Chapter 3. Homological Algebra 8 Chain and Cochain Complexes

The aim of this chapter is to introduce the fundamental results of homological algebra. Homological algebra appeared in the 1800’s and is nowadays a very useful tool in several branches of mathematics, such as algebraic topology, commutative algebra, algebraic geometry, and, of particular interest to us, group theory.

Throughout this chapter R denotes a ring, and unless otherwise speciﬁed, all rings are assumed to be unital and associative.

Reference: [Rot09] J. J. R, An introduction to homological algebra. Second ed., Universitext, Springer, New York, 2009. [Wei94] C. A. W, An introduction to homological algebra, Cambridge Studies in Advanced Math- ematics, vol. 38, Cambridge University Press, Cambridge, 1994.

8 Chain and Cochain Complexes Deﬁnition 8.1 (Chain complex)

(a) A chain complex (or simply a complex) of R-modules` is a sequence p ‚ ‚q“ ¨¨¨›Ñ ` ›Ñ1 ›Ñ ´ ›Ñ ¨ ¨ ¨ ˆ ˙ 1 1 ZC C C C P P p ´ q ˝ ` “ ‚ p ‚ ‚q R 1 1 where for each , C is an R-modules and Hom C C satisﬁes 0. We often write simply C instead of C . Z (b) The integer is called theP degree of the R-module C . Z (c) The R-linear‚ maps ( ) are called the diﬀerential maps“ . P † Z P § 0 (d) A complex C is called non-negative (resp. positive) if C 0, for all (resp. for all 0 ).

Notice that sometimes we will omit the indices and write for all diﬀerential maps, and thus the

31 Skript zur Vorlesung: Cohomology of Groups SS 2018 32 ˝ ` “ “ “ 2 § 1 condition 0 can be written as 0. If there is an integer N such that C 0 for all N, then we omit to write the zero modules and` zero maps` on the right-hand side of the complex: ¨¨¨›Ñ ` ›Ñ ` ›Ñ N 2 N 1 N 2 N 1 N C “ C C• Similarly, if there is an integer N such that C 0 for all N, then we omit to write the zero modules and zero maps on the left-hand side of the´ complex: ›Ñ ´ ›NÑ1 ´ ›Ñ ¨ ¨ ¨ N N N 1 N 2 C C C

1 p ‚ ‚q p ‚ ‚q Deﬁnition 8.2 (Morphism of complexes) Z ‚ ‚ ›Ñ ‚ ›Ñ P ˝ ` “ A1 morphism of complexesZ (or a chain map) between two chain complexes C and D , ` ˝ ` P 1 written : C D , is a familiy of R-linear maps : C D ( ) such that 1 1 for each , that is` such/ that` the/ following/ diagram´ commutes:/ ¨¨¨ 2 ` 1 ´ 1 ¨¨¨ ` 1 ´ 1 1 C ✏ 1 C✏ 1 C ✏ 1 `/ 1 ` / / 1 ´ / ¨¨¨ ` ´ ¨¨¨ 2 1 1 1 1 D D D

Notation. Chain complexes together with morphisms of chain complexes (and composition given by R degreewise composition of R-morphisms) form a category, which we will denote by Ch( Mod).

Definition 8.3 (Subcomplex1 / quotient complex) 1 Z ‚ p ‚ ‚q § P 1 1 Z p qÄ ´ P (a) A subcomplex C1 of a chain complex C is a family of R-modules 1Cã C ( ), such p ‚ ‚q ‚ Ñ ‚ 1 1 Z that C C for every . P In this case, C becomes a chain complex and the inclusion C C given by the 1 1 1 canonical‚ inclusion of C ‚ into C for each is a chain‚{ ‚ map. { Z 1 1 P { ›Ñ ´ { ´ (b) If C is a subcomplex of C , then the quotient complex C C is the familiy of R-modules C C 1 ( ) together with the differential maps : C1 C C C 1 uniquelyZ determined ‚ ‚ ›Ñ ‚{ ‚ P by the universal property›Ñ { of1 the quotient. In this case, the quotient map π : C C C defined for each by the canonical projection π : C C C is a chain map.

1 Definition‚ 8.4‚ (›KernelÑ ‚ / image / cokernel) p ‚ ‚q p ‚ ‚q Z Let : C D be‚ a morphism of chain complexes‚ between ‚C“pt and uDP ‚q. Then, Z 1 (a) the kernel of ‚ is the subcomplex of C‚ defined by ker ‚ “pt: ker u P ‚q ; (b) the image of is‚ the subcomplex of D defined by‚ Im“ ‚{: Im‚ ; and (c) the cokernel of is the quotient complex coker : D Im . Skript zur Vorlesung: Cohomology of Groups SS 2018 33

R With these notions of kernel and cokernel, one can show that Ch( Mod) is in fact an abelian category.

Deﬁnitionp ‚ 8.5‚q (Cycles, boundaries, homology) Let C be a chain complex of R-modules.“ p ‚q “ (a) An -cycle is an element of ker : `Z “C : p Z‚q. “ Z 1 (b) An -boundary is˝ an element` “ of Im Ñ : B ÑC :Ñ B .@ P 1 [Clearly, since 0, we have 0 B Z‚ C p ‚q “.]{ (c) The -th homologyZ module (or simply group) of C is H C : Z B . P p´q ›Ñ R R In fact, for each , H : Ch( Mod) Mod is a covariant additive functor (see Exercise 1, Exercise Sheet 6), which we deﬁne on morphisms as follows:

1 Lemma 8.6‚ ‚ ›Ñ ‚ p ‚ ‚q p ‚ ‚q ‚ Let : C D be a morphism of chain complexes between C and D . Then induces an R-linear map p ‚q p ‚q›Ñ p ‚q ` p ‚qﬁÑ p q` p ‚q H : H C H D Z P B C ˚ B D p ‚q

Z 1 for each P . To simplify, thisp ‚q map›Ñ isp often‚q{ denotedp ‚q by insteadp ‚q of›ÑH p ‚.q{ p ‚q 1 Proof : Fix , and let πp ‚:qZ ÄC p ‚q Z C B CP , resp. π : ˝Z Dp q“ Z´ ˝D pBq“D , be the canonical projections. p qP p` q ˘ 1 First, notice that Z C Z D because if Z , then 0. Hence, p ‚q Ä p ‚q P p ‚q “ ` p q P ` we have Z D . 1 ‚ ` ˘ p q“ ˝ ` p q“ ` ˝ ` p qP1 p ‚q 1 Similarly, we have B C B D . Indeed, if B C , then for some1 C , 1 1 ˝ and because is a chain map we have 1 B D . Therefore, by the universal property of the quotient, there exists a unique R-linear map π such 1 that the following diagram commutes: / / p ‚q p ‚q p ‚q{ π 3 p ‚q Z C Z 1D Z D B D ˝ π ✏ π p ‚q{ p ‚q 1 Z C B C p ‚q “ ˝ Set H : π . The claim follows. p ‚q It should be thought that the homology module` H/ C measures/ the "non-exactness" of the sequence ` 1 ´ Z 1 1 p´q P C C C Moroever, the functors H ( ) are neither left exact, nor right exact in general. As a matter of fact, using the Snake Lemma, we can use s.e.s. of complexes to produce so-called "long exact sequences" of R-modules. Skript zur Vorlesung: Cohomology of Groups SS 2018 34

/ ‚ / ‚ / / Theorem‚ 8.7 (Long‚ exact‚ sequence‚ in homology‚ ) ψ Let 0 C D E 0 be a s.e.s. of chain complexes. Then there is a long exact sequence ` / ˚ / ˚ / / ˚ / ˚ / ¨¨¨ 1 p ‚q p ‚q p ‚q ´ p ‚q ´ p ‚q ¨¨¨ δ ψ δ ψ Z 1 1 HP C Hp ‚qD›Ñ ´Hp E‚q H C H D 1 where for each , δ : H E H C is an R-linear map, called connecting homomorphism. ‚

Note: Here 0 simply denotes the zero¨¨¨› complexÑ ›Ñ, that›Ñ is the›Ñ complex ¨ ¨ ¨ 0 0 0 0 0 ‚

consisting of zero modules and zero morphisms. We often write simply 0 instead of 0 . ‚ ‚ ‚ Z P Proof : To simplify, we denote all diﬀerential maps of the three complexes C , D , E with the same letter , and we ﬁx . First, we apply the/ “non-snake”/ part of the/ Snake Lemma/ to the commutative diagram ψ 0 C D E 0 ✏ ✏ ✏ / ´ / ´ / / ´ 1 ´ 1 ´ ψ 1 1 1 0 C D E 0

and we obtain two exact sequences / / / p ‚q p ‚q p ‚q ψ 0 Z C Z D Z E ´ / ´ / / ´ { 1 ´ { 1 ´ { and ψ 1 1 1 C Im D Im E Im 0´ Shifting indices in both sequences we obtain similar sequences in degrees 1, and respectively. Therefore, we have a commutative diagram with/ exact rows of the/ form: / { { { ` ` ψ ` 1 1 1 C Im D Im E Im 0 ✏ ✏ ✏ / ´ / ´ / ´ p ‚q 1 ´ p ‚q 1 ´ p ‚q ψ 1 1 1 { 0` ›Ñ Z ´ p C‚q Z D Z E 1 ›Ñ ´1 ` Ñ where : C Im Z C is the unique R-linear map induced by the universal property of ‚ ‚ 1 1 the quotient by : C C (as Im ker by deﬁnition of a chain complex), and similarly for D and E . Therefore, the Snake Lemma yieldsp ‚q › theÑ existencep of‚q the connecting homomorphisms

“ p ‚q “ ´ p ‚q δ : ker E coker C loooomoooon loooooomoooooon 1 Z H E H C P

for each ` as/ well as the˚ required/ long˚ exact/ sequence: / ˚ / ˚ / ¨¨¨ 1 p ‚q p ‚q p ‚q ´ p ‚q ´ p ‚q ¨¨¨ δ ψ δ ψ 1 1 H“ C H“ D H“ E H“ C H“ D loomoon loomoon loomoon loooomoooon loooomoooon ker ker ker coker coker Skript zur Vorlesung: Cohomology of Groups SS 2018 35

We now describe some important properties of chain maps and how they relate with the induced morphisms in homology.

‚ ‚ ›Ñ ‚ p ‚q DeﬁnitionZ 8.8 (Quasi-isomorphism) P A chain map : C D is called a quasi-isomorphism if H is an isomorphism for all . ‚ ‚ Warning: A quasi-isomorphism does not imply that the complexes C and D are isomorphic as chain complexes. See Exercise 2, Sheet 5 for a counter-example.

In general complexes are not exact sequences, but if they are, then their homology vanishes, so that there is a quasi-isomorphism from the zero complex.

Exercise‚ [Exercise 3, Exercise Sheet 5] Z Let C be‚ a chain complex of R-modules. ProveP that TFAE: Z (a) C‚ is exact (i.e. exact at pC ‚q“for each );P (b) C is acyclic‚ ›,Ñ that‚ is, H C 0 for all ; (c) The map 0 C is a quasi-isomorphism.

1 ‚ ‚ ‚ ›Ñ ‚ p ‚ ‚q p ‚ ‚q Deﬁnition 8.9 (Homotopic chain maps / homotopy equivalence) Z t ›Ñ ` u P Two chain maps ψ : C D between chain complexes C and D are called 1 1 (chain) homotopic if there exists a familiy´ “ of R`-linear˝ ` maps´ ˝ : C D such that Z 1 1 P ψ

for each . / ` / / / ¨¨¨ ` 1 ´ ¨¨¨ 1 ´ 1´ ´ ´ ` ´ `C C C ´ ψ 1 ψ 1 1 ψ 1 ✏ y ✏ y ✏ / / 1 / / ¨¨¨ ` 1 1 ´ ¨¨¨ ` 1 1 D 1 D D ‚ „ ‚ ‚ ‚ ›Ñ ‚ In this case, we write ψ . ‚ ›Ñ ‚ ‚ ˝ ‚ „ ‚ ‚ ˝ ‚ „ ‚ Moreover, a chain map : C D is called a homotopy equivalence if there exists a chain C D map σ : D C such that„ σ id and σ id . Note: One easily checks that is an equivalence relation on the class of chain maps.

Proposition‚ ‚ 8.10‚ ›Ñ ‚ If ψ : C D are homotopic morphisms of chain complexes, then they induce the same morphisms in homology, that is Z p ‚q“ p ‚q p ‚q›Ñ p ‚q@P H H ψ : H C H D Skript zur Vorlesung: Cohomology of Groups SS 2018 36 Z P P p ‚q

1 1 Proof : Fix and let´ pZq“C . Then,` ` with´ the notationp q“ of` Deﬁnitionp q ` ´8.9, wep q haveP p ‚q ` ˘ ` 1 ˘ 1“ ψ 1 P 1p ‚q B D loooomoooon loooomoooon B D 0 ` p ‚qP p ‚q Hence, for every B C H C , we have p p ‚q´ p ‚qq p ` p ‚qq “ p ‚ ´ ‚qp ` p ‚qq “ ` p ‚q H p ‚q´H pψ ‚q” B C pH ‚q“ ψp ‚q B C 0 B D In other words H H ψ 0, so that H H ψ .

Remark 8.11 (Out of the scope of the lecture!) Homotopy of complexes leads to considering the so-called homomotopy category of R-modules, R denoted Ho( Mod), which is very useful in algebraic topology or representation theory of ﬁnite groups¨ for example. It is deﬁned as follows: “ R R ¨ The objects are the chain complexes, i.e. Ob“ Ho( Mod) {„Ob Ch( Mod) . Ho(RMod) Ch(RMod) The morphisms are given by Hom : Hom .

It is an additive category, but it is not abelian in general though. The isomorphisms in the homotopy category are exactly the classes of the homotopy equivalences.

Dualizing the objects and concepts we have deﬁned above yields the so-called "cochain complexes" and the notion of "cohomology".

Deﬁnition 8.12 (Cochain complex / cohomology)

(a) A cochain complex of‚ R-modules‚ is a sequence´ ´ ` p q“ ¨¨¨›Ñ ›Ñ1 ›Ñ ›Ñ ¨ ¨ ¨ ˆ 1 1 ˙

CZ C C C ` ` P P p q ˝ “ ‚ ‚ ‚ 1 1 p q R where for each , C is an R-module and Hom C C satisﬁes 0. We often write simply“C pinstead‚q “ of C . (b) The elements of Z :“ Z pC ‚q :“ ker ´ are the -cocycles. 1 (c) The elements of B : B C : Im are the -coboundaries‚ p ‚q “. { (d) The -th cohomology module (or simply group) of C is H C : Z B .

Similarly¨ to the case of chain complexes, we can deﬁne: p ‚ ‚q p ‚ ‚q ‚ ‚ ›Ñ ‚ ›Ñ Z ´ ´ ´ Z MorphismsP of cochain˝ complexes“ ˝(or simply cochainP maps) between two cochain complexes C and D ˜ , written 1 1: C 1 D , as a familiy of R-linear maps : C D ( ) such that ˜ for each , that is such that the following diagram Skript zur Vorlesung: Cohomology of Groups SS 2018 37

´ ´ ` / ´ / / ` / ¨¨¨ 2 1 1 ¨¨¨ commutes: ´ 1 ` 1 C ✏ C C ✏ ´ 1 ´ ✏ 1 ` / ´ / / ` / ¨¨¨ 2 1 1 ¨¨¨ ˜ 1 ˜ ˜ 1 ˜ ¨ D D D

¨ subcomplexes, quotient complexes;

¨ kernels, images, cokernels of morphisms of cochain complexes. Cochain complexes together with morphisms of cochain complexes (and composition given by degreewise composition of R-morphisms) form an abelian category, which we will denote by R CoCh( Mod).

Exercise: formulate these deﬁnitions in a formal way.

‚ ‚ ‚ / ‚ / ‚ / ‚ / ‚ Z Theorem 8.13 (Long exact sequence in cohomology) P ψ p ‚q›Ñ ` p ‚q Let 0 C D E 0 be a s.e.s. of cochain 1 complexes. Then, for each , there exists a connecting homomorphism δ : H E H C such that the following sequence is ` ˚ ˚ ˚ ˚ exact: / ‚ / ‚ / ‚ / ` ‚ / ` ‚ / ¨¨¨ 1 p q p q p q p q p q ¨¨¨ δ ψ δ 1 1 ψ H C H D H E H C H D

Proof : Similar to the proof of the long exact sequence in homology (Theorem 8.7). Apply the Snake Lemma.

9 Projective Resolutions Deﬁnition 9.1 (Projective resolution) Let M be an R-module. A projective resolution/ of M / is a non-negative/ complex of projective p ‚ ‚q“ ¨¨¨ R-modules 3 2 1 ` 2 1 0 ˘ P • P p ‚q“P { P – • 0 ‚ 0 1 which is exact at P for every 1 and such that H P P Im M. Moreover, if P is a free⇣R-module for every 0, then P is called a free resolution of M.

0 Notation: Letting ε : P M denote/ the quotient/ homomorphism,/ / / we/ have a so-called augmented complex ¨¨¨ 3 2 1 ε 2 1 0 P P P M 0 ⇣ ‚ associated to the projective resolution, and this augmented complex is exact. Hence we will also denote projective resolutions of M by P M.

Z Z Z /Z ¨ /Z Example 8 “ { The -module M admits the following projective resolution: 0 . Skript zur Vorlesung: Cohomology of Groups SS 2018 38

We now prove that projective resolutions do exist, and consider the question of how "unique" they are.

Proposition 9.2 Any R-module has a projective resolution. (It can even chosen to be free.) ⇣ – { Proof : We use the fact that every R-module is a quotient of a free R-module (Proposition⇣ 6.4). Thus there 0 0 Ñ 0 exists a free module P together with a surjective R-linear map ε : P M such that M P ker ε. { –1 1 1 0 Next, let P be a free R-module together with a surjective R-linear map : P ker ε P such that 1 1 / / / P ker ker ε : 1 < ε 1 0 P " " - P M 1

´ ´ ›Ñ ´ ker ε ⇣ 1 1 2 ´ Ñ ´ Inductively, assuming that the R-homomorphism : P P has already been deﬁned, then { – ´ 1 1 there exists a free R-module P and a surjective R-linear map : P ker P with 1 P ker ker . The claim follows.

1 Theoremp 9.3‚ ‚ (Liftingq p Theorem‚ ‚q ) Let P and Q be two non-negative• chain complexes such that 1. P‚ is a projective R-module for• every 0; p ‚q“ • ⇣ 1 ⇣ 2. Q is exactp at‚qQ for every 1p (that‚q is H Q 0, for all 1). p0 ‚q›Ñ0 p ‚q 0 0 ‚ ‚ ›Ñ ‚ Let ε : P H P and ε : Q H Q be the quotient homomorphims. 1 0 0 p ‚q“ ˝ “ ˝ If : H P H Q is an R-linear map, then there exists a chain map : P Q inducing ‚ 0 0 the given map in degree-zero homology, that is such that H and ε ε . Moreover, such a chain map is unique up to homotopy.‚

In the situation of the Theorem, it is said that 1 lifts .

›Ñ 0 Proof : Existence. Beacuse P is projective and ε is surjective, by definition (Def. 6.7), there exists an 0 0 0 R-linear map : P Q such that the/ following/ diagram/ commutes ¨¨¨ 1 p ‚q“ { ε 0 ö 0 0 1 P D H P P Im 1 ✏ 0 1 ✏ / / / 1 ¨¨¨ 1 p ‚q“ { ε 0 0 0 1 1 1 1 1 ˝ “ ˝ ˝ ˝ Q “ ˝ ˝ H “Q Q Imp ˝ qÑ “ 0 0 1 “ 1 0 1 1 that is ε ε . But then, ε 1 loomoonε 0, so that Im ker ε Im . Again 0 1 ›Ñ 1 ˝ “ ˝ by Definition 6.7, since P is projective and 1 is surjective onto its image, there exists an R-linear map 1 1 1 0 1 1 1 : P Q such that : / 1 ˝ 1 0 1 0 D P ö P 1 ✏ 0 ✏ / / 1 1 / 1 “ 1 0 Q 1 Im 1 ker ε inc Q Skript zur Vorlesung: Cohomology of Groups SS 2018 39 ›Ñ ‚ ‚ ›Ñ ‚ The morphisms : P Q are constructed similarly by induction on . Hence the existence of a chain map : P Q as required. ‚ ‚ ›Ñ ‚ ‚ „ ‚ ‚ ´ ‚ Uniqueness. For the uniqueness statement, suppose ψ : P Q also lifts the given morphism .We have to prove• that ψ“ (or equivalently´ that‚ ‚ψ›Ñis homotopic‚ to the zero chain map). “ ´ “ p ‚q´ p ‚q“ ´ “ ´ ›Ñ p ‚q ´ p ‚q›Ñ For each 0 set σ : ψ , so that σ : P Q is becomes a chain map. In particular 0 0 0 0 0 2 0 1 0 0 σ ψ H H ψ 0. Then we let :0 H Q and : H P Q be the zero maps. Therefore, in degree zero, we/ / have the following/ maps: p ‚q ε 0 0 0 P ´ H P ´ 0 ✏ x x 1 0 2 / / / 1 p ‚q 0 0 ε 0 1 Q H Q 0 “ ´ ˝ ` ˝ ´ ›Ñ ` 2 1 • ›Ñ ` where clearly 0 0 ε . This1 provides us with the starting point for constructing a homotopy ´ § §1 ´ ` ˝ ` ´ ˝ “ •´ 1 : P Q by induction on . So let 0 and suppose : P Q is already constructed 1 1 1 1 1 for each´ “2 p ‚q 1 and´ “ satisfiesp ‚q ´ “ “ ´ σ for“ each “ 1, and´ where“ “ we´ identify

1 0 1 0 2 2 0 1 0 1 1 1 P H P Q H Q´´P˝ 0 Q ε“ ε ` 0

1 1 1 1 1 Now, we check that the image˝ ´ of σ ´ ˝ “ is˝ contained´ ˝ in´ker˝ Im : 1 ` 1 ˘ “ ˝ ´p ´ ´ 1 ´ ˝ ´ q˝ σ 1 σ 1 2 1 “ ˝ σ ´ σ´ ˝ 1 “ ´ σ˝ ´σ ´ ˝ “ 1 1 σ ‚ σ 0 1 ` ›Ñ ` where the last-nut-one1 equality holds because both σ is a chain map. Therefore, again by Deﬁnition 6.7, ` ˝ “ ´ ´ ˝ 1 since P is projective and 1 is surjective onto its image, there exists an R-linear map : P Q 1 1 such that σ : / ´ / / ´ 1 ´ ¨¨¨ D ´ ´ 1 2 ´ ´ ˝ ´ ´ P 1 P 2 P 1 ✏ 1 ✏ w 2 ✏ x σ / w σ/ σ/ / ` 1 1 ´ 1 ´ ¨¨¨ ` ´ 1 1 2 Q 1 Q Q 1 Q 1 ´ “ “ ` ˝ ` ´ ˝ 1 Hence we have ψ σ 1 , as required.

As a corollary, we obtain the required statement on the uniqueness of projective resolutions:

1 ⇣ ⇣ Theorem‚ 9.4 (Comparison‚ Theorem) ‚ ‚ ε ε ‚ ‚ ›Ñ ‚ ‚ ‚ ›Ñ ‚ Let P M and Q M be two projective resolutions of an R-module M. Then P and Q are ‚ ˝ ‚ „ ‚ ‚ ˝ ‚ „ ‚ homotopy equivalent. More precisely, there exist chain maps : P Q and ψ : Q P P Q lifting the identity on M and such that ψ›Ñ Id and ψ Id . M ‚ ‚ ›Ñ ‚ Proof : Consider the identity morphism1 Id : M M. p ‚q“ ˝ “ ˝ ‚ ‚ ›Ñ ‚ By the Lifting Theorem, there exists a chain1 map : P Q , unique up to homotopy, such that 0 M M p ‚q“ 0 ˝ “ ˝ H Id and Id ε ε . Likewise, there exists a chain map ψ : Q P , unique up to 0 M M 0 homotopy, such that H Id and Id ε ε ψ . Skript zur Vorlesung: Cohomology of Groups SS 2018 40

/ / / / / / ¨¨¨ O ¨¨¨ O 1 O O ε 1 ö 0 ö ¨¨¨ D P D ¨¨¨ D P D D P D M 0 ✏ 1 ✏ 1 0 ✏ 0 M ✏ M ψ / / ψ / ψ / 1 Id / Id / ¨¨¨ 1 ¨¨¨ 1 ε 1 0 Q Q 1 Q M 0 ‚ ˝ ‚ ‚ p ‚q›Ñ p ‚q P ‚ ˝ ‚ M„ 0‚ 0 ‚ ˝ ‚ ‚ Now, ψ and Id are both chain maps that lift the identity map Id : H P H P . Therefore, p ‚q›Ñ p ‚q P Q by the uniqueness statement in the Lifting Theorem, we have ψ Id . Likewise, ψ and Id ‚ ˝ ‚ „ ‚ M 0 0 are both chain maps that lift the identity map Id : H Q H Q , therefore they are homotopic, Q that is ψ Id .

Another way to construct projective resolutions is given by the following Lemma, often called the Horse- shoe Lemma, because it requires to ﬁll in a horseshoe-shaped diagram:

1 / 1 / / 2 / 1 ⇣ 1 Lemma 9.5 (Horseshoe Lemma) ‚ 2 ε 1 2 ⇣ 2 2 ‚ Let 0 M M εM 0 be a short exact sequence of R-modules. Let P M be a resolution of M and P M be a projective resolution of M .

. . .✏ .✏ 1 2

P✏ 1 P✏1 1 2

1 2 ✏ 0 ✏0 / P 1 / / P 2 / ε ε ✏ ✏ 0 M M M 0

⇣ 001 2 Z ‚ – ‘ P • ε ‚ ‚ / 1 / / 2 / / 1 / 0 / 2 / Then, there exists a resolution P M of M such that P P P‚ for each‚ ‚ and‚ the s.e.s.‚ 1 π 1 ⇣ 1 0 M‚ M‚ M 0 lifts to a s.e.s. of chain complexes 0 P ‚ P P 0 ⇣ ε ‚ where and π are the canonicalε injection and projection. Moreover, if P M is a projective resolution, then so is P M.

Proof : Exercise 3, Exercise Sheet 6. [Hint: Proceed by induction on , and use the Snake Lemma.]

Finally, we note that dual to the notion of a projective resolution is the notion of an injective resolution:

Deﬁnition 9.6 (Injective resolution) Let M be an R-module. An injective resolution of M is a non-negative cochain complex of injective Skript zur Vorlesung: Cohomology of Groups SS 2018 41

‚ ‚ / / / R-modules p q“ 0 1 2 ¨¨¨ 0 1 2 ` ‚ ˘ I• I I p q“I { – 0 0 ã which is exact at I forÑ every 1 and such that H I ker 0 M. 0 Notation: Letting : M I denote the natural/ injection,/ / we have/ a so-called augmented complex 0 1 ¨¨¨ 0 1 2 M p ‚ I‚q I I ã ‚ Ñ associated to the injective resolution I , and this augmented complex is exact. Hence we will also denote injective resolutions of M by M I .

Similarly to projective resolutions, one can prove that an injective resolution always exists. There is also a Lifting Theorem and a Comparison Theorem for injective resolutions, so that they are unique up to homotopy (of cochain complexes).

10 Ext and Tor We now introduce the Ext and Tor groups, which are cohomology and homology groups obtained from applying Hom and tensor product functors to projective/injective resolutions. We will see later that Ext groups can be used in group cohomology to classify abelian group extensions.

⇣ Z Deﬁnition 10.1 (Ext-groups) ‚ P • ε 0 Let M and N be two left R-modules and let P M be a projective resolution of M. For , the n-th Ext-group of M and N is p q “ p ‚ q R ` R ˘ Ext MN : H Hom P N p ‚ q R that is, the -th cohomology group of the cochain complex Hom P N .

Recipe: ‚ 1. Choose a projective resolution P of M. p´ q R 2. Apply the left exact contravariant functor Hom/ N/ to the projective/ resolution ‚ “ ¨¨¨ 3 2 1 ` 2 1 0 ˘ P P P P

˚ ˚ ˚ to obtain a cochain complex / / / p q p q p q ¨¨¨ 1 2 3 R 0 R 1 R 3 Hom P N Hom P N Hom P N

of abelian groups (which is not exact in general).

3. Compute the cohomology of this new complex. p q First of all, we have to check that the deﬁnition of the abelian groups ExtR MN is independent from the choice of the projective resolution of M. Skript zur Vorlesung: Cohomology of Groups SS 2018 42

1 ⇣ ⇣ Proposition‚ 10.2 ‚ p ‚ q ε ε p ‚ q ` R ˘ If P M and Q M are two projective resolutions of M, then the groups H Hom P N ` R ‚ ˘ ‚ and H Hom Q N are (canonically) isomorphic, via the homomorphisms induced by the chain M maps between P and Q given by the Comparison Theorem‚ ‚ ›Ñ applied‚ to the‚ identity‚ ›Ñ morphism‚ Id . ‚ ˝ ‚ „ ‚ ‚ ˝ ‚ „ ‚ Proof : By the Comparison Theorem, there exist chain maps : P Q and ψ : Q P lifting the p´ q P Q identity on M and such that ψ Id and ψ Id . ˚ R ˚ Now, applying the functorp ‚ Homq›Ñ N pyields‚ q morphisms of cochainp ‚ complexesq›Ñ p ‚ q R R ˚ ˚ R ˚R ˚ ‚ ˝‚ „: Hom‚ Q N‚ ˝ ‚ „Hom‚ P N and ψ˝ : Hom„ P pN‚ q Hom ˝ Q „N p ‚ q ˚ R R Q P Hom P N Hom Q N Since ψ Id and ψ Id , it follows that ψ Id and ψ Id . ˚ But then, passing to cohomology, inducesp ‚ a groupq ›Ñ homomorphismp ‚ q ` R ˘ ` R ˘ ˚ ˚ : H Hom Q ‚N H Hom P N

˚ (see Exercise 1, Exercise Sheet 5). Sincep ‚ q is›Ñ unique up top homotopy,‚ q so is , and hence is unique because˚ homotopic˚ chain ˚ maps˚ induce the same morphisms˚ ˚ in cohomology.˚ ˚ Likewise, there is˚ a ‚ ˝ „ ` ˝ R „ ˘ ` ˝ R “ ˘ ˝ “ unique˚ homomorphism ψ : H Hom P N H Hom Q N of abelian groups induced by ψ . Finally, ψ Id and ψ Id imply that ψ Id and ψ Id. Therefore, and ψ are canonically deﬁned isomorphisms. Z Proposition 10.3 (Properties of ExtR ) P ° 1 2 1 2 0 Let MM pM andq–NN Np be qR-modules and let be an integer. The following holds: 0 R (a) ExtR MN Hom MN . ›Ñ

˚ 1 2 (b) Any morphism of R-modules α : M p M inducesq›Ñ a groupp homomorphismq 2 1 α : ExtR M N ExtR M N ›Ñ 1 2 (c) Any morphism of R-modules β˚ : N p N inducesq›Ñ a groupp homomorphismq 1 2 β : ExtR MN ExtR MN p q“ • (d) If P is a projective R-module, then ExtRp PNq“ 0 for all • 1. (e) If I is an injective R-module, then ExtR MI 0 for all 1. ⇣ ‚ p´ q ε ‚ R Proof : (a) Let P M be a projective resolution of M. Applying the left exact functor Hom N to ˚ ˚ ˚ the resolution P yields the cochain/ complex / / p q 1 p q 2 p q 3 ¨¨¨ R 0 R 1 R 3 Hom P N Hom P N Hom P N ˚ ˚ p q“ p ‚ q “ { – Therefore, 0 0 ` R ˘ ExtR MN H Hom P N ker 1 0 ker⇣1 ¨¨¨ ›Ñ1 ›Ñ ›Ñ ‚ ε ε 0 Now, the tail P M 0 of the augmented complex P M is an exact sequence ˚ of R-modules, so that the induced/ sequence ˚ / / p q p q 1 ¨¨¨ ε R R 0 0 Hom MN Hom P N Skript zur Vorlesung: Cohomology of Groups SS 2018 43 p q p q

R R 0 ˚ ˚ is exact at Hom MN and atp Hom q–P N and“ it follows– thatp q 0 R ˚ ExtR MN ker 1 Im ε Hom MN 1 ⇣ 1 ⇣ because‚ ε is injective. ‚ ε ε 1 1 1 ‚ ‚ 2›Ñ ‚ ‚ 2 (b) Let P M be a projective˚ resolution1 of M and P M be a projective˚ resolution of M . The p ‚ q›Ñ p ‚ q Lifting Theorem implies that α lifts to a chain map : P P . Then, induces a morphism R R of chain complexes : Hom ˚P N p Homq›ÑP Np and thenq induces a morphism in cohomology ˚ ˚ 2 1 • “ : ExtR M N ExtR M N ⇣ for each‚ 0 and we set α : . ‚ ε p ‚ q›Ñ p ‚ q (c) Let P M be a projective resolution of M. Then, there is a morphism of cochain complexes R ˚1 R 2 β : Hom P N Hom P N induced by β, which, in turn, induces a homomorphism of ⇣ / / / ‚ ¨¨¨ abelian groupsε β in cohomology. “ “ ›Ñ (d) Let P M be a projective resolution of M.Choose 0 0 P as a projective 0 1 P resolution of P (i.e. P : P, P 0, . . . ), augmented by the identity morphism Id : P P. Then the induced cochain complex is / / / p q ¨¨¨ 0 0 0 R p q“Hom PN • 0 0 ⇣ R ‚ p´ q so that clearlyε Ext PN 0 for each 1. R (e) Let P M be a projective resolution of M. Since I is injective, the functor Hom I is exact. ˚ ˚ ˚ Therefore the induced cochain complex/ / / p q 1 p q 2 p q 3 ¨¨¨ R 0 R 1 R 2 Hom P I Hom P I Hom P I

is exact and its cohomology is zero. The claim follows. Z Remark 10.4 P • p´ q ›Ñ p ´q ›Ñ 0 R R Using the proposition one can prove that for every , Ext N : Mod Ab is a R contravariant additive functor, and ExtR M : Mod Ab is a covariant additive functor.

Theorem 10.5 (Long exact sequences of Ext-groups) / / / / ψ 1 2 3 (a) Any s.e.s. 0 N N N 0 of R-modules induces a long exact sequence of abelian groups/ ˚ / ˚ / / / p q p q p q 0 p q 0 0 ψ 0 δ 1 / R 1 ˚ / R 2 ˚ / R 3 / `R 1 / 0 Ext pMN q Ext pMN q Ext pMN q Ext Mp N q ψ δ 1 1 2 3 1 ExtR MN ExtR MN ExtR MN ExtR MN / / / / α β 1 2 3 (b) Any s.e.s. 0 M M M 0 of R-modules induces a long exact sequence of ˚ abelian groups/ / ˚ / / / p q p q p q 1 p q β α δ 0 ˚ 0 ˚ 0 1 / R 3 / R 2 / R 1 / `R 3 / 0 Ext pM Nq Ext pM Nq Ext pM Nq Ext Mp N q β α δ 1 3 2 1 3 ExtR M N ExtR M N ExtR M N ExtR M N Skript zur Vorlesung: Cohomology of Groups SS 2018 44 ‚

Proof : (a) Let P be a projective resolution of M. Then there is an induced short exact sequence of cochain ‚ ‚ complexes / / / / p ‚ q p ‚ q p ‚ q ψ R 1 R 2 ZR 3 0 Hom P N Hom P N HomP • P N 0 0 because each module/ P is projective.˚ Indeed,/ at each degree˚ / this sequence/ is p q p q p q ψ R 1 R 2 R 3 0 Hom P N p Hom´q P N Hom P N 0 R obtained by applying the functor Hom P ˚ , which is exact as P is projective. It is then easily checked that this gives a s.e.s. of cochain complexes, that is that the induced diﬀerential maps commute with the induced homomorphisms . Thus, applying Theorem 8.13 yields the required long exact‚ sequence in cohomology. ‚ 1 ‚ 3 (b) Let P be a projective resolution of M and let Q be a projective resolution of M . By the 2 Horseshoe Lemma (Lemma 9.5), there exists a projective resolution R of M and a short exact / / / / sequence of chain complexes ‚ ‚ ‚

0 P R Q 0 • lifting the initial s.e.s. of R-modules. Since/ Q / is projective/ for each/ 0, the sequences

• 0 P R Qp´ q 0 / / R / / are split exact for each 0p. Thereforeq applyingp Homq N yieldsp a splitq exact s.e.s. R R R 0 • Hom Q N Hom R N Hom P N 0 / / / / for each for each 0. It followsp ‚ q that there is ap s.e.s.‚ q of cochain complexesp ‚ q R R R 0 Hom Q N Hom R N Hom P N 0

The associated long exact sequence in cohomology (Theorem 8.13) is the required long exact sequence.

p ´q p´ q The above results show that the Ext groups “measure” and "repair" the non-exactness of the functors R R Hom M and Hom N .

The next result is called “dimension-shifting” in the literature (however, it would be more appropriate to call it “degree-shifting”); it provides us with a method to compute Ext-groups by induction.

Remark 10.6 (Dimension shifting) Let N be an R-module and consider a s.e.s./ / / / α β 0 L P M 0 p q“ • of R-modules, where• is P projective (if M is given, take e.g. P free mapping onto M, with kernel L). Then ExtR PN 0 for all 1 and applying the long exact sequence of Ext-groups yields at ˚ ˚ each degree 1 an exact sequence/ of the form / ` / p q p q α δ 1 β 0 ExtR LN ExtR MN 0 Skript zur Vorlesung: Cohomology of Groups SS 2018 45

` where the connecting homomorphism δ isp thereforeq– forcedp toq be an isomorphism: 1 ExtR MN ExtR LN

Note that the same method applies to the second variable with a short exact sequence whose middle term is injective.

A consequence of the dimension shifting argument is that it allows us to deal with direct sums and products of modules in each variable of the Ext-groups. For this we need the following lemma:

Lemma 10.7

Consider the following commutative diagram1 / of R-modules/ 2 with/ exact rows: α β A✏ A✏ A 0 1 / / 2 /

B B ψ B 0 P p 2 2q ˝ “ ˝ R Then there exists a morphism Hom A B such that β ψ . Moreover, if and are isomorphisms, then so is .

Proof : Exercise 5, Exercise Sheet 7.

Proposition 10.8 (Ext and direct sums) t u P I (a) Let M be a family of R-modules and let N be an R-module. Then ˜ ¸ – p q@• P P à π ExtR M N ExtR M N 0 I I t u P I (b) Let M be an R-module and let N be a family of R-modules. Then ˜ ¸ – p q@• P P π π ExtR M N ExtR MN 0 I I

“ Proof : (a) Case 0. By Proposition 10.3(a) and the universal property of the direct sum (Proposi- tion 4.2), we have ˜ ¸ – ˜ ¸ – p q– p q 0 P P P P à R à π R π ExtR M N Hom M N Hom M N ExtR M N I • I P I I / / / / Now, suppose that 1 and choose for each I a s.e.s. of R-modules 0 L P M 0 where P is projective (e.g. choose/ P free/ with quotient isomorphic/ to M/ and kernel L ). These sequences induce a s.e.s. P P P À À À 0 I L I P I M 0 Skript zur Vorlesung: Cohomology of Groups SS 2018 46 • Case 1: We proceed by induction on . First, for n=1, using a long exact/ sequence of Ext-groups,/ we obtain a commutative/ diagram p P q p P q 0 p P q p P q δ 1 1 R À ö R À R À R À Hom –I P N Hom –I L N Ext I M N Ext I P N ✏ ✏ / / / P p q P p q P p q P p q 1 1 ± R ± R ± ± I Hom P N I Hom L N I ExtR M N I ExtR P N

with¨ the following properties: ¨ the morphisms of the bottom row are induced componentwise; ¨ “ both rows are exact; and the two vertical isomorphismsP are given byP the case 0. À Since P is projective for every I, so is I P , thus Proposition 10.3 yields ˜ ¸ – – p q@P 1 P P 1 R à π R Ext I P N 0 I Ext P N I

Therefore Lemma 10.7 yields ˜ ¸ – p q 1 P P 1 R à π R Ext I M N I Ext M N • p ´ q Now assume that 2 and assume that the claim holds for the 1 -th Ext-groups, that is ´ ´ ˜ ¸ – p q 1 P P 1 R à π R Ext I L N I Ext L N

Then, applying the Dimension Shifting argument yields ´ ˜ ¸ – ˜ ¸ 1 P P R à R à Ext I L N Ext I M N ´ p q– p q@P and 1 R R Ext L´N Ext M N I p q– p q so that P 1 P π R π R I Ext L N I Ext M N

Hence the required isomorphism ˜ ¸ – p q P P R à π R Ext I M N I Ext M N

(b) Similar to (a): proceed by induction and apply a dimension shift. (In this case, we use s.e.s.’s with injective middle terms.)

b ´ ´b To end this chapter, we introduce the Tor-groups, which “measure” the non-exactness of the functors R R M and N. Skript zur Vorlesung: Cohomology of Groups SS 2018 47

‚ DeﬁnitionZ 10.9 (Tor-groups) P • Let M be a right R-module and N be a left R-module. Let P be a projective resolution of N. For 0 , the n-th Tor-group of M and pN is q “ p b ‚q R R Tor MN : H M b P ‚ R that is, the -th homology group of the chain complex M P .

Proposition 10.10 1 2 3 1 2 3 Let MM M M bep right qR-modules and let NN N N be left R-modules. R (a) The groupp q–Tor MbN is independant of the choice of the projective resolution of N. R R (b) Tor0 p´MNq M N. R (c) Tor p N´q is an additive covariant functor. R (d) Tor pM ‘ is anq– additivep covariantq‘ functor.p q R R R 1 2 1 2 (e) Tor pM M‘ Nq–TorpM Nq‘TorpM Nq. R R R 1 2 1 2 (f) Tor MN N Tor MN Tor MN . p q“ • R (g) If either M or N is ﬂat (so in particular if either M or N is projective), then Tor MN 0 for all 1. / / / / α β 1 2 3 (h) Any s.e.s. 0 M M M 0 of right R-modules induces a long exact sequence / ` / ˚ / ˚ / / ¨¨¨ ` p q 1 p q p q p q ¨¨¨ R δ R α R β R δ 3 1 2 3 Tor/ 1 M N Tor/ M N b Tor/ M N b Tor/ M N / ¨¨¨ p q 1 b N b N b R δ α Id β Id 3 1 R 2 R 3 R Tor1 M N M N M N M N 0

of abelian groups./ / / / α β 1 2 3 (i) Any s.e.s. /0 N N ` / N 0 of˚ left/ R-modules induces˚ / a long exact sequence/ ¨¨¨ ` p q 1 p q p q p q ¨¨¨ R δ R α R β R δ 3 1 2 3 Tor/ 1 MN Tor/ MN b Tor/ MN b Tor/ MN / ¨¨¨ p q 1 b M b M b R δ Id α Id β 3 R 1 R 2 R 3 Tor1 MN M N M N M N 0

of abelian groups.

The proof of the above results are in essence similar to the proofs given for the Ext-groups.