V. v. Prasolov Elements of Homology Theory Graduate Studies in Mathematics 81 2007

V. V. Prasolov

Elements of Homology Theory

v. V. Prasolov

Graduate Studies in Mathematics

Volume 81

_._,i&. American Mathematical Society Providence, Rhode Island

David Cox (Chair) Walter Craig N. V. Ivanov

Steven G. Krantz

B. B. IIpaconoB

8JIEMEHThI TEOPI1I1 rOMOJIOrl1H MIIHMO, MocKBa, 2005

This work was originally published in Russian by MIIHMO under the title "8neMeHTLI TeOpH:" rOMOnOr"" 2005. The present translation was created under license for the American Mathematical Society and is published by permission.

Translated from the Russian by Olga Sipacheva

2000 Mathematics Subject ClasSIfication. Primary 55-01.

For additional information and updates on this book, visit www.ams.org/bookpages/gsm-81

Library of Congress Cataloging-in-Publication Data Prasolov, V. V. (Viktor Vasil'evich) [Elementy teorii gomologii. English] Elements of homology theory / V. V. Prasolov.

p. cm. - (Graduate studies in mathematics; v. 81) Includes bibliographical references and index. ISBN-13: 978-0-8218-3812-9 (alk. paper) ISBN-I0: 0-8218-3812-1 (alk. paper) 1. Homology theory. I. Title.

QA612.3.P73 2007 514'.23---dc22 2006047074

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Contents

Preface

Notation

Chapter 1. Simplicial Homology 1. Definition and Some Properties 2. Invariance of Homology 3. Relative Homology 4. Cohomology and Universal Coefficient Theorem 5. Calculations 6. The Euler Characteristic and the Lefschetz Theorem

Chapter 2. Cohomology Rings 1. Multiplication in Cohomology 2. Homology and Cohomology of Manifolds 3. The Kiinneth Theorem

Chapter 3. Applications of Simplicial Homology 1. Homology and Homotopy 2. Characteristic Classes 3. Group Actions 4. Steenrod Squares

Chapter 4. Singular Homology 1. Basic Definitions and Properties

vii

ix

1 1 6

12 21 35 51

59 59 69 95

111 111 131 173 184

195 195

-v

vi Contents

2. The Poincare and Lefschetz Isomorphisms for Topological Manifolds 227

3. Characteristic Classes: Continuation 252 Chapter 5. Cech Cohomology and de Rham Cohomology 263

1. Sheaf Cohomology 263 2. De Rham Cohomology 275 3. The de Rham Theorem 289

Chapter 6. Miscellany 301 1. The Alexander Polynomial 301 2. The Arf Invariant 317 3. Embeddings and Immersions 325 4. Complex Manifolds 339 5. Lie Groups and H-Spaces 344

Hints and Solutions 365

Bibliography 403

Index 411

Preface

This book is a natural continuation of the author's earlier book Elements of Combinatorial and Differential Topology (American Mathematical Society, Providence, Rl, 2006), which we refer to as Part I here. (A corrected Russian version of Part I is available at http://www.mccme.ru/prasolov.)

In Chapter 1, we define simplicial homology and cohomology and give many examples of their calculations and applications. At this point, the book diverges from standard modern courses in algebraic topology, which usually begin with introducing singular homology. Simplicial homology has a simpler and more natural definition. Moreover, it is simplicial homology that is usually involved in calculations. For this reason, we introduce sin-gular homology near the end of the book and use it only when it is indeed necessary, mainly in studying topological manifolds.

Homology and cohomology groups with arbitrary coefficients ale expres-sed in terms of integral homology by means of the functors Tor and Ext. The properties of these functors are very important for homology theory, so we discuss them in detail.

We first prove the Poincare duality theorem for simplicial (co ) homology. This proof applies only to smooth (to be more precise, triangulable) mani-folds. There is no triangularization theorem for topological manifolds, and the proof ofthe Poincare duality theorem for them uses, of necessity, singular (co) homology. This proof is given in Chapter 4; it is very cumbersome.

Chapter 2 considers an important algebraic structure on cohomology, the cup product of Kolmogorov and Alexander. It is particularly useful in the case of manifolds. Multiplication in cohomology is related to many topo-logical invariants of manifolds, such as the intersection form and signature.

-vii

viii Preface

One possible approach to constructing multiplication in cohomology is based on a theorem of Kiinneth, which expresses the (co)homology of X x Y in terms of those of X and Y and is of independent interest.

Chapter 3 is devoted to various applications of (simplicial) homology and cohomology. Many of them are related to obstruction theory. One of such applications is the construction of the characteristic classes of vector bundles. Other approaches to constructing characteristic classes (namely, the universal bundle and axiomatic approaches) are also discussed. Then, we consider the (co ) homological properties of spaces with actions of groups; we construct transfers and Smith's exact sequences. We conclude the chap-ter with constructing Steenrod squares, which generalize multiplication in cohomology.

In Chapter 4, we define singular (co)homology and describe some of its applications; in particular, we obtain certain properties of characteristic classes. (Technically, it is more convenient to prove them by using singular cohomology, although the assertions themselves can be stated for simplicial cohomology. )

Chapter 5 considers yet another approach to constructing cohomology theory, namely, Cech cohomology and de Rham cohomology, which are closely related to each other. For the de Rham cohomology, we prove the Poincare duality theorem. Then, we carryover the construction of de Rham, which was originally introduced for smooth manifolds, to arbitrary simplicial complexes.

The final Chapter 6 is devoted to various applications of homology the-ory, largely to the topology of manifolds. We begin with a detailed account of the Alexander polynomials, which we construct by using the homology of cyclic coverings; the Arf invariant is also considered. Then, we prove the strong Whitney embedding theorem. We also give a formula for calculating the Chern classes of complete intersections and discuss some homological properties of Lie groups and H -spaces.

The book contains many problems (with solutions) and exercises. The problems are based on the materials of topology seminars for second-year students held by the author at the Independent University of Moscow in 2003.

The basic notation, as well as theorems and other assertions, of Part I are mostly used without explanations; in some cases, we give references to the corresponding places in Part I.

This work was financially supported by the Russian Foundation for Basic Research (project nos. 05-01-01012a, 05-01-02805-NTsNIL_a, and 05-01-02806-NTsNIL....a) .

Notation

Hk(Xj G), the k-dimensional homology group of X with coefficients in Gj Hk(Xj G), the k-dimensional cohomology group of X with coefficients in Gj Hom(A, B), the group of homomorphisms A --+ Bj A B, the tensor product of the Abelian groups A and Bj Tor(A, B), see p. 29j Ext (A, B), see p. 29j Coker a, the cokernel of the homomorphism a (see p. 15)j [Mn], the fundamental class of the manifold Mnj X(X), the Euler characteristic of Xj AU), the Lefschetz number of the map Ij u(M4n), the signature of the manifold M4nj Ek, the k-dimensional trivial vector bundlej Wk({), the kth Stiefel Whitney class of the bundle {j Ck({), the kth Chern class of the bundle {j Pk({), the kth Pontryagin class of the bundle {j Sqi, the Steenrod square.

-ix

Chapter 1

Simplicial Homology

1. Definition and Some Properties

The homology groups of a topological space X can be defined in several ways; different definitions are equivalent only for sufficiently good l spaces. The simplest definition is that of simplicial homology. Unfortunately, it has the essential drawback of not being invariant; to be more precise, the proof of its invariance requires some effort. (By invariance here we mean that the homology groups of homeomorphic spaces are isomorphic.) However, the main ideas of homology theory are most transparent at the level of simplicial homology; for this reason, we begin with a detailed discussion of simplicial homology.

1.1. Definition of Homology Groups. Let K be a simplicial complex (see Part I, p. 99). We assume that all of its simplices are oriented, i.e., the vertices of each simplex are ordered (the orientations of different simplices are not assumed to be compatible). We denote a simplex with vertices ao, al, ... , an (in this order) by lao, al,"" an]. Two simplices lao, al, ... , an] and [au(O) , au(l),"" au(n)] are said to have the same orientation ifsignu = 1; if sign u = -1, then these simplices have opposite orientations.

We define the boundary of a simplex as

8[0,1, ... ,n] = L(-I)i[O, ... ,i, ... ,n], where [0, ... , i, ... , n] = [0, ... , i-I, i + 1, ... , n]. At this point, we begin to consider formal sums of simplices. To be more precise, we declare simplices with the same vertices and orientations to be equivalent and consider the

1 Most of the spaces that arise in topology from geometry are sufficiently good, as opposed to those arising from functional analysis.

-1

2 1. Simplicial Homology

set of equivalence classes of simplices. Moreover, we assume that the sum of two simplices with the same vertices and opposite orientations is o. Thus, we can denote simplices with opposite orientations by ~ and -~.

It is convenient to denote the simplex [a, bJ by an arrow from its starting vertex a to the end vertex b. In this notation, the boundary of the simplex [0,1, 2Jlooks as shown in Figure 1; this fully agrees with thE' intuitive idea of a boundary.

2

Figure 1. The boundary of the simplex [0,1,2]

The following assertion is very important for homology theory.

Theorem 1.1. For any simplex~, OO~ = o.

Proof. Let i > j. The simplex [ ... ,], ... , i, ... J occurs in the expression for 00[0, . .. , nJ two times: in (-I)io [ ... , i, ... J with the sign (_I)i+i and in (-I)io[ ... ,J, ... J with the opposite sign (_I)i+i-l. 0

We could consider not only linear combinations of simplices with integer coefficients but also finite2 sums of the form E ai~~' where each ai is an element of some Abelian group G and each ~~ is a simplex of dimension k. The expression E ai~~ is called an n-chain (with coefficients in the group G). Chains can be added to each other; therefore, they form an Abelian group. The group of k-chains is denoted by Ck(K; G). For brevity, we often denote this group by Ck(K) or simply Ck.

We have defined the map 0 for simplices. Extending it by linearity, we obtain a map Ok: Ck --+ Ck-l, which we call the boundary homomorphism. For the O-simplex ~o, we set 8o~o = o.

A chain c E Ck is called a cycle if OkC = 0, i.e., c E Ker Ok. The group of k-dimensional cycles is denoted by Zk. A chain C E Ck is said to be a boundary if c = Ok+Ic' for some chain c' E CH1, i.e., C E Imok+I. The group of k-dimensional boundaries is denoted by Bk. It follows from 00 = a that Bk c Zk; therefore, we can consider the quotient group Hk(K) = Zk/ B k. its elements are equivalence classes of cycles: cycles are equivalent if their

2For infinite sums, we could not define the homomorphism a in the case wn re one simplex is a face for infinitely many simplices.

1. Definition and Some Properties 3

difference is a boundary; such cycles are said to be homologous. The group Hk(K) is called the k-dimensional simplicial homology group of the complex K. To indicate the group of coefficients, we use the notation Hk(K; G). Remark. For the first time, "homological numbers" were mentioned in the works of Riemann (1857) and Betti (1871). The correct definition was given by Poincare. He realized that, in defining chains, not only the geo-metric shapes of simplices but also their multiplicities should be taken into account. Initially, homology groups (with coefficients in Z) were described in terms of their ranks (Betti numbers) and torsion numbers. In the 1926 short paper [97], Emmy Noether made the important observation that the equivalence classes of cycles can be regarded as groups.

It can be seen from the definition that if a simplicial complex K contains no k-simplices, then Hk(K) = a.

Let us calculate homology groups for some simplicial complexes.

Example 1. If K consists of n isolated points, then Ho(K; G) = ~ and Hk(K; G) = a for k ~ 1.

n

Proof. Clearly, in the situation under consideration, we have Co = G EEl ... EEl G, Ck = a for k ~ 1, Zo = Co, and Bo = a. 0

Example 2. Let Sl be the simplicial complex that is the boundary of the 2-simplex [0,1,2]. Then HO(Sl; G) = HI (Sl; G) = G. Proof. Clearly, B1 = 0, whence HI = Zl. Consider an arbitrary I-chain

c = ao[l, 2] + ad2, 0] + a2[a, 1] E C l , where ao, aI, a2 E G. We have

oc = (al - a2)[a] + (a2 - ao)[l] + (ao - al)[2]. Therefore, Zl consists of chains of the form a[l, 2] + a[2, 0] + ala, 1], where a E G.

The equalities o(a[l, 2] + a[2, 0]) = a[a] - a[l] and o(a[2, 1] + a[l, 0]) = a[a]- a[2] show that any a-chain has the form a[a] (up to a boundary). On the other hand, if

a[a] = oc = (al - a2)[a] + (a2 - ao)[l] + (ao - at}[2], then a2 = ao and ao = a1; hence a = al - a2 = a. Thus, the group HO(Sl; G) = Col Eo is isomorphic to G. 0

The argument which we used to calculate HO(Sl) applies to any con-nected simplicial complex.


Theorem 1.2. If K is a connected simplicial complex, then Ho(K; G) = G. Proof. Arbitrary vertices [m] and [n] can be joined by I-simplices [m, ill, [iI, i2], ... , [ik' n]; therefore,

a[n] - arm] = 8(a[m, ill + ali!, i2] + ... + a[ik' n]). This means that, up to a boundary, any O-chain has the form arm], where [m] is a fixed vertex. It remains to verify that if the chain arm] is a boundary, then a = O.

Suppose that

arm] = 8(Laa[ia,ja]) = Laa[ja]- Laa[ia]. a a a

The sum of coefficients on the right-hand side vanishes. Therefore, a = O. [] Exercise. Suppose that S2 is represented as the boundary of the simplex [0,1,2,3]. Prove that Ker 82 consists of chains of the form a8[D, 1,2,3] and Ker8l = Im~.

1.2. Chain Complexes. A chain complex is defined as a family of Abelian groups Ck and homomorphisms 8k: Ck --+ Ck-l satisfying the relations 8k8k+1 = O. If Ck = a for k < 0, then the chain complex is said to be nonnegative.

An Abelian group F is free if it contains a set of elements {fa} such that any f E F has a unique representation in the form f = naJal + ... +na/cfa/c , where na E Z and all fal" .. , falc are distinct. The set {fa} is called a basis of the group F. A free Abelian group can be equivalently defined as a (finite or infinite) direct sum of copies of Z. If all of the groups Ck are free, then the chain complex is said to be free.

For any chain complex C .. , we can consider the homology groups Hk(C,,) = Ker8k/Im8k+1'

A chain map of chain complexes C .. and C~ is a family of homomorphisms CPk: Ck --+ C~ satisfying the condition ~CPk = CPk-18k for each k. Each chain map CPk: Ck --+ C~ induces a family of homomorphisms cp .. : Hk(C,,) --+

Hk(C~) for which (cp'I/J) .. = CP .. 'I/J ... Every simplicial map f: K --+ L induces the map A: Ck(K) --+ Ck(L)

defined by

I ([ ]) = {[f(ao), . .. , f(ak)] Jk ao,,ak a if f(ad -=I !(aj) for i -=I j, if !(ai) = !(aj) for some i -=I j.

If the dimension of the simplex with vertices !(ao), ... , !(ak) is equal to k or less than k - 1, then, obviously, 8~A = A-18k, i.e., the map IS chain (in

1. Definition and Some Properties 5

the latter case, both sides of this equality vanish). Suppose that I(ao) = I(at) and the points I(at), . .. , I(ak) are different. In this case, we have !k([ao, ... , ak]) = 0, and !k-I(8[ao, ... , ak]) = [/(at), !(a2), ... , I(ak)]-[/(ao), l(a2), ... , I(ak)] = 0, because I(ao) = I(at).

Thus, any simplicial map I: K -+ L induces a homomorphism I.: Hk(K)[O] -+ Hk(L)j the identity map induces the identity homomorphism, and (lg). = I.g.

If I, g: K -+ L are simplicial maps and there exists a family of homo-morphisms Dk : Ck(K) -+ CkH(L) satisfying the conditions

8kH Dk + Dk-18k = gk - A, then such a family D is called a chain homotopy between I and g.

The notion of chain homotopy has a geometric origin. Namely, suppose that I and 9 are maps joined by a homotopy H, and these maps (including the homotopy) are simplicial. Then the homotopy H determines the chain homotopy that assigns the (k + I)-chain H(/:l.k) to each k-simplex /:l.k. This chain is the curvilinear prism swept out by the image of the simplex during the homotopy. The boundary of the prism consists of the bases I(/:l.k) and g(/:l.k) and the lateral surface H(8/:l.k). Taking orientations into account, we obtain precisely the expression for the chain homotopy:

8H(/:l.k) - H(8/:l.k) = g(/:l.k) - I(/:l.k). Theorem 1.3 (on chain homotopy). II simplicial maps I, g: K -+ L are chain homotopic, then I. = g .

Proof. Suppose that Zk E Ck(K) and 8kzk = O. Then gk(Zk) - Ik(Zk) = 8kHDkZk + Dk- 18kZk = 8kH(Dkzk),

Le., the cycles gk(Zk) and A(Zk) are homologous. 1.3. Homology of Simplices and of Their Boundaries.

Theorem 1.4. II k ~ 1, then Hk(/:l.n) = o.

o

Proof. Note that 6,n = [b, aI, . .. , an] is a cone (over /:l.n-l = [al, ... , an]). To each simplex [ail' ... ,ai",] we assign the simplex [b, ail' ... ,ai",]. Extend-ing this map by linearity, we get a homomorphism Ck_l(/:l.n-l) -+ Ck(6,n); we denote the image of Ck-l by [b, Ck-l]. It is easy to verify that 8[b, Ck] = Ck - [b, 8ck] for k ~ 1 and 8[b, eo] = eo - E(eo)b, where E(E npaip) = E np.

Any chain Zk E Ck(6,n) has a unique representation in the form Zk = Ck + [b,dk-l], where Ck E Ck(6,n-l) and dk- 1 E Ck_l(/:l.n-l). Suppose that


8Zk = O. Then 8Ck+dk-l-[b,8dk-l] = 0 for k > 1 and act +cio-c:(do)b = 0 for k = 1. In both cases, 8Ck + dk-l = 0; therefore,

alb, Ck] = Ck - [b,8ck] = Ck + [b, dk-l] = Zk, i.e., any cycle Zk E Ck(.6.n ) with k ~ 1 is a boundary. o Corollary. Let a.6.n be the simplicial complex consisting of all simplices in .6.n except.6.n itself. Then H n _ 1 (8.6.n ) = G (ifn ~ 2) and Hk(8.6.n ) = 0 for 0 < k < n - 1. Proof. Up to the dimension n - 1, the homology groups of a.6.n coincide with those of .6.n . The complex 8.6.n contains no n-simplices; therefore, lman = 0, which means that Hn _ 1 (8.6.n ) = Ker8n - 1 . But Hn(.6.n ) = 0; hence Ker 8n - 1 = lm~, where ~ is the differential in the chain complex for .6. n. If.6. n = [0, 1, ... ,n], then 1m ~ consists of elements of the form

a(l:~o(-I)i[o, ... ,i, ... ,n]), where a E G. 0 2. Invariance of Homology

First, we prove a theorem about acyclic supports, which allows us to prove the chain homotopy of two chain maps in many cases. Then, we twice apply this theorem in different situations to prove the topological invariance of homology. Finally, using the same theorem, we prove the homotopy invariance of homology.

We can define homology with coefficients in any Abelian group G. But the case most important for applications is the one in which G is the additive group of some commutative ring with identity (e.g., G = Z, Q, or Zp); on the other hand, some important facts about homology (and especially cohomology) groups can be proved only for such groups of coefficients. For this reason, in what follows, we usually assume the group of coefficients to be the additive group of some commutative ring with identity.

2.1. Acyclic Supports. A simplicial complex is said to be acyclic if its homology coincides with that of the singleton. For example, the cone over any simplicial complex is acyclic; this is proved in precisely the same way as the acyclicity of simplices (see Theorem 1.4 on p. 5; the proof of this theorem uses only the representation of a simplex as a cone).

Suppose we have a chain Ck = l:ak.6.f E Ck(L). We refer to any sub complex L' c L containing all the simplices .6.~ as a support of Ck.

We say that a chain map cP is augmentation-preserving3 if CPO (2: ai.6.?) = l: bj.6.~, where l: ai = 2: bj .

3The origin of this term is explained on p. 17.

2. Invariance of Homology 7

Theorem 1.5 (on acyclic supports). Let 'Pk, 1/;k: Ck(K) -+ Ck(L) be aug-mentation-preserving chain maps. Suppose that to any simplex ~ C K there corresponds a complex L(~) c L so that the following conditions are satisfied:

(1) if~' c ~, then L(~') c L(~)j (2) the complex L(~) is acyclicj (3) the complex L(~k) is a support for both chains 'Pk(~k) and 1/;k(~k).

Suppose also that the coefficient group G is the additive group of a ring with identity. Then the maps 'Pk and 1/;k are chain homotopicj in particular, 'P. = 1/; .

Proof. We construct a chain homotopy Dk by induction on k. First, sup-pose that k = O. Let ~o be a vertex of K. The complex L(~O) is a support for the chains 'Po(~O) and 1/;0(~0). Since the maps 'PO and 1/;0 are augmentation-preserving, it follows that ('Po -1/;o)(a~O) = Ebi~?' where E bi = O. In the acyclic complex L(~O), the chain E bi~?' where E bi = 0, is the boundary of some chainj therefore, ('Po - 1/;o)(a~O) = 81Do(a~0), where Do(a~O) is a I-chain for which L(~O) is a support. The equality

Do(a~O) + Do(b~O) = Do(a + b)~O) may be false. To make it true, we choose an identity element 1 in the ring G, fix a I-chain Do(1 ~O), and set

Do(a~O) = aDo(I ~O). In what follows, we denote chains of the form 1 ~ simply by ~ (note that this notation makes no sense if G is an arbitrary group of coefficients).

Suppose that the required homomorphisms Do, ... , Dk-l are already constructed and each L(~i) is a support for the chain Di(~i). The only requirement to the homomorphism Dk is that 8k+lDk(~) = Ck for any k-simplex ~ in K, where Ck = 'Pk(~) -1/;k(~) - Dk-18k(~)' All simplices

8k(~) are contained in ~j hence L(~) is a support for the chain 8k(~), and therefore L(~) is a support for the chain Dk-18k(~)' Thus, L(~) is a support for Ck, and

8kCk = (8kk - 8k'Pk - 8kDk--18k)(~) = (8kk - 8k'Pk - (1/;k-18k - 'Pk-18k - Dk-28k-18k))(~) = O.

We have assumed that k ~ 1. The acyclicity of the complex L(~) implies Hk(L(~)) = O. Hence there exists a chain Dk(~) for which L(~) is a support and 8k+l D k(6.) = Ck. 0

2.2. Topological Invariance of Homology. In this section, we prove that any continuous map f: IKI -+ ILl of simplicial complexes induces a homomorphism f.: Hk(K) -+ Hk(L). Moreover, (lg). = f.g., which


implies the isomorphism of the homology groups of homeomorphic simpli-cial complexes. To define f*, we use the simplicial approximation theorem (see Part I, p. 105). Theorem 1.6. Suppose that the coefficient group G is the additive group of a ring with identity. Let K' be the barycentric subdivision of a simplicial complex K. Then the homology groups Hk(K') and Hk(K) are isomorphic.

Proof. First, consider the simplicial map j: K' - K defined as follows. If K = ~k = [0,1, ... , kJ, then any simplicial map K' - K can be defined by assigning the vertices (labels) 0,1, ... , k to the vertices of the complex K'. To each vertex of K' which is the barycenter of some simplex [io, ... , i p ] we assign one of the vertices io, ... , i p An example of such an assignment is given in Figure 2. It is easy to show that precisely one k-simplex of K' has complete set of labels, and the orientation of this simplex coincides with that of K. Indeed, take the barycenter of K and consider the face whose vertices have labels different from those of the barycenter; take the barycenter of this face, and so on.

Figure 2. A set of labels

For general simplicial complexes, the map j: K' - K is defined simi-larly. Note that it is a simplicial approximation of the identity map. The map j induces a chain map jk: Ck(K') - Ck(K).

Now, let us define a chain map ik: Ck(K) - Ck(K'); it is not induced by a simplicial map. We set io(v) = v and idvo, VI] = [b, VI]- [b, vol, where b is the barycenter of the simplex [vo, VI]. Formally, the definition of i 1 can be written as il(~I) = [b,ioo~I]. For k > 1, we set ik(~k) = [b,ik_IO~k], where b is the barycenter of the complex ~k. It remains to verify that ik is a chain map, i.e., Okik = ik-IOk. Clearly,

okik(~k) = 0k[b, ik_lOk~k] = ik_lOk(~k) - [b, Ok_Iik_IOk~k]. Therefore, if Ok-lik-I = ik-20k-l, then Ok-lik-lOk = ik-20k-lfh = and Okik = ik-lOk

The map j takes exactly one k-simplex from the baryc{; tric subdivision of ~k to ~k and preserves its orientation; the remaining k simplices are


mapped to simplices of dimension less than k. This implies that ikik is the identity map, and for a support of the chain ikik(fl.'), where fl.' is any sim-plex from K', we can take the barycentric subdivision of the simplex A from K that contains A'. The barycentric subdivision of A is the cone over 8Aj therefore, the barycentric subdivision of a simplex is an acyclic simplicial complex. Thus, the maps ikik and idck(K') have a common acyclic support. Clearly, the map ioio takes vertices to verticesj hence, it is augmentation-preserving. Therefore, ikik is chain homotopic to the identity.

Thus, the induced homomorphisms i*i* and i*i* are the identity maps, and hence the groups Hk(K') and Hk(K) are isomorphic. 0 Remark 1. The map i*: Hk(K) - Hk(K') has a simple geometric meaning at the level of chains; namely, to every simplex it assigns the sum of simplices into which it decomposes under the barycentric subdivision. The inverse map i*: Hk(K') - Hk(K) is canonical4 only at the level of homology. Remark 2. It is easy to prove Theorem 1.6 without appealing to acyclic supports. Indeed, it suffices to verify that i*i* is the identity map. But all simplices of the barycentric subdivision of A must have the same coefficients in any cyclej therefore, the restriction of ikik to Zk is the identity.

Let I: / K/ - / L / be a continuous map of simplicial complexes. We define a homomorphism 1*: Hk(K) - Hk(L) as follows. Take the nth barycentric subdivision K(n) and consider a simplicial approximation 'P: K(n) - L of I. We set 1* = 'P*iin), where iin ): Hk(K) - Hk(K(n)) is the canonical isomorphism. We must verify that this map is well defined, i.e., if tP: K(m) -L is another simplicial approximation of I, then 'P*iin) = tP*iim ). First, suppose that m = n.

Theorem 1.7. II 'P, tP: K - L are simplicial approximations 01 a map I: /K/-/L/, then 'P* = tP* Proof. We use only the following property of the simplicial maps 'P and tP Let ao, .. , ak be the vertices 01 some simplex in K. Then the points

4We say that a map is canonical if it is uniquely determined by construction or for some other reasons. For example, any group isomorphism Z:I -+ Z2 is canonical. Any ring isomorphism Z -+ Z is canonical also, but a group isomorphism Z -+ Z is canonical only if it is assumed to take 1 to 1 rather than to -l.

In linear algebra, the most important example of a noncanonical isomorphism is the isomor-phism between a linear space V and its dual space V. This isomorphism depends on the choice of a basis in V.

In topology, noncanonical isomorphisms often arise between bundles because the fibers of bundles are homeomorphic but the homeomorphisms are not canonical; the canonicity of the homeomorphisms between fibers means the triviality of the bundle. Therefore, the homotopy and homology groups of different fibers are isomorphic, but the isomorphisms are not always canonical because the products of the base and the fibers may be twisted.


cp(ao), . .. , cp(ak), .,p(ao), ... , .,p(ak) are the vertices of a simple:/J in L. To prove this assertion, take a point x E int[ao, ... , ak]. The point f(x) is interior for a unique simplex [bo, . .. ,bzl in L. According to the definition of a simplicial approximation, we have cp(x), .,p(x) E [bo, ... , bzl. Therefore, since the maps cp and .,p are simplicial, it follows that the simplices cp([ao, ... , ak]) and .,p([ao, .. . , ak]) belong to [bo, . .. ,bd. The maps cpo,.,po: Co(K) -+ Co(L) are augmentation-preserving. According to the acyclic support theorem, the maps CPk and .,pk are chain homotopic. D

Now, let us prove the equality cp.i~n) = .,p.i~rn) for n =1= m. For definite-ness, suppose that n > m. We can take K(rn) (instead of K) for the initial complex and consider its pth barycentric subdivision for p = n - m. Then the required equality takes the form cp.i~P) = .,p .

In the proof of Theorem 1.6, we constructed a map j: K(l) -+ K, which is a simplicial approximation of the identity map. Recall that a composition of simplicial approximations is a simplicial approximation of a composition (see Part I, the corollary of Theorem 3.17 on p. 104). Therefore, the identity map has a simplicial approximation jCP): K(p) -+ K; the maps j't) and i~P) are mutually inverse. Since .,pj(p) is a simplicial approximation of I, it follows that cp. = .,p.j(p), i.e., cp.i(p) = .,p . Theorem 1.8. If I: IKI -+ ILl and g: ILl -+ IMI are continuous maps, then (gf). = g.I. Proof. First, we construct a simplicial approximation .,p: L' -+ M of g, and then, a simplicial approximation cp: K' -+ L' of I. Let jK: K' -+ K and jL: L' -+ L be simplicial approximations of the identity maps. These maps form the commutative diagram

IKI ~ ILl ~ IMI rjK rjL

IK'I ~ IL'I ~ IMI Since .,pcP is a simplicial approximation of gl, we have

(gf). = (.,pcp).(j~)-l = .,p.CP.(j~)-l. The maps jLcp and .,p are simplicial approximations of f and g, resppctively; hence I. = jfcp.(j!


Proof. If f: IKI -+ ILl and g: ILl -+ IMI are mutually inverse continuous maps, then f. and g. are mutually inverse homomorphisms of homology groups. 0

As an application of the topological invariance of homology, we give yet another proof of the nonexistence of a retraction of the ball onto its boundary.

Theorem 1.9. There exists no continuous map r: nn -+ sn-l = ann such that rex) = x for all x E sn-l. Proof. Suppose that r: nn -+ sn-l is a continuous map for which rlsn 1 = idsn-l. Consider the inclusion i: sn-l ~ nn. The composition ri is the identity mapj therefore, r.i.: Hn - 1 (sn-l j Z) -+ H n - l (sn-lj Z) is the iden-tity map Z -+ Z (or Z ffi Z -+ Z ffi Z, if n = 1). On the other hand, the image of i.: Hn-l (sn-lj Z) -+ H n- l (nnj Z) is zero (or contained in Z, if n = 1) because Hn_1(nn j Z) = 0 for n ~ 2 and Ho(nl; Z) = Z. 0 2.3. Homotopy Invariance of Homology.

Theorem 1.10. Homotopic maps of simplicial complexes induce the same map on homology.

Proof. Let /0, II: IKI -+ ILl be homotopic simplicial maps of simplicial complexes K and L. This means that there is a continuous map F: IKI x I -+ ILl for which FIIKlx{o} = fo and FIIKlx{l} = II. The Cartesian product IKI x I can be turned into a simplicial complexj moreover, we can assume that IKI x {O} and IKI x {I} are endowed with the same simplicial structure and for any simplex Ll in K, the set Ll x I is a sub complex in IKI x I (see Part I, p. 130). Consider the inclusions io: K -+ IKI x {O} ~ IK1 x I and i 1 : K -+ IKI x {1} ~ IKI x I. Clearly, fo = Fio and h = Fil. Therefore, it is sufficient to verify that io. = it. (we use the equalities fo. = F.io. and h .. = F.it.).

Let Llk be a simplex in K. The chains io(Llk) and il(Llk) have the common support Llk x I. This support is acyclic because it is homeomorphic to nk+l R:: Llk+l. 0

Corollary. If simplicial complexes K and L are homotopy equivalent, then their homology groups are isomorphic.

As an application of the homotopy invariance of homology, we give yet another proof of dimension invariance (one proof was given in Part I on pp.60-61). Theorem 1.11. If n i- m, then the spaces lRn and]Rm are not homeomor-phic.


Proof. For definiteness, suppose that n > m ~ 1; in particular, n ~ 2. If the spaces lRn and lRm are homeomorphic, then the spaces lRn \ {O} '" sn-l and lRm \ {O} '" sm-l are homeomorphic as well; hence they are homotopy equivalent. Thus, Z = Hn_l(sn-l; Z) = Hn_l(sm-l; Z) and n - 1 > 0; therefore, m = n. This contradiction completes the proof. 0

3. Relative Homology

Let K be a simplicial complex and L c K a sub complex. Then Ck(L) c Ck(K), and we can consider the quotient group Ck(K, L) = Ck(K)/Ck(L); the elements of this group are called relative k-chains. The homomor-phism 8k: Ck(K) -+ Ck-1(K) induces a homomorphism 8k : Ck(K,L) -+ Ck-l(K, L) with the same property 88 = o. Therefore, we can again take the groups Zk(K,L) = Ker8k and Bk(K,L) = Im8k+l and consider the quotient group Hk(K, L) = Zk/ B k, which is called the k-dimensional rela-tive homology group.

Example 3. If each path-connected component of a complex K contains at least one point of a sub complex L, then Ho(K, L) = o. Proof. Let us join a vertex x E K with some vertex y E L by a polygonal line 0: formed by edges. Then 810: = [x] - [y] '" [x]; therefore, Bo ::J Zoo D Exercise. Prove that Ho(K, L; G) = G EB EB G, where n is the number

--------n

of path-connected components of K containing no points of L.

Problem 1 (excision isomorphism). Suppose that K U M is a simplicial complex, K and M are its sub complexes, and L is a sub complex in K. Prove that Hk(K, L) ~ Hk(K U M, L U M) for all k.

A map of simplicial pairs J: (K,L) -+ (K',L') induces the following homomorphism J.: H.(K, L) -+ H.(K', L') of relative homology groups. First, we take the map Ck(K)/Ck(L) -+ Ck(K')/ JCk(L), and then, using the inclusion JCk(L) c Ck(L'), apply the canonical projection Ck(K')/ JCk(L) -+ (Ck(K')/ JCk(L/(Ck(L')/ JCk(L ~ Ck(K')/Ck(L').

As a result, we obtain a chain map J#: C.(K, L) -+ C.(K', L'). It induces a homomorphism between the homology groups.

3.1. Exact Homology Sequences of Pairs. The inclusion i: L -+ K induces a homomorphism i.: Hk(L) -+ Hk(K). Moreover, any absolute cycle can be regarded as a relative cycle; thus, we have a homomorphism p.: Hk(K) -+ Hk(K, L). Let us construct a connecting homomorphism 8.: Hk(K, L) -+ Hk-l(L). Take a relative cycle Zk E Ck(K, L) and its absolute representative Zk; this means that Zk E Ck(K) and "'k = Zk+Ck(L).

3. Relative Homology 13

By assumption, (AZk = 0, i.e., OkZk E Ck-1(L). The map 0.: Hk(K, L) -+ H k-l (L) acting as Zk ........ OkZk is well defined because if Yk E Ck (L ), then Ok(Zk + Yk) = OkZk + OkYk rv OkZk' Theorem 1.12. The sequence of homomorphisms

... -- Hk(L) ~ Hk(K) ~ Hk(K,L) ~ Hk-l(L) -- ... is exact.

Proof. (1) 1m i. C Kerp . Any absolute cycle Zk E Ck(L) corresponds to the zero relative cycle.

(2) Kerp. C Imi . Suppose that an absolute cycle Zk E Ck(K) corre-sponds to a relative cycle homologous to zero. Then Zk = z~ + zZ, where

z~ E Ck(L) and zZ = OZk+1 for Zk+1 E Ck+l(K). Therefore, the cycle Zk is homologous to the cycle z~ E Ck (L ).

(3) Imp. C Kero . Suppose that Zk E Zk(K) and Zk = Zk + Ck(L). Then Zk ........ OkZk = O.

(4) Kero. C Imp . Suppose that Zk = Zk + Ck(L) and OkZk = O. Then for a representative of the relative cycle Zk we can take the absolute cycle Zk.

(5) Imo. C Keri . If Zk-l = OkZk E Ck-l(L), then the cycle Zk-l is homologous to zero in K.

(6) Keri. elmo . If Zk-l E Ck-1(L) and Zk-l = OkZk, where Zk E Ck(K), then Zk-l = O.(Zk + Ck(L)). 0

Relative homology groups can be expressed in terms of absolute groups; namely, for k ~ 2, we have

where C L is the cone over L. The first isomorphism is obvious even at the level of relative chains. The second follows from the exact sequence

Hk(CL) ~ Hk(K U CL) ~ Hk(K U CL, CL) ~ Hk_1(CL) of a pair. Since the cone CL is contractible, it follows that Hk(CL) Hk_l(CL) = 0; therefore, the middle map is an isomorphism. Note that the complex K U CL is homotopy equivalent to K/ L because the complex CL is contractible and, hence, K U L r-.J (K U CL)/CL ~ K/ L. Example 4. If n ~ 1, then

Hk(Dn, sn-l; G) = {oG for k = n, for k 1= n.


Proof. The exact sequence

Hk(Dn) ---+ Hk(Dn,sn-l) ---+ Hk_1(sn-1) ---+ Hk_l(Dn) shows that Hk(Dn,sn-l) ~ Hk_l(sn-l) for k ~ 2. For k = 0, the required assertion follows from Example 3. For k = 1, we have one of the exact sequences

and o ---+ HI (Dl, SO) ---+ G EB G ---+ G ---+ O.

Here the map G - G is an isomorphism and G EB G - G has the form (a, b) 1--+ a + b. The former has zero kernelj the kernel of the latter consists of elements of the form (a, -a) and, therefore, is isomorphic to G. 0 Remark. The difficulties involved in the consideration of the case k = 1 can be avoided by turning to reduced homology, which is defined on p. 17.

A generalization of exact homology sequences is the following algebraic construction. Suppose we have a short exact sequence of chain maps

O C ' i. C p. e" 0 ---+ ---+ ---+ .---+

(for example, C~ = Ck(L), Ck = Ck(K), and C; = Ck(K)/Ck(L)). Then we can define a connecting homomorphism a.: Hk(C:) - Hk-l(C~), Namely, take zZ E C; for which a;zZ = O. We have zZ = PkCk for some Ck E Ck, and o = azzz = Pk(akCk). Therefore, akCk = ik-ldk_1 for some dk_1 E C~_l' We set a.z; = dk - I It can be proved in precisely the same way as for exact sequences of pairs that this map is well defined at the homology level and the sequence

... ~ Hk(C~) ~ Hk(C.) ~ Hk(C:) ~ Hk-I(C~) ~ ... is exact.

Example 5. Let K be a simplicial complex, and let 0 - G' - G - Gil - 0 be a short exact sequence of Abelian groups. Then we have a short exact sequence of chain complexes

0---+ Ck(Kj G') ---+ Ck(Kj G) ---+ Ck(Kj Gil) ---+ O. The connecting homomorphism {3.: Hk(Kj Gil) - Hk-1(Kj G) is called the Bockstein homomorphism.

The most interesting examples of Bockstein homomorphisms can be obtained from the exact sequences


which determine homomorphisms Hk(Kj Zm) -+Hk-l(Kj Z) and Hk(Kj Zm) -+ Hk-I(KjZm).

The cokernel of a homomorphism a: A -+ A' is defined as Coker a = A'/Ima.

Problem 2. Given a commutative diagram of Abelian groups with exact rows

0----+ A --+ B --+ C --+ 0

la Ip l~ o ---+ A' ---+ B' ---+ C' ---+ 0,

prove that there is an exact sequence

o -+ Ker a -+ Ker {1 -+ Ker'Y -+ Coker a -+ Coker {1 -+ Coker'Y -+ O.

For a triple of simplicial complexes LI C L2 C K, we can construct an exact homology sequence of the triple

as follows. The isomorphism

determines a short exact sequence

This short exact sequence of chain complexes induces the required exact sequence of homology groups.

In dealing with exact sequences, the following assertion is often useful.

Theorem 1.13 (Steenrod's five lemma). Suppose that

is a commutative diagram of Abelian groups with exact rows. If 'PI, 'P2, 'P4, and 'P5 are isomorphisms, then 'Pa is an isomorphism.

Proof. Consider the groups A~ = A2/Imal, A~ = Kera4, B~ = B2/Im{11, and B~ = Ker {14. It is sufficient to prove the required assertion for the


simpler diagram

, 02 03, o ~ A2 ~ A3 ~ A4 ------+ 0

1 ~2 I 1 ~3, 1 ~4 , fJ2 fJ3 B' 0 0~B2~B3~ 4~ ,

in which cP~ and cP~ are isomorphisms. Clearly, Ker CP3 C Ker(.B~CP3) = Ker(cp~Q~) = Ker(Q~) = Im(Q~)j

therefore, Kercp3 ~ Ker(cp3Q~) = Ker(.B~cp~) = 0, i.e., CP3 is a monomor-phism. Moreover, Im(.B~CP3) = Im(cp~Q~), where Q~ is an epimorphism and cP~ is an isomorphism. Hence .B~CP3 is an epimorphism, and B3 = 1m CP3 + Ker .B~. Since

Ker.B~ = Im.B~ = Im(.B~cp~) = Im(cp3Q~) C Imcp3, it follows that B3 = 1m CP3, i.e., CP3 is an epimorphism. D

Remark. The five lemma remains valid if the diagram is commutative up to sign, i.e., CP2Ql = .BICPl, etc. Indeed, the proof involves only the groups Ker and 1m, which do not change under the replacement of cP by -cpo

Using the five lemma, we can easily prove the following theorem.

Theorem 1.14. Suppose that f: (K,L) -+ (K',L') is a map of pairs for which the induced maps K -+ K' and L -+ L' are homotopy equivalences. Then f.: Hk(K, L) -+ Hk(K', L') is an isomorphism. Proof. Consider the following commutative diagram with exact rows:

Hk(L) ~ Hk(K) ~ Hk(K, L) ~ Hk-l(L) ~ Hk-l(K) 'lisO . '1 iso '1? 'liSO . 'liSO

Hk(L') ~ Hk(K') ~ Hk(K', L') ~ Hk-l(L') ~ Hk-l(K'). The five lemma implies that the middle vertical arrow is an isomorphism. D

Problem 3. Suppose that 02 03 0~A2~A3~A4~0

1~2 fJ2 1~3 fJ3 1~4 0~B2~B3~B4~0

is a commutative diagram of Abelian groups with exact rows. (a) Prove that if CP2 and CP4 are monomorphisms, then C(3 is a monomor-

phism.


(b) Prove that if r.p2 and r.p4 are epimorphisms, then r.p3 is an epimorphism. Problem 4. Given a commutative diagram

with exact rows, prove that r.p2 induces an isomorphism

: Ker(r.p3Q2)/(KerQ2 + Kerr.p2) --+ (Imr.p2 nlmlh)/Im(r.p2Qd. Problem 5 ([83]). (a) Given two commutative diagrams satisfying the assumptions of Steenrod's five lemma in which the respective homomor-phisms, except for the isomorphisms r.p~ = TIl and r.p~ = T/2, coincide, prove that

for any x E A3 . (b) Suppose that the diagrams in (a) are diagrams of rings and their

homomorphisms. Prove that the isomorphisms TIl and Tl2 in these diagrams can be different if and only if there exists a nontrivial additive homomor-phism 6: A3 --+ A3 such that dQ2 = 0, Q3d = 0, and d(xy) = (dX)Y + xdy + (dX)(dy) for all x, y E A 3 .

(c) Give an example of two diagrams satisfying the assumptions of (a) for which the homomorphisms TIl and T/2 are different.

3.2. Reduced Homology. The statements and proofs of many theorems can be simplified by considering the reduced homology groups ih(K) instead of the homology groups Hk(K) themselvesj below, we give two equivalent definitions of reduced homology groups.

Definition 1. The reduced homology group Hk(K) is the kernel of the homo-morphism p*: Hk(K) --+ Hk(*), where p: K --+ * is the map from the com-plex K to the singleton.

Definition 2. Let us replace the map 80: CoCK) --+ 0 by : CoCK) --+ Z, where (v) = 1 for each vertex v. If Cl E CI(K), then alCI = OJ therefore, for the new chain complex, we can also define the reduced homology group Hk(K).

The map is called an augmentation. This term is used because we augment the chain complex


to the chain complex

... ---+ Cl(K) ~ Co(K) -=--. C-l(K) ---+ 0, where C_l(K) = Z.

Definitions 1 and 2 are equivalent because Hk(K) = Hk(K) for k ~ 1 and the map Po: Co (K) --+ Co (*) = Z coincides with c.

For an arbitrary map i: * --+ X, we have pi = id.; therefore,

Ho(K) = Imp. ffi Keri. = Ho(K) ffi Z. Thus, Ho(K) ~ Ho(K, *).

For reduced homology, exact sequences of pairs are also defined. Example 4 shows that Hk(Dn, sn-l) ~ Hk(sn) for all k. We have already shown that Hk(K, L) ~ Hk(K U CL) for k ~ 2. Using

reduced homology, we can also show that Hk(K, L) ~ Hk(KUCL) {or k = 0 and 1.

Exercise. Given a simplicial (possibly, disconnected) complex K embedded in R n , prove that Hi(Rn,K) ~ Hi l(K) for i ~ 2 and HI(Rn,K) ~ Ho(K). Moreover, a path between points of K represents the zero element of the group HI (Rn , K) if and only if these points belong to the same connected component of K.

3.3. The Mayer-Vietoris Sequence. The exact sequence of Mayer-Viet oris relates the homology groups of the union of two simplicial com-plexes Ko and K l , of their intersection, and of these complexes themselves. In the simplest case, where Ko and Kl consist of finitely many points, we can use the well-known inclusion-exclusion formula N = No + Nl - NOl, where N is the number of points in Ko U K l , No and Nl are the numbers of points in Ko and K 1 , respectively, and NOl is the number of points in Ko n K 1 . In the language of exact sequences, this formula can be written as

0---+ Ho(Ko n KI) ---+ Ho(Ko) ffi Ho(Kd ---+ Ho(Ko U KI) ---+ O. Another simple example of an exact sequence of this form arises in the

homology of the wedge K = Ko V Kl of two simplicial complexes. In this case, we have the isomorphism Ck(K) ~ Ck(Ko) ffi Ck(KI) for k ~ 1 already at the level of chains. Moreover, the kernel of the boundary homomorphism on Cl(K) is the direct sum of the kernels of the boundary homomorphisms on Cl(Ko) and C1(Kt). Therefore, for any k ~ 1, we have the canonical isomorphism Hk(K) ~ Hk(Ko) ffi Hk(KI).

These examples suggest the existence of an exact sequence

0- Hk(Ko n Kt) ---+ Hk(Ko) ffi Hk(Kd ---+ Hk(Ko U kl) ---+ o.


However, representing the circle S1 as the union of two arcs Ko and K1 with two-point intersection, we see that no such exact sequence exists for k = 1. In reality, the k-dimensional homology of a complex Ko U Kl depends not only on the k-dimensional homology ofthe complexes Ko n K1, K o, and K1 but also on their (k - 1 )-dimensional homology; the correct exact sequence is as follows.

Theorem 1.15 (Mayer Vietoris [84, 142]). Suppose that K is a simplicial complex, Ko and Kl are subcomplexes of K such K = Ko U K 1, and L = Ko n K 1 . Then there is an exact sequence

... --+ Hk(L) --+ Hk(Ko) ffi Hk(KJ) --+ Hk(K) --+ Hk-l(L) --+ .. , . Proof. The Mayer Vietoris sequence arises from the exact sequence

(1) 0 --+ C.(L) (jOl-jI), C.(Ko) ffi C.(KJ) ~ C.(K) --+ 0, which we describe below. The complex C.(Ko) ffi C.(Kd consists of the groups Ck(Ko) ffi Ck(K1 ), and the boundary homomorphism in it is the direct sum of the boundary homomorphisms; i.e., a( co, c1) = (acO, ac1 ). The maps jO/: L --+ KO/ and iO/: KO/ --+ K are the natural embeddings.

Let us show that (1) is exact. The map (jo, -jl) is, obviously, a mono-morphism. It follows from K = Ko U Kl that (io, i1) is an epimorphism. Indeed, take c = L ai~~ E Ck(K) and let cO be the sum of all terms for which ~f C Ko. Then Kl is a support of the chain c - co, and (io, i1)(CO, c - cO) = cO + (c - cO) = c.

It remains to verify exactness in the middle term. The image of the map (jo, -JI) consists of chains of the form (c, -c), where c is a chain in L. The kernel of (io, iJ) consists of chains of the form (c, -c), where c is a chain in KonKl =L.

Clearly, the homology of the chain complex C.(Ko) ffi C.(K1) is isomor-phic to H.(Ko) ffi H.(K1). 0

Mayer Vietoris exact sequences exist also for reduced homology. Indeed, setting (jo, -jl)(n) = (n, -n) and (io, iJ)(m, n) = m + n, we obtain the commutative diagram

with exact rows. Using the Mayer Vietoris exact sequence for reduced homology, we can

easily calculate the homology groups of a suspension (see Part I, p. 130).


Theorem 1.16 (the suspension isomorphism). Let K be a simplicial com-plex, and let EK be the suspension over K. Then, for any k ~ 1, there exists a canonical isomorphism Hk(EK) ~ Hk-l(K). Proof. We can represent EK in the form Ko U Kl, where Ko and Kl are cones over K and Ko n Kl = K. Let us write the Mayer Vietoris sequence for reduced homology:

Hk(Ko) ffi Hk(Kl ) --+ Hk(EK) --+ Hk-l(K) --+ Hk-l(Ko) ffi Hk-l(Kl ). The spaces Ko and Kl are contractible; therefore, the first and last terms in this sequence are trivial, and the middle map is an isomorphism. 0

Remark. We use reduced homology because for ordinary homology, the case k = 1 must be considered separately.

Problem 6. Calculate the homology of the torus T2.

Problem 7. Prove that Hk(SP x sq, SP V sq) ~ Hk(Sp+q) for all k ~ 1. Problem 8. Calculate the homology of the space SP x sq.

Problem 9. (a) Calculate the homology of the complement of a knot in S3 (the answer does not depend on the knot).

(b) Consider a link with n connected components in S3. Calculate the homology of the complement of this link (the answer depends only on n). Problem 10. Let M n be a smooth manifold. Prove that Hk(Mn\int Dn) ~ Hk(Mn ) for 1 ~ k ~ n - 2. (Here Dn is a ball contained in some chart of Mn.) Problem 11. Suppose that K is a simplicial complex and C = {L l , ... , Ln} is a family of its sub complexes such that they cover K and all their inter-sections are acyclic. Prove that the homology groups of the complex K are isomorphic to those of the nerve6 of the cover C.

The Mayer Vietoris theorem has the following generalization (concern-ing the relative M ayer- Vietoris sequence). Theorem 1.17. For any subcomplexes Lo C Ko and Ll C Kl of a simplicial complex K, there is an exact sequence

--+ Hk(Ko n Kl, Lo n Lt) --+ Hk(Ko, Lo) ffi Hk(Kl, L l ) --+ Hk(Ko U Kl, Lo U Lt) --+ Hk-l(Ko n Kl, Lo n Ll) --+ .

6The definition of the nerve of a cOVer is given in Part I on p. 108.

4. Cohomology and Universal Coefficient Theorem 21

Proof. The relative Mayer Vietoris sequence arises from the short exact sequence (2)

G.(Ko n KI) (jo,-il) G.(Ko) G.(KI) (io,h) G.(Ko U KI) 0 o ---+ --__ I $ -- ---+ G.(Lo n Ll) G.(Lo) G.(LI) G.(Lo U Ld

defined as follows. For each quadruple of Abelian groups HI C H2, G 1 C G2, where HI C Gl and H2 C G2 , the canonical map Gd HI -+ G2 / H2 of their quotients is defined. The maps io, il and io, il are such canonical maps.

If c E G.(Lo) n G.(Ld, then c E G.(Lo n Ld. Therefore, (jo, -it) is a monomorphism.

We show that (io, id is an epimorphism. Take c= E ai6.~ EG.(KoUKd and let cO be the sum of all terms for which 6.~ C Ko. Then the chain c - cO is contained in C.(Kl)' Therefore, the pair (cO, c - cO) can be associated with an element of the middle group in (2). The image of this element under the map (io, it) coincides with c (mod G.(Lo U Ll))'

It remains to verify the exactness in the middle term. The image of the map (jo, -il) consists of relative chains (c (mod G.(Lo)), -c (mod G.(Lt))), where c is a chain in KonKo. The kernel of (io, il) consists of relative chains (cO (mod G.(Lo)), c1 (mod G.(Ll))), where cO + c1 E G.(Lo U Ll)' Clearly, the image is contained in the kernel. Consider an element of the kernel. Let ifJ be the sum of all terms in the chain cO + c 1 that are contained in G. (Lo), and let c1 = - (cO + & + cl ). Then cl E G. (L1 ) j therefore, for c we can take cO + & = _ (c1 + (1 ). 0

4. Cohomology and Universal Coefficient Theorem

4.1. Cohomology. Suppose that K is a simplicial complex, G an Abelian group, and Gk(K) = Ck(Kj Z). A homomorphism ck: Ck(K) -+ G is called a k-dimensional cochain with coefficients (or values) in G. The group of k-dimensional cochains is denoted by Gk(Kj G) = Hom(Gk(K), G).

The text in small print below refers to infinite simplicial complexes. It can be skipped at the first reading.

For any family of Abelian groups {Go}, the direct sum ffio Go and the Carte-sian product ITo Go are defined. Both groups consist of indexed sets (go) under componentwise addition. The difference between them is that the group ffi", Go consists of the sets (go) in which go "10 only for finitely many indices a, whereas the group ITo Go consists of all such sets. For finite families of groups, Cartesian products coincide with direct sums.

Suppose that the k-simplices in a complex K are indexed by a. According to the definition of groups of chains, we have Ck(Kj G) = ffio Go, where Go ~ G. But Ck(Kj G) = ITo Go for Go ~ G. Indeed, a cochain is a function on a set of


simplices with values in G. To each simplex ~~ any value can be assigned; infinitely many values may be nonzero.

For the group of chains a direct sum rather than a Cartesian product is taken because boundary homomorphisms cannot be defined for Cartesian products. The reason for this is that the direct sums of free Abelian groups are free, while the Cartesian products are not.

Suppose that G is the additive group of a ring with identity. Then to each simplex ~~ we can assign the k-cochain (~~). which takes the value 1 at ~~ and vanishes at all other k-simplices. Therefore, any k-cochain can be written as a formal sum L:go~~)*, where go< E G. This sum can have infinitely many terms.

Take Ck E Ck(K) and ck E Ck(K; G). We denote the value of the homomorphism ck at the element Ck by (ck, Ck). Such a pairing allows us to associate with the boundary operator 0: Ck+1(K) --+ Ck(K) the dual operator c5: Ck (K; G) --+ Ck+ 1 (K; G) defined by

(c5ck,Ck+1) = (Ck,OCk+1)' Note that the operator c5 increases dimension, whereas 0 decreases it.

Exercise. Prove that the value of a cochain c5ck at the simplex [vo, . .. , Vk+1] equals

k+I

L(ck, [vo, ... , Vi,"" Vk+1])' i=O

Remark. Sometimes, the operator c5: Ck(K; G) --+ CHI (K; G) is defined by

(c5ck,Ck+1) = _(-l)k(ck,oCk+1)' This choice of sign is based on the following convention suggested by Mac-Lane: when the n- and m-dimensional symbols are transposed, the right-hand side must be multiplied by (-1 )mn; the maps 0 and c5 are then assigned the dimensions -1 and + 1. We do not adopt this convention for two reasons. First, we use the duality between maps 0 and c5 very often, and the intro-duction of an additional sign would make it somewhat awkward. Second, to determine how the sign changes under transpositions of symbols, nontrivial considerations are needed every time.

The equality 00 = 0 implies c5c5 = O. Therefore, the cohomology groups Hk(K; G) = Zk(K; G)j Bk(K; G) and HO(K; G) = ZO(K; G)

are defined; here Zk and Bk+1 are the kernel ann. the image of the homo-morphism c5: Ck --+ Ck+1, respectively. The elements of the groups Zk and Bk are called cocycles and co boundaries.

Example 6. If K is a connected simplicial complex, then HO(K; G) = G.


Proof. Letel = [VO,VIJ. Then8q = [VIJ-[VOJ. Hence (deO,el) = (eO,8el) = (cO, [VI]) - (cO, [vo]). Thus, the equality deo = 0 means that the co chain cO takes the same value at any two vertices joined by an edge. For a connected simplicial complex K, this means that the cochain cO takes the same value at all vertices. Therefore, ZO(Kj G) = G. D Theorem 1.18. If G is the additive group of a field F, then Hk(Kj G) is the dual space of Hk(Kj G).

Proof. The groups Ck (K j G) and Ck (K j G) are linear spaces over the field F, and C k is the dual space of Ck. The map 8 is linear, and the map d is its dual. Therefore, we must prove that if A: U - V and B: V - W are linear maps for which BA = 0, then Ker A* / 1m B* is the dual space of Ker B/lmA.

The linear functions on Ker B can be interpreted as linear functions on V considered up to functions of the form g(Bx)j the functions g(Bx) constitute the space 1m B*. The linear functions on Ker B / 1m A can be interpreted as the linear functions on Ker B satisfying the condition f(Ax) = 0 for all Xj the functions for which j(Ax) = 0 form the space Ker A*. D Corollary. If G is the additive group of a field F and one of the linear spaces Hk(K; G) and Hk(K; G) over F is finite-dimensional, then Hk(K; G) ~ Hk(K; G)i this isomorphism is not canonical.

Any simplicial map cp: K - L induces a homomorphism cp*: Hk(Lj G) - Hk(K; G) acting in the opposite direction. Indeed, each cochain ek E Ck(L; G) corresponds to the cochain cpkek E Ck(K; G) defined by cpkck(6.) = ek (cp(6.)).

The reduced cohomology groups are defined by replacing the augmenta-tione: Co(K) -Zbythedualmap: G-CO(K;G),where(co) =ge(CO).

The groups of relative cochains are defined as

Ck(K, Lj G) = Hom(Ck(K, L); G). Eaeh group of relative cochains is a subgroup in the group of absolute eochains. It consists of the eochains vanishing on Ck(L). Recall that the relative chain group is a quotient of the absolute chain group. Under duality, quotient groups correspond to subgroups.

In the relative case, the coboundary and boundary homomorphisms are dual also. The relative cohomology groups are defined in a natural way.

Exercise. Prove that if each connected component of IKI contains at least one connected component of ILl. then HO(K, L; G) = O.


The short exact sequence

(3) induces the dual sequence

(4) 0- Ck(LjG) L Ck(KjG) L Ck(K,LjG) - O. This sequence is exact as well, but proving this requires some effort. More-over, the proof of the exactness of the sequence (4) uses a special fea-ture of the sequence (3), namely, its split property. An exact sequence a -+ A :!..... B :!!... c -+ a is said to be split if it satisfies any of the three equivalent conditions in Lemma 1.1.

Lemma 1.1. Let a -+ A :!..... B :!!... c -+ a be an exact sequence. Then the following conditions are equivalent:

(5) (a) this sequence has the forrn1

a -----+ A .2...... A ffi C ~ C -----+ 0, where i is the natural embedding and p is the natural projectionj

(b) the homomorphism cp has a left inverse, i.e., there exists a homo-morphism 4>: B -+ A for which 4>cp = idA j

(c) the homomorphism,p has a right inverse, i.e., there exists a homo-morphism \lI: C -+ B for which ,p\ll = ide.

Proof. Clearly, conditions (b) and (c) hold for any sequence of the form (5)j to see this, it suffices to set 4>(a, e) = a and \lI(c) = (0, e). It remains to verify that if (b) or (c) holds, then the sequence has the form (5).

Suppose that 4>cp = idA. Let us show that B = Imcp ffi Ker4>. Any element b E B can be represented as b = cp4>(b) + (b-cp4>(b)), where cp4>(b) E 1m cp and b - cp4>(b) E Ker 4>. Moreover, if b E 1m cp n Ker 4>, then b = cp(a) and a = 4>(b) = 4>cp(a) = aj hence b = O.

Suppose that ,pw = ide. Let us show that B = Ker.,p ffi 1m \lI. Any element b E B can be represented as b = (b - \lI.,p{b)) + \lI,p{b), where b - w1j;(b) E Ker,p and W,p(b) Elm \lI. Moreover, if bE Ker,p n 1m W, then b = w(e) and a = ,pCb) = ,p\I!(e) = ej hence b = O. 0

7Formally, this means that there exists an isomorphism /: B --+ AEBe for which the follOwing diagram is commutative:


Problem 12. Prove Lemma 1.1 using the five lemma.

Problem 13. Let p: E --+ B be a fibration8 with fiber F. (a) Prove that if there exists a section s: B --+ E (this means that

po s = idB is the identity map on B), then 7rn(E) ~ 7rn(B) E9 7rn(F). (b) Prove that if there exists a retraction r: E --+ F, then 7r n (E) ~

7rn(B) E9 7rn(F). (c) Prove that if the fiber F is contractible in the space E, then 7rn(B) ~

7rn(E) E9 7rn-l(F). It is easy to verify that if the group C is free, then the exact sequence

o --+ A ~ B !t C --+ 0 is split. Indeed, the map '1}1: C --+ B can be constructed as follows. Take a basis in C; for each basis element c, we set w(c) = b, where b is any element of 1/J-l(c).

Since the group Ck(K, L) is free, it follows that the exact sequence (3) is split.

Theorem 1.19. (a) If a sequence A ~ B ~ C --+ 0 is exact, then so is the dual sequence

- Hom(A, G) ...!!.- Hom(B, G) .- Rom(C, G) .- o.

(b) If a sequence 0 --+ A ~ B !t C --+ 0 is exact and split, then so is the dual sequence

- 0.- Hom(A, G) ...!!.- Hom(B, G) .- Hom(C, G) .- O.

Proof. (a) First, let us show that Kef'l,b = O. Suppose that C E Ker, i.e., 0= ifi(c) = Co1/J. This means that c(1/J(b)) = 0 for all b E B. By assumption, 1/J is an epimorphism; therefore, c = O.

~ow, let us show that 1m = Ker rj;. The equality 1/J 0 cp = 0 implies 'l: 0 1/J = 0, whence 1m ifi c Ker rj;. Suppose that b E Ker rj;, i.e., 0 = rj;(b) = b 0 cpo _This means that b(lm cp) = 0; therefore, b induces a homomor-phism b/ : B / 1m cp --+ G. The homomorphism 1/J induces an isomorphism 1/J/: B / 1m cp --+ C. Consider the homomorphism b' (1/J') -1 E Hom( C, G). Clearly,

8The definition of a fibration is given in Part I on p. 162.


because the diagram

B

Yl~ G +-- B / 1m r.p ~ C

is commutative. (b) Let cI>: B -+ A be a homomorphism for which cI>r.p = idA. Then the

composition ~ -Hom(A,G) -+ Hom(B,G) ~ Hom(A,G)

is the identity. Therefore, .:p is an epimorphism, i.e., the dual sequence is exact. Clearly, the homomorphism


groups Hom(Ck(K), G) are replaced by CkG, where is the tensor product of Abelian groups, which is defined as follows. Suppose that A and B are Abelian groups, F(A, B) is the free Abelian group with basis A x B, and R(A, B) is the subgroup in F(A, B) generated by the elements of the form (a + a', b) - (a, b) - (a', b) and (a, b + b') - (a, b) - (a, b'). Then A B = F(A, B)I R(A, B). The coset containing an element Ca, b) is denoted by ab. Less formally, A B is the group with generators a b and defining relations (a + a') b = a b + a' b and a (b + b') = a b + a b'. Remark. The group A B may contain an element al bl + ... + ak bk that cannot be represented in the form a b.

It is easy to show that Z G ~ Gj the isomorphism is defined by n 9 1-+ ng. Indeed, n 9 = 1 9 + ... + 1 9 = 1 (ng). Therefore, any element of the group Z G ~ G has the form 1 gj there are no relations between these elements.

A little more involved argument proves that Zn G ~ GlnG. Indeed, any element of the group Zn G has the form 1 g, and 1 ng = O. In particular, Zn Zm ~ Zd, where d = GCD(n,m). Moreover, Zn Q = o.

The very definition of tensor product readily implies the following two assertions.

1. !fa map ep: AxB -- C is bilinear, i.e., ep(a+a', b) = ep(a, b) +ep(a', b) and ep(a, b + b') = ep(a, b) + ep(a, b'), then there exists a homomorphism tj;: A B -- C for which tj;(a b) = ep(a, b).

2. For any two homomorphisms ep: A -- A' and 'If;: B -- B', there exists a homomorphism ep : A B -- A' B' for which (ep )(a b) -ep(a) (b). Indeed,

(ep )[(a + a') b - a b - a' b] = [ep(a) + ep(a')] (b) - ep(a) (b) - ep(a') (b) = o.

For the tensor product, the following theorem is validj it is dual to Theorem 1.19.

Theorem 1.20. (a) If a sequence A ~ B :!:... C -- 0 is exact, then so is the sequence

AG~BG~CG--O, where 1 = idG.

(b) If a sequence 0 -- A ~ B :!:... C -- 0 is exact and split, then so is the sequence

0---+ AG ~ BG ~ CG ---+ o.


Proof. (a) First, let us show that 1jJ 1 is an epimorphism and its kernel coincides with Ker1jJ G. The homomorphism 1jJ 1 induces a homomor-phism

-----1jJ 1: B G / (Ker 1jJ G) -+ C G. -----It is sufficient to verify that 1jJ 1 is an isomorphism. Consider the map

'It: CxG -+ BG/(Ker1jJG) defined by 'It(c,g) = bg+Ker1jJG, where 1jJ(b) = c. This map is well defined because if 1jJ(b') = c, then b 9 - b' 9 = (b-b')g, where b-b' E Ker1jJ. The map 'It is bilinear; therefore, it induces a homomorphism ~: C G -+ B G / (Ker 1jJ G). The maps 1 and ~ are mutually inverse.

Thus, Ker(1jJl) = Ker1jJG = ImcpG = Im(cpl); the last equality holds because the group Im( cp 1) is generated by all elements of the form cp(a) g.

(b) Let ~: B -+ A be a homomorphism for which iPcp = idA. Then (~ 1) 0 (cp 1) = iPcp 1 = idA ide = idAe;

therefore, cp 1 is a monomorphism, and the map iP 1 splits the exact sequence of tensor products. 0

Exercise. Prove that the sequence 0 -+ Z Z2 -+ Q Z2 induced by the exact sequence 0 -+ Z -+ Q is not exact.

4.3. The Groups Tor and Ext. The homology and cohomology groups with arbitrary coefficients can be expressed in terms of integral homol-ogy groups. The expressions include operations Tor and Ext, which assign Abelian groups Tor(A, B) and Ext(A, B) to pairs of Abelian groups A and B. In cmputations, it is sufficient to know Tor(A, B) and Ext(A, B) for A, B = Z, Zm, Q, and JR.

The groups Tor(A, B) and Ext(A, B) are defined as follows. The Abelian group A can be specified by generators and defining relations. This means that there exists an exact sequence

(6) i p o -+ R -+ F -+ A -+ 0, where F and R are free Abelian groups (the group F is defined by generators and R, by relations; the group R is free because it is a subgroup of a free group). The exact sequence (6) is called a free resolution of the Abelian group A. The exact sequence (6) induces exact sequences

RB ~ FB ~ AR ---+ 0 and

Hom(R, B) ~ Hom(F, B) J- Hom(A, B) ~ o.

4. Cohomology and Universal CoefIicient Theorem 29

We set Tor(A, B) = Ker(i 1) and Ext(A, B) = Cokeri' = Hom(R, B)/Imi' (note that both definitions use only the map i: R --+ F). Thus, the groups Tor(A, B) and Ext(A, B) complete the exact sequences under consideration to the exact sequences

il pl o --. Tor(A, B) --. R B - F B ----+ A B --. 0 and

o +-- Ext(A, B) +-- Hom(R, B) ~ Hom(F, B) ...L Hom(A, B) +-- O. First, we must check that the groups Tor(A, B) and Ext(A, B) thus

defined do not depend on the choice of a free resolution.

Lemma 1.2. (a) For an arbitrary homomorphism cP: A --+ A' of Abelian groups A and A', any free resolutions of A and A' can be completed to a commutative diagram

i P O~R~F~A~O

l~l l~o l~ i' p' O~R'~F'~A'~O.

(b) If tPo and tPl are other homomorphisms completing the free resolu-tions to a commutative diagram, then there exists a homomorphism Do: F --+ R' for which i'Do = CPo - tPo and Doi = CPI - tPl .

Proof. (a) Take bases {Ta.} and {ft3} in the free Abelian groups Rand F. The map p' is an epimorphism; therefore, F' contains elements f~ such that P'(J~) = cPp(Jt3). For the basis elements, we set cpo(Jt3) = f~; then, we extend CPo to a group homomorphism F --+ F'. We have p' CPo = cpp.

By assumption, pi = 0; hence p'cpoi(ra) = cppi(Ta) = O. It follows from Ker p' = 1m i' that R' has an element T~ for which i'(T~) = CPOi(Ta). We set CPI(Ta ) = r~. Then i'CPI = cpoi.

(b) Choose a basis {fa} in F. By assumption, p'(cpo-tPo)(fa) = cpp(Ja)-cpp(Ja) = O. Therefore, R' has an element {T~} for which i'(T~) = (cpo -Wo)(fa). We set Do(fa) = r~. Then i'Do = cpo - tPo, whence i'Doi = (cpo - tPo)i = i'(CPl -WI)' The map i is a monomorphism; therefore, Doi = CPI - tPl. 0

The awkward formulation of Lemma 1.2 has a very natural interpretation in the language of homological algebra. Namely, consider the sequence of homomorphisms


as a chain complex C. (here Co = F, Cl = R, and Ck = 0 for k > 1). Lemma 1.2 asserts that any homomorphism cp: A --+ A' induces a chain map C. --+ C~, which is unique up to chain homotopy. Thus, the induced homology map H.(C.) --+ H.(C~) is uniquely determined. Moreover, the corresponding homology group homomorphisms H .. (C. G) --+ H .. (C~ G) and H.(Hom(C~, G)) --+ H.(Hom(C., G)) are uniquely determined as well. It is these homomorphisms that we are interested in because HI (C. G) = Tor(A, G) and HI (Hom(C., G)) = Ext(A, G). Note also that Ho(C. G) = FG/RG ~ AG and Ho(Hom(C.,G)) ~ Hom(A,G).

Now, the relations (id). = id and (cp'I/J). = CP.'I/J. imply that the group Tor(A, B) is well defined. For Ext(A, B), the proof is similar; the only difference is that (cp'I/J). = 'I/J .. CP. for this group.

Let us calculate the groups Tor(A, B) and Ext(A, B) in simple cases. Note that if the group A is free, then Tor(A, B) = 0 and Ext(A, B) = O. In particular, Tor(Z, B) = 0 and Ext(Z, B) = O.

For the group A = Zn, consider the free resolution

xn o ----+ Z ----+ Z ----+ Zn ----+ o.

First, we calculate Tor(Zn, B). Any element of the group Z B can be represented in the form 1 b. We must calculate the kernel of the map 1 b 1--+ n b = 1 nb. Clearly, it consists of the elements 1 b for which nb = O. Thus, the group Tor(Zn, B) is isomorphic to Ker(B ~ B). In particular, Tor(Zn, Z) = 0 and Tor(Zn, Zm) = Z(n,m)'

Now, we calculate Ext(Zn, B). Clearly, Hom(Z, B) ~ B, and to multi-plication by n in Z corresponds the homomorphism B --+ B that takes each b to nb. Therefore, Ext(Zn, B) = B/nB. In particular, Ext(Zn, Z) = Zn and Ext(Zn, Zm) = Z(n,m)'

Problem 16. Let A = Zk EB T, where T is a finite Abelian group. Prove that Ext (A, Z) ~ T. Theorem 1.21. (a) If A is an Abelian group such that Ker(A ~ A) = 0 for any n E N, then Tor(A, B) = 0 for any group B.

(b) If B is an Abelian group such that nB = B for any n E N, then Ext(A, B) = 0 for any group A.

Proof. (a) Let A' be an arbitrary finitely generated subgroup in A. Then A' is a free Abelian group; therefore, for any free resolution

. p o --+ R ~ F --+ B ----+ 0,


we have the commutative diagram

l'i A' R ----t A' F

(7) At~AL in which the map l' i is a monomorphism. The vertical arrows in this diagram are monomorphisms too because Rand F are free groups. Let us prove that the map 1 i is a monomorphism. Take any element w = al n + ... + ak rk in A R. The group A' can be chosen so as to contain the elements al, ... , ak. Since the diagram (7) is commutative and, moreover, l' i and the right vertical arrow are monomorphisms, it follows that (1 i)(w) 1= o.

(b) Any free resolution i 'P A O~R~F~ ~O

determines a homomorphism Hom(F, B) -+ Hom(R, B)j we have to prove that this is an epimorphism. Let 'P: R -+ B be a homomorphism. Take y E F \ R and extend 'P to the group generated by Rand y as follows:

(1) if my R for all mEN, then we set cjS(y) = OJ (2) if my E R for some mEN, then we choose the least number n E N

with this property and put cjS(y) = b, where b is the element of B such that nb = 'P(ny).

An extension of'P over the entire group F can be constructed by induc-tion (if the group F/ R is not finitely generated, then the induction is trans-finite). 0 Corollary. If G = Q, JR, or C (we regard these fields as groups under addition), then Tor(G, B) = 0 and Ext(A, G) = 0 for any Abelian groups A andB.

In calculating homology and cohomology groups with coefficients in G by using the universal coefficient theorem, the groups Tor(A, G) and Ext(A, G) arise. If G = Q, JR, or C, then we have Tor(A, G) = 0 because Tor(A, B) ~ Tor(B, A) for any Abelian groups A and B. Note that Ext does not have this propertyj for example, Ext(Z, Zn) = 0 and Ext(Zn, Z) = Zn. Problem 17. Prove that Tor(A, B) ~ Tor(B, A), and the isomorphism is canonical.

An Abelian group T is said to be periodic if for any t E T, there exists a positive integer n for which nt = o.


Problem 18. Prove that if Ext(T, Z) = 0 and T is a periodic group, then T=O.

Problem 19. Prove that if a sequence 0 - A :!.... B:!!.....C is exact, then so is the sequence

- ~ 0----+ Hom(G,A) ~ Hom(G, B) ----+ Hom(G,C) for any Abelian group G.

An Abelian group G is said to be divisible if for any positive integer n, the map G - G defined by 9 1-----+ ng is an epimorphism.

Problem 20. Let 0 - A:!.... B:!!.....C - 0 be an exact sequence. (a) Prove that, for any free Abelian group F, the sequence

- ~ 0----+ Hom(F, A) ~ Hom(F, B) ----+ Hom(F, C) ----+ 0 is exact.

(b) Prove that, for any divisible group G, the sequence - ~

0-- Hom(A, G) ~ Hom(B,G) -- Hom(C,G) -- 0 is exact.

A free resolution of an Abelian group A is often called a projective res-olution because there is the dual notion of injective resolution; an injective resolution of an Abelian group A is an exact sequence 0 - A - G - H - 0 in which G and H are divisible Abelian groups.

Any Abelian group A has an injective resolution. Indeed, any quotient of a divisible group is divisible. Therefore, it is sufficient to embed A in a divisible group G. Suppose that A = F / R, where F is a free group. The group F is a direct sum of a number of copies of Z; hence it is a subgroup of the group G defined as the direct sum of the same number of copies of Q. The group G has a subgroup corresponding to R. The quotient group G / R contains A as a subgroup.

The group Ext(A, B) can be defined not only by using a projective res-olution of A but also by using an injective resolution of B. Namely, let o - B - G - H - 0 be an injective resolution of the group B. Accord-ing to Problem 19, the induced sequence 0 - Hom(A, B) _ Hom(A, G) -Hom(A, H) is exact. Therefore, it can be completed to an exact sequence of the form

o ----+ Hom(A, B) ----+ Hom(A, G) ----+ Hom(A, H) ----+ Ext(A, B) ----+ o. Problem 21. Prove that &t(A, B) ~ Ext(A, B).


Problem 22. Given an exact sequence of Abelian groups 0 - A _ B -C - 0, prove that, for any Abelian group X, there are exact sequences

o ---+ Hom(X, A) ---+ Hom(X, B) ---+ Hom(X, C) ---+ Ext (X, A) -+ Ext(X, B) ---+ Ext(X, C) ---+ 0

and

0-- Ext(A, X) -- Ext(B, X) -- Ext(C, X) -- Hom(A,X) -- Hom(B,X) -- Hom(C,X) -- O.

4.4. Universal Coefficient Theorem. Homology and cohomology with arbitrary coefficients can be expressed in terms of integral homology. Name-ly, the following theorem is valid.

Theorem 1.22 (universal coefficient formulas). For any Abelian group G, there are exact sequences

o ---+ Hk(K) G ---+ Hk(Kj G) ---+ Tor(Hk_l (K), G) -+ 0 and

0-- Hom(Hk(K), G) -- Hk(KjG) -- Ext(Hk-l(K),G) -- 0, where Hk(K) = Hk(Kj Z). These exact sequences splitj hence

and Hk(Kj G) ~ Hom(Hk(K), G) ffi Ext (Hk- 1 (K), G)

(the isomorphisms are not canonical). Proof. Consider the exact sequence of chain complexes

It is assumed that Ck = Ck(Kj Z), etc. The group B k - 1 is freej therefore, tensoring with G, we obtain the exact

sequence Z G il8il Bl8il o ---+ k --+ Ck G ---+ B k- 1 G ---+ o.


Any short exact sequence of chain complexes induces an exact sequence of homology groups. Taking into account the fact that the homology groups of a chain complex with zero boundary homomorphisms coincide with the chain groups, we obtain the exact sequence

-----+ Bk G -----+ Zk G -----+ Hk(C. G) -----+ Bk 1 G -----+ Zk-l G -----+ It follows directly from the definition of connecting homomorphisms in exact sequences of homology groups that the map Bk G --+ Zk G in this sequence is induced by the embedding j: Bk --+ Zk. Thus, we obtain the exact sequence

1811 0--+ ZkG/BkG -----+ Hk(C.G) -----+ Ker(Bk-lG ~ Zk-lG) --+ O. Here Zk G/ Bk G = Hk G and Hk(C. G) = Hk(Kj G). More-

1811 over, Ker(Bk-1 G ~ Zk-l G) = Tor(Hk_1(K), G) because any exact sequence

o -----+ B k-l -----+ Z k-l -----+ H k-1 -----+ 0 is a free resolution for the group H k - 1 = H k - 1 (Kj Z).

Similarly, for cohomology, we have the exact sequence

0-- Hom(Hk(K) , G) -- Hk(KjG) -- Ext(Hk_1(K),G) -- O. It remains to show that these exact sequences split. In the exact sequence

of homology groups, at the level of cycles, the homomorphism 'P: Hk(K) G --+ Hk(KjG) has the structure 'P((Eni~f) g) = Enig~f. It is required to construct a homomorphism ~: Hk(Kj G) --+ Hk(K) G for which ~'P = id. The exact sequence

i 8 o -----+ Z k -----+ Ck -----+ B k-l -----+ 0

splits because the group Bk-1 is free. Suppose that I: Ck --+ Zk is a splitting map, i.e., Ii = idz". Restricting the composition

Ck G ~ Zk G -----+ Hk G to the cycles of the complex C. G, we obtain a map that induces a homo-morphism ~: H k (K j G) --+ H k (K) G since I 1 takes boundaries to boundaries. Clearly, if Eni~f is a cycle in Ck, then ~'P((Eni~n g) =

~(E nig~n = (E ni~~) g. For cohomology, the fact that the exact sequpnce is split follows from

Theorem 1.19, (b). 0 Corollary. IfG = Q, JR, ore, then Hk(KjG) ~ Hk(K)G and Hk(KjG) ~ Hom(Hk(K), G).

5. Calculations 35

In calculating the groups Tor and Ext, the following fairly obvious iso-morphisms are useful:

Ext(Al EB A2, B) ~ Ext (AI, B) EB Ext(A2' B), Tor(Al EB A2 , B) ~ Tor(A1 , B) EB Tor(A2 , B).

Remark. For infinite families of Abelian groups, we have

Ext (~A.,B) = I] Ext (A., B), Tor (~A.,B) = ~Tor(A.,B),

Ext ( A, II B) = II Ext(A, Ba ), a a

Problem 23. Prove that if H1(KjZ) ~ ZrEBT1, where Tl is a finite group, then HI(Kj Z) ~ zr. Problem 24. Suppose that Hi(KjZ) = zn, EB11 and Hi(KjZ) = zm, EBr, where 11 and r are finite groups. Prove that mi = ni and r ~ 11-1' Problem 25. Derive the assertion of Problem 24 directly from the definition of homology and cohomology groups, without using the universal coefficient theorem.

Problem 26. Let X and Y be finite simplicial complexes, and let f: X -- Y be a continuous map. Prove that if the map f.: HI (X) -- HI (Y) is zero, then so is f*: Hl(y) __ Hl(X).

5. Calculations

5.1. Fundamental Classes. Many homological properties of manifolds are consequences of the fact that for an orient able closed manifold Mn, the group Hn(Mnj Z) is nontrivial and generated by one homology classj a similar assertion is true for the Z2-homology of an arbitrary closed manifold. Pseudomanifolds9 have the same homological properties.

Theorem 1.23. Let M n be a pseudomanifold, and let G be any Abelian group.

(a) If Mn ha.~ nonempty boundary, then Hn(Mnj G) = O. (b) If Mn is closed, then

if Mn is orientable, H (Mn'G) ~ {G n, x2 Ker( G --+ G) if Mn is nonorientable.

9The definition of a pseudomanifold was given in Part I on p. 109. Recall that any smooth manifold can be triangulated, and a triangulated manifold is a pseudomanifold. Therefore, all theorems on pseudomanifolds presented below are also valid for smooth manifolds.


Proof. The group Hn(Mn; G) consists of chains en = E ai~f with ai E G for which 8en = o. Suppose that simplices ~f and ~j have a common face ~~-1. Since any pseudomanifold must be unramified, it follows that

~~-1 is a face for only these two simplices; therefore, it is contained in 8en with coefficient ai aj. The coefficient is (ai - aj) if the orientations of the simplices ~~ and ~j are compatible, and (ai + aj) otherwise. The coefficient ai aj can vanish only if ai = aj. Therefore, according to the strong connectedness condition, if 8en = 0, then en = E a~~, where the summation is over all n-simplices.

If a simplex ~n-l belongs to the boundary of M n , then it is contained in 8en with coefficient a. Therefore, 8en = 0 implies a = 0; i.e., Hn(Mn; G) = o for any coefficient group G.

Changing (if necessary) the orientations of some simplices, we can assume that en = E a~f. If the pseudomanifold Mn is closed and 2a i- 0, then 8en = 0 if and only if the orientations of all simplices ~f are compatible; if 2a = 0, then 8en = O. Therefore, if M n is orientable, then Hn(Mn; G) ~ G, and if M n is nonorientable, then Hn(Mn; G) ~ Ker(G ~ G). 0

For any closed pseudo manifold M n , the homology class of the cycle E ~f in Hn(Mn; Z2) is called the fundamental class of Mn (more formally, this cycle should be written as E l~f, where 1 E Z2). The term fundamen-tal class is also used for the homology class of the cycle E ~f in Hn(Mn; Z) provided that the closed pseudomanifold Mn is oriented (the orientations of all simplices ~f are compatible with that of Mn). The fundamental class of a pseudomanifold M n is denoted by [Mn]; for the coefficient group Z2, the notation [Mnh is sometimes used. Problem 27. Prove that if Mn is a closed orient able manifold, then the assertion of Problem 10 on p. 20 remains valid for k = n - 1. Is it essential that the manifold M n is closed? orientable?

For pseudomanifolds with boundary, the same argument proves that

if Mn is orient able, if Mn is nonorientable.

Indeed, if a simplex ~n-l belongs to 8Mn , then its contribution to the rela-tive chain is zero. The definition of fundamental classes in relative homology is the same as in absolute homology.

Let Mn and Nn be closed oriented pseudomanifolds. A map f: Mn -+ N n takes the fundamental class [Mn] to an element of the group HnCNn; Z), and any element of this group has the form k[Nn], where k E Z. The integer degf for which (degJ)[Nn] = f*([Mnj) is called the degree of the map f. If

5. Calculations 37

the pseudomanifolds M n and N n are closed Cbut not necessarily orient able ), then the degree of f modulo 2 can be defined. Remark. This definition is equivalent to that given in Part I on p. 111, where the degree of a map was defined as the number of preimages of points counted with signs.

If M n and Nn are pseudomanifolds with boundary, then we can also define the degree (or degree modulo 2) ofa map f: (Mn, aMn) -+ (Nn, aNn) by considering the image of the fundamental class [Mn] in the group Hn(Nn, aNn;Z) (or in Hn(Nn,aNn;Z2))' Theorem 1.24. Let i.: HncaWn+l) -+ HnCwn+1) be the homomorphism induced by the natural embedding. If the pseudomanifold W n +1 is orientable and its boundary awn+l is connected, then i", = 0 for the coefficient groups Z and Z2, and if it is nonorientable, then i", = 0 for the coefficient group Z2.

Proof. The group HnCalvn+l) is cyclic; therefore, it suffices to show that the fundamental cycle [awn+l] is the boundary of some cycle in W n+1 . Clearly, it is the boundary of the fundamental cycle [Wn+l]. If the pseudo-manifold Wn+l is nonorientable, then its fundamental cyclt' is defined only for the coefficient group Z2. 0

Corollary. II I: awn+l -+ M n is the restriction of some continuous map F: Wn+l -+ Mn, the pseudomanilold wn+l is orientable, and its boundary awn+l is connected, then deg I = O.

Proof. The homomorphism f",: Hncawn+l) -+ HnCMn) can be represented as the composition f", = F",i",; hence f", = O. 0 Problem 28. Let f: sn -+ sn be a map, and letE/: ESn -+ Esn be the map whose restriction to sn x it} coincides with I for each t. Prove that deg I = deg Ef by using the suspension isomorphism.

5.2. Cellular Homology. If a simplicial complex K is endowed with the structure of a CW-complex Csee Part I, p. 118) X and each skeleton Xk is a subcomplex in K, then the integral homology of K can be calculated by using the chain complex C. in which every Ck is the free group whose gen-erators are in one-to-one correspondence with the k-cells of the complex X. This substantially facilitates calculating the homology because the number of cells in X may be many times smaller than the number of simplices in K. For example, a minimal triangulation of the sphere sn contains 2n+l_l simplices (of all dimensions), while a minimal cell decomposition consists of only two cells (n-dimensional and zero-dimenSional).


Lemma. For i i= k, Hi(Xk,Xk- 1) = 0, and Hk(Xk,Xk- 1) is the free Abelian group whose generators are in one-to-one correspondence with the k-cells of x.

Proof. Any chain Ci E Ci (X k , X k-1) has a unique representation in the form of the (finite) sum of chains Cia:, where the {e~} are the k-cells of X. Moreover, 8Ci = 'E 8~a:. Any group Ci(e~, 8e~) has the form Ci(Dk, Sk-l) for some triangulation of the disk Dk. Therefore, Hi (Xk, Xk-1) is the direct sum of the group~ H l (Dk,Sk-1). 0

The exact sequence

Hi+l(Xk+1,Xk) ____ Hi(Xk) ____ Hi(Xk+1) ____ Hi(Xk+1,Xk) for the pair (Xk+1,Xk) shows that Hi(Xk) ~ Hi(Xk+1) for i f. k,k + 1. Hence Hk_l(Xk) ~ Hk_1(Xk+1) !:>! Hk_1(Xk+2) ~ .. , ~ Hk-1(X).

Let Ck = Hk(Xk,Xk 1). Consider the map 8k : Ck -+ Ck- 1 defined as follows. We arrange the exact sequences for the pairs (Xk,Xk- 1 ) and (X k - 1 , X k - 2 ) along horizontal and vertical lines:

The map 8k: Ck -+ Ck-l is the composition of the horizontal and vertical arrows. The group Hk(Xk) can be regarded as a subgroup in Ck. Moreover, we can identify Hk(Xk) with Ker 8k because if ep = hep' for a monomorphism h, then Kercp = Kerep'. Identifying Hk_l(Xk - 1 ) with Ker8k_1 in a similar way, we obtain Hk _ 1(X k ) ~ Ker8k-I!Im8k.

The map 8k : Ck -+ Ck-l acts as follows. Suppose that to a generator c~ E Ck corresponds the cell e~. Consider an absolute chain [e~] representing the relative fundamental class (e~, 8e~). At the level of H k-l ()( k-l ), we have 8[e~] = L: na:,6[e~-l]. Thus, 8~ = L: na:,6c~-l. The number na:,6 is equal to the degree of the map Sk-l -+ e~-l /8e~-1 = Sk-l induced by the characteristic map of the cell e~.

The equality 88 = 0 follows from 88[e~1 = O.

5. Calculations 39

The homology with coefficients in Z2 is calculated similarlyj in this case, the degree modulo 2 should be considered.

As an example, we calculate the homology of closed two-dimensional surfaces. We have already calculated the groups Ho and H 2 j let us calculate HI Example 7. HI (nT2) = z2n and HI (nT2j Z2) = z~n (nT2 is a special case of the surface 8 2 #pT2 #q K #r p2 defined in Part I on p. 142). Proof. Consider the standard representation of nT2 by using a 4n-gon. For this representation, the complex for calculating the cellular homology has the form

z ~ z2n ~ Z --. o. Here 81 = 0 because there is one zero-dimensional cell, and (h = 0 because each I-cell occurs two times with opposite orientations in the boundary of the 2-ce11.

For the coefficient group Z2, the calculation is similar. o

For a sphere with n handles, the 2n cycles generating the one-dimensional homology group can be chosen as shown in Figure 3. To prove this, we must

Figure 3. Basis cycles

verify that these cycles are homologous to the sides of the 4n-gon from which the sphere with handles is obtained. First, consider a handle from which a disk is removed (see Figure 4). Clearly, to the cycles in Figure 3 corre-spond those in Figure 5; the latter cycles are homologous to the sides of the polygon.

Example 8. HI (mP2) = zm-l ED Z2 and HI (mP2 j Z2) = Zr. Proof. The complex for calculating the cellular homology of mP2 has the form

Z ~ zm ~ Z --. o. We again have 81 = 0, but this time, (h(!) = (2, ... ,2) because each I-cell occurs two times with the same orientation in the boundary of the 2-ce11. We


Figure 4. BlIBis cycles on a handle

Figure 5. BlIBis cycles on the polygon

must take the quotient group of zm modulo the subgroup generated by the element (2, ... ,2). Consider el - (1,0, ... ,0), ... , em-l = (0, ... ,0,1,0), and em = (1, ... ,1). We have

The condition (al, ... , am) E Imih is equivalent to al - am = 0, ... , am-l-am = 0 and am == 0 (mod 2). Therefore, the quotient group zm / 1m ih is isomorphic to zm-l EB Z2; the group zm-l is generated by el. ... , em-I. and the group Z2 is generated by em. As before, for the coefficient group Z2, we have ih = O. 0

Cycles Q and (3 generating the integral homology group for the Klein bot-tle are shown in Figure 6. At the level of homology, we have 2(3 = 0 because the boundary of the Mobius band hatched in Figure 6a is homologous to 2(3.

Problem 29. Calculate the homology groups of closed two-dimensional surfaces with coefficients in Zp for p f:. 2. Pr

V. v. Prasolov Elements of Homology Theory Graduate Studies in Mathematics 81 2007

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