: If K ∩C contained two vertices y ,y(cid:5), it would be possible to find a i i i reduced cycle in T by following a reduced edge path in K from y to y(cid:5) i i i followed by the reduced edge path in C from y(cid:5) to y that does not pass i i through (cid:22)w1,w 1 (cid:5)(cid:23). (There must be at least one vertex in common because Γ is connected.) Now define a strong deformation retraction of Γ onto C as follows: On eachK ,itisastrongdeformationretractionofK ontoy ,whichexistsby i i i Problem 7-13; and on C it is the identity. The resulting map is continuous by the gluing lemma, and shows that Γ(cid:25)S1. It remains to show that the path class [f1] is a generator of π1(Γ,v). Let z be any vertex in C. The path a that starts at z and traverses each edge of C in order at constant speed is clearly path homotopic to the standard generator of S1 ≈ C (or its inverse). Choosing any path b from z to v yields an isomorphism Φ b : π1(Γ,z)→π1(Γ,v) as in Theorem 7.11. Thus a generator of π1(Γ,v) is Φ b [a]=[b−1· a · b]. Since b−1· a · b is a path that goes from v to w1, traverses (cid:22)w1,w 1 (cid:5)(cid:23), and returns to v, it is homotopic to f1. (Remember that the path class of f1 is independent of which paths we choose from v to w1 and w 1 (cid:5).) This completes the proof in the case n=1. Now suppose n > 1, and assume that the theorem is true whenever there are fewer than n edges in the complement of a maximal tree. We will apply the Seifert–Van Kampen theorem in the following way. For each i = 1,…,n, choose a point x in the interior of the edge (cid:22)w ,w(cid:5)(cid:23) (Figure i i i 10.3(b)).LetU =Γ(cid:3){x1,…,x n−1 }andV =Γ(cid:3){x n }.BothU andV are openinΓ,andjustasbeforeitiseasytoconstructdeformationretractions to show that U ∩V (cid:25)T, U (cid:25)T ∪(cid:22)w ,w(cid:5)(cid:23), and V (cid:25)Γ(cid:3)Int(cid:22)w ,w(cid:5)(cid:23). By n n n n theinductivehypothesis,π1(V,v)=(cid:22)[f1],…,f n−1andπ1(U,v)=(cid:22)f n . Since U ∩ V is simply connected, Corollary 10.3 shows that π1(Γ,v) is isomorphic to the free product of these two free groups, which is the free group on [f1],…,[f n ] as claimed.

Applications 217 Second Special Case: One Simply Connected Set The other special case of the Seifert–Van Kampen theorem we will use is thatinwhichoneoftheopensets,sayU,issimplyconnected.Inthatcase, diagram (10.1) simplifies considerably. Because the top group π1(U,q) is trivial,boththehomomorphismsi∗ andk∗ aretrivial,andthefreeproduct in the middle reduces to π1(V,q). Moreover, the homomorphism Φ is just equaltol∗,andthemapF isjustequaltoj∗,sotheentirediagramcollapses to π1(U ∩V,q)−j→∗ π1(V,q)−l→∗ π1(X,q). The conclusion of the theorem reduces immediately to the following corol- lary. Corollary 10.8. Assume the hypotheses of the Seifert–Van Kampen the- orem, and suppose in addition that U is simply connected. Then inclusion l: V (cid:9)→X induces an isomorphism (cid:30) π1(X,q) ∼ =π1(V,q) j∗π1(U ∩V,q), where j∗π1(U ∩V,q) is the normal closure of j∗π1(U ∩V,q) in π1(V,q). If the fundamental groups of V and U ∩V have finite presentations π1(V,q) ∼ =(cid:22)β1,…,β n |σ1,…,σ s (cid:23), π1(U ∩V,q) ∼ =(cid:22)γ1,…,γ p |τ1,…,τ t (cid:23), then π1(X,q) has the presentation π1(X,q) ∼ =(cid:22)β1,…,β n |σ1,…,σ s ,v1,…,v p (cid:23), where v a is an expression for j∗γ a ∈π1(V,q) in terms of {β1,…,β m }. Wewillgivetwoapplicationsofthiscorollary.Thefirstistogiveanother proof that Sn is simply connected when n≥2. Another proof of Theorem 8.7. As in the previous proof, we take Sn = U ∪V with U = Sn(cid:3){N} and V = Sn(cid:3){S}, both of which are simply connected because they are homeomorphic toRn. Moreover, U∩V is path connected because it is homeomorphic to Rn(cid:3){0}. Corollary 10.8 to the Seifert–Van Kampen theorem says that for any point q ∈U ∩V, π1(Sn,q) is isomorphic to a quotient of π1(V,q) by a certain subgroup. But this quotient is trivial because π1(V,q) is itself trivial. The next proposition will allow us to compute the fundamental groups of all compact surfaces. Now it will become clear why we chose similar notations for surface presentations and group presentations.

218 10. The Seifert–Van Kampen Theorem v0 a i g q c P 0 a i π (cid:7)a i v(cid:7) M g(cid:7) q(cid:7) FIGURE 10.4. Proof of Proposition 10.9. Proposition 10.9. Let M be a 2-manifold determined by a polygonal surface presentation (cid:22)a1,…,a n | W(cid:23) with one face, in which all ver- tices are identified to a single point. Then π1(M) has the presentation (cid:22)a1,…,a n |W(cid:23). Proof. First we set up some notation (see Figure 10.4). Let P be a regular polygon in the plane with 2n sides, centered at the origin, and let π: P → M denote the quotient map determined by the given presentation. Set U = IntP and V = P (cid:3){0}, and let U (cid:7) = π(U), V (cid:7) = π(V) ⊂ M. Since U and V are saturated open sets, the restrictions of π to U and V are (cid:7) (cid:7) (cid:7) (cid:7) quotientmaps,andtheirimagesU,V areopeninM.Moreover,U,V,and U (cid:7) ∩V (cid:7) are all path connected because they are images of path connected sets in P. Choose a base point q ∈U ∩V lying on the line segment from 0 tothevertexv0 onthey-axis,andletq(cid:7)=π(q)∈U (cid:7)∩V (cid:7) .Finally,letv(cid:7)∈M denote the single vertex (the image of all the vertices of P). In general, we will use symbols without tildes to denote sets, points, or paths in P, and the same symbols with tildes to denote their images in M. The restriction of π to U is a one-to-one quotient map and therefore a homeomorphism. Since U is convex, it is simply connected, and therefore (cid:7) U is simply connected as well. Thus we will be able to use Corollary 10.2 oncewefindpresentationsforthefundamentalgroupsofV (cid:7) andU (cid:7)∩V (cid:7) .The (cid:7) details are a bit involved, but the basic idea is just that V is homotopy

Applications 219 equivalent to a bouquet of circles, so π1(V (cid:7) ,q(cid:7)) is a free group on the n generators determined by a1,…,a n ; and U (cid:7)∩V (cid:7) is homotopy equivalent to acircle,soj∗π1(U (cid:7)∩V (cid:7) ,q(cid:7))istheinfinitecyclicgroupgeneratedbytheword W in these generators. Consider first V (cid:7) . Observe that its preimage V ⊂P is homotopy equiva- lentto∂P,bythestraight-linehomotopyH thatmoveseachpointradially outward to the boundary. The map π×Id: P ×I → M ×I is a quotient map by Lemma 4.35 (or just by the closed map lemma). Since V ×I is a saturated open subset of P ×I, the restriction of π×Id to it is a quotient map. Because π ◦H respects the identifications of this quotient map, it (cid:7) (cid:7) descends to a strong deformation retraction H of V onto π(∂P): V ×I H (cid:2) V π×Id π (cid:5) (cid:5) V (cid:7) ×I (cid:7) (cid:2) V (cid:7) . H On ∂P, π identifies the edges in pairs and identifies all the vertices to a point, so π(∂P) is homeomorphic to a bouquet of n circles, one for each labela .(Toverifythis,youcan,forexample,constructquotientmapsfrom i 2n disjoint intervals to S1∨···∨S1 and to π(∂P), both of which make the (cid:7) same identifications.) Therefore, π1(V,q) is a free group on n generators. We need to identify these generators explicitly. For each i, let (cid:7)a denote i apathinM thattraversesthesingleedgelabeleda (theimageunderπ of i twoedgesofP)intheindicateddirection.SinceallverticesofP projectto v(cid:7), (cid:7)a i is a loop based at v(cid:7). From Example 10.6, π1(S1∨···∨S1,v(cid:7)) is freely generatedby[(cid:7)a1],…,[(cid:7)a n ].Theisomorphismπ1(S1∨···∨S1,v(cid:7))→π1(V (cid:7) ,v(cid:7)) induced by inclusion takes these generators to themselves, thought of as loops in V (cid:7) , so π1(V (cid:7) ,v(cid:7)) is the free group (cid:22)[(cid:7)a1],…,(cid:7)a n . We are really interested in the base point q(cid:7), not v(cid:7). Let g be the radial straight-line path from q to v0, let g(cid:7) = π ◦ g, a path from q(cid:7) to v(cid:7), and consider the loops (cid:7)a(cid:5) based at q(cid:7), defined by i (cid:7)a(cid:5) =g(cid:7) · (cid:7)a · g(cid:7)−1. i i These are just the images of the loops (cid:7)a under the isomorphism from i π1(V (cid:7) ,v(cid:7)) to π1(V (cid:7) ,q(cid:7)) provided by the path g(cid:7)−1 as in Theorem 7.11, so we conclude that π1(V (cid:7) ,q(cid:7)) is the free group (cid:22)[(cid:7)a(cid:5) 1 ],…,(cid:7)a(cid:5) n . Next we turn to U (cid:7)∩V (cid:7) . Since π is one-to-one on U ∩V, U (cid:7)∩V (cid:7) is home- omorphic to U ∩V = IntP (cid:3){0}. Let C be the circle centered at 0 and (cid:7) passing through q, and let C be its image in M. (We may assume that q is chosen close enough to 0 that C is contained in the interior of P.) A de- formation retraction of IntP (cid:3){0} onto C is easily constructed by moving each point radially inward or outward toward C at constant speed. Thus

220 10. The Seifert–Van Kampen Theorem π1(U (cid:7)∩V (cid:7) ,q(cid:7)) is infinite cyclic, generated by the path class of any loop that (cid:7) goes around C once. Let c be a loop that goes counterclockwise around C, and let (cid:7)c = π ◦c, a loop in U (cid:7) ∩V (cid:7) based at q(cid:7). Then π1(U (cid:7) ∩V (cid:7) ,q(cid:7)) is the infinite cyclic group (cid:22)(cid:7)c. Now Corollary 10.8 to the Seifert–Van Kampen theorem says that π1(M,q(cid:7)) ∼ = (cid:22)[(cid:7)a(cid:5) 1 ],…,[(cid:7)a(cid:5) n ] | b(cid:23), where b is an expression for j∗[(cid:7)c] ∈ π1(V (cid:7) ,q(cid:7)) in terms of the generators [(cid:7)a(cid:5)],…,[(cid:7)a(cid:5) ]. So to complete the proof, we need 1 n only find such an expression. Thekeyobservationisthatg(cid:7)−1· (cid:7)c · g(cid:7),aloopbasedatv(cid:7),ispathhomotopic inV (cid:7) totheloopW (cid:8) obtainedfromW byreplacingeacha by(cid:7)a .Toseethis, i i we first work upstairs in P: Let H be the strong deformation retraction of V onto ∂P, and set F(s,t)=H(g−1· c · g(s),t). The map F is a homotopy from g−1· c · g to a path that goes once around (cid:7) ∂P in the clockwise direction, and it descends to a homotopy in V from g(cid:7)−1· (cid:7)c · g(cid:7)to W (cid:8) . It follows that (cid:7)c∼g(cid:7) · g(cid:7)−1· (cid:7)c · g(cid:7) · g(cid:7)−1 ∼g(cid:7) · W (cid:8)· g(cid:7)−1 ∼W (cid:8)(cid:5), where W (cid:8)(cid:5) is obtained from W by replacing each a by (cid:7)a(cid:5); the last equiva- lence follows by inserting g(cid:7)−1· g(cid:7)between each pair i of sym i bols in the word (cid:8) W. Thus we have shown that the generator [(cid:7)c] of π1(U (cid:7)∩V (cid:7) ,q(cid:7)) is mapped by inclusion to [W (cid:8)(cid:5)] ∈ π1(V (cid:7) ,q(cid:7)), where [W (cid:8)(cid:5)] is obtained from W by replacing each a i by [(cid:7)a(cid:5) i ]. By Corollary 10.2, therefore, π1(M,q(cid:7)) has the presentation (cid:22)[(cid:7)a(cid:5)],…,[(cid:7)a(cid:5) ] | W (cid:8)(cid:5); relabeling the symbols a instead of [(cid:7)a(cid:5)] yields the 1 n i i presentation given in the statement of the proposition. Example 10.10. Using the results of this section, we have the following presentations for the fundamental groups of compact, connected surfaces: (a) π1(S2) ∼ =(cid:22) (cid:23) (the trivial group). (b) π1(T2#···#T2) ∼ =(cid:22)β1,γ1,…,β n ,γ n |β1γ1β 1 −1 γ 1 −1···β n γ n β n −1γ n −1 =1(cid:23). (c) π1(P2#···#P2) ∼ =(cid:22)β1,…,β n |β 1 2···β n 2 =1(cid:23). In particular, for the torus this gives π1(T2) ∼ = (cid:22)β,γ | βγ = γβ(cid:23), which agreeswiththeresultwederivedearlier.Inthecaseoftheprojectiveplane, this gives π1(P2) ∼ =(cid:22)β |β2 =1(cid:23)∼ =Z/(cid:22)2(cid:23).

Proof of the Theorem 221 Proof of the Theorem In this section we prove the Seifert–Van Kampen theorem (Theorem 10.1) and its corollary about finite presentations (Corollary 10.2). Proof of Theorem 10.1. Becausewewillbeconsideringpathsandtheirho- motopy classes in various spaces, for this proof we will refine our notation to specify explicitly where homotopies are assumed to lie. If a and b are pathsinX thathappentolieinoneofthesubsetsU,V,orU∩V,wewill use the notation a∼b, a∼b, a ∼ b, a∼b U V U∩V X toindicatethataispathhomotopictobinU,V,U∩V,orX,respectively. Wewrite[a] U forthepathclassofainπ1(U,q),andsimilarlyfortheother sets.Thus,forexample,ifaisaloopinU∩V,thehomomorphismsinduced by the inclusions i: U ∩V (cid:9)→U and k: U (cid:9)→X can be written i∗([a] U∩V )=[a] U , k∗([a] U )=[a] X . We will have to consider two different types of products: path class mul- tiplication within any one fundamental group, and word multiplication in the free product group. As usual, we will denote path and path class mul- tiplication by a dot, as in · · [a] [b] =[a b] . U U U To emphasize the distinction between the two products, we denote multi- plication in the free product group by an asterisk, so, for example, · [a] U ∗[b] U ∗[c] V =[a b] U ∗[c] V ∈π1(U,q)∗π1(V,q). Then the map Φ: π1(U,q)∗π1(V,q)→π1(X,q) can be written Φ([a1] U ∗[a2] V ∗···∗[a m−1] U ∗[a m ] V ) · · · · =k∗[a1] U l∗[a2] V ··· k∗[a m−1] U l∗[a m ] V (10.3) · · · · =[a1] X [a2] X ··· [a m−1] X [a m ] X · · · · =[a1 a2 ··· a m−1 a m ] X . LetN denotethenormalclosureofF(π1(U∩V,q))inπ1(U,q)∗π1(V,q). We need to prove three things: (1) Φ is surjective; (2) N ⊂KerΦ; and (3) KerΦ⊂N.

222 10. The Seifert–Van Kampen Theorem (cid:12) (cid:13) (cid:12) (cid:13) (cid:12) (cid:13) a i+1 a i a i−1 n n n a i+1 a i h i+1 h h i−1 i a U V q FIGURE 10.5. Proof that Φ is surjective. Step 1: Φ is surjective. Let a: I →X be any loop in X based at q. By the Lebesgue number lemma, we can choose n large enough that a maps each subinterval [(i−1)/n,i/n] either into U or into V. (This is why it is importantthatthesetsU andV beopen.)Lettinga denotetherestriction i of a to [(i−1)/n,i/n] (reparametrized so that its parameter interval is I), the path class of a in X factors as · · [a] X =[a1 ··· a n ] X . The problem with this factorization is that the paths a are not loops in i general. To remedy this, for each i = 1,…,n−1, choose a path h from i q to a(i/n) (Figure 10.5). If a(i/n) ∈ U ∩V, choose h to lie entirely in i U ∩V; otherwise, choose it to lie in whichever set U or V contains a(i/n). (This is why the sets U, V, and U ∩V must all be path connected.) Then set (cid:7)a i = h i−1 · a i · h − i 1 (where we let h0 and h n be the constant loop c q ), so that each (cid:7)a is a loop based at q and lying entirely in either U or V. It i follows easily that a also factors as · · [a] X =[(cid:7)a1 ··· (cid:7)a n ] X . Now consider the element γ =[(cid:7)a1] U ∗[(cid:7)a2] V ∗···∗[(cid:7)a n ] V ∈π1(U,q)∗π1(V,q),

Proof of the Theorem 223 wherewechooseeitherU orV foreach(cid:7)a dependingonwhichsetcontains i its image. Then as in (10.3) above, · · Φ(γ)=[(cid:7)a1 ··· (cid:7)a n ] X =[a] X . This proves that Φ is surjective. Step 2: N ⊂KerΦ. If we can show that F(π1(U∩V,q)) is contained in KerΦ,thenitsnormalclosureN willbecontainedinKerΦaswellbecause KerΦ is normal. Let [a] U∩V ∈π1(U ∩V,q) be arbitrary. Then Φ◦F([a] U∩V )=Φ((i∗[a] U∩V )−1∗(j∗[a] U∩V )) =Φ([a−1] ∗[a] ) U V =[a−1· a] X =1. Step 3: KerΦ⊂N. This is the crux of the proof. Let γ =[a1] U ∗[a2] V ∗···∗[a k ] V ∈π1(U,q)∗π1(V,q) be an arbitrary element of the free product, and suppose that Φ(γ) = 1. Using (10.3) again, this means that · · [a1 ··· a k ] X =1, which is equivalent to · · a1 ··· a k ∼c q . X We need to show that γ ∈N. · · LetH: I×I →X bethepathhomotopyfroma1 ··· a k toc q inX.By the Lebesgue number lemma again, we can subdivide I×I into squares of side1/nsothatH mapseachsquareS =[(i−1)/n,i/n]×[(j−1)/n,j/n] ij either into U or into V. Let v denote the image under H of the vertex (i/n,j/n); and let a ij ij denote the restriction of H to the horizontal line segment [(i−1)/n,i/n]× {j/n},andb therestrictiontotheverticalsegment{i/n}×[(j−1)/n,j/n], ij both suitably reparametrized on I (see Figure 10.6). TherestrictionofH tothebottomedgeofI×I,wheret=0,isequalto · · the path product a1 ··· a k . By taking n to be a sufficiently large power of 2, we can ensure that the endpoints of the paths a in this product are i

224 10. The Seifert–Van Kampen Theorem a1n … a in … a nn H a ij b i−1,j b ij a i,j−1 S ij q a10 … a i0 … a n0 FIGURE 10.6. Proof that KerΦ⊂N. of the form i/n, so the path obtained by restricting H to the bottom edge of the square can also be written · · · · · · · · H0 ∼a1 ··· a k ∼(a10 ··· a p0) ··· (a r0 ··· a n0). In the free product, this means that · · · · γ =[a10 ··· a p0] U ∗···∗[a r0 ··· a n0] V . We would like to factor this in the free product as [a10] U ∗[a20] U ∗··· and soforth.Butthesepathsarenotloopsbasedatq,sowecannotyetusethis relation directly. This is easy to fix as in Step 1: For each i and j, choose a path h from q to v , staying in U∩V if v ∈U∩V, and otherwise in U ij ij ij or V; if v happens to be the base point q, choose h to be the constant ij ij loop c . Then define loops q (cid:7)a ij =h i−1,j · a ij · h − ij 1 , (cid:7) b ij =h i,j−1 · b ij · h − ij 1 , (10.4) each of which lies entirely in U or V. Then γ can be factored as γ =[(cid:7)a10] U ∗[(cid:7)a20] U ∗···∗[(cid:7)a n0] V . (10.5) The main idea of the proof is this: We will show that modulo N, the expression (10.5) for γ can be replaced by the corresponding expression obtained by restricting H to the top edge of the first row of squares: γ ≡[(cid:7)a11] U ∗···∗[(cid:7)a n1] V (mod N).

Proof of the Theorem 225 Repeating this argument, we move up to the next row, and so forth by induction, until we obtain γ ≡[(cid:7)a1n ] U ∗···∗[(cid:7)a nn ] V (mod N). But the entire top edge of I ×I is mapped by H to the point q, so each (cid:7)a is equal to the constant loop c , and this last product is equal to the in q identity. This shows that γ ∈N, completing the proof. Thus we need to prove the following inductive step: Assuming by induc- tion that γ ≡[(cid:7)a1,j−1] U ∗···∗[(cid:7)a n,j−1] V (mod N), (10.6) we need to show that γ is equivalent modulo N to the same expression with j−1 replaced by j. First we observe the following simple fact: Suppose a is a loop in U∩V. Then [a] and [a] are in the same coset in the free product modulo N, U V because [a] V ∗N =[a] U ∗([a] − U 1∗[a] V )∗N =[a] U ∗F([a] U∩V )∗N =[a] U ∗N. Since N is normal, this also implies x∗[a] ∗y∗N = x∗[a] ∗N ∗y = U U x∗[a] ∗N ∗y = x∗[a] ∗y∗N for any x,y in the free product. Thus, V V as long as we are computing modulo N and a is a loop in U ∩V, we can freely interchange [a] with [a] wherever either appears. U V Consider a typical square S , and suppose for definiteness that H maps ij S into V. The boundary of S , traversed clockwise starting at the lower ij ij · · −1· −1 left corner, is mapped to the path (b i−1,j a ij ) (b ij a i,j−1 ). By Lemma 7.12, this means that a i,j−1 ∼b i−1,j · a ij · b − ij 1 . (10.7) V Using the definition (10.4) of the loops (cid:7)a and (cid:7) b , (10.7) yields ij ij (cid:7)a i,j−1 =h i−1,j−1 · a i,j−1 · h − i,j 1 −1 ∼h i−1,j−1 · b i−1,j · a ij · b − ij 1· h − i,j 1 −1 (10.8) V ∼(cid:7) b i−1,j · (cid:7)a ij ·(cid:7) b − ij 1 , V since the interior factors of h ij and h i−1,j and their inverses cancel out. Now start with the expression (10.6) for γ. For each factor [(cid:7)a i,j−1] U , check whether the square S above it is mapped into U or V. If it is ij

226 10. The Seifert–Van Kampen Theorem mapped into V, then (cid:7)a i,j−1 must map into U∩V, and we can replace this factor by [(cid:7)a i,j−1] V modulo N. Correct each factor whose square maps into U similarly. By (10.8),wecanreplaceeachsuchfactor[(cid:7)a i,j−1] V by[ (cid:7) b i−1,j ] V ∗[(cid:7)a ij ] V ∗ (cid:7) −1 [b ] , and similarly for the factors in U. Thus ij V γ ≡[ (cid:7) b0,j ] U ∗[(cid:7)a1j ] U ∗[ (cid:7) b1j ] − U 1∗···∗[ (cid:7) b n−1,j ] V ∗[(cid:7)a nj ] V ∗[ (cid:7) b nj ] − V 1 (mod N) ≡[(cid:7)a1j ] U ∗···∗[(cid:7)a nj ] V (mod N). (cid:7) Here we have used the facts that the interior b factors all cancel each ij (cid:7) (cid:7) (cid:7) (cid:7) other out (replacing [b ij ] U by [b ij ] V when necessary), and b0j and b nj are both equal to the constant loop c . This completes the inductive step and q thus the proof. Proof of Corollary 10.2. To simplify the notation, let A and B denote the free groups (cid:22)α1,…,α m (cid:23) and (cid:22)β1,…,β n (cid:23), respectively, and write R = {ρ1,…,ρ r },S ={σ1,…,σ s },andG={u − 1 1 v1,…,u− p 1v p },allconsidered as subsets of A∗B. (Note that the relators τ in U ∩V do not enter into i ∼ either the statement of the corollary or its proof.) Then π1(U,q) = A/R ∼ and π1(V,q) = B/S by hypothesis. We will consider these isomorphisms as identifications, so Φ is a map from (A/R) ∗ (B/S) to π1(X,q). Let π: A∗B → (A/R)∗(B/S) denote the homomorphism induced by the projections A→A/R and B →B/S as in Problem 9-4. TheSeifert–VanKampentheoremsaysthatΦissurjectiveandhaskernel equal to F(π1(U ∩V,q)), which clearly contains π(G) by our choice of the set G. In fact, it is equal to π(G): To see this, we need only verify that π(G) includes all of F(π1(U ∩V,q)), which is to say every element of (cid:12)the form (i∗γ)−1 (cid:13)(j∗γ) for γ ∈ π1(U ∩V,q). Consider the quotient group (A/R)∗(B/S) /π(G).BydefinitionofG,eachelementoftheformu −1 v i i projects to 1 in this quotient, which is to say that u and v project to i i the same element. Given any γ ∈π1(U ∩V,q), express it as a word in the generatorsγ1,…,γ p ;theni∗γ andj∗γ willbeexpressedbythesameword, but with γ replaced by u in one case and v in the other. This shows that i i i i∗γ andj∗γ projecttothesameelementinthequotient,whichmeansthat (i∗γ)−1(j∗γ)∈π(G). The composition A∗B →π (A/R)∗(B/S)→Φ π1(X,q) is also surjective. The corollary will follow from the first isomorphism the- orem (Theorem A.13) once we show that its kernel is exactly R∪S∪G. Clearly,RandS arecontainedinKer(Φ◦π).Also,π(G)⊂F(π1(U∩V,q) byourchoiceofG,andthusπ(G)⊂KerΦ,whichmeansthatG∈Ker(Φ◦

Distinguishing Manifolds 227 π) as well. Therefore, the kernel of Φ◦π is a normal subgroup containing R∪S∪G, so R∪S∪G⊂Ker(Φ◦π). To prove the reverse inclusion, suppose w ∈ A∗B is a word such that Φ◦π(w) = 1. This means that π(w) ∈ KerΦ = π(G) = π(G). (The last equality follows because the homomorphic image of a normal subgroup under a surjective map is normal; see Exercise A.25 in the Appendix.) In other words, π(w)=π(g), where g is some element of G. This just means that wg−1 ∈Kerπ, which by Problem 9-4 is equal to R∗S (thought of as a subgroup of A∗B). This can be rewritten as w =hg for some h∈R∗S, g ∈G. Since clearly R∪S∪G must contain R∗S and G, this means that w ∈R∪S∪G, and the proof is complete. Distinguishing Manifolds Nowwearefinallyinapositiontofillthegapinourclassificationofsurfaces by showing that the different surfaces on our list are actually topologically distinct. We will do so by showing that their fundamental groups are not isomorphic.Eventhisisnotcompletelystraightforward,becauseitinvolves solvingtheisomorphismproblemforcertainfinitelypresentedgroups.But inthiscasewecanreducetheproblemtoamuchsimplerprobleminvolving abelian groups. GivenagroupG,thecommutatorsubgroupofG,denotedby[G,G],isthe normalclosureofthesetofallelementsoftheformαβα−1β−1forα,β ∈G. Clearly,thequotientgroupG/[G,G]isalwaysabelian,andthecommutator subgroup is trivial if and only if G itself is abelian. This quotient group is denoted by Ab(G) and called the abelianization of G. It is clear that isomorphic groups have isomorphic abelianizations. Ab(G) is the “largest” abelian quotient of G, or equivalently the largest abelian homomorphic image of G, in the sense that any other homomorphism into an abelian group factors through the abelianization, as the following characteristic property shows. Theorem 10.11 (Characteristic Property of Abelianizations). For any abelian group H and any homomorphism ϕ: G → H, there exists a unique homomorphism ϕ(cid:7): Ab(G) → H such that the following diagram commutes: ϕ(cid:2) G H (cid:3)(cid:4) π (cid:3)ϕ(cid:7) (cid:5)(cid:3) Ab(G). Exercise 10.1. Prove Theorem 10.11. Itisrelativelyeasytocomputetheabelianizationsofoursurfacegroups.

228 10. The Seifert–Van Kampen Theorem Theorem 10.12. The fundamental groups of compact surfaces have the following abelianizations: Ab(π1(S2))={1}; Ab(π1( (cid:31) T2#· · ! ·#T " 2)) ∼ =Z2n; n Ab(π1( (cid:31) P2#· · ! ·#P " 2)) ∼ =Zn−1×Z/(cid:22)2(cid:23). n Proof. The case of the sphere is obvious, so consider first an orientable surface of genus n, and let G=(cid:22)β1,γ1,…,β n ,γ n |β1γ1β 1 −1 γ 1 −1···β n γ n β n −1γ n −1(cid:23) be the fundamental group. Define a map ϕ: Ab(G)→Z2n as follows. Let e =(0,…,1,…,0)∈Z2n (1 in the ith place), and set i ϕ(β i )=e i , ϕ(γ i )=e i+n . Thought of as a map from the free group (cid:22){β ,γ }(cid:23) into Z2n, this sends i i the element β1γ1β 1 −1 γ 1 −1···β n γ n β n −1γ n −1 to (0,…,0), so it descends to a homomorphism from G to Z2n. By the characteristic property of the abelianization, it also descends to a homomorphism (still denoted by ϕ) from Ab(G) to Z2n. To go back the other way, define ψ: Z2n →Ab(G) by (cid:16) [β ], 1≤i≤n, ψ(e )= i i [γ i−n ], n+1≤i≤2n, where the brackets on the right-hand side denote the equivalence class in Ab(G), and extend it to be a homomorphism. It is easy to check that ϕ and ψ are inverses of each other. Next consider the connected sum of projective planes, and let H = (cid:22)β1,…,β n | β 1 2···β n 2(cid:23). Write f for the nontrivial element of Z/(cid:22)2(cid:23), and define ϕ: Ab(H)→Zn−1×Z/(cid:22)2(cid:23) by (cid:16) e , 1≤i≤n−1; ϕ(β )= i i f −e n−1 −···−e1, i=n. As before, ϕ(β2···β2) = (0,…,0) by direct computation (noting that 1 n f+f =0),soϕdescendstoAb(H).Thehomomorphismψ: Zn−1×Z/(cid:22)2(cid:23)→ Ab(H) defined by ψ(e i )=[β i ], ψ(f)=[β1 ···β n ] is easily verified to be an inverse for ϕ.

Distinguishing Manifolds 229 Corollary 10.13. Anycompact,connectedsurfaceishomeomorphictoex- actly one of the surfaces S2, T2#···#T2, or P2#···#P2. Proof. First note that the sphere cannot be homeomorphic to a connected sumoftoriorprojectiveplanes,becauseonehastrivialfundamentalgroup andtheotherdoesnot.Next,ifM isaconnectedsumofprojectiveplanes, thenAb(π1(M))containsanontrivialtorsionelement,whereastheabelian- ized fundamental groups of connected sums of tori are torsion free. There- fore, no connected sum of projective planes can be homeomorphic to a connected sum of tori. If M is a connected sum of n tori, then its abelian- ized fundamental group has rank 2n. Thus the genus (i.e., the number of tori in the connected sum) can be recovered from the fundamental group, so the genus of an orientable surface is a topological invariant. Similarly, a connected sum of n projective planes has abelianized fundamental group of rank n−1, so once again the genus is a topological invariant. Now we can tie up the loose ends regarding the combinatorial invariants we discussed at the end of Chapter 6. Recall that a compact 2-manifold is said to be orientable if it admits an oriented presentation. Corollary 10.14. A connected sum of projective planes is not orientable. Proof. By the argument in Chapter 6, if a manifold admits an oriented presentation, then it is homeomorphic to a sphere or a connected sum of tori. The preceding corollary showed that a connected sum of projective planes is not homeomorphic to any of these surfaces. Corollary 10.15. The Euler characteristic of a surface presentation is a topological invariant. Proof. Suppose P and Q are polygonal surface presentations such that |P| ≈ |Q|. Each of these presentations can be transformed into one of the standard ones by elementary transformations, and since the surfaces rep- resented by different standard presentations are not homeomorphic, both presentations must reduce to the same standard one. Since the Euler char- acteristic of a presentation is unchanged by elementary transformations, the two presentations must have had the same Euler characteristic to be- gin with. Because of this corollary, we can define the Euler characteristic of a compact surface M, denoted by χ(M), to be the Euler characteristic of any presentation of that surface.

230 10. The Seifert–Van Kampen Theorem Problems 10-1. Compute the fundamental group of each of the following spaces. (To “compute” a fundamental group means to give a presentation of the group together with a specific loop representing each generator.) (a) A closed disk with two points removed. (b) The projective plane with two points removed. (c) The connected sum of n tori with one point removed. (d) The connected sum of n tori with two points removed. 10-2. Give a purely algebraic proof that the groups (cid:22)α,β | αβαβ−1(cid:23) and (cid:22)ρ,γ |ρ2γ2(cid:23)areisomorphic.[Hint:LookattheKleinbottleforinspi- ration.] 10-3. Letnbeanintegergreaterthan2.Constructapolygonalpresentation whosegeometricrealizationhasafundamentalgroupthatiscyclicof order n. 10-4. SupposeM and N arecompact2-manifolds.Showthatanytwocon- nected sums of M and N are homeomorphic. 10-5. Compute the fundamental group of the complement of the three co- ordinate axes in R3. [Hint: This space is homotopy equivalent to the 2-sphere with six points removed.] 10-6. If M is a connected manifold of dimension at least 3 and q ∈ M, show that π1(M (cid:3){q}) ∼ =π1(M). 10-7. Let M and N be connected n-manifolds, n ≥ 3. Prove that the fun- damental group of M #N is isomorphic to π1(M)∗π1(N). 10-8. LetP bethepolyhedronofafinitesimplicialcomplexK,andassume that P is connected. Show that the fundamental group of P has a presentation with the same number of generators as π1(K(1)) and one relation for each 2-simplex in K. [Hint: First treat the case of a 2-dimensional complex by induction on the number of 2-simplices. Then use a similar argument to show that inclusion K(j) (cid:9)→ K(j+1) induces a fundamental group isomorphism for j ≥2.] 10-9. Let Γ be a finite connected graph. The Euler characteristic of Γ is χ(Γ)=V−E,whereV isthenumberofverticesandE isthenumber of edges. Show that the fundamental group of Γ is a free group on 1−χ(Γ)generators.Concludethatχ(Γ)isahomotopyinvariant,i.e., that homotopy equivalent graphs have the same Euler characteristic. [Hint:FirstshowbyinductiononthenumberofedgesthattheEuler characteristic of a finite tree is 1.]

Problems 231 0 2 4 … 2n−2 FIGURE 10.7. The space Xn of Problem 10-10. 10-10. Let X be the union of n circles of radius 1 centered at the points n {0,2,4,…,2n−2} in C, which are pairwise tangent to each other along the x-axis (Figure 10.7). (Note that X2 is homeomorphic to the figure eight space.) Prove that π1(X n ,1) is a free group on n generators, and describe explicit loops representing the generators. 10-11. Show that a connected, compact surface M is nonorientable if and only if it contains a subset homeomorphic to the M¨obius band. 10-12. Let X ⊂R3 be the union of the unit 2-sphere with the line segment {(0,0,z) : −1 ≤ z ≤ 1}. Compute π1(X,N), where N = (0,0,1) is the north pole, giving explicit generator(s). 10-13. Show that abelianization defines a functor from GROUP to AB. (You have to decide what the induced homomorphisms are.) 10-14. Given a group G, show that Ab(G) is the unique group that satisfies the characteristic property expressed in Theorem 10.11. 10-15. IfG1 andG2 aregroups,showthatAb(G1 ∗G2) ∼ =Ab(G1)⊕Ab(G2). Conclude as a corollary that the abelianization of a free group on n generators is free abelian of rank n.

11 Covering Spaces Sofarwehavedevelopedtwogeneraltechniquesforcomputingfundamental groups.Thefirstishomotopyequivalence,whichcanoftenbeusedtoshow thatonespacehasthesamefundamentalgroupasasimplerone.Thiswas used in Chapter 7, for example, to show that every contractible space is simply connected, and that the fundamental group of the punctured plane is infinite cyclic. The second is the Seifert–Van Kampen theorem, which wasusedinChapter10tocomputethefundamentalgroupsofspheresand compact surfaces. Theonlyotherfundamentalgroupwehavecomputedisthatofthecircle, for which we used a technique that at first glance might seem to be rather ad hoc. The strategy for computing π1(S1,1) in Chapter 8 was to use the properties of the exponential quotient map ε to show that any loop based at 1 in the circle lifts to a path in R that ends at an integer; this integer is called the winding number of the loop, and different loops are path homotopicifandonlyiftheyhavethesamewindingnumber.Anotherway to express this result is that lifting provides a one-to-one correspondence between the fiber of ε over 1 and the fundamental group of the circle. The main ingredients in the proof were the three “lifting properties” of the circle: the path lifting property (Lemma 8.3), the homotopy lifting property(Lemma8.4),andtheuniqueliftingproperty(Lemma8.2).These, in turn, followed from the basic fact that every point in the circle has an evenly covered neighborhood. In this chapter we introduce a far-reaching generalization of these ideas, and show how the same techniques can be applied to a broad class of topological spaces. This brings us to the last major subject in the course:

234 11. Covering Spaces (cid:7) X p q X FIGURE 11.1. An evenly covered neighborhood of q. covering spaces and covering maps. A covering map is a particular type of quotient map that has many of the same properties as the exponential quotient map. A careful study of covering maps will enable us to compute and analyze many more fundamental groups. Definitions and Basic Properties Let X (cid:7) and X be topological spaces, and let p: X (cid:7) → X be a continuous map. A subset U ⊂ X is said to be evenly covered by p if U is connected and open, and each component of p−1(U) is an open set that is mapped homeomorphically onto U by p (Figure 11.1). We usually visualize p−1(U) as a “stack of pancakes” that are projected down onto U by p. It is easy to see that any connected open subset of an evenly covered set is evenly covered. A covering map is a continuous surjective map p: X (cid:7) → X such that X (cid:7) is path connected and locally path connected, and every point q ∈ X has an evenly covered neighborhood. If p: X (cid:7) → X is a covering map, we call (cid:7) X a covering space of X, and X the base of the covering. (Some authors define covering spaces more generally, omitting the re- (cid:7) quirement that X be locally path connected or path connected, or even sometimes omitting any connectivity requirement at all. In that case vari-

Definitions and Basic Properties 235 ous connectivity hypotheses have to be added to the theorems below. We havechosentoincludethesehypothesesinthedefinitionofcoveringmaps, because most of the interesting results require them, such as the lifting criterion, the covering group structure theorem, and the classification of covering spaces, and this frees us from having to remember which connec- tivity hypotheses are necessary for which theorems. In any case, connected manifolds and most interesting spaces built from them will always satisfy the hypotheses.) Example 11.1. Theexponentialquotientmapε: R→S1 givenbyε(x)= e2πix is a covering map; this is the content of Lemma 8.5. Example 11.2. The nth power map p : S1 → S1 given by p (z) = zn is n n alsoacoveringmap.Foranyz0 ∈S1,thesetU =S1(cid:3){−z0 }haspreimage equalto{z ∈S1 :zn (cid:14)=−z0 },whichhasncomponents,eachofwhichisan open arc mapped homeomorphically by p onto U. n Example 11.3. Define E: Rn →Tn by E(x1,…,x n )=(ε(x1),…,ε(x n )), where ε is the exponential quotient map of Example 11.1. Since a product of covering maps is a covering map (Problem 11-2), E is a covering map. Example 11.4. Define a map π: Sn →Pn (n≥1) by sending each point x in the sphere to the line through the origin and x, thought of as a point in Pn. Then π is a covering map (Problem 11-1), and the fiber over each point of Pn is a pair of antipodal points {x,−x}. Exercise 11.1. Let Xn be the union of n circles in C as described in Problem 10-10. Define a map p: X3 → X2 by letting A, B, and C denote thecirclescenteredat0,2,and4,respectively(seeFigure11.2),anddefining ⎧ ⎪⎨z, z∈A; p(z)= 2−(z−2)2, z∈B; ⎪⎩ 4−z, z∈C. (In words, p is the identity on A, wraps B twice around itself, and reflects C onto A). Show that p is a covering map. Lemma 11.5 (Elementary Properties of Covering Maps). Every coveringmapisalocalhomeomorphism,anopenmap,andaquotientmap. A one-to-one covering map is a homeomorphism. Exercise 11.2. Prove Lemma 11.5. Itisimportanttorealizethatasurjectivelocalhomeomorphismmaynot be a covering map, as the next example shows.

236 11. Covering Spaces A B C FIGURE 11.2. Two views of the map of Exercise 11.1. Example 11.6. LetX (cid:7) betheinterval(0,2)⊂R,anddefinef: X (cid:7) →S1by f(x) = e2πix (Figure 11.3). Then f is a local homeomorphism (because it istherestrictionofthecoveringmapε),andisclearlysurjective.However, f is not a covering map, as is shown in the following exercise. Exercise 11.3. Prove that the map f in the preceding example is not a covering map by showing that the point 1 ∈ S1 has no evenly covered neighborhood. Recall from Chapter 8 that a local section of a continuous map is a continuous right inverse defined on some open subset. Lemma 11.7 (Existence of local sections). Let p: X (cid:7) → X be a cov- ering map. Given any evenly covered open set U ⊂ X, any q ∈ U, and any q(cid:7) 0 in the fiber over q, there exists a local section σ: U →X (cid:7) such that σ(q)=q(cid:7) 0. Proof. LetU (cid:7) 0 bethecomponentofp−1(U)containingq(cid:7) 0.Sincetherestric- tion of p to U (cid:7) 0 is a homeomorphism, we can just take σ =(p| (cid:7) )−1. U0 Proposition 11.8. For any covering map p: X (cid:7) → X, the cardinality of the fibers p−1(q) is the same for all fibers. Proof. IfU isanyevenlycoveredopensetinX,eachcomponentofp−1(U) contains exactly one point of each fiber. Thus, for any q,q(cid:5) ∈U, there are one-to-one correspondences p−1(q)↔{components of p−1(U)}↔p−1(q(cid:5)), which shows that the number of components is constant on U. It follows that the set of points q(cid:5) ∈X such that p−1(q(cid:5)) has the same cardinality as p−1(q) is open.

Definitions and Basic Properties 237 (cid:7) X f FIGURE 11.3. A surjective local homeomorphism that is not a covering map. Now choose any point q ∈X, and let A be the set of points in X whose fibershavecardinalityequaltothatofp−1(q).ThenAisopenbytheabove argument, and X (cid:3)A is open because it is a union of open sets (one for each cardinality not equal to that of p−1(q)). Since X is connected and A is not empty, A must be equal to X. If p: X (cid:7) → X is a covering map, the cardinality of any fiber is called the number of sheets of the covering. For example, the nth power map of Example 11.2 is an n-sheeted covering; the map π: Sn → Pn of Example 11.4isatwo-sheetedcovering;andtheexponentialquotientmapε: R→S1 has countably many sheets. Lifting Properties The key technical tools for working with covering spaces are the following three lifting properties, which are straightforward generalizations of the ones we proved for the circle in Chapter 8. In fact, the proofs for the circle apply almost verbatim to these more general propositions. We sketch the proofs in streamlined form here; if you remember the arguments given in Chapter 8, you can safely skip these proofs. If p: X (cid:7) → X is a covering map and ϕ: B → X is any continuous map, a lift of ϕ is a continuous map ϕ(cid:7): B →X (cid:7) such that p◦ϕ(cid:7)=ϕ: (cid:7) X (cid:3)(cid:4) ϕ(cid:7)(cid:3) p (cid:3) (cid:5) (cid:2) B ϕ X.

238 11. Covering Spaces Proposition 11.9 (Unique Lifting Property). Let p: X (cid:7) → X be a covering map. Suppose B is connected, ϕ: B → X is continuous, and ϕ(cid:7) 1,ϕ(cid:7) 2: B →X (cid:7) are lifts of ϕ that agree at some point of B. Then ϕ(cid:7) 1 ≡ϕ(cid:7) 2. Proof. As in the proof of Lemma 8.2, it suffices to show that S={b∈B : ϕ(cid:7) 1(b)=ϕ(cid:7) 2(b)} is open and closed in B. For any b ∈ S, let U ⊂ X be an evenly covered neighborhood of ϕ(b), and let U α be the component of p−1(U) containing ϕ(cid:7) 1(b)=ϕ(cid:7) 2(b). On the neighborhood V =ϕ(cid:7)− 1 1 (U α )∩ϕ(cid:7)− 2 1 (U α ) of b, ϕ=p◦ϕ(cid:7) 1 =p◦ϕ(cid:7) 2. Since p is injective on U α , this means ϕ(cid:7) 1 =ϕ(cid:7) 2 on V, so S is open. On the other hand, for b(cid:14)∈S, if U is an evenly covered neighborhood of ϕ(b),therearedisjointcomponentsU1,U2ofp−1(U)containingϕ(cid:7) 1(b),ϕ(cid:7) 2(b) suchthatpisahomeomorphismfromeachU i toU.LettingV =ϕ(cid:7)− 1 1 (U1)∩ ϕ(cid:7)− 2 1 (U2),weconcludethatϕ(cid:7) 1 (cid:14)=ϕ(cid:7) 2 onV,whichshowsthatSisclosed. Proposition 11.10 (Path Lifting Property). Letp: X (cid:7) →X beacov- ering map. Suppose f: I →X is any path, and q(cid:7) 0 ∈X (cid:7) is any point in the fiber of p over f(0). Then there exists a unique lift f (cid:7) : I → X (cid:7) of f such that f (cid:7) (0)=q(cid:7) 0. Proof. By the Lebesgue number lemma, n can be chosen large enough that p maps each subinterval [k/n,(k+1)/n] into an evenly covered open subset of X. Starting with f (cid:7) (0)=q(cid:7) 0, f (cid:7) is defined inductively by choosing an evenly covered neighborhood U containing f[k/n,(k +1)/n], a local k sectionσ : U →X (cid:7) suchthatσ (f(k/n))=f (cid:7) (k/n),andsettingf (cid:7) =σ ◦f k k k k on [k/n,(k+1)/n]. Because p◦f (cid:7) = (p◦σ )◦f = f, this is indeed a lift, k and it is unique by the unique lifting property. Proposition 11.11 (Homotopy Lifting Property). Let p: X (cid:7) → X be a covering map. Suppose f0,f1: I → X are path homotopic, and f (cid:7) 0,f (cid:7) 1: I →X (cid:7) are lifts of f0 and f1 such that f (cid:7) 0(0)=f (cid:7) 1(0). Then f (cid:7) 0 ∼f (cid:7) 1. Proof. If H: f0 ∼f1 is a path homotopy, by the Lebesgue number lemma we can choose n large enough that H maps each square of side 1/n into an evenly covered open set. Labeling the squares S = [i/n,(i+1)/n]× ij (cid:7) [j/n,(j+1)/n], we define a lift H of H square by square along the bottom row, then along the next row, and so on by induction as in the proof of Lemma 8.3. On each square S , set H (cid:7) = σ◦H, for an appropriate local ij (cid:7) section σ chosen so that the new definition of H matches the previous one atthecornerpoint(i/n,j/n).Onalinesegmentwheretwosuchdefinitions overlap, both the old and new definitions are lifts of the path obtained by restricting H to that segment, starting at the same point. Thus they are equal by the unique lifting property. On the left-hand and right-hand edges of I×I, where s=0 or s=1, H (cid:7) (cid:7) is a lift of the constant loop and therefore constant. The restriction H0 to

Covering Maps and the Fundamental Group 239 (cid:7) thebottomedgewheret=0isaliftoff0 startingatf0(0),andthereforeis (cid:7) (cid:7) (cid:7) (cid:7) equal to f0; and similarly H1 =f1. Thus H is the required path homotopy (cid:7) (cid:7) between f0 and f1. The next result is an immediate corollary of the homotopy lifting prop- erty. Corollary 11.12 (The Monodromy Theorem). Let p: X (cid:7) → X be a (cid:7) covering map. Suppose f0 and f1 are path homotopic paths in X, and f0, (cid:7) (cid:7) (cid:7) f1 are lifts of them starting at the same point. Then f0(1)=f1(1). Covering Maps and the Fundamental Group Ournexttheoremcharacterizesthefundamentalgrouphomomorphismin- duced by a covering map. Theorem 11.13 (Injectivity Theorem). Let p: X (cid:7) →X be a covering map. For any point q(cid:7) ∈ X (cid:7) , the induced homomorphism p∗: π1(X (cid:7) ,q(cid:7)) → π1(X,p(q(cid:7))) is injective. Proof. Suppose [f] ∈ π1(X (cid:7) ,q(cid:7)) is in the kernel of p∗. This means that p∗[f] = [c q ], where q = p(q(cid:7)), or in other words, p ◦ f ∼ c q in X. By the homotopy lifting property, therefore, any lifts of p◦f and c that start q at the same point must be path homotopic in X (cid:7) . Now, f is a lift of p◦f starting at q(cid:7), and the constant loop c q(cid:7) is a lift of c q starting at the same point; therefore, f ∼c q(cid:7) in X (cid:7) , which means that [f]=1. This theorem shows that the fundamental group of a covering space can be identified with a subgroup of the fundamental group of the base. We call this the subgroup induced by the covering. Example 11.14. Let p: X3 → X2 be the covering map of Exercise 11.1, and choose 1 as base point in both X3 and X2. To compute the subgroup induced by p, we need to compute the action of p on the generators of π1(X3,1). Let a, b, c be loops that go once counterclockwise around each circle A, B, and C, starting at 1, 1, and 3, respectively; and let b1 and b2 be the lower and upper halves of b, so b1 is a path from 1 to 3, b2 is a path · from 3 to 1, and b∼b1 b2. Using the result of Problem 10-10, π1(X3,1) is · · −1 the free group on [a], [b], and [b1 c b 1 ], and π1(X2,1) is the free group on [a] and [b]. The images of these generators under p∗ are [a], [b]2, and [b] · [a]−1· [b]−1, so the subgroup induced by p is the subgroup of (cid:22)[a],b generated by these three elements. Asourfirstsignificantapplicationofthetheorywegiveageneralsolution to the lifting problem for covering maps: This is the problem of deciding, givenacontinuousmapϕ: Y →X,whetherϕadmitsaliftϕ(cid:7)toacovering

240 11. Covering Spaces (cid:7) space X of X. The following theorem reduces this topological problem to an algebraic problem. Theorem 11.15 (Lifting Criterion). Suppose p: X (cid:7) → X is a covering map.LetY beaconnectedandlocallypathconnectedspace,andletϕ: Y → X be a continuous map. Given any points y0 ∈ Y and q(cid:7) 0 ∈ X (cid:7) such that p(q(cid:7) 0) = ϕ(y0), ϕ has a lift ϕ(cid:7): Y → X (cid:7) satisfying ϕ(cid:7)(y0) = q(cid:7) 0 if and only if the subgroup ϕ∗π1(Y,y0) of π1(X,ϕ(y0)) is contained in p∗π1(X (cid:7) ,q(cid:7) 0). Proof. The necessity of the algebraic condition is easy to prove (and, in fact, does not require any connectivity assumptions about Y). If ϕ(cid:7) satis- fies the conditions in the statement of the theorem, the following diagram commutes: π1(X (cid:7) ,q(cid:7) 0) (cid:3)(cid:4) (cid:3) (cid:3) ϕ(cid:7) ∗(cid:3) p∗ (cid:3) (cid:3) (cid:5) (cid:2) π1(Y,y0) ϕ∗ π1(X,ϕ(y0)). Therefore, ϕ∗π1(Y,y0)=p∗ϕ(cid:7) ∗π1(Y,y0)⊂p∗π1(X (cid:7) ,q(cid:7) 0). To prove the converse, we will “lift ϕ along paths” using the path lifting property. If ϕ(cid:7)does exist, it will have the following property: For any point y ∈Y andanypathf fromy0 toy,ϕ(cid:7)◦f isaliftofϕ◦f startingatq(cid:7) 0,and ϕ(cid:7)(y) is equal to the terminal point of this path. We use this observation to define ϕ(cid:7): Namely, for any y ∈Y, choose a path f from y0 to y, let ϕ (cid:4)◦f be the (unique) lift of the path ϕ◦f to a path in X (cid:7) starting at q(cid:7) 0, and set ϕ(cid:7)(y)=ϕ (cid:4)◦f(1). We need to show two things: (1) ϕ(cid:7) is well-defined, independently of the choice of the path f; and (2) ϕ(cid:7) is continuous. Then it is immediate from the definition that p◦ϕ(cid:7)(y) = p◦ϕ (cid:4)◦f(1) = ϕ◦f(1) = ϕ(y), so ϕ(cid:7) is a lift of ϕ. Claim 1: ϕ(cid:7) is well-defined. Suppose f and f(cid:5) are two paths from y0 to y (Figure 11.4). Then f(cid:5)· f−1 is a loop based at y0, so ϕ∗[f(cid:5)· f−1]∈ϕ∗π1(Y,y0)⊂p∗π1(X (cid:7) ,q(cid:7) 0). This means that [ϕ◦(f(cid:5)· f−1)]=[p◦g] for some loop g in X (cid:7) based at q(cid:7) 0. Thus we have the following path homotopy in X: p◦g ∼ϕ◦(f(cid:5)· f−1)=(ϕ◦f(cid:5)) · (ϕ◦f)−1,

Covering Maps and the Fundamental Group 241 (cid:7) X ϕ(cid:7)(y) ϕ (cid:4)◦f g q(cid:7) 0 ϕ (cid:5)◦f(cid:5) ϕ(cid:7) p f Y ϕ(y) ϕ ϕ◦f X y p◦g y0 ϕ(y0) ϕ◦f(cid:5) f(cid:5) FIGURE 11.4. Proof that ϕ(cid:7)is well-defined. which implies (p◦g) · (ϕ◦f)∼(ϕ◦f(cid:5)). Bythemonodromytheorem,theliftsofthesetwopathsstartingatq(cid:7) 0 have the same terminal points. Since the lift of p◦g is g, which starts and ends at q(cid:7) 0, this implies ϕ (cid:5)◦f(cid:5)(1)=g · ϕ (cid:4)◦f(1)=ϕ (cid:4)◦f(1), so ϕ(cid:7)is well-defined. Claim 2: ϕ(cid:7) is continuous. Before proving this, we will show that ϕ(cid:7) has one important property of a continuous map: It takes path connected sets to path connected sets. Let V ⊂ Y be path connected, and y1,y2 ∈ V be arbitrary. There is a path f in Y from y0 to y1, and a path g in V from · y1 to y2 (Figure 11.5); by definition, ϕ(cid:7) maps the path f g to the lift of · (ϕ◦f) (ϕ◦g). In particular, the lift of ϕ◦g is a path from ϕ(cid:7)(y1) to ϕ(cid:7)(y2) that is contained in ϕ(cid:7)(V). This proves that ϕ(cid:7)(V) is path connected. To prove that ϕ(cid:7) is continuous, it suffices to show that each point in Y has a neighborhood on which ϕ(cid:7) is continuous. Let y ∈ Y be arbitrary, let (cid:7) U beanevenlycoveredneighborhoodofϕ(y),andletU bethecomponent of p−1(U) containing ϕ(cid:7)(y) (Figure 11.6). If V is the path component of ϕ−1(U) containing y, the argument above shows that ϕ(cid:7)(V) is a connected

242 11. Covering Spaces (cid:7) X ϕ(cid:7)(V) ϕ(cid:7)(y1) ϕ(cid:7)(y2) Y ϕ(cid:7) q(cid:7) 0 f y0 g V y1 y2 FIGURE11.5.Proofthatϕ(cid:7)takespathconnectedsetstopathconnected sets. subset of p−1(U), and must therefore be contained in U (cid:7) . Since Y is locally path connected, V is open and thus is a neighborhood of y. Let σ: U →U (cid:7) be the local section of p taking ϕ(y) to ϕ(cid:7)(y), so p◦σ is the identity on U. The following equation holds on V: p◦ϕ(cid:7)=ϕ=p◦σ◦ϕ. Both ϕ(cid:7) and σ ◦ϕ map V into U (cid:7) , where p is injective, so this equation implies ϕ(cid:7)=σ◦ϕ on V, which is a composition of continuous maps. The following corollaries are immediate. Corollary 11.16. If p: X (cid:7) → X is a covering map and Y is a simply connected and locally path connected space, every continuous map ϕ: Y → X has a lift to X (cid:7) . Given any point y0 ∈ Y, the lift can be chosen to take y0 to any point in the fiber over ϕ(y0). Corollary 11.17. Suppose p: X (cid:7) →X is a covering map and X (cid:7) is simply connected. For any connected and locally path connected space Y, a con- tinuous map ϕ: Y → X has a lift to X (cid:7) if and only if ϕ∗ is the trivial homomorphism for any base point y0 ∈ Y. If this is the case, then the lift can be chosen to take y0 to any point in the fiber over ϕ(y0). Example 11.18. Consider the n-sheeted covering of the circle given by the nth power map p : S1 → S1. It is easy to check that the subgroup of n π1(S1,1) induced by p n is the cyclic subgroup generated by [α]n. Thus, for

Covering Maps and the Fundamental Group 243 (cid:7) X ϕ(cid:7)(y) U (cid:7) ϕ(cid:7) σ p Y ϕ X y ϕ(y) U V FIGURE 11.6. Proof that ϕ(cid:7)is continuous. any integer m, there is a continuous map f making the diagram S1 (cid:3)(cid:4) (cid:3) f (cid:3) p n (cid:3) (cid:5) (cid:2) S1 S1 p m commute if and only if m = nk for some integer k. If this is the case, the lift sending 1 to 1 is obviously f =p . k Dependence on Base Points It is important to remember that in general, the subgroup induced by a covering depends not only on the covering but also on the choice of base point q(cid:7)∈ X (cid:7) . As the next theorem shows, the subgroup may change when we change base point, but it can change only in a very limited way. Theorem 11.19 (Conjugacy Theorem). Let p: X (cid:7) →X be a covering map. For any q ∈X, as q(cid:7)varies over the fiber p−1(q), the set of subgroups p∗π1(X (cid:7) ,q(cid:7))⊂π1(X,q) is exactly one conjugacy class. Proof. First we will show that given any q(cid:7),q(cid:7)(cid:5) ∈ p−1(q), the subgroups p∗π1(X (cid:7) ,q(cid:7)) and p∗π1(X (cid:7) ,q(cid:7)(cid:5)) are conjugate. Let g(cid:7)be a path in X (cid:7) from q(cid:7)to

244 11. Covering Spaces q(cid:7)(cid:5) g(cid:7) X (cid:7) q(cid:7) f p q p◦f g X FIGURE 11.7. Proof of the conjugacy theorem. q(cid:7)(cid:5), and let g = p◦g(cid:7), which is a loop in X based at q (Figure 11.7). We have four maps π1(X (cid:7) ,q(cid:7)) Φ g(cid:7)(cid:2) π1(X (cid:7) ,q(cid:7)(cid:5)) p∗ p∗ (cid:5) (cid:5) π1(X,q) (cid:2) π1(X,q), (11.1) Φ g where Φ g(cid:7)[f] = [g(cid:7)−1] · [f] · [g(cid:7)], and Φ g is defined similarly. This diagram commutes, because p∗Φ g(cid:7)[f]=p∗[g(cid:7)−1· f · g(cid:7)] =[p◦(g(cid:7)−1· f · g(cid:7))] =[(p◦g(cid:7)−1) · (p◦f) · (p◦g(cid:7))] =[g−1] · p∗[f] · [g] =Φ g p∗[f]. This means that Φ g maps the subgroup p∗π1(X (cid:7) ,q(cid:7)) into p∗π1(X (cid:7) ,q(cid:7)(cid:5)), so wecanreplacethebottomtwogroupsin(11.1)bytheimagegroupsunder

Covering Maps and the Fundamental Group 245 p∗ to obtain π1(X (cid:7) ,q(cid:7)) Φ g(cid:7)(cid:2) π1(X (cid:7) ,q(cid:7)(cid:5)) p∗ p∗ (cid:5) (cid:5) p∗π1(X (cid:7) ,q(cid:7)) Φ (cid:2) p∗π1(X (cid:7) ,q(cid:7)(cid:5)). (11.2) g Now, in this diagram, both vertical maps are isomorphisms, and the top mapisanisomorphismbyTheorem7.11.Thismeansthatthebottommap Φ is an isomorphism as well. But Φ is exactly the map that sends any g g subgroup onto its conjugate by [g]−1, so this shows that p∗π1(X (cid:7) ,q(cid:7)) and p∗π1(X (cid:7) ,q(cid:7)(cid:5)) are conjugate subgroups. Conversely, let q(cid:7)∈ p−1(q), and suppose H is any subgroup of π1(X,q) conjugate to p∗π1(X (cid:7) ,q(cid:7)). This means that there is some element [g] ∈ π1(X,q) such that H = Φ g (p∗π1(X (cid:7) ,q(cid:7))). If we let g(cid:7) be the lift of g start- ing at q(cid:7), and q(cid:7)(cid:5) = g(cid:7)(1), the above construction shows that p∗π1(X (cid:7) ,q(cid:7)(cid:5)) = Φ g (p∗π1(X (cid:7) ,q(cid:7)))=H. Thereisanimportantspecialcaseinwhichthesubgroupp∗π1(X (cid:7) ,q(cid:7))does notdependonthechoiceofbasepoint.Wesaythatthecoveringp: X (cid:7) →X isnormalifp∗π1(X (cid:7) ,q(cid:7))isanormalsubgroupofπ1(X,p(q(cid:7)))foreachq(cid:7)∈X (cid:7) . Thismeans,inparticular,thatforanyfixedq ∈X thesubgroupp∗π1(X (cid:7) ,q(cid:7)) is independent of the choice of base point q(cid:7)in the fiber over q, because the onlysubgroupconjugatetoanormalsubgroupisitself.In fact,asthenext lemma shows, as long as the induced subgroup is normal for one choice of q(cid:7)∈X (cid:7) , it is normal for all of them. Lemma 11.20. Let p: X (cid:7) → X be a covering map, and suppose the sub- group induced by p is normal for one point q(cid:7)∈X (cid:7) . Then p is normal. Proof. Let q(cid:7),q(cid:7)(cid:5) be two points of X (cid:7) , and let q =p(q(cid:7)), q(cid:5) =p(q(cid:7)(cid:5)). Let g(cid:7)be a path from q(cid:7)to q(cid:7)(cid:5), and set g = p◦g(cid:7), which is a path from q to q(cid:5). If we replace q by q(cid:5) in the bottom right corner of diagram (11.1), the diagram still commutes, and the top and bottom rows are still isomorphisms. It follows from the commutativity of the diagram that Φ g takes p∗(π1(X (cid:7) ,q(cid:7))) to p∗(π1(X (cid:7) ,q(cid:7)(cid:5))). Since an isomorphism takes normal subgroups to normal subgroups, the result follows. Next we show that there is a natural right action of the fundamental group of the base on the fiber of any covering space. Recall from Chapter 3 that given an action of a group Γ on a set F, the orbit of a point x∈F is the set of all images of x under elements of the group (for a right action, · thisistheset{x γ :γ ∈Γ}),andtheactionissaidtobetransitiveifeach orbit is all of F.

246 11. Covering Spaces g(cid:7)(1) (cid:7) f(1) g(cid:7) q(cid:7) (cid:7) f p q g f FIGURE 11.8. Proof that the fundamental group acts on the fiber. Theorem 11.21 (Action of the Fundamental Group on a Fiber). Suppose p: X (cid:7) → X is a covering map and q ∈ X. There is a transitive right action of π1(X,q) on the fiber p−1(q), given by q(cid:7) · [f] = f (cid:7) (1), where f (cid:7) is the lift of f starting at q(cid:7)∈p−1(q). Proof. If q(cid:7)is any point in the fiber over q, any path f starting at q has a lift to a path f (cid:7) starting at q(cid:7)by the path lifting property. The monodromy (cid:7) theorem guarantees that the endpoint f(1) depends only on the path class · of f; therefore, q(cid:7) [f] is well-defined. To see that this is a group action, we need to check two things: · (a) q(cid:7) [c ]=q(cid:7); q · · · · (b) (q(cid:7) [f]) [g]=q(cid:7) ([f] [g]). For(a),justobservet · hattheconstantloopc q(cid:7)istheuniqueliftofc q starting at q(cid:7), and therefore q(cid:7) [c q ]=c q(cid:7)(1)=q(cid:7). To prove the composition property (cid:7) (b),supposef andg aretwoloopsbasedatq.Letf betheliftoff starting at q(cid:7), so that q(cid:7) · [f] = f (cid:7) (1). Now, if g(cid:7) is the lift of g starting at f (cid:7) (1), then by definition, (q(cid:7) · [f]) · [g]=g(cid:7)(1) (Figure 11.8). On the other hand, f (cid:7)· g(cid:7)is

The Covering Group 247 · clearly the lift of f g starting at q(cid:7). This means that · · · · q(cid:7) ([f] [g])=q(cid:7) [f g] =(f (cid:7)· g(cid:7))(1) =g(cid:7)(1). Finally, to prove that the action is transitive, just note that any two points q(cid:7),q(cid:7)(cid:5) in the fiber over q are joined by a path f (cid:7) because X (cid:7) is path connected. Setting f =p◦f (cid:7) , it is immediate that f (cid:7) is the lift of f starting at q(cid:7), and therefore q(cid:7) · [f]=q(cid:7)(cid:5). Corollary 11.22. Let p: X (cid:7) → X be a covering map, and suppose X (cid:7) is simply connected. The number of sheets of the covering is equal to the car- dinality of the fundamental group of X. Proof. Choose a base point q ∈ X and a point q(cid:7)in the fiber over q, and consider the map π1(X,q)→p−1(q) given by f→q(cid:7) · [f]. It is surjective because the action of the fundamental group is transitive. To show that it is injective, suppose that q(cid:7) · [f] = q(cid:7) · [g]. This means that the lifts f (cid:7) and g(cid:7)starting at q(cid:7)end at the same point. Since X (cid:7) is simply connected, f (cid:7)∼g(cid:7), and therefore [f]=p∗[f (cid:7) ]=p∗[g(cid:7)]=[g]. Example 11.23. Since π: Sn → Pn is a two-sheeted covering and Sn is simply connected, Corollary 11.22 shows that π1(Pn) is a two-element group, which must therefore be isomorphic to Z/(cid:22)2(cid:23). Corollary 11.24. If X is a simply connected space, any covering map p: X (cid:7) →X is a homeomorphism. (cid:7) Proof. The injectivity theorem shows that X is also simply connected. Thus Corollary 11.22 shows that the cardinality of the fibers is 1, so p is a one-to-one covering map and therefore a homeomorphism. The Covering Group In this section we introduce the group of covering transformations of a covering space, and explore its relation to the fundamental groups of the base and the covering space.

248 11. Covering Spaces Suppose p: X (cid:7) →X is a covering map. A homeomorphism ϕ: X (cid:7) →X (cid:7) is called a covering transformation if p◦ϕ=p: (cid:7) ϕ (cid:2) (cid:7) X X p(cid:6) (cid:3)p (cid:6)(cid:7) (cid:3)(cid:9) X. Coveringtransformationsarealsovariouslyknownasdeck transformations or automorphisms of the covering. Let C (X (cid:7) ) denote the set of all covering transformations of X (cid:7) with re- p specttop.Itiseasytoverifythatthecompositionoftwocoveringtransfor- mations, the inverse of a covering transformation, and the identity map of X (cid:7) are all covering transformations; thus C (X (cid:7) ) is a group, called the cov- p (cid:7) ering group or the automorphism group of the covering. It acts on X (on the left) in a natural way, and the definition of covering transformations implies that each orbit is a subset of a single fiber. Example 11.25. For the covering ε: R → S1, the integral translations x (cid:10)→ x+k for k ∈ Z are easily seen to be covering transformations. More generally,foranyintegers(k1,…,k n ),thetranslation(x1,…,x n )(cid:10)→(x1+ k1,…,x n +k n )isacoveringtransformationofE: Rn →Tn.Wewillprove below that these are all of them. Example 11.26. If π: Sn → Pn is the covering map of Example 11.4, then the antipodal map A(x) = −x is a covering transformation. We will see shortly that C (Pn) is the two-element group {Id,A}. π Proposition 11.27 (Properties of the Covering Group). Let p: X (cid:7) →X be a covering map. (a) If two covering transformations agree at one point, they are identical. (b) The covering group acts freely and continuously on X (cid:7) : If ϕ(q(cid:7)) = q(cid:7) for some q(cid:7)∈X (cid:7) , then ϕ=Id(cid:7). X (c) For any q ∈X, each covering transformation permutes the points of the fiber p−1(q). (d) ForanyevenlycoveredopensetU ⊂X,eachcoveringtransformation permutes the components of p−1(U). Proof. Note that a covering transformation ϕ is, in particular, a lift of p: (cid:7) X (cid:3)(cid:4) (cid:3) ϕ p (cid:3) (cid:3) (cid:5) (cid:7) (cid:2) X p X.

The Covering Group 249 Thus (a) follows from the unique lifting property. The covering group acts continuously because each covering transformation is continuous by definition; the fact that it acts freely follows from (a) by comparing ϕ with the identity. Part (c) follows from the fact that if q(cid:7)∈ p−1(q), then p(ϕ(q(cid:7))) = p(q(cid:7)) = p, so ϕ takes the fiber over q to itself; since the same is trueofϕ−1,ϕactsasapermutationofthefiber.Toprove(d),letU bean evenlycoveredopenset,andletU beacomponentofp−1(U).Sinceϕ(U ) α α is a connected subset of p−1(U), it must be contained in a single compo- nent; applying the same argument to ϕ−1 shows that ϕ(U ) is exactly a α component. Example 11.28. Consider again the covering group of ε: R → S1. Let ϕ∈C (R)bearbitrary.Ifwesetn=ϕ(0),thenbothϕandthetranslation ε x (cid:10)→ x+n are covering transformations taking 0 to n, and are therefore equal by Proposition 11.27(a). Thus the covering group of ε: R → S1 is equal to Z acting on R by integral translations. By a similar argument, the covering group of E: Rn → Tn is Zn acting by translations, and the covering group of π: Sn → Pn is equal to the cyclic group of order 2 generated by the antipodal map. (cid:7) Because of Proposition 11.27, the action of the covering group on X is completely determined by the action on any fiber. However, unlike the action of the fundamental group on a fiber that we defined in Theorem 11.21,theactionofthecoveringgrouponfibersisnottransitiveingeneral. It is often useful to have a criterion for deciding when two points in a fiber are in the same orbit. Proposition 11.29 (Orbit Criterion). Let p: X (cid:7) → X be a covering map. (a) If q(cid:7), q(cid:7)(cid:5) ∈ X (cid:7) are two points in the same fiber p−1(q), there exists a covering transformation taking q(cid:7) to q(cid:7)(cid:5) if and only if the induced subgroups p∗π1(X (cid:7) ,q(cid:7)) and p∗π1(X (cid:7) ,q(cid:7)(cid:5)) are equal. (b) C (X (cid:7) ) acts transitively on each fiber if and only if p is a normal p covering. Proof. Ifthereexistsϕsuchthatϕ(q(cid:7))=q(cid:7)(cid:5),thenϕ∗: π1(X (cid:7) ,q(cid:7))→π1(X (cid:7) ,q(cid:7)(cid:5)) is an isomorphism, so p∗π1(X (cid:7) ,q(cid:7)) = p∗ϕ∗π1(X (cid:7) ,q(cid:7)) = p∗π1(X (cid:7) ,q(cid:7)(cid:5)). Con- versely, if the two subgroups are equal, then the lifting criterion yields a lift p(cid:7): X (cid:7) → X (cid:7) satisfying p◦p(cid:7)= p and p(cid:7)(q(cid:7)) = q(cid:7)(cid:5). Reversing the roles of q(cid:7) and q(cid:7)(cid:5), we get a lift p(cid:7)(cid:5) satisfying p(cid:7)(cid:5)(q(cid:7)(cid:5)) = q(cid:7). To show that p(cid:7) and p(cid:7)(cid:5) are inverses of each other, note that p(cid:7)(cid:5)◦p(cid:7)and Id(cid:7) are lifts of p taking q(cid:7) X to itself, and thus are equal by the unique lifting property, and similarly p(cid:7)◦p(cid:7)(cid:5) =Id(cid:7). Therefore, p(cid:7)is the required covering transformation. X Nowsupposepisnormal.Thismeansthatforanyq(cid:7),q(cid:7)(cid:5) inthesamefiber, p∗π1(X (cid:7) ,q(cid:7))=p∗π1(X (cid:7) ,q(cid:7)(cid:5)),sobypart(a)thereisacoveringtransformation

250 11. Covering Spaces taking q(cid:7)to q(cid:7)(cid:5). Conversely, if C (X (cid:7) ) acts transitively on the fiber p−1(q), p the groups p∗π1(X (cid:7) ,q(cid:7)) coincide for all q(cid:7)∈p−1(q), which is to say that p is normal. The next theorem is the central result concerning the relationship be- tweencoveringspacesandfundamentalgroups.Itgivesanexplicitformula for the covering group in terms of the fundamental groups of the covering space and the base, and can be used to compute the fundamental groups of certain spaces from properties of their coverings. For normal coverings, the theorem simply says that the covering group is isomorphic to the quo- tientofthefundamentalgroupofthebasebythesubgroupinducedbythe covering.Thestatementforgeneralgroupsissomewhatmorecomplicated, and involves the following algebraic notion: If G is a group and H ⊂ G is a subgroup, the normalizer of H in G, denoted by N(H), is the set of all elements γ ∈G such that γ−1Hγ =H. The normalizer N(H) is easily seen to be a subgroup of G; it is in fact the largest subgroup in which H is normal. Theorem 11.30 (Covering Group Structure Theorem). Suppose p: X (cid:7) → X is a covering map and q(cid:7) ∈ X (cid:7) . The covering group C (X (cid:7) ) is p isomorphic to the quotient N(p∗π1(X (cid:7) ,q(cid:7))) . p∗π1(X (cid:7) ,q(cid:7)) The isomorphism is induced by the map α: N(p∗π1(X (cid:7) ,q(cid:7))) → C p (X (cid:7) ) that · sends [f] to the unique covering transformation ϕ taking q(cid:7)to q(cid:7) [f]. Beforewegivetheproofofthisimportanttheorem,whichisrathertech- nical, let us derive some immediate consequences to illustrate its utility. Corollary 11.31 (Normal Case). If p: X (cid:7) → X is a normal covering, q(cid:7)∈X (cid:7) , and q =p(q(cid:7)), then C p (X (cid:7) ) ∼ =π1(X,q)/p∗π1(X (cid:7) ,q(cid:7)). Corollary 11.32 (Simply Connected Case). If p: X (cid:7) → X is a cov- ering map and X (cid:7) is simply connected, then for any q(cid:7)∈ X (cid:7) the map α of Theorem 11.30 is an isomorphism from π1(X,q) to C p (X (cid:7) ), where q =p(q(cid:7)). Example 11.33. Since the covering group of ε: R → S1 is infinite cyclic and R is simply connected, Corollary 11.32 yields another proof that the fundamental group of the circle is infinite cyclic. In fact, if you look back carefullyattheproofinChapter8,youwillseethatthisisreallythesame proof we gave there, decorated with some fancier terminology. Example 11.34. Because the covering group of π: Sn → Pn is the two- element group {Id,A}, Corollary 11.32 gives another proof that π1(Pn) ∼ = Z/(cid:22)2(cid:23).

The Covering Group 251 Proof of the covering group structure theorem. Write H = p∗π1(X (cid:7) ,q(cid:7)) ⊂ π1(X,q). We will show that α: N(H) → C p (X (cid:7) ) is a surjective homomor- phismwhosekernelisH;bythefirstisomorphismtheorem,thisprovesthe result. Let [g] ∈ N(H) be arbitrary, and let q(cid:7)(cid:5) = q(cid:7) · [g] as defined in Theorem 11.21. Recall that q(cid:7)(cid:5) is the terminal point of the lift g(cid:7) of g starting at q(cid:7). The key claim is that there exists a covering transformation ϕ ∈ C (X (cid:7) ) p such that ϕ(q(cid:7))=q(cid:7)(cid:5). By the orbit criterion, to prove the claim it suffices to show that p∗π1(X (cid:7) ,q(cid:7)(cid:5)) = p∗π1(X (cid:7) ,q(cid:7)). Let Φ g(cid:7): π1(X (cid:7) ,q(cid:7)) → π1(X (cid:7) ,q(cid:7)(cid:5)) be the isomor- phism determined by the path g(cid:7) as in Theorem 7.11. From the commu- tative diagram (11.2) in the proof of the conjugacy theorem, we conclude that p∗π1(X (cid:7) ,q(cid:7)(cid:5))=p∗Φ g(cid:7)π1(X (cid:7) ,q(cid:7)) =Φ g p∗π1(X (cid:7) ,q(cid:7)) =[g]−1· H · [g]=H =p∗π1(X (cid:7) ,q(cid:7)). Thus there exists a covering transformation ϕ such that ϕ(q(cid:7)) = q(cid:7)(cid:5); it is necessarily unique by Proposition 11.27(a). Define α[g]=ϕ. To show that α is a homomorphism, let [g1],[g2] ∈ N(H), and write α[g ]=ϕ , so that ϕ is a covering transformation satisfying ϕ (q(cid:7))=g(cid:7)(1). i i i i i · · Let ϕ12 = α[g1 g2], so ϕ12(q(cid:7)) = g(cid:5) 1 g2(1). We need to show that ϕ12 = ϕ1 ◦ϕ2. It suffices to show that these two covering transformations agree at one point, so let us show that ϕ12(q(cid:7)) = ϕ1 ◦ ϕ2(q(cid:7)), or equivalently · g(cid:5) 1 g2(1)=ϕ1(g(cid:7) 2(1)). Now, the lift g(cid:7) 2 of g2 is a path in X (cid:7) starting at q(cid:7). Because p◦ϕ1 = p, the image ϕ1 ◦g(cid:7) 2 of g(cid:7) 2 under ϕ1 is also a lift of g2, but this one starts at · ϕ1(q(cid:7)) = g(cid:7) 1(1) (Figure 11.9). Thus the path product g(cid:7) 1 (ϕ1 ◦g(cid:7) 2) makes · sense, and is the lift of g1 g2 starting at q(cid:7). In summary, · ϕ12(q(cid:7))=g(cid:5) 1 g2(1) · =g(cid:7) 1 (ϕ1 ◦g(cid:7) 2)(1) =ϕ1 ◦g(cid:7) 2(1) =ϕ1(ϕ2(q(cid:7))), which was to be proved. To show that α is surjective, let ϕ ∈ C (X (cid:7) ) be arbitrary, let q(cid:7)(cid:5) = ϕ(q(cid:7)), p andletg(cid:7)beapathinX (cid:7) fromq(cid:7)toq(cid:7)(cid:5).Theng =p◦g(cid:7)isaloopinX.Moreover,

252 11. Covering Spaces ϕ1 ◦g(cid:7) 2 ϕ1 (cid:7) X g(cid:7) 1 q(cid:7) g(cid:7) 2 g1 q g2 X FIGURE 11.9. Proof that α is a homomorphism. the orbit criterion shows that p∗π1(X (cid:7) ,q(cid:7)) = p∗π1(X (cid:7) ,q(cid:7)(cid:5)) because q(cid:7) and q(cid:7)(cid:5) are in the same orbit. On the other hand, from (11.2), p∗π1(X (cid:7) ,q(cid:7)(cid:5)) = Φ g p∗π1(X (cid:7) ,q(cid:7)). Thus Φ g p∗π1(X (cid:7) ,q(cid:7))=p∗π1(X (cid:7) ,q(cid:7)), which is to say that [g]∈ N(H), and the construction above gives α[g]=ϕ. Finally,weneedtoshowthatKerα=H.Let[g]∈N(H),letg(cid:7)bethelift ofgstartingatq(cid:7),andwriteϕ=α[g].Thenϕistheidentitytransformation if and only if ϕ(q(cid:7)) = g(cid:7)(1) = q(cid:7), which means that g(cid:7) is a loop in X (cid:7) ; so ϕ is the identity if and only if [g]=[p◦g(cid:7)]=p∗[g(cid:7)] for some [g(cid:7)]∈π1(X (cid:7) ,q(cid:7)), i.e., [g]∈H.

Problems 253 Problems 11-1. Prove that for any n ≥ 1 the map π: Sn → Pn defined in Example 11.4 is a covering map. 11-2. Show that a finite product of covering maps is a covering map: If p :X (cid:7) →X are covering maps for i=1,…,n, then so is the map i i i p1 ×···×p n :X (cid:7) 1 ×···×X (cid:7) n →X1 ×···×X n . 11-3. Suppose p: X (cid:7) →X is a covering map. (cid:7) (a) If X is an n-manifold and X is Hausdorff, show that X is an n-manifold. (cid:7) (b) If X is an n-manifold, show that X is an n-manifold. 11-4. Suppose p: X (cid:7) →X is a covering map and X is a compact manifold. (cid:7) Show that X is compact if and only if p is a finite-sheeted covering. 11-5. Let S be the following subset of C2: S ={(z,w):w2 =z, w (cid:14)=0}. (It is the graph of the two-valued complex square root “function” described in Chapter 1, with the origin removed.) Show that the projection π1: C2 → C onto the first coordinate restricts to a two- sheeted covering map p: S →C(cid:3){0}. 11-6. Show that there is a two-sheeted covering of the Klein bottle by the torus. (cid:8) 11-7. LetM,M,andN beconnectedmanifoldsofdimensionnandsuppose p: M (cid:8)→M is a k-sheeted covering map. Show that there exists a k- (cid:8) sheetedcoveringofM#N bytheconnectedsumofM withk copies ofN.[Hint:ChoosetheballtobecutoutofM tolieinsideanevenly covered neighborhood.] 11-8. Showthateverynonorientablecompactsurfaceofgenusnhasatwo- sheetedcoveringbyanorientableoneofgenusn−1.[Hint:UseProb- lem 11-7 and induction.] 11-9. Show that a proper local homeomorphism between connected, path connected, and locally compact Hausdorff spaces is a covering map. 11-10. A continuous map f: S1 →S1 is said to be odd if f(−z)=−f(z) for all z ∈ S1, and even if f(z) = f(−z) for all z ∈ S1. Show that every odd map has odd degree, as follows.

254 11. Covering Spaces (a) Let p2: S1 → S1 be the two-sheeted covering map of Exam- ple 11.2. If f is odd, show that there exists a continuous map g: S1 → S1 such that degf = degg and the following diagram commutes: f(cid:2) S1 S1 p2 p2 (cid:5) (cid:5) (cid:2) S1 g S1. (b) Ifdegf iseven,showthat g liftstoamap g(cid:7): S1 →S1 suchthat p2 ◦g(cid:7)=g. (c) Show that g(cid:7)◦p2 and f are both lifts of g ◦p2 that agree at either (1,0) or (−1,0), so they are equal everywhere; derive a contradiction. 11-11. Show that every even map f: S1 →S1 has even degree. 11-12. Provethehamsandwichtheorem:Iftwopiecesofbreadandonepiece of ham are placed arbitrarily in space, all three pieces can be cut in half with a single slice of the knife. (If you do not like ham, you may wishtosubstitutetofu.)Moreprecisely,giventhreedisjoint,bounded, connected open subsets U1,U2,U3 ⊂ R3, there exists a plane that simultaneously bisects all three, in the sense that the plane divides R3 into two half spaces H+ and H− such that for each i, U ∩H+ i has the same volume as U ∩H−. [Hint: For any ω ∈ S2, show that i thereareuniquerealnumbers(λ1,λ2,λ3)suchthattheplanethrough λ ω and orthogonal to ω bisects U . If there does not exist a plane i i bisecting all three sets, define a map F: S2 →S1 by F(ω)= (cid:3) (λ1 −λ2)+i(λ2 −λ3) . (λ1 −λ2)2+(λ2 −λ3)2 ShowthatF iscontinuous,andF◦ιS1 contradictstheresultofProb- lem11-10,whereιS1: S1 (cid:9)→S2 istheinclusionmap.Youmayassume thatthereisavolumefunctionVolassigninganonnegativerealnum- ber to each open set in R3 and satisfying the following properties: The volume of a set is unchanged by translations or rotations; the volumes of balls, cylinders, and rectangular solids are given by the usual formulas; and if U ⊂V then Vol(U)≤Vol(V).] 11-13. Let p: X3 →X2 be the covering map of Exercise 11.1. (a) Determine the covering group C p (X3). (b) Determine whether p is a normal covering.

Problems 255 FIGURE 11.10. The covering map of Problem 11-14. (c) For each of the following maps f: S1 →X2, determine whether f has a lift to X3 taking 1 to 1. i. f(z)=z. ii. f(z)=z2. iii. f(z)=2−z. iv. f(z)=2−z2. 11-14. LetX4betheunionoffourcirclesdescribedinProblem10-10,andlet p: X4 → X2 be the covering map indicated schematically in Figure 11.10. Answer the questions of Problem 11-13 for this covering. 11-15. Let E be the figure eight space of Example 7.21, and let X be the unionofthex-axiswithinfinitelymanyunitcirclescenteredat{2πk+ i:k ∈Z}.Letp: X →EbethemapthatsendseachcircleinX onto the upper circle in E by translating in the x-direction and sends the x-axis onto the lower circle by x(cid:10)→ieix−i. You may accept without proof that p is a covering map. (a) Identifythesubgroupp∗π1(X,0)ofπ1(E,0)intermsofthegen- erators for π1(E,0). (b) Compute the covering group C (X). p (c) Determine whether p is a normal covering. 11-16. This problem shows that the hypothesis that Y is locally path con- nected is necessary for the lifting criterion (Theorem 11.15) to hold. Let X be the topologist’s sine curve (Example 4.10), and let Y be the union of X with a path in the plane from (1,sin1) to (0,1) that intersects X only at those two points. (a) Show that Y is simply connected.

256 11. Covering Spaces (b) Show that there is a map f: Y →S1 that has no lift to R. 11-17. Suppose X is a compact polyhedron and p: X (cid:7) → X is a covering map. (cid:7) (a) ShowthatX andX admittriangulationssuchthatpisinduced by a simplicial map. [Hint: Use barycentric subdivision.] (b) SupposeK,K(cid:7) arefinitecomplexessuchthat|K|=X,|K(cid:7)|=X (cid:7) , and p is induced by a simplicial map from K(cid:7) to K. If p is an n-sheeted covering, show that χ(K(cid:7) )=nχ(K).

12 Classification of Coverings The main thrust of the preceding chapter was to learn about fundamental groups by studying covering maps. In this chapter we reverse the process andexplorewhatthereistobelearnedfromthefundamentalgroupabout the existence and uniqueness of covering spaces. The key idea is provided bytheconjugacytheoremoftheprecedingchapter:Eachcoveringspaceof X determines a conjugacy class of subgroups in the fundamental group of X. Webeginwiththeuniquenessquestion.Inthefirstsectionofthechapter we define isomorphisms of covering spaces, and show that two covering spaces are isomorphic if and only if they induce the same conjugacy class of subgroups. Thenweaddresstheexistencequestion.Theultimategoalistoshowthat forasufficientlynicespaceX (anyconnectedmanifold,forexample),every conjugacyclassofsubgroupsofπ1(X,q)correspondstosomecovering.This isaccomplishedinseveralstages.FirstweshowthatX hasauniquesimply connected covering space, called its “universal covering space.” Then we show how to construct coverings as quotients of a given space by certain group actions. The d´enouement is the last theorem of the chapter, which putstogetheralltheprecedingresultstogiveacompleteclassificationofall coverings of X up to isomorphism: They are in one-to-one correspondence with conjugacy classes of subgroups of the fundamental group of X. We illustratethetheorybydeterminingtheuniversalcoveringspacesofallthe compact surfaces and classifying all the coverings of the torus.

258 12. Classification of Coverings (cid:7) (cid:7) X1 f (cid:7) (1) X2 ϕ q(cid:7) (cid:7) f g(cid:7) q(cid:7) 1 q(cid:7) 2 p1 p2 q X f FIGURE 12.1. Proof that a covering homomorphism is surjective. Covering Homomorphisms In this section we examine the question of how to tell when two covering spacesare“thesame.”Asusual,weconsidertwocoveringsthesameifthey are related by a suitable isomorphism. We begin by defining some terms. Let X be a space, and let p1: X (cid:7) 1 →X, p2: X (cid:7) 2 →X be two coverings of X. A covering homomorphism from p1 to p2 is a continuous map ϕ: X (cid:7) 1 → X (cid:7) 2 such that p2 ◦ϕ=p1: (cid:7) ϕ (cid:2) (cid:7) X1 X2 p1 (cid:6) (cid:6)(cid:7) (cid:3)(cid:9) (cid:3)p2 X. A covering homomorphism that is also a homeomorphism is said to be an isomorphismofcoverings.Itiseasytoseethatinthiscasetheinversemap is also a covering homomorphism. We say two coverings are isomorphic if there is an isomorphism between them; this is an equivalence relation on the set of coverings of X. Note that an isomorphism from a covering to itself is just a covering transformation. Aninterestingfeatureofcoveringhomomorphismsisthattheyarethem- selves covering maps, as the following lemma shows. Lemma 12.1. Let p1: X (cid:7) 1 → X and p2: X (cid:7) 2 → X be coverings of X, and let ϕ be a covering homomorphism from p1 to p2. Then ϕ is a covering map. Proof. First we show that ϕ is surjective. Let q(cid:7)∈X (cid:7) 2 be arbitrary. Choose some q(cid:7) 1 ∈ X (cid:7) 1, and let q(cid:7) 2 = ϕ(q(cid:7) 1) ∈ X (cid:7) 2, q = p1(q(cid:7) 1) = p2(q(cid:7) 2) ∈ X (Figure 12.1). There is a path g(cid:7) in X (cid:7) 2 from q(cid:7) 2 to q(cid:7). Let f = p2 ◦g(cid:7), which is a

Covering Homomorphisms 259 −1 p (V) 1 ϕ−1(U) ϕ U α −1 p (V) 2 p1 q(cid:7) U p2 q V FIGURE 12.2. An evenly covered neighborhood of q(cid:7). (cid:7) path in X starting at q, and let f be the unique lift of f to a path in X (cid:7) 1 starting at q(cid:7) 1. Consider now the path ϕ◦f (cid:7) in X (cid:7) 2. Its initial point is ϕ◦f (cid:7) (0) = ϕ(q(cid:7) 1) = q(cid:7) 2, and it satisfies p2 ◦ϕ◦f (cid:7) = p1 ◦f (cid:7) = f, so ϕ◦f (cid:7) is theliftoff toX (cid:7) 2 startingatq(cid:7) 2.Bytheuniqueliftingproperty,thismeans that ϕ◦f (cid:7) =g(cid:7), so ϕ(f (cid:7) (1))=g(cid:7)(1)=q(cid:7), which shows that ϕ is surjective. Toshowthatϕisacoveringmap,letq(cid:7)∈X (cid:7) 2 bearbitrary;letq =p2(q(cid:7))∈ X; let U1, U2 ⊂ X be neighborhoods of q that are evenly covered by p1 and p2, respectively; and let V be the component of U1 ∩U2 containing q. Thus V is a neighborhood of q that is evenly covered by both p1 and p2. Let U be the component of p −1 (V) containing q(cid:7). We need to show that 2 the components of ϕ−1(U) are mapped homeomorphically onto U by ϕ. −1 Consider the restrictions of p1 and ϕ to the “stack of pancakes” p 1 (V) −1 (Figure 12.2). Since U is both open and closed in p (V), it follows that 2 ϕ−1(U) is both open and closed in p −1 (V), and is thus a union of compo- 1 nents. On any such component U , the following diagram commutes: α U α (cid:6) ϕ (cid:6)(cid:7) p1 U (cid:5)(cid:3)(cid:9) (cid:3)p2 V.

260 12. Classification of Coverings Since p1 and p2 are homeomorphisms in this diagram, so is ϕ. Thekeytodeterminingwhentwocoveringspacesareisomorphicistode- cidewhentherearecoveringhomomorphismsbetweenthem.Thisquestion is answered by the following theorem. Theorem 12.2 (Covering Homomorphism Criterion). Suppose p1: X (cid:7) 1 → X and p2: X (cid:7) 2 → X are two coverings of X, and q(cid:7) 1 ∈ X (cid:7) 1, q(cid:7) 2 ∈ X (cid:7) 2 are base points such that p1(q(cid:7) 1) = p2(q(cid:7) 2) = q ∈ X. There exists a covering homomorphism from p1 to p2 taking q(cid:7) 1 to q(cid:7) 2 if and only if p1∗ π1(X (cid:7) 1,q(cid:7) 1)⊂p2∗ π1(X (cid:7) 2,q(cid:7) 2). Proof. A covering homomorphism from p1 to p2 can also be viewed as a lift of p1: (cid:7) X2 (cid:3)(cid:4) ϕ (cid:3) p2 (cid:3) (cid:5) (cid:7) (cid:2) (12.1) X1 p1 X. Thusboththenecessityandthesufficiencyofthesubgroupconditionfollow from the lifting criterion (Theorem 11.15). Example 12.3. Let p : S1 →S1 be the nth power map defined in Exam- n ple 11.2. The subgroup of π1(S1,1) induced by p n is the cyclic subgroup generated by [α]n (Example 11.18). By the covering homomorphism crite- rion, there is a homomorphism from p to p if and only if m is divisible m n by n; the homomorphism in that case is just p . m/n Example 12.4. ConsiderthefollowingtwocoveringsofT2:E: R2 →T2is thecoveringofExample11.3(theproductoftwocopiesofε: R→S1);and p: S1×R→T2 is given by p(z,y)=(z,ε(y)). Writing π1(T2) ∼ =(cid:22)β(cid:23)×(cid:22)γ(cid:23), we see that E∗π1(R2) is trivial, while p∗π1(S1×R)=(cid:22)β(cid:23)×{1}. Therefore, there exists a covering homomorphism from E to p. (Why do the base points not matter?) It is easy to check that ϕ(x,y) = (ε(x),y) is such a homomorphism. Thefollowingtheoremcompletelysolvestheuniquenessquestionforcov- ering spaces up to isomorphism. Theorem 12.5 (Covering Isomorphism Theorem). Two coverings p1: X (cid:7) 1 → X and p2: X (cid:7) 2 → X are isomorphic if and only if for some q ∈X and base points q(cid:7) 1 ∈p − 1 1 (q) and q(cid:7) 2 ∈p − 2 1 (q), the induced subgroups p1∗ π1(X (cid:7) 1,q(cid:7) 1) and p2∗ π1(X (cid:7) 2,q(cid:7) 2) are conjugate in π1(X,q). If this is the case, these subgroups are conjugate for every such q, q(cid:7) 1, and q(cid:7) 2.

The Universal Covering Space 261 Proof. If there exists an isomorphism ϕ: X (cid:7) 1 → X (cid:7) 2, choose q(cid:7) 1 ∈ X (cid:7) 1 ar- bitrarily and set q(cid:7) 2 = ϕ(q(cid:7) 1). The covering homomorphism criterion ap- plied to ϕ and ϕ−1 guarantees that the two subgroups p1∗ π1(X (cid:7) 1,q(cid:7) 1) and p2∗ π1(X (cid:7) 2,q(cid:7) 2) are contained in each other, so they are equal. Thus by the conjugacy theorem (Theorem 11.19), the subgroups associated with any other choices of base points in the same fibers are conjugate. Conversely,supposethetwosubgroupsareconjugateforsomechoiceofq, q(cid:7) 1, and q(cid:7) 2. By the conjugacy theorem, we can change to a new base point q(cid:7) 2 (cid:5) ∈ X (cid:7) 2 such that p2∗ π1(X (cid:7) 2,q(cid:7) 2 (cid:5)) = p1∗ π1(X (cid:7) 1,q(cid:7) 1). Then by the covering homomorphism criterion there exist homomorphisms ϕ from p1 to p2 and ψ from p2 to p1, with ϕ(q(cid:7) 1) = q(cid:7) 2 (cid:5) and ψ(q(cid:7) 2 (cid:5)) = q(cid:7) 1. The composite map ψ◦ϕ is a covering transformation of p1 that fixes q(cid:7) 1, so it is the identity. Similarly, ϕ◦ψ is the identity, so ϕ is the required isomorphism. The Universal Covering Space When the results of the preceding section are applied to simply connected covering spaces, they yield some extremely useful results. Proposition 12.6 (Properties of Simply Connected Coverings). (a) Let p: X (cid:7) → X be a covering map with X (cid:7) simply connected. If p1: X (cid:7) 1 →X is any covering, there exists a covering map p(cid:7): X (cid:7) →X (cid:7) 1 such that the following diagram commutes: (cid:7) X (cid:6) p(cid:7) (cid:6)(cid:7) p X (cid:7) 1 (cid:5)(cid:3)(cid:9) (cid:3)p1 X. (b) Anytwosimplyconnectedcoveringsofthesamespaceareisomorphic. Proof. Since the trivial subgroup is contained in every other subgroup, part (a) follows from the covering homomorphism criterion and the fact that every covering homomorphism is a covering map. Part (b) follows immediately from the covering isomorphism theorem. Part (a) of this proposition says that a simply connected covering space covers every other covering space of X. Because of this, any covering of X (cid:7) byasimplyconnectedspaceX (whichisuniqueby(b))iscalledauniversal (cid:7) covering, and X is called the universal covering space of X. Example 12.7. Theuniversalcoveringspaceofthen-torusisRn,because weconstructedacoveringmapE: Rn →TninExample11.3.Theuniversal covering space of Pn is Sn, by the covering map π of Example 11.4.

262 12. Classification of Coverings As the next theorem shows, every “reasonable” space, including every manifold, has a universal covering space. We say that a space X is lo- cally simply connected if it admits a basis of simply connected open sets. Clearly,alocallysimplyconnectedspaceislocallypathconnected,because simply connected sets are path connected. Any manifold is locally simply connected, because it has a basis of Euclidean balls. Theorem 12.8 (Existence of the Universal Covering Space). Ev- ery connected and locally simply connected topological space (in particular, every connected manifold) has a universal covering space. Proof. To get an idea how to proceed, suppose for a moment that X does have a universal covering p: X (cid:7) → X. The key fact is that once we choose base points q(cid:7) 0 ∈ X (cid:7) and q0 = p(q(cid:7) 0) ∈ X, the fiber p−1(q) over any q ∈ X is in one-to-one correspondence with path classes from q0 to q. To see why, define a map E from the set of such path classes to p−1(q) by sending [f] to the terminal point of the lift of f starting at q(cid:7) 0. Since lifts of homotopic pathshavethesameterminalpointbythemonodromytheorem,E iswell- defined. E is surjective, because given any q(cid:7)in the fiber over q, there is a path f (cid:7) from q(cid:7) 0 to q(cid:7), and then p◦f (cid:7) is a path from q0 to q whose lift ends at q(cid:7). Injectivity of E follows from the fact that X (cid:7) is simply connected: If (cid:7) (cid:7) f1, f2 are two paths from q0 to q whose lifts f1,f2 end at the same point, then f (cid:7) 1 and f (cid:7) 2 are path homotopic, and therefore so are f1 = p◦f (cid:7) 1 and f2 =p◦f (cid:7) 2. Now let X be any space satisfying the hypotheses of the theorem, and choose any base point q0 ∈X. Guided by the observation in the preceding (cid:7) paragraph, we define X to be the set of path classes of paths in X starting at q0, and define p: X (cid:7) → X by p[f] = f(1), which is well-defined because (cid:7) path homotopic paths have the same terminal point. We will prove thatX has the required properties in a series of steps. Step1:TopologizeX (cid:7) .WedefineatopologyonX (cid:7) byconstructingabasis. Forany[f]∈X (cid:7) andanysimplyconnectedopensetU ⊂X containingf(1), define the set [f · U]⊂X (cid:7) by · · [f U]={[f a]:a is a path in U starting at f(1)}. · Let B denote the collection of all such sets [f U]; we will show that B is a basis.First,sinceX islocallysimplyconnected,forany[f]∈X (cid:7) thereexists · a simply connected open set U containing f(1), and clearly [f] ∈ [f U]. Thus the union of all the sets in B is X (cid:7) . Tochecktheintersectioncondition,suppose[h]∈X (cid:7) isintheintersection · · · · oftwobasissets[f U],[g V]∈B.Thismeansthath∼f a∼g b,where a is a path in U and b is a path in V (Figure 12.3). Let W be a simply connected neighborhood of h(1) contained in U ∩V (such a neighborhood · existsbecauseX hasabasisofsimplyconnectedopensets).If[h c]isany · · · · · · elementof[h W],then[h c]=[f a c]∈[f U]becausea cisapathin

The Universal Covering Space 263 V W b X g c h a q0 U f · FIGURE 12.3. Proof that the collection of sets [f U] is a basis. · · · · · U.Similarly,[h c]=[g b c]∈[g V].Thus[h W]isabasissetcontained · · in [f U]∩[g V], which proves that B is a basis. From now on, we endow X (cid:7) with the topology generated by B. Step 2: X (cid:7) is path connected. Let [f] ∈ X (cid:7) be arbitrary. We will show that there is a path in X (cid:7) from q(cid:7) 0 to [f], where q(cid:7) 0 =[c q0 ]. For any 0≤t≤1, define f : I →X by t f (s)=f(ts), t so f t is a path in X from q0 to f(t). Then define f (cid:7) : I →X (cid:7) by (cid:7) f(t)=[f ]. t Clearly, f (cid:7) (0)=[f0]=q(cid:7) 0, and f (cid:7) (1)=[f1]=[f]. So we need only show that (cid:7) (cid:7) f is continuous; for this it suffices to show that the inverse image under f of any basis open set [h · U] ⊂ X (cid:7) is open. Let t0 ∈ I be a point such that f (cid:7) (t0) ∈ [h · U] (Figure 12.4). This means that f t0 ∼ h · c for some path c lying in U, and in particular that f(t0) = f t0 (1) ∈ U. For any 0 ≤ t ≤ 1, define a path f by t0t f t0t (s)=f(t0+s(t−t0)). · This path just follows f from f(t0) to f(t), so f t0 f t0t is easily seen to be path homotopic to f . t

264 12. Classification of Coverings U h X c f(t) f f(t0) q0 (cid:7) FIGURE 12.4. Proof that X is path connected. By continuity of f, there is some δ > 0 such that f(t0 −δ,t0+δ) ⊂ U. If t∈(t0 −δ,t0+δ), then · · · f ∼f f ∼h c f , t t0 t0t t0t from which it follows that f (cid:7) (t)=[f ]=[h · c · f ]∈[h · U]. t t0t Thisshowsthatf (cid:7)−1[h · U]containstheset(t0 −δ,t0+δ),sof (cid:7) iscontinuous. Step 3: p is a covering map. Let U ⊂X be any simply connected open set. We will show that U is evenly covered. Choose any point q1 ∈ U. We begin by showing that p−1(U) is the · disjointunionofthesets[f U]as[f]variesoverallthedistinctpathclasses · f(cid:2)rom q0 to q1. It is obvious from the definition of p that p[f U] ⊂ U, so [f · U]⊂p−1(U). Conversely, if [g]∈p−1(U), then g(1)=p[g]∈U, so [f] there is a path b in U from(cid:2)g(1) to q1, and [g]=[g · b · b−1]∈[(g · b) · U]. This proves that p−1(U)= [f · U]. [f] This shows, in particular, that p is continuous: X has a basis of simply connected open sets, and the inverse image under p of any such set is a union of basis sets and therefore open. And p is clearly surjective, because each q ∈X is equal to p[g] for any path g from q0 to q. · Next we show that p is a homeomorphism from each set [f U] to U. It is surjective because for each q ∈U there is a path a from f(1) to q in U, soq =p[f · a]∈p[f · U].Toseethatitisinjective,let[g],[g(cid:5)]∈[f · U],and supposep[g]=p[g(cid:5)],orinotherwords,g(1)=g(cid:5)(1)(Figure12.5).Thenby definition of [f · U], g ∼f · a and g(cid:5) ∼f · a(cid:5) for some paths a,a(cid:5) in U from f(1) to g(1). Since U is simply connected, a ∼ a(cid:5) and therefore [g] = [g(cid:5)]. Finally,pisanopenmapbecauseittakesbasisopensetstoopensets,and · therefore p: [f U]→U is a homeomorphism.

The Universal Covering Space 265 U g a X f a(cid:5) g(cid:5) q0 · FIGURE 12.5. Proof that p is injective on [f U]. · Eachset[f U]isopenbydefinition,andeachispathconnectedbecause (cid:7) itishomeomorphictothepathconnectedsetU.ItfollowsthatX islocally path connected. To complete the proof that p is a covering map, we need to show that for any two paths f and f(cid:5) from q0 to q1, the sets [f · U] and [f(cid:5) · U] are either equal or disjoint. If they are not disjoint, there exists [g] ∈ [f · U]∩[f(cid:5) · U], so g ∼ f · a ∼ f(cid:5) · a(cid:5) for paths a, a(cid:5) in U from q1 to g(1). Since U is simply connected, a ∼ a(cid:5), which implies f ∼ f(cid:5) and therefore [f · U]=[f(cid:5)· U]. Step 4: X (cid:7) is simply connected. Suppose F: I →X (cid:7) is any loop based at q(cid:7) 0. Let f = p◦F, so F is a lift of f. If we write f (cid:7) (t) = [f t ] as in Step 3, then p◦f (cid:7) (t) = p[f t ] = f t (1) = f(t), so f (cid:7) is also a lift of f starting at q(cid:7) 0. (cid:7) By the unique lifting property, F =f. Since F is a loop, [c q0 ]=q(cid:7) 0 =F(1)=f (cid:7) (1)=[f1]=[f], so f is null homotopic. By the homotopy lifting property, this means that F is null homotopic as well. A careful study of this proof shows that it does not really need the full strengthofthehypothesisthatX islocallysimplyconnected.Eachtimewe usethefactthataloopinasmallopensetU ⊂X isnullhomotopic,allwe really need to know is that it is null homotopic in X. For this reason, it is traditionaltomakethefollowingdefinition:AspaceX issemilocallysimply connected if it admits a basis of open sets U with the property that every loopinU isnullhomotopicinX.Itcanbeshownthataconnected,locally path connected space admits a universal covering space if and only if it is semilocallysimplyconnected(see[Mas89]or[Sie92]).Sinceourmotivation

266 12. Classification of Coverings for studying the fundamental group is to understand manifolds, we have no need for this extra generality. Once you have understood the proof of the existence of the universal coveringspaceofaspaceX,youshouldforgetthecomplicatedconstruction (cid:7) (cid:7) of X in terms of path classes, and just think of X as a simply connected space with a covering map to X. The uniqueness theorem tells us that all (cid:7) the relevant properties of X can be derived from these facts. Proper Group Actions ThenextstepinclassifyingcoveringsistostartwithaspaceY anddevelop atechniqueforconstructingspacescoveredbyY.Inthenextsectionwewill apply this to the universal covering space in order to derive a classification theorem for coverings of a given space X. To get an idea how to construct spaces covered by Y, let us suppose p: X (cid:7) → X is a normal covering. (The restriction to normal coverings will notbealimitationintheend:Forreasonsthatwillsoonbecomeapparent, the construction in this section will produce only normal coverings, but in the next section we will be able to use them to produce all coverings of a given space.) As we observed in the previous chapter, the covering group C (X (cid:7) ) acts p (cid:7) continuouslyandfreelyonX (ontheleft).Theorbitcriterion(Proposition 11.29) says that C (X (cid:7) ) acts transitively on each fiber when p is normal, so p the identifications made by p are exactly those determined by the equiva- lence relation x ∼ y if and only if y = ϕ(x) for some ϕ ∈ C (X (cid:7) ). Since p p is a quotient map by Lemma 11.5, X is homeomorphic to the orbit space determined by the left action of C (X (cid:7) ) on X (cid:7) (see Chapter 3). p NowletY beanyspace,andsupposewearegivenaleftactionbyagroup Γ on Y. Our aim in this section is to describe conditions under which the quotient map π: Y → Y/Γ onto the orbit space is a covering map whose covering group is Γ. Note that this construction can produce only normal coverings, because Γ acts transitively on the fibers of any orbit space by definition. Not every group action yields a covering map, of course. Clearly, the actionmustbecontinuousandfree(Proposition11.27(b)).Moreover,every pointofacoveringspaceY hasaneighborhood(oneofthe“pancakes”over an evenly covered open set) whose images under the covering group are all disjoint, which places a strong restriction on the actions we can consider. A simple condition that will guarantee that a group action has the req- uisite properties in all cases of interest to us is the following. A continuous actionofatopologicalgroupΓonaspaceY issaidtobeproperifthemap · Γ×Y → Y ×Y given by (g,y) (cid:10)→ (y,g y) is a proper map, i.e., if the inverse image of any compact set under this map is compact.

Proper Group Actions 267 Properactionshavethefollowingusefulalternativecharacterizations,at least for discrete group actions on locally compact Hausdorff spaces, the onlytypewewillbeconcernedwith.Foranyg ∈ΓandanysubsetK ⊂Y, · · we let g K ={g y :y ∈K}. Proposition 12.9. For a discrete group Γ acting on a locally compact Hausdorff space Y, the following are equivalent: (a) Γ acts properly. · (b) For any compact set K ⊂Y, K∩(g K)=∅ for all but finitely many g ∈Γ. (c) For every y,y(cid:5) ∈ Y, there exist neighborhoods U of y and U(cid:5) of y(cid:5) such that U ∩(g · U(cid:5))=∅ for all but finitely many g ∈Γ. Proof. We will show (a) =⇒ (b) =⇒ (c) =⇒ (a). Assume first that the action of Γ is proper, and let Φ: Γ×Y → Y ×Y denote the proper map · Φ(g,y)=(y,g y). Given any compact set K ⊂Y, the set Φ−1(K×K)={(g,y)∈Γ×Y :y ∈K, g · y ∈K} iscompact.ThusitsprojectionontoΓiscompactandthereforefinite.But · this projection includes all g ∈Γ such that K∩(g K)(cid:14)=∅, so this proves (a) =⇒ (b). Now suppose (b) holds. Because Y is locally compact Hausdorff, any pointsy,y(cid:5) ∈Y haveprecompactneighborhoodsU andU(cid:5),respectively.Let · K bethecompactsetU∪U(cid:5).Thenthesetofg ∈ΓsuchthatK∩(g K)=∅ is finite, which implies (c). Finally, if (c) holds, let L be an arbitrary compact subset of Y ×Y. For any(y,y(cid:5))∈L,chooseneighborhoodsU ofyandU(cid:5)ofy(cid:5)asin(c),soU×U(cid:5) isaneighborhoodof(y,y(cid:5)).Thesetofsuchproductneighborhoodsas(y,y(cid:5)) ranges over L is an open cover of L, so finitely many such neighborhoods U (g 1 − × 1· U U 1 (cid:5) i (cid:5) , ) . (cid:14)= .., ∅ U } m is × fi U ni m (cid:5) te. c L ov e e t r S L = . F S o 1 r ∪ e · a · c · h ∪ i S , m th a e n s d et K S = i = π1( { L g ) ∈ ⊂ Γ Y. : T U h i e ∩ n it is straightforward to check that Φ−1(L) is contained in the compact set S×K.SinceY isHausdorff,LisclosedinY×Y;andsinceΦiscontinuous, Φ−1(L) is a closed subset of a compact set and thus compact. Corollary 12.10. IfadiscretegroupΓactsfreelyandproperlyonalocally compact Hausdorff space Y, every point y ∈Y has a neighborhood U such · that U ∩(g U)=∅ unless g =1. Proof. Taking y = y(cid:5) in Proposition 12.9(c), we obtain neighborhoods U and U(cid:5) of y such that U ∩(g · U(cid:5)) = ∅ except for finitely many group elements 1,g1,…,g m ∈Γ. Since the action is free and Y is Hausdorff, for

268 12. Classification of Coverings each g there are disjoint neighborhoods W of y and W(cid:5) of g · y. Let i i i i U (cid:7) =U ∩U(cid:5)∩W1 ∩(g 1 −1· W 1 (cid:5))∩···∩W m ∩(g m −1· W m (cid:5) ). (cid:7) We will show that U has the required properties. First consider g = g for some i. If y ∈ U (cid:7) ⊂ g −1· W(cid:5), then g · y ∈ W(cid:5), i i i i i whichisdisjointfrom W andthereforefromU (cid:7) .ThusU (cid:7)∩(g · U (cid:7) )=∅.On i i the other hand, if g ∈ Γ is not the identity and not one of the g ’s, then i for any y ∈ U (cid:7) ⊂ U(cid:5), we have g · y ∈ g · U(cid:5), which is disjoint from U and (cid:7) therefore also from U. Agroupactionpossessingthepropertyexpressedinthiscorollary,orthat expressed in part (c) of Proposition 12.9, or something closely related to these(dependingonwhomyouread)hastraditionallybeencalledproperly discontinuous. This is a particularly unfortunate term, because the group actionsweareinterestedinareallcontinuous,sooneisforcedtospeakofa “continuous properly discontinuous action.” We will avoid the problem by workingonlywithproperactions,whichhavemanyimportantapplications intopologyandgeometry,andarequitesufficientaslongasweconfineour attention to locally compact Hausdorff spaces. Theorem 12.11. LetY beaconnected,locallypathconnected,locallycom- pact Hausdorff space (for example, a connected manifold), and suppose a discrete group Γ acts continuously, freely, and properly on Y. Then Y/Γ is Hausdorff, the quotient map π: Y → Y/Γ is a normal covering map, and C (Y)=Γ, considered as a group of homeomorphisms of Y. π Proof. Clearly, π is surjective and continuous. In fact, it is an open map, for the following reason: If U ⊂Y is open, then π−1(π(U)) is the union of · all sets of the form g U as g ranges over Γ. This is a union of open sets and therefore open, so π(U) is open. Toshowthat π isacoveringmap,let y ∈Y,andchooseaneighborhood U ofy asinCorollary12.10.LetV (cid:7) ⊂U bethecomponentofU containing (cid:7) y; clearly, V still has the property that its images under Γ are disjoint. Let (cid:7) V =π(V), which is open in Y/Γ because π is an open map. Now, π−1(V) is the union of the disjoint connected open sets g · V (cid:7) for g ∈ Γ, so to show that π is a covering it remains only to show that π is a homeomorphism from each such set onto V. Because for each g ∈ Γ, g: V (cid:7) →g · V (cid:7) is a homeomorphism and the diagram (cid:7) g (cid:2) · (cid:7) V g V π(cid:6) (cid:3)π (cid:6)(cid:7) (cid:3)(cid:9) V commutes, it suffices to show that π: V (cid:7) → V is a homeomorphism. It is surjective,continuous,andopen;anditisinjectivebecauseπ(v)=π(v(cid:5))for

Proper Group Actions 269 v,v(cid:5) ∈V (cid:7) impliesv(cid:5) =g · vforsomeg ∈Γ,sov =v(cid:5) becauseV (cid:7)∩(g · V (cid:7) )=∅ when g (cid:14)=1. This proves that π is a covering map. · · Ifg ∈Γ,thenx(cid:10)→g xisacoveringtransformation,sinceπ(g x)=π(x) bydefinition;thusΓ⊂C (Y).Byconstruction,Γactstransitivelyoneach π fiber,soπ isanormalcovering.Ifϕisanycoveringtransformation,choose y ∈ Y and let y(cid:5) = ϕ(y). Then there is some g ∈ Γ such that g · y = y(cid:5); · since ϕ and x (cid:10)→ g x are covering transformations that agree at a point, they are equal. Thus Γ is the full covering group. · ToshowthatthequotientspaceisHausdorff,letΦ(g,y)=(y,g y)asin the proof of Proposition 12.9. Since Φ is proper, it is closed by Proposition 4.32, so Φ(Γ×Y) is a closed subset of Y ×Y. Let x,x(cid:5) ∈Y/Γ be distinct points. Choosing y,y(cid:5) ∈ Y such that π(y) = x and π(y(cid:5)) = x(cid:5), the fact that x(cid:14)=x(cid:5) means that (y,y(cid:5))(cid:14)∈Φ(Γ×Y). Therefore, (y,y(cid:5)) has a product neighborhood U ×U(cid:5) ⊂ Y ×Y that is disjoint from Φ(Γ×Y). Since π is open, π(U) and π(U(cid:5)) are neighborhoods of x and x(cid:5), respectively. Any point z ∈ π(U)∩π(U(cid:5)) would satisfy z = π(v) = π(v(cid:5)) for some v ∈ U and v(cid:5) ∈ U(cid:5). But this would mean that v(cid:5) = g · v for some g ∈ Γ, so (v,v(cid:5)) = (v,g · v) ∈ Φ(Γ×Y), which contradicts the fact that U ×U(cid:5) is disjoint from the image of Φ. Thus π(U)∩π(U(cid:5)) = ∅, which shows that Y/Γ is Hausdorff. (cid:8) Corollary 12.12. Let M be a connected n-manifold on which a dis- (cid:8) crete group Γ acts continuously, freely, and properly. Then M/Γ is an n-manifold, and the quotient map π: M (cid:8) → M (cid:8) /Γ is a normal covering map. Proof. We know from Theorem 12.11 that π is a normal covering map and (cid:8) M/Γ is Hausdorff, and therefore it is a manifold by Problem 11-3(a). Example 12.13 (Lens Spaces). ByidentifyingR4 withC2 intheobvi- ous way, we can consider S3 as the following subset of C2: S3 ={(z1,z2)∈C2 :|z1 |2+|z2 |2 =1}. Fix a pair of relatively prime integers 1 ≤ m < n, and define an action of Z/(cid:22)n(cid:23) on S3 by (cid:12) (cid:13) k · (z1,z2)= e2πik/nz1,e2πikm/nz2 . It can easily be checked that this action is free, and it is proper because Z/(cid:22)n(cid:23) is a finite group (Problem 12-6). The orbit space S3/(Z/(cid:22)n(cid:23)) is thus a compact 3-manifold whose universal covering space is S3 and whose fun- damentalgroupisisomorphictoZ/(cid:22)n(cid:23).Thismanifold,denotedbyL(n,m), is called a lens space.

270 12. Classification of Coverings A particularly important example of a free proper action arises when we consider a topological group G and a discrete subgroup Γ (that is, a subgroup that is a discrete subspace). Recall from Chapter 3 that left translationdefinesaleftactionofΓonGwhosequotientisthecosetspace G/Γ. Proposition 12.14. Let Γ be a discrete subgroup of a connected, locally path connected, locally compact Hausdorff topological group G. Then Γ acts freely and properly on G by left translations, so the quotient map π: G → G/Γ is a normal covering map. Proof. As we observed in Example 3.35(e), G acts freely on itself, so the restrictionofthisactiontoΓiscertainlyfree.Wewillshowthattheaction is proper by showing that it satisfies property (b) of Proposition 12.9. Let K ⊂ G be any compact set. If γ ∈ Γ is an element such that K ∩ γK (cid:14)= ∅, then there exist g1,g2 ∈ K such that g1 = γg2, which is to say γ ∈ KK−1 = {g1g 2 −1 : g1,g2 ∈ K}. This set KK−1 is compact because it is the image of K ×K under the continuous map from G×G to G given by (g1,g2)(cid:10)→g1g 2 −1 . Because Γ is discrete, there can be only finitely many elements of Γ in KK−1. Corollary 12.15. SupposeGandH areconnected,locallypathconnected, locally compact Hausdorff topological groups, and ϕ: G→H is a surjective continuous homomorphism with discrete kernel. If ϕ is an open or closed map, then it is a normal covering map. Proof. Let Γ = Kerϕ. By the preceding proposition, the quotient map π: G → G/Γ is a normal covering map. The assumption that ϕ is either open or closed implies that it is a quotient map, and by the first isomor- phism theorem the identifications made by ϕ are precisely those made by π. Thus the result follows from the uniqueness of quotient spaces. Example 12.16 (Coverings of the Torus). For any integers a,b,c,d such that ad−bc (cid:14)= 0, consider the map p: T2 → T2 given by p(z,w) = (zawb,zcwd). This is easily seen to be a surjective continuous homomor- phism,anditisaclosedmapbytheclosedmaplemma.Onceweshowthat it has discrete kernel, it will follow from the preceding corollary that it is a normal covering map. (cid:12) Le(cid:13)t A denote the invertible linear transformation of R2 whose matrix is a b . Then we have a commutative diagram c d A(cid:2) R2 R2 E E (cid:5) (cid:5) T2 (cid:2) T2 (12.2) p

Proper Group Actions 271 where E(x,y) = (e2πix,e2πiy) is the universal covering map of the torus. To identify Kerp, note that p◦E(x,y)=(1,1) ⇐⇒ E◦A(x,y)=(1,1) ⇐⇒ A(x,y)∈Z2 ⇐⇒ (x,y)∈A−1(Z2), where A−1(Z2) denotes the additive subgroup {A−1(m,n) : (m,n) ∈ Z2} of R2. Because E is surjective, this shows that Kerp=E◦A−1(Z2). SinceA−1 hasrationalentries,itfollowseasilythateachelementofKerp has finite order in T2. Moreover, since Z2 is generated (as a group) by the two elements (1,0) and (0,1), Kerp is generated by their images under E◦A−1. An abelian group that is generated by finitely many elements of finite order is easily seen to be finite; in particular, it is discrete. Application: Universal Coverings of Higher Genus Surfaces As another application of the theory of proper group actions, we will show that the unit disk B2 ⊂ C is the universal covering space of all the ori- entable surfaces of genus n ≥ 2. The construction is rather involved, so we will describe the main steps and leave some of the details for you to work out. Some of these steps can be done a bit more straightforwardly if you know a little about Riemannian metrics and their geodesics, but we will not assume any such knowledge. We will, however, assume a passing acquaintance with complex analysis, at least enough to understand what it means for a function to be complex analytic. We begin by describing a special metric on the disk. For z1,z2 ∈ B2, define (cid:20) (cid:21) −1 2|z1 −z2 |2 d(z1,z2)=cosh 1+ (1−|z1 |2)(1−|z2 |2) . This is a metric, called the hyperbolic metric. (The only property of a metricthatisnotstraightforwardtocheckisthetriangleinequality;away to prove it is indicated in Problem 12-8.) The disk with this metric, called the hyperbolic disk, is one model of non-Euclideanplanegeometry.The“straightlines”inthisgeometry,called hyperbolic geodesics,aretheintersectionswiththediskofEuclideancircles andlinesmeetingtheunitcircleorthogonally(Figure12.6).(Alinesegment throughtheorigincanbethoughtofasthelimitingcaseofacirculararcas theradiusgoestoinfinity.)Itiseasytocheckthat“twopointsdeterminea line”:Thatis,givenanytwopointsinthedisk,thereisauniquehyperbolic geodesic passing through both points.

272 12. Classification of Coverings 0 FIGURE 12.6. Hyperbolic geodesics. The most interesting feature of the hyperbolic metric is that it is pre- served by a transitive group action. Let α and β be complex numbers with |α|2−|β|2 >0, and define αz+β ϕ(z)= . (12.3) βz+α A straightforward calculation shows that ϕ is continuous, takes the disk to itself, and preserves the hyperbolic metric in the sense that d(ϕ(z1),ϕ(z2)) = d(z1,z2) for all z1,z2 ∈ B2. Any such map is called a M¨obius transformation of the disk, and the set M of all such maps is a group under composition, called the M¨obius group of the disk. It can be (cid:12) (cid:13) identified with the group of matrices of the form αβ and so is a topo- β α logical group acting continuously on B2. M¨obius transformations take geodesics to geodesics, as can be seen by substituting ϕ(z) for z in the equation defining a circle or line intersecting theboundaryofthediskorthogonally,andnotingthatitreducestoanother equation of one of the same types. In fact, the same computation shows that a Mo¨bius transformation takes the intersection of the disk with any Euclidean circle or line to another set of one of the same forms. One special case worth noting is that any rotation of the disk z (cid:10)→ eiθz is a Mo¨bius transformation with α = eiθ/2 and β = 0, so the hyperbolic metricisinvariantunderrotations.In fact,anyM¨obiustransformationthat takes the origin to itself must be of this form, because (12.3) reduces to ϕ(z)=(α/α)z inthatcase.Observealsothatthehyperbolicdistancefrom theorigintozdependsonlyon|z|,soanymetricballB (0)abouttheorigin r is actually a Euclidean disk centered at 0, and its boundary is a Euclidean circle.SinceM¨obiustransformationspreservehyperbolicdistanceandtake circles to circles, it follows that every metric ball is a Euclidean disk. (Its Euclidean center may not be the same as its hyperbolic center, however). ItalsofollowsthatthehyperbolicmetricgeneratestheEuclideantopology.

Proper Group Actions 273 The left action of M on the disk defined by (12.3) is transitive because any z0 ∈B2 is carried to 0 by the M¨obius transformation z−z0 ϕ(z)= . (12.4) 1−z0z In fact, more is true: Given any two pairs of points z0,z1 and z 0 (cid:5),z 1 (cid:5) such that d(z0,z1) = d(z 0 (cid:5),z 1 (cid:5)), there is a unique Mo¨bius transformation taking z0 to z 0 (cid:5) and z1 to z 1 (cid:5) (and therefore taking the geodesic segment joining z0,z1 to the one joining z 0 (cid:5),z 1 (cid:5)). To prove this, let ψ = ρ◦ϕ, where ϕ is the transformation (12.4) and ρ is a rotation moving ϕ(z1) to the positive x-axis, so that ψ takes z0 to 0 and z1 to some λ > 0. Similarly, there is a transformation ψ(cid:5) taking z(cid:5) to 0 and z(cid:5) to λ(cid:5) > 0. Since M¨obius trans- 0 1 formations preserve distances, λ and λ(cid:5) are at the same distance from 0 along the positive x-axis and therefore must be equal, so ψ(cid:5)−1 ◦ψ is the transformation we seek. It is unique because if γ is any M¨obius transfor- mation taking z0 to z 0 (cid:5) and z1 to z 1 (cid:5), the composition ψ(cid:5) ◦γ ◦ψ−1 fixes 0 and therefore must be a rotation, and since it also fixes λ, it must be the identity, which implies γ =ψ(cid:5)−1◦ψ. Each M¨obius transformation ϕ is complex analytic with nowhere van- ishing derivative. Multiplication by the complex derivative ϕ(cid:5)(z0) defines a linear map from C to C, which can be interpreted geometrically as the action of ϕ on tangent vectors to curves: For any differentiable pa- rametrized curve f: (−ε,ε) → B2 with f(0) = z0, the chain rule gives (ϕ◦f)(cid:5)(0) = ϕ(cid:5)(z0)f(cid:5)(0). Thus ϕ acts on tangent vectors by multiplying them by the nonzero complex number ϕ(cid:5)(z0), and since all tangent vectors are rotated through the same angle, every Mo¨bius transformation is con- formal, meaning it preserves angles between tangent vectors. (We will also be considering angles between geodesics, by which we always mean angles between their tangent vectors.) In particular, if ϕ(z) = eiθz is rotation through an angle θ, then ϕ(cid:5)(0) = eiθ rotates tangent vectors through the same angle. It follows that the only M¨obius transformation that fixes the origin and fixes the direction of a tangent vector at the origin is the iden- tity. In fact, a Mo¨bius transformation that fixes any point and a tangent direction at that point must be the identity, because conjugation with a transformationtakingthefixedpointto0yieldsatransformationthatfixes 0 and a tangent direction at 0. NowletM beacompactorientablesurfaceofgenusn≥2.Wewillshow that there is a discrete subgroup Γ ⊂ M acting freely and properly on B2 suchthatM ishomeomorphictoB2/Γ.ItfollowsfromTheorem12.11that the universal covering space of M is B2. Recall from Chapter 6 the standard polygonal presentation of M as a quotient of a polygonal region with 4n sides whose edges are identified in pairs. We will realize M as a quotient of a compact region in B2 bounded byageodesicpolygon,thatis,theunionoffinitelymanygeodesicsegments. We begin by constructing a 4n-sided geodesic polygon whose edges have

274 12. Classification of Coverings θ P θ ≈π−π/2n. θ ≈0. θ =π/2n. FIGURE12.7.Geodesicpolygonswithinteriorangles0<θ<π−π/2n. equal lengths and meet at equal angles (a regular geodesic polygon). Start with 4n points (z0,z1,…,z4n = z0) equally spaced on some circle about the origin. Because the hyperbolic metric is invariant under rotations, the geodesic segments joining z j and z j+1 for j = 0,…,4n−1 all have the same length and meet at equal angles, so their union is a regular geodesic polygon. As the radius of the circle goes to zero, these geodesics approach linesegmentsthroughtheorigin,anddefinesmallregulargeodesicpolygons whoseinterioranglesareveryclosetowhattheywouldbeintheEuclidean case, namely π −π/2n (see Figure 12.7). As the points get farther from the origin, the arcs become nearly tangent to each other, defining geodesic polygons with interior angles very near zero. By continuity, somewhere in between there is a polygon whose interior angles are exactly θ = π/2n. (Note that this does not work when n = 1, so we cannot construct a covering of the torus in this manner.) Let P be the compact subset of B2 consisting of this regular geodesic polygon together with the bounded component of its complement. Choose one vertex v0, and label the edges a1,b1,a − 1 1 ,b − 1 1 ,…,a n ,b n ,a− n 1,b− n 1 in counterclockwise order starting from v0. (See Figure 12.8, but ignore the −1 vertex labels other than v0 for now.) For each edge pair a j ,a j , there is −1 a unique Mo¨bius transformation α that takes the edge labeled a onto j j the one labeled a , with the initial vertex of one going to the initial vertex j −1 of the other. Similarly, let β be the transformation taking b to b and j j j respecting the initial and terminal vertices. Let Γ ⊂ M be the subgroup generated by {α ,β : j = 1,…,4n}. We will call the generators α , β , j j j j and their inverses edge pairing transformations. One important property of the edge pairing transformations is easy to verify: If σ is any edge pairing transformation, then P ∩σ(P) consists of exactly one edge of P. To see why, suppose σ takes an edge e to another edgee(cid:5).Thenclearly,P∩σ(P)containse(cid:5).Notethatthecomplementofany geodesic in B2 has exactly two components, which we may call the sides of thegeodesic.BecauseP isconnectedandliesononesideofeachofitsedges, thesameistrueofσ(P).Usingconformalityandfollowingwhatσdoestoa vector that is perpendicular toe and points into P, it is easy to check that

Proper Group Actions 275 v0 v3 a1 b1 v2 α1 −1 a 1 v1 β1 −1 b 1 v4 FIGURE 12.8. Edge pairing transformations. σ(P)liesontheoppositesideofe(cid:5) fromP,andthereforeP∩σ(P)consists of exactly the edge e(cid:5). Because P is obviously homeomorphic to a regular Euclidean polygon, the quotient of P by the identifications determined by the edge pairing transformations is homeomorphic to M. Let p: P → M denote the quotient map. Theorem 12.17. The group Γ is discrete and acts freely and properly on B2, and the quotient B2/Γ is homeomorphic to M. The restriction of this quotient map to P is p. Proof. Thefirstthingwewillproveisthattheedgepairingtransformations satisfythesamerelationasthegeneratorsofthefundamentalgroupof M: α1 ◦β1 ◦α 1 −1◦β 1 −1◦···◦α n ◦β n ◦α n −1◦β n −1 =Id. (12.5) Actually, it will be more convenient to prove the equivalent identity ob- tained by inversion: β n ◦α n ◦β n −1◦α n −1◦···◦β1 ◦α1 ◦β 1 −1◦α 1 −1 =Id. (12.6) To simplify the notation, let us write the sequence of transformations on the left-hand side of (12.6) as σ4n ◦···◦σ2 ◦σ1. −1 By definition, σ1 = α 1 takes v0, the initial vertex of the edge labeled −1 a1, to the initial vertex of the edge labeled a 1 . If we label the vertices in

276 12. Classification of Coverings V3 3θ V0 V2 2θ α α−1 β−1 V1 θ β 4θ V4 FIGURE12.9.ImagesofavectorV0 underedgepairingtransformations. counterclockwiseorderstartingfromv0 asv0,v3,v2,v1,v4 asinFigure12.8, itiseasytocheckonestepatatimethatσ j takesv j−1tov j forj =1,…,4. Since v4 is also the initial vertex of the edge labeled a2, we can continue by induction to number all the remaining vertices v5 through v4n = v0 in suchawaythatσ j (v j−1)=v j .Inparticular,σ4n ◦···◦σ2 ◦σ1(v0)=v0.To show that this composition is the identity, it suffices to show that it fixes a tangent direction at v0. For any vertex v , we will measure angles of vectors at v from the edge j j adjacent to v j in the counterclockwise direction (so we measure from a1 at −1 v0,fromb 1 atv1,etc.).Positiveangleswillalwaysbeunderstoodtomean counterclockwise rotation from that edge. Let θ =π/2n be the measure of the interior angles of P. Let V0 be a nonzero vector that makes an angle of 0 at v0 (see Figure 12.9), and for j = 1,…,4n let V j be the image of V0 under σ j ◦···◦σ1, so that σ j takes V j−1 to V j . We will prove the following claim: For each j, the angle of V at v is jθ. For j =0 this is immediate from the definition j j −1 −1 of V0. For j = 1, note that σ1 = α 1 takes a1 to a 1 , and therefore takes −1 V0 toavectorV1 thatpointsinthedirectionofa 1 ,whichmakesanangle −1 θ with b . Next, since M¨obius transformations preserve angles, the image 1 −1 V2 of V1 under σ2 = β 1 makes an angle θ with b1, which is the same as −1 an angle 2θ with a 1 . A similar analysis shows that the angles of V3 and V4 are 3θ and 4θ, respectively, and the claim is then proved for all j by

Proper Group Actions 277 induction. In particular, the angle of V4n is 4nθ =2π, so V4n points in the same direction as V0. This completes the proof of (12.5). Now we have to prove that Γ is discrete and acts freely and properly on B2. It seems to be impossible to prove this by directly analyzing the action of Γ, so instead we resort to a rather circuitous trick due originally toPoincar´e.Wewillconstruct“byhand”acoveringspaceofM thatought to be its universal covering space, as a union of infinitely many copies of P—one for each element of π1(M)—with “adjacent” copies glued together bytheidentificationsdeterminedbytheedgepairingtransformations.Only later will we show that this space is homeomorphic to B2, and therefore is simply connected and so is in fact the universal covering space. Let G be the abstract group with presentation (cid:22)α1,β1,…,α n ,β n | α1β1α 1 −1 β 1 −1···α n β n α n −1β n −1(cid:23), which is isomorphic to π1(M). Let ∼ be the equivalence relation on G×P generated by all relations of the form (g,σ(z)) ∼ (gσ,z), where σ is an edge pairing transformation and both z (cid:8) and σ(z) are points in∂P. Give G the discrete topology, and letM denote the quotient space G×P/∼. We will denote the equivalence class of (g,z) in M (cid:8) by [g,z], and the quotient map by π: G×P →M (cid:8) . Left translation in the G factor defines a natural continuous action of G on G×P. This respects the identifications made by π, so it descends to a continuous action of G on M (cid:8) , satisfying g(cid:5)· [g,z] = [g(cid:5)g,z]. This action is free, because (g(cid:5)g,z)∼(g,z) only when g(cid:5) =1. The subset P (cid:7) = π({1}×P) = {[1,z] : z ∈ P} of M (cid:8) is homeomorphic to P (why?), and M (cid:8) is the union of the sets g · P (cid:7) = {[g,z] : z ∈ P} as g (cid:8) ranges over G. Each of these sets is a homeomorphic copy of P in M, and thecopiesg · P (cid:7) andg(cid:5)· P (cid:7) intersectinanedgepreciselywheng andg(cid:5) differ by a single edge pairing transformation. Since there are only finitely many · (cid:7) suchtransformations,thismeansinparticularthateachsetg P intersects only finitely many others. Because ∼ identifies only points (g,z) with z ∈ ∂P, the fiber of π over any point [g0,z0] for z0 ∈ IntP consists of exactly one point (g0,z0) ∈ G×P. If z0 is in ∂P but is not a vertex, then z0 lies on one edge, and there is exactly one edge pairing transformation σ that identifies that edge with another edge; thus the fiber over [g0,z0] is exactly two points (g0,z0) and (gσ−1,σ(z0)). If z0 is a vertex of P, then by the argument at the beginning of the proof there is a sequence of edge pairing transformations σ1,…,σ4n (possibly a cyclic permutation of the sequence we considered earlier) such that the points z j =σ j ◦···◦σ1(z0) are the vertices of P, so −1 −1 −1 the fiber over [g0,z0] consists of the 4n points (g0σ 1 ,z1), (g0σ 1 σ 2 ,z2), … , (g0σ 1 −1···σ 4 − n 1 ,z4n )=(g0,z0).

278 12. Classification of Coverings There is a natural continuous map p(cid:7): M (cid:8) → M given by p(cid:7)[g,z] = p(z), obtained from p◦π2 by passing to the quotient: G×P π2(cid:2) P π p (cid:5) (cid:5) (cid:8) (cid:2) M p(cid:7) M. Clearly, p(cid:7) is surjective, because p(cid:7)(P (cid:7) ) = M. It is a quotient map for the following reason: If U ⊂M (cid:8) is an open set that is saturated with respect to p(cid:7),thenπ−1(U)⊂G×P isopenandsaturatedwithrespecttop(cid:7)◦π =p◦π2, and since p◦π2 is a quotient map, it follows that p(cid:7)(U)=p(cid:7)◦π(π−1(U))= p◦π2(π−1(U)) is open. You can check that the fibers of p(cid:7)are precisely the (cid:8) (cid:8) orbitsofGinM,sowecanidentifyM withtheorbitspaceM/G.Wewish to show that p(cid:7)is actually a covering map. To show that p(cid:7)is a covering, by Theorem 12.11 it suffices to show that (cid:8) M is connected, locally path connected, locally compact, and Hausdorff, (cid:8) and that the action of G on M is proper. Connectedness is easy: If σ is an edge pairing transformation taking edge e to edge e(cid:5), then the connected sets P (cid:7) and σ · P (cid:7) have the points [1,σ(z)]=[σ,z] in common for z ∈e, so P (cid:7)∪(σ · P (cid:7) ) is connected. By induction, any set of the form P (cid:7)∪(σ1 · P (cid:7) )∪ ···∪(σ m ···σ1) · P (cid:7) is connected. Since M (cid:8) is the union of all such sets, and (cid:7) (cid:8) they all have points of P in common, M is connected. (cid:8) ToprovetheotherpropertiesofM,wefirstneedtointroducesomemore maps. Let τ: G → Γ be the homomorphism that sends each generator α i orβ toitself(thoughtofasanelementofΓ),whichiswell-definedbecause i (12.5) holds in Γ. The map G × P → B2 defined by (g,z) (cid:10)→ τ(g)z is continuous and respects the identifications made by ∼, so it descends to a continuous map δ: M (cid:8)→B2 given by δ[g,z]=τ(g)z. It takes the action of G on M (cid:8) over to the action of Γ on B2, in the sense that · δ(g x)=τ(g)◦δ(x). (12.7) ThemostimportantfeatureofM (cid:8) isthateveryx∈M (cid:8) hasaneighborhood U with the following properties: (i) U is mapped homeomorphically by δ onto a closed hyperbolic ball B (δ(x))⊂B2. ε (ii) δ(U)=B (δ(x)). ε (iii) U intersects the sets g · P (cid:7) for only finitely many g ∈G. We will call any such set U a regular hyperbolic neighborhood of x.

Proper Group Actions 279 Fromtheexistenceofregularhyperbolicneighborhoodsitfollowsimme- diately that

• M (cid:8) islocallycompactandlocallypathconnected,becauseeachregular hyperbolic neighborhood has these properties.
• M (cid:8) is Hausdorff: Let x,x(cid:5) ∈ M (cid:8) , and let U, U(cid:5) be regular hyperbolic neighborhoodsofthem.Ifx(cid:5) (cid:14)∈U,thenshrinkingU abitifnecessary we may assume x(cid:5) (cid:14)∈ U, so that U and U(cid:5) (cid:3)U are disjoint neigh- borhoods of x and x(cid:5). On the other hand, if x(cid:5) ∈U, then the inverse imagesunderδ| ofdisjointneighborhoodsofδ(x)andδ(x(cid:5))areopen U sets separating x and x(cid:5).
• The action of G on M (cid:8) is proper: With x, x(cid:5), U, and U(cid:5) as above, there can be at most finitely many g ∈G such that U∩(g · U(cid:5))(cid:14)=∅, because U and U(cid:5) intersect only finitely many of the sets g · P (cid:7) . Thus, to complete the proof that p(cid:7)is a covering map, we need only prove the existence of a regular hyperbolic neighborhood of each point. (cid:8) Let x=[g0,z0] be an arbitrary point of M. The fiber over x consists of finitelymanypointsoftheform(g j ,z j ),wherez j =σ j ◦···◦σ1(z0)forsome (possibly empty) sequence of edge transformations σ1,…,σ j and g j = g0σ 1 −1···σ j −1 . (The fiber contains one, two, or 4n such points depending onwhetherz0 isaninteriorpoint,anedgepoint,oravertex.)Chooseε>0 smaller than half the distance from z0 to any edge that does not contain z0. Let W ⊂ G×P be the union of the sets {g j }×(B ε (z j )∩P), and let U = π(W). Because W is a saturated open set, U is a neighborhood of x inM (cid:8) .Similarly,W istheunionofthesets{g }×(B (z )∩P),asaturated j ε j · (cid:7) closed set, so π(W)=U. Clearly, U intersects g P for only finitely many g. To complete the proof that U is a regular hyperbolic neighborhood, we need to show that δ is a homeomorphism from U to B ε (z0) taking U to B ε (z0). Since the diagram δ (cid:2) U B (δ(x)) ε g τ(g) (cid:5) (cid:5) · (cid:2) · g U B (δ(g x)) δ ε commutes for each g ∈ G and the vertical maps are homeomorphisms, it suffices to prove this for x=[1,z0]∈P (cid:7) . We consider three cases. Case I: z0 ∈ IntP. In this case, U ⊂ P (cid:7) , and it is immediate from the definitions that δ is one-to-one on U, δ(U) = B ε (z0), and δ(U) = B ε (z0). Since U is the image under π of a compact set, it is compact, so δ: U → B ε (z0) is a homeomorphism by the closed map lemma.

280 12. Classification of Coverings z0 {1}×P (cid:7) δ P z1 {σ−1}×P FIGURE 12.10. Hyperbolic neighborhood of an edge point. Case II: z0 ∈ ∂P, but z0 is not a vertex. Let e0 denote the edge con- taining z0. By our choice of ε, B ε (z0)∩P contains the entire portion of B ε (z0) lying on one side of e0 (Figure 12.10). There is one edge pairing transformation σ that takes e0 to another edge e1, and thus takes z0 to z1 =σ(z0)∈e1. As a Mo¨bius transformation of B2, σ takes B ε (z0) home- omorphically onto B ε (z1). Since B ε (z0)∩P and σ−1(B ε (z1)∩P) lie on opposite sides of e0, B ε (z0)=(B ε (z0)∩P)∪σ−1(B ε (z1)∩P). Then δ(U)=δ (cid:7) (W)=(B ε (z0)∩P)∪σ−1(B ε (z1)∩P)=B ε (z0). The restriction of δ to U is one-to-one, takes U onto B ε (z0), and as before is a homeomorphism by the closed map lemma. Case III:z0 is a vertex of P.Thenδ(U)=δ (cid:7) (W)istheunionofthesets δ (cid:7) ({σ −1···σ −1}×(B (z )∩P))=σ −1◦···◦σ −1 (B (z )∩P), 1 j ε j 1 j ε j wherez1,…,z4n aretheverticesofP.Toseewhatthesesetsare,lookback attheproofof (12.5);fromthatanalysis,itfollowsthatσ −1◦···◦σ −1 maps 1 j z j to z0 and maps B ε (z j )∩P to the sector of B ε (z0) lying between the geodesics passing through z0 at angles −jθ and (−j+1)θ (Figure 12.11). These sectors fit together to make up the entire closed ball B ε (z0), and δ maps U bijectively to B ε (z0). As above, it is a homeomorphism by the closed map lemma.

Proper Group Actions 281 δ(U) P FIGURE 12.11. Hyperbolic neighborhood of a vertex point. This completes the proof of the existence of hyperbolic neighborhoods andthustheproofthatp(cid:7): M (cid:8)→M isacoveringmap.Tofinishtheproofof thetheorem,wewillshowthatδ: M (cid:8)→B2 isalsoacoveringmap.SinceB2 is simply connected, this implies that δ is a homeomorphism. The theorem follows from this, as we now show. First, τ: G→Γ is a group isomorphism: It is surjective because it takes generators of G to generators of Γ; and it is injective because if τ(g)=Id, then for any x ∈ M (cid:8) we have δ(g · x) = τ(g)δ(x) = δ(x), which implies · g x=xandthereforeg =1becauseGactsfreely.Itfollowsthattheaction ofΓonB2isequivalenttothatofGonM (cid:8) underthehomeomorphismδ,and thequotientmapB2 →B2/Γisequivalenttothecoveringmapp(cid:7): M (cid:8)→M. Therefore, the action of Γ on B2 is free and proper, and the restriction of thecoveringmaptoP isp(cid:7)◦δ−1| =p.ToseethatΓisadiscretesubgroup P of M, suppose γ → γ in Γ. By continuity γ z → γz for any z ∈ B2, and i i setting g =τ−1(γ ), g =τ−1(γ), and x=δ−1(z) we obtain g · x→g · x. i i i Since the g ’s are covering transformations, this can happen only if g =g i i (and therefore γ =γ) for all sufficiently large i. i Toshowthatδ isacovering,weneedthefollowingadditionalfactabout regularhyperbolicneighborhoods:There exists some r >0 such that every point x ∈ M (cid:8) has a regular hyperbolic neighborhood U whose closure is x mapped homeomorphically by δ onto B (δ(x)). To prove this, let K ⊂ M (cid:8) r (cid:7) · (cid:7) denote the union of P together with its images σ P under the 4n edge i pairing transformations σ . Since K is compact, so is its image δ(K)⊂B2, i and it is easy to see that δ(K) contains a neighborhood of P. As U ranges over regular hyperbolic neighborhoods of points in K, the sets δ(U) form an open cover of δ(K). Let c be a Lebesgue number for this cover, and choose r < c small enough that for each z ∈ P the hyperbolic ball B (z) r is contained in δ(K). This means that for every z ∈ P, there is a regular

282 12. Classification of Coverings (cid:7) P P U δ(U) x δ(x) x0 U δ(x0) x0 FIGURE 12.12. Finding regular hyperbolic balls of fixed radius. hyperbolic neighborhood U of some point x∈K such that B (z)⊂δ(U). r For each x0 ∈ P (cid:7) , choose a regular hyperbolic neighborhood U of some x ∈ K such that B r (δ(x0)) is contained in δ(U) (Figure 12.12), and let U x0 = (δ| U )−1(B r (δ(x0)); then δ: U x0 → B r (δ(x0)) is the restriction of a homeomorphism and hence is itself a homeomorphism. Since δ is injective o o t n h P e (cid:7) r a x nd ∈ δ M (cid:8) (x , 0) th ∈ er δ e (U is x0 s ) o , m U e x0 g is ∈ th G e d su e c si h re t d ha n t ei g gh · b x orh ∈ oo P (cid:7) d , o s f o x w 0 e . F c o a r n a s n e y t U =g−1· U · . x g x We can now prove that δ is a covering map. First we need to show that (cid:8) it is surjective. If it were not, the image set δ(M) would have a boundary point z0 ∈ B2. There is some point z ∈ δ(M (cid:8) ) whose distance from z0 is less than r/2. But then z = δ(x) for some x ∈ M (cid:8) , and δ(U ) = B (z), x r which is a neighborhood of z0. This contradicts the assumption that z0 is a boundary point of the image. For any z0 ∈ B2, we will show that B r/2(z0) is evenly covered. Let V beacomponentofδ−1(B r/2(z0))inM (cid:8) .SinceM (cid:8) islocallypathconnected, V is open. We need to show that δ: V → B r/2(z0) is a homeomorphism. Choose x∈V, set z =δ(x), and let σ =(δ| )−1: B (z)→U . Now, σ(B r/2(z0)) is a connected subset of U δ x−1(B r/ r 2(z0)) tha x t contains a point x in common with V, so it must be contained in V. This implies, for any z(cid:5) ∈B r/2(z0), that δ(σ(z(cid:5)))=z(cid:5), so δ: V →B r/2(z0) is surjective. On the other hand, ∂B r (z) is disjoint from B r/2(z0) by the triangle inequality.Sinceδ takes∂U to∂B (z),itfollowsthat∂U ∩V =∅.Now, x r x V ∩U is open in M (cid:8) and therefore open in V, and V ∩U = V ∩U is x x x closed in V. Since V is connected, V ∩U is all of V, which means that x V ⊂U . Thus δ| is the restriction of a homeomorphism, so it is injective x V and open, and therefore δ: V →B r/2(z0) is a homeomorphism.

The Classification Theorem 283 Corollary 12.18. Let M be a compact surface. The universal covering space of M is homeomorphic to (a) S2 if M ≈S2 or P2; (b) R2 if M ≈T2 or P2#P2; (c) B2 if M is any other surface. Proof. ThiswasprovedforalltheorientablesurfacesandP2inthischapter. If M is a connected sum of n≥2 projective planes, then by Problem 11-8 M hasatwo-sheetedcoveringbytheorientablesurfaceN ofgenusn−1.If (cid:8) (cid:8) M istheuniversalcoveringspaceofM,thenM alsocoversN byCorollary 12.6(a), so M and N have the same universal covering space. Note that R2 and B2 are homeomorphic, so up to topological equiva- lence there are only two simply connected 2-manifolds that cover compact surfaces. It is useful, however, to distinguish the two cases because of the differentcharacteroftheircoveringtransformations.Forexample,thecov- ering transformations for the torus are all translations of the plane that preserve the Euclidean metric, while for the higher genus orientable sur- faces they are M¨obius transformations. The Classification Theorem In this section we assemble the results of this chapter to come up with a complete classification of coverings of a given space. The idea is that every covering of X is itself covered by the universal covering space, and intermediatecoveringscanbebuiltfromtheuniversalcoveringasquotients by suitable group actions. Theorem 12.19 (Classification of Coverings). LetX beaconnected, locally simply connected, locally compact Hausdorff space (for example, any connected manifold), and let q ∈ X be any base point. There is a one-to- one correspondence between isomorphism classes of coverings of X and conjugacy classes of subgroups of π1(X,q). The correspondence associates each covering p(cid:5): X(cid:5) →X with the conjugacy class of its induced subgroup. Proof. The covering isomorphism theorem shows that there is at most one isomorphism class of coverings corresponding to any conjugacy class of subgroups, so all we need to show is that there is at least one. Let H ⊂ π1(X,q) be any subgroup in the given conjugacy class. Let p: X (cid:7) → X be the universal covering of X, and choose a base point q(cid:7)∈p−1(q). Then the simply connected case of the covering group structure theorem (Corollary 11.32) shows that π1(X,q) is isomorphic to the covering group C p (X (cid:7) ),

284 12. Classification of Coverings under the map α: π1(X,q)→C p (X (cid:7) ) that sends [f] to the unique covering transformation ϕ taking q(cid:7)to q(cid:7) · [f]. Let H (cid:7) =α(H)⊂C (X (cid:7) ). p Since C (X (cid:7) ) acts freely and properly on X (cid:7) , it follows easily that H (cid:7) does p too.SoletX(cid:5) denotethequotientspaceX (cid:7) /H (cid:7) andπ: X (cid:7) →X(cid:5) thequotient map;byTheorem12.11,π isanormalcoveringmap.Moreover,p: X (cid:7) →X is constant on the fibers of π (since they are contained in the fibers of p), so p descends to a continuous map p(cid:5): X(cid:5) → X such that the following diagram commutes: (cid:7) X (cid:6) π (cid:6)(cid:7) p X(cid:5) (cid:3)p(cid:5) (cid:5)(cid:3)(cid:9) X. We have to show that p(cid:5) is a covering map. Let q1 ∈ X be arbitrary, let U be a neighborhood of q1 that is evenly covered by p, and let U(cid:5) be any component of p(cid:5)−1(U). To show that p(cid:5) is a covering map, it suffices to show that U(cid:5) is mapped homeomorphically onto U by p(cid:5). Because X(cid:5) is locally path connected, U(cid:5) is open and closed in p(cid:5)−1(U). Thusπ−1(U(cid:5))isopenandclosedinπ−1(p(cid:5)−1(U))=p−1(U),whichimplies that it is a union of components of p−1(U). If U (cid:7) is any such component, the following diagram commutes: (cid:7) U (cid:6) π (cid:6)(cid:7) p U(cid:5) (cid:3)p(cid:5) (cid:5)(cid:3)(cid:9) (12.8) U. In this diagram, p = p(cid:5)◦π is a homeomorphism, so π is injective on U (cid:7) . If π(U (cid:7) )(cid:14)=U(cid:5), then (cid:5) π(π−1(U(cid:5)))= π(ϕ(U (cid:7) ))=π(U (cid:7) )(cid:14)=U, ϕ∈H (cid:7) which contradicts the fact that π: X (cid:7) →X(cid:5) is surjective. Thus π: U (cid:7) →U(cid:5) is bijective, and because it is an open map, it is a homeomorphism. Since p and π are homeomorphisms in (12.8), so is p(cid:5).

The Classification Theorem 285 ϕ(q(cid:7)) (cid:7) X ϕ q(cid:7) (cid:7) f π p X(cid:5) q(cid:5) f p(cid:5) X q p(cid:5)◦f FIGURE 12.13. Proof of the classification theorem. The last step is to show that p(cid:5) ∗ π1(X(cid:5),q(cid:5)) = H for some q(cid:5) ∈ X(cid:5) such that p(cid:5)(q(cid:5))=q. Let q(cid:5) =π(q(cid:7)) and consider the following diagram: π1(X(cid:5),q(cid:5)) α(cid:2)(cid:5) C π (X (cid:7) ) p(cid:5) ι ∗ (cid:5) (cid:5) π1(X,q) α (cid:2) C p (X (cid:7) ), where α(cid:5) and α represent the isomorphisms given by the covering group structure theorem, and ι is inclusion. If this diagram commutes, we are done, because then p(cid:5) ∗ π1(X(cid:5),q(cid:5))=α−1◦ι◦α(cid:5)(π1(X(cid:5),q(cid:5))) =α−1◦ι(C (X (cid:7) )) π =α−1(H (cid:7) )=H. Toseethatitcommutes,let[f]∈π1(X(cid:5),q(cid:5))bearbitrary,andletϕ=α(cid:5)[f], soϕtakesq(cid:7)toq(cid:7) · [f]=f (cid:7) (1),wheref (cid:7) istheliftoff toapathinX (cid:7) starting at q(cid:7)(Figure 12.13). Then ι◦α(cid:5)[f]=ϕ, thought of as an element of C (X (cid:7) ). p On the other hand, α◦p(cid:5)[f] = α[p(cid:5)◦f] is the transformation ψ ∈ C (X (cid:7) ) ∗ p taking q(cid:7)to p (cid:4)(cid:5)◦f(1). Now, p◦f (cid:7) = p(cid:5) ◦π ◦f (cid:7) = p(cid:5) ◦f, so f (cid:7) is the lift of p(cid:5)◦f starting at q(cid:7), which implies that p (cid:4)(cid:5)◦f(1) = f (cid:7) (1). Thus ϕ = ψ and the diagram commutes.

286 12. Classification of Coverings We end with a pair of interesting and representative examples. Example 12.20 (Coverings of Lens Spaces). By the preceding the- orem, the coverings of the lens space L(n,m) are in one-to-one corre- spondence with subgroups of Z/(cid:22)n(cid:23). (Since Z/(cid:22)n(cid:23) is abelian, each con- jugacy class contains precisely one subgroup.) Since every subgroup of a cyclic group is cyclic (Exercise A.27), the only possibilities for subgroups G ⊂ π1(L(n,m)) are cyclic groups of order p where p is a factor of n. In each such case, a covering of L(n,m) is obtained by restricting the action of Z/(cid:22)n(cid:23) on S3 to G, and mapping the resulting quotient space down to L(n,m)bysendingeachG-equivalenceclasstoitsZ/(cid:22)n(cid:23)-equivalenceclass. If n = pq for positive integers p and q, let G ⊂ Z/(cid:22)n(cid:23) be the cyclic sub- group of order p generated by (the coset of) q. It is easy to check from the definitions that S3/G = L(p,m), and we obtain a q-sheeted covering L(p,m) → L(n,m). These are the only coverings of the lens spaces up to isomorphism. Our last application will be to classify all the coverings of the torus up to isomorphism. Proposition 12.21 (Classification of Torus Coverings). Every cov- ering of T2 is isomorphic to precisely one of the following: (a) the universal covering E :R2 →T2; (b) the coverings p:S1×R→T2 by p(z,y)=(zaε(y)b,zbε(y)−a), where (a,b) are integers with a≥0 and b>0 if a=0; (c) the coverings p:T2 →T2 by p(z,w)=(zawb,wc), where (a,b,c) are integers with 0≤b0. Proof. Note that all of these maps are coverings: the universal cover by Example11.3;themapsinpart(b)byProblem12-7;andthoseinpart(c) by Example 12.16. Let us use q = (1,1) ∈ T2 as base point, and represent π1(T2,q) as the productgroup(cid:22)β(cid:23)×(cid:22)γ(cid:23),whereβ andγ arethepathclassesofthestandard generatorofπ1(S1,1)inthefirstandsecondfactors,respectively.Thenthe map (m,n)(cid:10)→βmγn is an isomorphism of Z2 with π1(T2,q). The classification theorem says that isomorphism classes of coverings of T2 are in one-to-one correspondence with subgroups ofπ1(T2,q) under the correspondence that matches a covering p : X → T2 with the subgroup induced by p. So we begin by showing that each subgroup of Z2 is one and only one of the following: (i) the trivial subgroup; (ii) infinite cyclic subgroups generated by (a,b) satisfying the conditions of (b) above;