NOTES ON CHAIN COMPLEXES

NOTES ON CHAIN COMPLEXES ANDEW BAKE These notes are intended as a very basic introduction to (co)chain complexes and their algebra, the intention being to point the beginner at some of the main ideas which should be further studied by in depth reading. An accessible introduction for the beginner is [2]. An excellent modern reference is [3], while [1] is a classic but likely to prove hard going for a novice. Most introductory books on algebraic topology introduce the language and basic ideas of homological algebra. 1. Chain complexes and their homology Let be a ring and Mod the category of right -modules. Then a sequence of -module homomorphisms L f M g N is exact if Ker g = Im f. Of course this implies that gf =. A sequence of homomorphisms fn+1 M n f n Mn 1 f n 1 is called a chain complex if for each n, f n f n+1 = or equivalently, Im f n+1 Ker f n. It is called exact or acyclic if each segment f n+1 M n+1 M n f n Mn 1 is exact. We write (M, f) for such a chain complex and refer to the f n as boundary homomorphisms. If a chain complex is finite we often pad it out to a doubly infinite complex by adding in trivial modules and homomorphisms. In particular, if M is a -module we can view it as the chain complex with M = M and M n = whenever n. It is often useful to consider the trivial chain complex = ({}, ). Given a complex (M, f), we define its homology to be the complex (H (M, f), ) where H n (M, f) = Ker f n / Im f n+1. A morphism of chain complexes h: (M, f) (N, g) is a sequence of homomorphisms h n : M n N n for which the following diagram commutes. M n h n f n Mn 1 h n 1 N n g n Nn 1 Notice that if u Ker f n we have g n (h n (u)) =, while if v M n+1, h n (f n+1 (v)) = g n+1 (h n+1 (v)). Together these allow us to define for each n a homomorphism h : H n (M, f) H n (N, g); h (u + Im f n+1 ) = h n (u) + Im g n+1. Date: [6/4/29]. 1

2 ANDEW BAKE It is easy to check that if j : (L, l) (M, f) is another morphism of chain complexes then (hj) = h j and for the identity morphism id: (M, f) (M, f) we have id = id. This shows that each H n is a covariant functor from chain complexes to -modules. From now on we will always write a complex as (M, d) where the boundary d is really the collection of boundary homomorphisms d n : M n M n 1 which satisfy d n 1 d n = ; we often symbolically indicate these relations with the formula d 2 =. Given a morphism of chain complexes h: (L, d) (M, d) we may define two new chain complexes Ker h = ((Ker h), d) and Im h = ((Im h), d), where (Ker h) n = Ker h: L n M n, (Im h) n = Im h: L n M n. The boundaries are the restrictions of d to these. A cochain complex is a collection of -modules M n together with coboundary homomorphisms d n : M n M n+1 for which d n+1 d n =. The cohomology of this complex is (H (M, d), ) where H n (M, d) = Ker d n / Im d n 1. 2. The homology long exact sequence Let h: (L, d) (M, d) and k : (M, d) (N, d) be morphisms of chain complexes and suppose that is short exact, i.e., (L, d) h (M, d) k (N, d) Ker h =, Im k = (N, d), Ker k = Im h. Theorem 2.1. There is a long exact sequence of the form H n+1 (N, d) +1 H n (L, d) h H n (M, d) k H n (N, d) H n 1 (L, d) Furthermore, given a commutative diagram of short exact sequences h (L, d) (M, d) (N, d) p q r (L, d) h (M, d) k k (N, d)

NOTES ON CHAIN COMPLEXES 3 there is a commutative diagram H n+1 (N, d) +1 r H n (L, d) H n+1 (N p, d) +1 H n (L, d) h H n (M, d) q h H n (M, d) k H n (N, d) r H n 1 (L, d) k H n (N p, d) H n 1 (L, d) Proof. We begin by defining : H n (N, d) H n 1 (L, d). Let z + Im d n+1 H n (N, d) = Ker d n / Im d n+1. Then since k : M n N n is epic, we may choose an element w M n for which k(w) = z. Then d n (w) M n 1 and kd n (w) = d n k(w) = d n (z) =. So d n (w) Ker k. By exactness, d n (w) = h(v) for some v L n 1. Notice that hd n 1 (v) = d n 1 h(v) = d n 1 d n (w) =. But h is monic, so this gives d n 1 (v) =. Hence v Ker d n 1. We take (z + Im d n+1 ) = v + Im d n H n 1 (L, d). It is now routine to check that for z Im d n+1 we have (z + Im d n+1 ) = + Im d n, hence is a well defined function. Exactness is verified by checking on elements. For cochain complexes we get a similar result for a short exact sequence of cochain complexes but the are replaced by homomorphisms (L, d) h (M, d) k (N, d), δ n : H n (N, d) H n+1 (L, d). Example 2.2. Consider the following exact sequence of chain complexes of Z-modules (written vertically). H (L ) H (M ) H (N ) = 2 = 1 Z 2 Z Z/2 = Z 2 2 2 Z Z/2 = 1

4 ANDEW BAKE Taking homology we obtain the long exact sequence H (L ) H (M ) H (N ) = 1 Z/2 = Z/2 From this we conclude that Z/2 = Z/2 : H 1 (N ) = Z/2 H (L ) = Z/2 is an isomorphism. 3. Tensor products, free resolutions and Tor Let M be any right -module and N be left -module. Then we can form the tensor product M N which is an abelian group. It is also an -module if is commutative. The definition involves forming the free Z-module F (M, N) with basis consisting of all the pairs (m, n) where m M and n ; then M N = F (M, N)/S(M, N), where S(M, N) F (M, N) is the subgroup generated by all the elements of form (m 1 + m 2, n) (m 1, n) (m 2, n), (m, n 1 + n 2 ) (m, n 1 ) (m, n 2 ), (mr, n) (m, rn), where m, m 1, m 2 M, n, n 1, n 2 N, r. We usually denote the coset of (m, n) by m n; such elements generate the group M N. The tensor product M N has an important universal property which characterizes it up to isomorphism. Write q : M N M N for the quotient function. Let f : M N V be a function into an abelian group V which satisfies f(m 1 + m 2, n) = f(m 1, n) + f(m 2, n), f(m, n 1 + n 2 ) = f(m, n 1 ) + f(m, n 2 ), f(mr, n) = f(m, rn) for all m, m 1, m 2 M, n, n 1, n 2 N, r, there is a unique homomorphism f : M N V for which f = f q. M N q f V! f M N Proposition 3.1. If f : M 1 M 2 and g : N 1 N 2 are homomorphisms of -modules, there is a group homomorphism for which f g : M 1 N 1 M 2 N 2 f g(m n) = f(m) g(n).

NOTES ON CHAIN COMPLEXES 5 Proposition 3.2. Given a short exact sequence of left -modules there is an exact sequence g 1 g 2 N 1 N2 N3, 1 g 1 1 g 2 M N 1 M N 2 M 3. N Because of this property, we say that M ( ) is right exact. Obviously it would be helpful to understand Ker 1 g 1 which measures the deviation from left exactness of M ( ). Let be a ring. A right -module F is called free if there is a set of elements {b λ : λ Λ} F such that every element x F can be uniquely expressed as x = λ Λ b λ t λ for elements t λ. We say that the b λ form a basis for F over. We can make a similar definition for left modules. For example, for n 1, n = {(t 1,..., t n ) : t 1,..., t n } is free on the basis consisting of the standard elements e 1 = (1,,..., ),..., e n = (,...,, 1). This works whether we view n as a left or right module. An exact complex (3.1) F k F 1 F M is called a resolution of M. Here we view M as the ( 1)-term and as the ( 2)-term. If each F k is also free over then it is called a free resolution of M. Every M admits such a free resolution. Given such a free resolution F M of a right -module M, and a left -module N, we can form a new complex F N, where the boundary maps are obtained by tensoring those of F with the identity on N an taking F N rather than using the original map F M. Here F n N is in degree n. We define It is easy to see that Tor n (M, N) = H n (F N). Tor (M, N) = M N. Of course we could also form a free resolution of N, tensor it with M and then take homology. Theorem 3.3. Tor has the following properties. i) Tor (M, N) can be computed by using free resolutions of either variable and the answers agree up to isomorphism. ii) Given -module homomorphisms f : M 1 M 2 and g : N 1 N 2 there are homomorphisms (f g) = f g : Tor n (M 1, N 1 ) Tor n (M 2, N 2 ) generalizing f g : M 1 N 1 M 2 N 2. iii) For a free right/left -module P /Q and n > we have Tor n (P, N) = = Tor n (M, Q).

6 ANDEW BAKE iv) Associated to a short exact sequence of right -modules M 1 M 2 M 3 there is a long exact sequence Tor n+1(m 3, N) Tor n (M 1, N) Tor n (M 2, N) Tor n (M 3, N) Tor n 1(M 1, N) M 1 N M 2 N M 3 N and associated to a short exact sequence of left -modules N 1 N 2 N 3 there is a long exact sequence Tor n+1(m, N 3 ) Tor n (M, N 1 ) Tor n (M, N 2 ) Tor n (M, N 3 ) Tor n 1(M, N 1 ) M N 1 M N 2 M N 3 Corollary 3.4. Let Q be a left -module for which Tor n (M, Q) = for all n > and M. Then for any exact complex (C, d), the complex (C Q, d 1) is exact, and H n (C Q, d 1) = H n (C, d) Q. An -module M for which Tor n (M, N) = for all n > and left -module N is called flat. Given a module M, it is always possible to find a resolution F M for which each F k is flat. Then we also have Proposition 3.5. If F M is a flat resolution, then Tor n (M, N) = H n (F N, d 1). 4. The Künneth Theorem Suppose that (C, d) is a chain complex of free right -modules. For any left -module N we have another chain complex (C N, d 1) with homology H (C N, d 1). We would like to understand the connection between this homology and H (C, d) N.

NOTES ON CHAIN COMPLEXES 7 Begin by taking a free resolution of N, F N. For each n the complex C n F C n N is still exact since C n is free. The double complex C F has two compatible families of boundaries, namely the horizontal ones coming from the boundaries maps d tensored with the identity, d 1, and the vertical ones coming from the identity tensored with the boundary maps δ of F, 1 δ. We can take the two types of homology in different orders to obtain H v (H h (C F )) = H v = Tor (H (C, d), N), H h (H v (C F )) = H h (C N)) = H (C N, d 1). In general, the precise relationship between these two involves a spectral sequence, however there are situations where the relationship is more direct. Suppose that N has a free resolution of the form F 1 F N. This will always happen when = Z or any (commutative) pid and for semi-simple rings. Then for any right -module M and n > 1, Tor n (M, N) =. Now consider what happens when we tensor C with such a resolution. We obtain a short exact sequence of chain complexes C F 1 C F C N and on taking homology, an associated long exact sequence as in Theorem 2.1. H n+1 (C N) +1 H n (C F 1 ) H n (C F ) Because F and F 1 are free, Corollary 3.4 gives in each case so our long exact sequence becomes H n (C F i ) = H n (C ) F i, H n+1 (C N) +1 H n (C ) F 1 H n (C ) The segment F H n (C ) F 1 H n (C ) H n (C N) H n 1 (C F 1 ) H n (C N) H n 1 (C ) F 1 F

8 ANDEW BAKE is part of the complex used to compute Tor (H n (C ), N) and we actually have the sequence in which Tor 1 (H n (C ), N) H n (C ) F 1 H n (C ) F Tor (H n (C ), N) Tor (H n (C ), N) = H n (C ) N. For : H n (C N) H n 1 (C F 1 ) we find that Ker = Tor (H n (C ), N), Im = Tor 1 (H n 1 (C ), N). Theorem 4.1 (Künneth Theorem). Let (C, d) be a chain complex of free right -modules and N a left -module. For each n there is an exact sequence H n (C ) N H (C N) Tor 1 (H n 1 (C ), N). 5. Projective and injective resolutions, Hom and Ext Let M, N be two right -modules. Then we define Hom (M, N) = {h : M N : h is a homomorphism of -modules}. If is commutative then Hom (M, N) is also an -module, otherwise it may only be an abelian group. If f : M M and g : N N are homomorphisms of -modules, then there are functions f : Hom (M, N) Hom (M, N); f h = h f, g : Hom (M, N) Hom (M, N ); g h = g h. These are group homomorphisms and homomorphisms of -modules if is commutative. Proposition 5.1. Let M, N be right -modules. (a) Given a short exact sequence of -modules the sequence f 1 f 2 M 1 M2 M3, Hom (M 3, N) f 2 Hom (M 2, N) f 1 Hom (M 1, N) is exact. (b) Given a short exact sequence of -modules the sequence is exact. g 1 g 2 N 1 N2 N3, Hom (M, N 1 ) g 1 Hom (M, N 2 ) g 2 Hom (M, N 3 ) These result show that Hom (, N) and Hom (M, ) are left exact. An -module P is called projective if given an exact sequence M f N and a (not usually unique) homomorphism p: P N, there is a homomorphism p: P M for which

NOTES ON CHAIN COMPLEXES 9 f p = p. P p p M N f In particular, every free -module is projective. It is easy to see that if P is projective then Hom (P, ) is right exact. Now suppose that P M is a resolution of M by projective modules (for example, each P n could be free). Then for any N we can form the cochain complex Hom (P, N) whose n-th term is Hom (P n, N). The n-th cohomology group of this is Ext n (M, N) = H n (Hom (P, N)). It turns out that this is independent of the choice of projective resolution of M. Notice also that Ext (M, N) = Hom (M, N). An -module J is called injective if given an exact sequence K g L and a homomorphism q : K J, there is a (not usually unique) homomorphism q : L J for which qg = q. K L g q It is easy to see that if J is injective, then Hom (, J) is right exact. If N J is an exact cochain complex in which each J n is injective (i.e., an injective resolution of M) then we may form the cochain complex Hom (M, J ) whose n-term is Hom (M, J n ). The n-th cohomology group of this is q J rext n (M, N) = H n (Hom (M, J )). It turns out that this is independent of the choice of injective resolution of N. Notice also that rext (M, N) = Hom (M, N). Proposition 5.2. For -modules M, N there is a natural isomorphism rext n (M, N) = Ext n (M, N). eferences [1] H. Cartan & S. Eilenberg, Homological Algebra, Princeton University Press (1956). [2] J. J. otman, An introduction to Homological Algebra, Academic Press (1979). [3] C. A. Weibel, An Introduction to Homological Algebra, Cambridge University Press (1994).