# are additive in each variable. Explicitly, the condition on composition means that given a diagram

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1 1. Abelian categories Most of homological algebra can be carried out in the setting of abelian categories, a class of categories which includes on the one hand all categories of modules and on the other hand categories of (abelian) sheaves. In this section we give a quick introduction to abelian categories. There are three layers of structure, and we discuss each in succession Pre-additive categories. In a general category C the homsets C(A, B) are simply sets; however, in many specific cases we find that these sets carry some additional structure. For example, consider the category Pos of posets and order-preserving functions. For two posets A, B, the homset Pos(A, B) is again an ordered set if we let f g f(x) g(x) for all x A. Similarly, in the category Rel of sets and relations we may define an ordering on Rel(A, B) by saying that R S if and only if R (as a subset of A B) is contained in S, i.e. if arb implies asb. Also, in the category of abelian groups, the homset AbGrp(A, B) is more than a mere set: it is itself an abelian group under the operation of pointwise addition. More generally, in any category of modules we have that R Mod(A, B) is an abelian group. The phenomenon that categories may have homsets with additional structure goes by the name of enrichment: we say that the category Rel is enriched in posets, or that the category of R-modules is enriched in abelian groups. We will not go into the theory of enriched categories in general (a standard reference is Max Kelly s book Enriched Category Theory ) but only concentrate on the abelian group case. Definition 1.1. A category C is called pre-additive (equivalently: enriched in abelian groups) if for each pair of objects A, B of C the homset C(A, B) is an abelian group; moreover, it is required that the composition functions are additive in each variable. C(B, C) C(A, B) C(A, C) Explicitly, the condition on composition means that given a diagram we have A f B g h C k D (g + h)f = gf + hf; k(g + h) = kg + kh. Note that pre-additivity is really an extra piece of structure on the category: in particular, a category may be pre-additive in different ways. To see this, take a category with two objects A, B and let Hom(A, A) = 0 = Hom(B, B), and Hom(A, B) a 4-element set. Then since we have Z/4Z = Z/2Z Z/2Z there are two different abelian group structures on Hom(A, B). Examples 1.2. (1) As mentioned earlier, any category of R-modules is pre-additive when we define a pointwise group structure on the homsets. In particular the category of abelian groups is pre-additive. 1

4 Abelian categories. The final step to make is to add to the notion of additive category some exactness properties: Definition 1.8 (Abelian Category). An additive category C is called abelian if: (i) every map in C has a kernel; (ii) every map in C has a cokernel; (iii) every monomorphism in C is the kernel of its cokernel; (iv) every epimorphism in C is the cokernel of its kernel. Note that this definition is again self-dual so that C is abelian if and only if C op is. The conditions on monomorphisms and epimorphisms imply that these are regular; in particular, an abelian category is balanced, in the sense that any morphism which is at the same time mono and epi is an isomorphism. Examples 1.9. (1) The leading example of an abelian category is R Mod for any ring R. (2) If C is abelian then so is C D for any D. (3) If C is abelian and D is (pre-)additive then the category of additive functors D C is again abelian. (Thus the fact that R Mod is abelian is a consequence of the fact that this category is of the form AbGrp R and that AbGrp is abelian.) (4) The category of finite abelian groups is abelian. (5) The category of free abelian groups is additive but not abelian (check!). We begin with a few easy consequences of the definition. Lemma In an abelian category, every morphism factors as an epimorphism followed by a monomorphism. Moreover, such factorization is essentially unique. Proof. Consider a morphism f : A B. First form the kernel Ker(f) A. Next, take the cokernel of the result, as to obtain A e Coker(Ker(f)). Since fi = 0, the universal property of the cokernel induces a factorization f = me where m : Coker(Ker(f)) B. Usually, one writes Im(f) for Coker(Ker(f)). It is now straightforward to check that m is a monomorphism. Moreover, given any other factorization f = m e with m monic, e epi, we get a diagram Im(f) e m A B e m D where the dotted arrow arises from the universal property of the cokernel. Using the universal property of e one obtains an arrow in the opposite direction and it is readily seen that these are inverse. i

6 6 m To see that Hom R (A, ) is left exact, consider a kernel Ker(f) M f m N. We wish to show that Hom R (A, M) Hom R (A, Ker(f)) is the kernel of Hom R (A, f). Thus take p : A M. Then f (p) = 0, that is, pf = 0, if and only if p factors through Ker(f), if and only if p is of the form m (q). However, Hom R (A, ) is typically not right exact. Indeed, consider an epimorphism M e N. To say that e : Hom R (A, M) Hom R (A, N) is surjective is to say that every f : A N is in the image of e, i.e. is of the form f = ef for some f : A M. See the diagram Thus, we see: M f e A f N. Proposition For a fixed R-module A, the (covariant) hom-functor Hom R (A, ) : R Mod AbGrp is left exact. Moreover, it is exact if and only if A is projective. We may dualize all of the above, and consider, for a fixed R-module B, the (contravariant!) functor Hom R (, B) : (R Mod) op AbGrp. Using the fact that kernels in C op are cokernels in C, one finds that Hom R (, B) preserves kernels but in general not cokernels. Therefore: Proposition For a fixed R-module B, the contravariant hom-functor Hom R (, B) : (R Mod) op AbGrp is left exact. Moreover, it is exact if and only if B is injective. Exercise 11. Consider the forgetful functor R Mod AbGrp. Is this functor left exact? Right exact? Exercise 12. Suppose F : C D is an additive functor which has a left adjoint. Prove that F is left exact. Formulate and derive a dual statement The Freyd-Mitchell embedding theorem. Categories of modules are not only the motivating example of abelian categories, but they are also a universal example, in the sense that for most questions about abelian categories can be settled in categories of modules. More concretely, there are several related embedding theorems (for the reader interested in more information on this, see Peter Freyd s book on Abelian Categories, freely available online). We state one useful version here: Theorem Let C be a (small) abelian category. Then there exists a ring R and a full embedding C R Mod. This embedding preserves all abelian structure (i.e. is additive and preserves kernels and cokernels). So, every small abelian category is a full abelian subcategory of a category of modules. The power of this result lies in the fact that we can now reduce

7 problems about abelian categories in general to problems about categories of modules. More precisely, in order to show that a given statement holds in an arbitrary abelian category it suffices to show that it holds in categories of modules. As an illustration, consider Lemma 1.10: we gave a general arrow-theoretic proof of this, but in a category of modules there is a more intuitive proof using elements and functions (by using the usual surjectiveinjective factorization of a homomorphism). The embedding theorem now says that it would have been sufficient to prove the lemma only for categories of modules. There are a few things to keep in mind: first, one cannot use the theorem to generalize every possible statement about categories of modules to arbitrary abelian categories. For example, consider the statement Abelian categories are complete. Certainly categories of modules are complete, but not every abelian category is (for example, the category of finite abelian groups is not). The problem here is that the statement is a statement about arbitrarily large collections of objects and arrows, while the embedding theorem only applies to small abelian categories. By contrast, a statement like In an abelian category the pullback of an epimorphism is again an epimorphism works fine, since this involves only finitary data about objects and arrows in the category. In short, the embedding theorem works if you have a statement which is really a statement about small subcategories of abelian categories. Note that given a small set of objects and arrows of a given abelian category we may always consider the abelian subcategory generated by these objects and arrows, and that this subcategory is again small. Exercise 13. Give examples of statements which are true in some abelian categories but not in all categories of modules. Exercise 14. Give an example of an abelian category which does not have enough projectives. Why does the embedding theorem not apply? 7

9 This is called the n-th homology object of C. In a category of modules, this is thus a quotient module; in a general abelian category it may be constructed as the cokernel of the inclusion Im(d n+1 ) Ker(d n ). Since we have an object H n (C) for each n Z, H(C) = (H n (C)) n Z is itself a Z-graded object of C. The following notation and terminology is standard, especially in categories of modules: Z n = Z n (C) is the object Ker(d n ) C n, and is called the object of n-cycles of C; B n = B n (C) is the object Im(d n+1 ) C n, and is called the object of n-boundaries of C; in concrete cases we call elements of Z n n-cycles and elements of B n n-boundaries. If two elements of Z n differ by a boundary (i.e. if a b B n for a, b Z n ) then we say that a and b are homologous. Finally, the elements of H n (C) = Z n (C)/B n (C) are called homology classes of C. Examples of chain complexes will be given below. For now, we note the following: given any composable pair of maps A p B q C, we say that this diagram is exact at B if Ker(q) = Im(p). In a chain complex C, we only have Im(d n+1 ) Ker(d n ), so C is not necessarily exact at C n. The homology object H n (C) measures how far C is from being exact at C n. This explains the following definition and terminology: Definition 2.2 (Exact Complex, Acyclic Complex). A chain complex C n is exact if it is exact at each C n. It is called acyclic if H n (C) = 0 for all n. Clearly a chain complex is exact if and only if it is acyclic. The term acyclic stems from the fact that if the homology H n (C) (cycles modulo boundaries) vanishes, then all cycles are killed by some boundary. Next, we need to organize chain complexes into a category. Definition 2.3 (Chain Map). A morphism of chain complexes from (C, d) to (D, d ) is a graded morphism f : C D which commutes with the differentials. Explicitly, for each n Z, the following diagram is required to commute: C n f n d D n d C n 1 fn 1 D n 1. Morphisms of chain complexes are usually simply called chain maps. It is clear that such maps compose and that the degreewise identity is a chain map. Thus we obtain a category Ch(C) of chain complexes in C and chain maps between them. Exercise 15. Show that the category Ch(C) is a full subcategory of the diagram category C Zop where Z is the poset of integers regarded as category. The following exercise collects a few elementary facts about boundaries, cycles and homology: Exercise 16. Let (C, d) be a chain complex. Show that (i) the graded object Z = (Z n ) of cycles can be made into a subcomplex of C; 9

10 10 (ii) the graded object B = (B n ) of boundaries can be made into a subcomplex of C; (iii) the graded object H = (H n ) can be made into a complex, but has trivial differential. Exercise 17. Let (C, d) be a chain complex of R-modules. Suppose c n are elements of R, and define a new differential by letting d n = c n d n. Check that (C, d ) is again a chain complex. Now suppose that every c n {1, 1}. How does the homology of (C, d ) relate to that of (C, d)? Moreover, the condition on morphisms that they commute with the differentials ensures the following: Exercise 18. Any chain map f : C D sends cycles to cycles and boundaries to boundaries, and induces morphisms in homology H n (f) : H n (C) H n (D). As a consequence of the preceding exercise, we find that the assignment C H n (C) is a functor Ch(C) C for each n Z. This is useful for various reasons; for example, one now knows that if two chain complexes are isomorphic, then they have isomorphic homology. Example 2.4 (Simplicial homology). The following example of a chain complex stems from topology. Consider the CW-complex X which looks like That is, X is obtained by glueing three copies of the unit interval into a triangle. We think of this space as consisting of three vertices (0-cells), labelled a, b, c and three edges (1-cells), labelled f, g, h and no higher cells. Thus the picture becomes: f a b h g c We now build a chain complex S(X) from X. We define S(X) 0 to be the free abelian group on the generating set {a, b, c} and we let S(X) 1 be the free abelian group on the generating set {f, g, h}. We let S(X) be 0 in all other degrees. (We say that the chain complex S(X) is concentrated in degrees 0, 1). Writing a, b, c for the free abelian group on {a, b, c} and so on, we thus obtain 0 f, g, h d a, b, c 0 0 S(X) 1 d S(X) 0 0 We must now define a differential d : S(X) 1 S(X) 0 ; it suffices to specify d(f), d(g) and d(h), since S(X) 1 is free on {f, g, h}. Define d(f) = b a, d(g) = c b and d(h) = c a. Note how this chain complex captures

11 the combinatorial information about the space X; the differentials encode the way the 1-cells are glued together. Now we may compute the homology groups of S(X); in degree 0, we find (check) that the generator a is not in the image of d, and thus d is not surjective. However, d(f) and d(g) are linearly independent so we may describe the image of d as the subgroup of a, b, c generated by (b a) and (c b). Therefore, the group H 0 (S(X)) = a, b, c /Im(d) is isomorphic to Z. In degree 1, we find that Ker(d) contains a non-zero element, namely f + g h. It is easily seen that any other element of Ker(d) is a multiple of this element, so that H 1 (S(X)) = Ker(d) is the free abelian group on one generator f + g h. For those acquainted with singular homology groups this will not come as a surprise, since H 1 (S 1 ) = Z. Exercise 19. Consider this time, instead of the open triangle in the example above, a solid triangle Y, to be thought of as three vertices, three 1-cells, and one 2-cell. Adapt the construction of a chain complex from the example above to this situation. What are the 1-boundaries? What is the homology? Do the same for the solid tetrahedron and the hollow tetrahedron (union of four triangles). In practice, one often deals with complexes which are zero in all negative degrees (we say that such a complex is concentrated in positive degrees, or that it is a non-negative complex). To indicate this, we write C n C n 1 C 1 C 0 0 Similarly, a complex which does not have nonzero terms above dimension n will be written 0 C n C n 1. We conclude this section with some structural properties of the construction of Ch(C) from C. Most details will be left as exercises. First, the category Ch(C) inherits all limits and colimits which happen to exist in C (and since C was assumed to be abelian, this includes biproducts, zero object, kernels and cokernels). This is proved as follows: the category Ch(C) is a full subcategory of a category of diagrams in C. We know that this category of diagrams inherits all these limits and colimits, since they may be computed pointwise. It now suffices to check that the category Ch(C) inherits these constructions. Exercise 20. Verify this. Also verify that every monic in Ch(C) is a kernel, and every epi is a cokernel. Conclude that Ch(C) is abelian. This allows for the following formulation: Proposition 2.5. The construction C Ch(C) is a functor from the category of abelian categories and additive functors to itself. By the previous exercise, we only need to verify functoriality; we leave it as an exercise to the reader to check that an additive functor F : C D induces an additive functor Ch(F ) : Ch(C) Ch(D). 11

12 Cohomology. We shall now briefly indicate how the above concepts dualize. First of all, a cochain complex in C is a diagram C n 1 dn 1 C n d n C n+1 dn+1 such that d 2 = 0. The elements of the objects C n (again, if we are in a category of modules) are referred to as n-cochains, those of Ker(d n ) as n-cocycles, and those of Im(d n 1 ) as n-coboundaries. Moreover, we define the n-th cohomology object of C to be the quotient H n (C) = Ker(d n )/Im(d n 1 ). Again we usually denote cochain complexes by (C, d), or by (C n, d n ). A morphism of cochain complexes f : (C, d) (D, d ) (usually called a cochain map) is a graded map f n : C n D n commuting with differentials. Any morphism of cochain complexes sends cocycles to cocycles and coboundaries to coboundaries, and so induces a morphism H n (f) : H n (C) H n (D) in cohomology. We denote by CoCh(C) the category of cochain complexes in C and cochain maps. Each H n ( ) is a functor CoCh(C) C. The following proposition is easily obtained by dualizing results in the previous section: Proposition 2.6. The category CoCh(C) is abelian. The assignment C CoCh(C) is a functor from the category of abelian categories and additive functors to itself. Exercise 21. Prove the proposition by indicating how to dualize the results about chain complexes. The following exercise gives an idea of how cochain complexes can arise, namely by taking a chain complex and applying a contravariant hom-functor: Exercise 22. Let (C, d) be a chain complex in C, and let B be an object of C. Verify that we have a cochain complex whose degree n object is Hom(C n, B), and whose codifferentials are given by composition with the differentials of C. Also show that this process is contravariantly functorial in C, so gives rise to a functor Hom(, B) : (Ch(C)) op CoCh(C). Also show that the process is covariantly functorial in B Homotopy. We will now turn to the question of when two chain complexes have the same (isomorphic) homology, as well as the question of when two chain maps induce the same map in homology. Because of the functoriality results of the first section, we know that isomorphic chain complexes have isomorphic homology; however, the converse does not hold as witnessed by the example after this definition. Definition 2.7 (Quasi-isomorphism). A morphism f : C D of chain complexes is a quasi-isomorphism if H n (f) is an isomorphism for all n.

13 Here is an example of a quasi-isomorphism (which is not an isomorphism). Consider 13 (1) 0 Z 4 Z d Z/2Z 0 f 2 f 1 f Z/4Z d Z/2Z 0 The differential d is the quotient n n(mod2), as is d. The first homology of the top complex is 2Z/4Z = Z/2Z. All other homology groups of the top complex are trivial. The second complex also has homology Z/2Z in degree one and has trivial homology in all other degrees. The chain map f is defined by f 1 (n) = n(mod2), and f 2 = 1. It is now easily verified that f 1 induces an isomorphism in homology. Clearly, f is itself not an isomorphism of chain complexes. In general it is not easy to determine whether a given map is a quasiisomorphism. Also, the question whether H n (f) = H n (g) (or, equivalently, when f g induces the zero map in homology) is a hard one. Therefore we shall now investigate a stronger notion which is both better behaved and more managable in practice. Definition 2.8. Let f, g : (C, d) (D, d ) be chain maps. A homotopy from f to g is a degree 1 map α n : C n D n+1 such that f n g n = d n+1α n + α n 1 d n. In this case, we write f g and say that the maps f and g are homotopic. Note the parallel with the notion of homotopy between continuous functions of topological spaces. Diagrammatically, a homotopy α from f to g may be pictured as d d C n+1 C d n d C n 1 α n+1 f g α n f g α n 1 f g d d D n+1 d D n d D n 1 Example 2.9 (Simplicial Homotopy). Let us give an example of homotopy which will make clear how this notion is related to the notion of homotopy between continuous maps of spaces. Consider the following two spaces: the first is the unit interval I, with endpoints labelled 0, 1 and edge labelled i. The second is the solid triangle Y, whose vertices we label a, b, c, whose edges we label f, g, h and whose 2-cell is labelled γ. To these spaces we associate chain complexes S(I) and S(Y ), respectively. S(Y ) is the following complex: 0 γ d 2 f, g, h d 1 a, b, c 0 0 S(Y ) 2 d 2 S(Y ) 1 d 1 S(Y ) 0 0

14 14 Here, the map d 1 is defined as in example 2.4, while d 2 (γ) = g h + f. The complex S(I) is: 0 i d 0, S(I) 1 d S(I) 0 0 with differential d(i) = 1 0. We now may consider two maps from I to Y, the first sending i to f and the second sending i to h. These maps induce chain maps φ, ψ : S(I) S(Y ) in the obvious manner. We wish to construct a homotopy α from φ to ψ. In degree 0, define α 0 (0) = 0, and α 0 (1) = g. Then certainly we have dα 0 (0) = 0 = a a = φ 0 (0) ψ 0 (0), and also dα 0 (1) = d( g) = dg = b c = φ 0 (1) ψ 0 (1). In degree 1, let α 1 (i) = γ. Then dα 1 (i) + α 0 d(i) = dγ + α 0 (1 0) = (g h + f) g = f h = φ 1 (i) ψ 1 (i). Thus, α is indeed a homotopy from α to β. Thus, the situation is very similar to that in algebraic topology, where one has two relations on continuous functions: the first is that of being homotopic (this is the stronger relation) and the second is that of inducing the same homomorphism in singular homology. In fact, the ambitious reader may wish to verify how the passage from topological spaces to chain complexes sends homotopic maps of spaces to homotopic chain maps. The following exercises develop a few elementary facts about the homotopy relation: Exercise 23. Let F : C D be an additive functor and let f g in C. Then F f F g in D. Exercise 24. The relation is an equivalence relation on the set of chain maps from C to D. Moreover, the homotopy relation is compatible with composition on both ends, in that f g implies fh gh and kf kh. Finally, is compatible with the additive structure, in that f f, g g implies f + g f + g. The previous exercise in effect tells us that is a congruence on the category of chain complexes. Therefore, we may consider the quotient category Ch(C)/, usually called the homotopy category. Its objects are chain complexes, but morphisms are now homotopy classes of chain maps. The homotopy category is additive. Note that there is a canonical additive quotient functor Ch(C) Ch(C)/. Exercise 25. Show that the quotient functor Q C : Ch(C) Ch(C)/ has the following universal property: for any additive functor F : Ch(C) D which identifies homotopic maps, there is a unique additive functor ˆF : Ch(C)/ D such that ˆF Q C = F. In particular, conclude that Q C is natural in C.

15 Returning to the original question about when two chain maps induce the same map in homology, we now give the main observation about chain homotopy: Lemma If f g then H n (f) = H n (g). Proof. We reason with modules (the reader may wish to construct an elementfree proof). Of course it is sufficient to show that H n (f g) = 0, and that in turn is established by showing that fc gc is a boundary for all n-cycles c. So consider c Z n (C), and let α be a homotopy from f to g, where f, g : (C, d) (D, d ). Then fc gc = (f g)(c) = d α(c) + αd(c), since α is a homotopy. Now d(c) = 0 since c was assumed to be a cycle, so (f g)(c) = d α(c), and therefore it (f g)(c) is a boundary as required. The converse is false: consider the following complex: 15 (2) 0 Z 2 Z Z/2Z 0 This complex (call it C) is exact, so has trivial homology. Therefore any two maps from C to itself will induce the same map in homology. Consider the two maps 1, 0 : C C, the identity and the zero map. We claim that there cannot exist a homotopy between them. Indeed, a homotopy α would in particular give a map α 0 : Z/2Z Z which would be a section of the quotient Z Z/2Z, which cannot exist. From the notion of chain homotopy we may also define the following. First, say that a chain map f : C D is null-homotopic if f 0 (where 0 : C D is the zero map). Then we may say that a chain complex C is contractible if the identity map 1 : C C is null-homotopic. The homotopy from 1 to 0 is then called a contracting homotopy for C. Clearly, if C is contractible, then H n (C) = 0 for all n. Exercise 26. Verify this. Lastly, We say that a chain map f is a homotopy equivalence if it has a homotopy inverse, i.e. a map g in the opposite direction for which fg 1 and gf 1. We may define two chain complexes C, D to be homotopyequivalent if there exists a homotopy equivalence between them. Note that the above example 2 of an acyclic complex is in fact an example of a complex for which the map C 0 is a quasi-isomorphism but not a homotopy equivalence. In particular, C is not contractible. Therefore, contractibility is a stronger notion than exactness/acyclicity. Exercise 27. Show that f is a homotopy equivalence if and only if it gets sent by Q C to an isomorphism in the homotopy category. Also show that two complexes are homotopy equivalent if and only if they become isomorphic in the homotopy category. Exercise 28. A chain complex C is called split if there is a degree +1 map s from C to itself such that dsd = d. Show that any contractible complex is split. Show that any complex of vector spaces is split. Also give an example of an exact complex which is not split.

16 16 3. Derived functors We now turn to the key construction in homological algebra, namely that of the derived functors of an additive functor. The crucial idea is that of a resolution of an object, which is a special kind of complex. The first section describes resolutions in general, as well as a method for constructing them. Then in the second section we prove the technically important comparison theorem, which tells us that, up to homotopy, taking resolutions is a functorial process. With that in place, we can define derived functors and study some of their properties Resolutions. Resolutions are a way of replacing, or approximating, an object in an abelian category by a complex of better-behaved objects. This idea is well-known from, say, group theory, where one often wants to have, for an abelian group A, a free presentation of the group. Explicitly, this consists of giving a free abelian group F (X), where X is a set of generators, and a subgroup R of F (X) (which is then again free) of relations. These assemble to form an exact sequence (3) 0 R F (X) A 0 and we may think of this sequence as providing us with a replacement, or resolution, of A by free objects. Definition 3.1 (Resolution, Projective Resolution). Let A be an object in C. A (left) resolution of A is an exact chain complex of the form P n P n 1 P 1 P 0 ɛ A 0. ɛ We write P A for such a resolution; the map ɛ is called the augmentation map. A resolution P ɛ A of A is called projective if each P i is a projective object. Clearly, the sequence (3) is a projective (even free) resolution of the abelian group A. Thus, we have shown that every abelian group admits a projective resolution. There are a few things to be noted about the definition. First of all, one may consider the complex P of projectives in its own right, i.e. as P n P n 1 P 1 P 0 0. Of course, this complex P is no longer exact at the zero term, but it is still exact for every n > 0. However, we have H 0 (P ) = A, and therefore to say that P ɛ A is a projective resolution of A is the same as saying that P is a complex of projectives which is exact for all n > 0 and for which H 0 (P ) = A. Exercise 29. Check this. Exercise 30. Yet another way of defining a projective resolution of A is the following: a projective resolution of A is a complex P of projective objects, together with a homotopy equivalence P A, where A is the complex whose only non-zero term is A in degree 0. We shall now show how to obtain projective resolutions of objects. Clearly, for an object A to admit a projective resolution, it is necessary that it can be covered by a projective object P 0. The following lemma tells us that if

17 every object can be covered by a projective, then we can construct projective resolutions. Proposition 3.2. Let C be an abelian category with enough projectives. Then every object admits a projective resolution. Proof. Let A be an object. We construct a resolution P A inductively. ɛ To define P 0 A choose any projective cover of A. Clearly this will make ɛ P 0 A 0 exact at A. Next, assume we have constructed P n 1 d n 1 d P n 2 P 1 ɛ 1 P0 A, and that we wish to define P n. First note that if we took the kernel d n 1 Ker(d n 1 ) P n 1 P n 2, we would have exactness at P n 1. However we cannot let P n = Ker(d n 1 ) because it need not be projective. Thus we cover it by a projective and take that to be P n : ɛ 17 d n:=i ne n d P n n 1 P n 1 P n 2 e n i n Ker(d n 1 ) Since e n is epi, we have that Im(d n ) = Im(i n e n ) = Im(i n ) = Ker(d n 1 ) so the complex is exact at P n 1 as needed. Note that the example of abelian groups at the beginning of the section may be seen as arising in this way: the process halts after one step, after taking the kernel of the projective cover F (X) A. This is special to abelian groups, because of the fact that a subgroup of a free abelian group is again free. In some cases, even shorter resolutions may be formed: consider a field R and an R-module A (i.e. a vector space over R). Then A is itself projective, being free, and thus a projective resolution of A would be 0 P 0 = A 1 A 0. The projective dimension of a ring R is the (supremum over all modules of the) minimum number n needed to form a projective resolution 0 P n P 1 P 0 A 0 of an R-module A. Thus, fields have projective dimension 0, while Z has projective dimension 1. Other rings however, need not have finite projective dimension. Exercise 31. Consider the ring R = Z/8Z. Then Z/4Z is an R-module. Note that R is itself projective (even free) as R-module. Show that the following is a projective resolution of Z/4Z: Such a resolution is called periodic. 4 R 2 R 4 R Z/4Z 0. We now briefly discuss the dual notion of injective resolution.

18 18 Definition 3.3 (Injective Resolution). Let A be an object in C. A (right) resolution of A is an exact cochain complex of the form 0 A η I 0 I 1 I n 1 I n. η We write A I for such a resolution; the map η is called the co-augmentation map. A resolution A η I of A is called injective if each I i is an injective object. Note that this is equivalent to having a complex I which is exact at all n > 0 and for which H 0 (I) = A. Proposition 3.2 dualizes: Proposition 3.4. If C is an abelian category with enough injectives, then every object admits an injective resolution. As an example of an injective resolution, consider an abelian group A. We know that we may embed this group in an injective (=divisible) group I 0, say by taking I 0 = a A Q/Z. Then let I 1 be the cokernel of this embedding; because quotients of divisible groups are again divisible, I 1 is injective, so we obtain a resolution 0 A I 0 I 1 0. The injective dimension of a ring R is defined in a way dual to that of projective dimension. The above argument shows that the injective dimension of Z is 1. The injective dimension of a field is The comparison theorem. Suppose that we are in an abelian category with enough projectives. We know now that every object admits a projective resolution. However, the way we constructed this resolution was by choosing projective covers at each degree, so there is in general no reason why this resolution is in any way canonical, or unique. In this section we prove a key technical result which allows us to express in what way different projective resolutions are related. As a consequence of the result, we obtain the important fact that projective resolutions are unique up to homotopy equivalence. Before we prove the comparison theorem, we need a lemma which will facilitate the reasoning: Lemma 3.5. Let P be a projective object, let C be an exact chain complex and let f : P C n be a map. Then if d n f = 0, there exists a map f : P C n+1 such that d n+1 f = f. Proof. Factor d n+1 as C n+1 Im(d n+1 ) = Z n (C) C n. Because of the universal property of the kernel of d n, the assumption that d n f = 0 implies that f factors through Z n (C). But now we may use the fact that P is projective to obtain the desired factorization: P f f C n+1 Z n (C) C n

19 Theorem 3.6 (Comparison Theorem). Let A, B be objects, let P 19 ɛ A and Q ɛ B be projective resolutions and let f : A B be a map. Then there exists a chain map f : P Q commuting with the augmentation maps. Moreover, such chain map is unique up to homotopy. In a picture: P d n P n 1 P 1 ɛ P 0 A 0 f n f n 1 f 1 f 0 f d Q n Q n 1 Q 1 ɛ Q 0 B 0 Proof. A chain map f is constructed inductively: for f 0, use the fact that P 0 is projective and ɛ epi. Inductively, assume that we have already constructed f 0,..., f n 1 and that we wish to construct f n : P n Q n. Observe that the composite d f n 1 d : P n Q n 2 is zero. Therefore lemma 3.5 applies to f n 1 d : P n Q n 1, and we take f n to be a lift of f n 1 d. We now need to show that if f, g : P Q both are liftings of f, then f g. Equivalently, we may show that the difference h = f g is nullhomotopic. To this end, we construct a homotopy α, again by induction. For the base case, we need α 0 : P 0 Q 1 such that d α 0 = h 0 = f 0 g 0. Note that the map h 0 satisfies ɛ h 0 = ɛ (f 0 g 0 ) = ɛ f 0 ɛ g 0 = f ɛ f ɛ = 0. Therefore, lemma 3.5 applies and we may let α 0 be a lift of h 0. Inductively, assume that we have already constructed α 0,..., α n 1 such that h n 1 = α n 2 d + dα n 1 and that we wish to construct α n : P n Q n+1 such that h n = d α n + α n 1 d, i.e. d α n = h n α n 1 d. That means that the desired α n is a lift of the map h n α n 1 d, and we wish to use lemma 3.5 once more to define this lift. So we compute d (h n α n 1 d) = d h n d α n 1 d = d h n (h n 1 α n 2 d)d by IH = d h n h n 1 d + α n 2 dd = d h n h n 1 d since dd = 0 = 0 since d h = hd Corollary 3.7. Any two projective resolutions of A are homotopy equivalent. Proof. If P A and Q A are both projective resolutions of A, then we may apply the comparison theorem to the identity on A in two ways, the first resulting in a chain map f : P Q and the other in g : Q P. The composite gf : P P need not be the identity, but by the second part of the comparison theorem is homotopy equivalent to the identity. Similarly, fg 1 Q. In particular, any two projective resolutions of A have the same homology. Exercise 32. Illustrate the construction of the comparison map for the case where f : A B is the abelian group map Z/2Z 2 Z/4Z using free resolutions of A, B as in the example at the beginning of the section.

20 20 As expected, everything dualizes. We formulate the comparison theorem and its corollary for injective resolutions. Theorem 3.8 (Comparison Theorem). Let A, B be objects, let A η I and B η J be injective resolutions and let f : A B be a map. Then there exists a cochain map f : I J commuting with the co-augmentation maps. Moreover, such cochain map is unique up to homotopy. Corollary 3.9. Any two injective resolutions of A are homotopy equivalent Derived functors. We now have everything set up to introduce the central concept of left- and right derived functors. Definition 3.10 (Left Derived Functors). Let F : C D be a right exact functor between abelian categories. We define a sequence of functors L i F : C D for i 0 by, for A an object of C: (L i F )(A) = H i (F P ) where P A is a projective resolution of A. Thus to compute L i F (A) we perform the following steps: (1) Choose a projective resolution P A of A. (2) Take the complex of projectives P and apply F to it. This will give a chain complex in D: F P n+1 F P n F P n 1 F P 1 F P 0 0. Note that because F is not necessarily exact, the complex F P will not be acyclic in general. (3) Take the i-th homology of F P. Note that the corollary to the comparison theorem guarantees that, up to isomorphism, L i F (A) does not depend on a choice of resolution P. Next, the comparison theorem also gives for free that L i F is functorial. Moreover, since homology functors are additive, each of the derived functors is also additive. In the definition of L i F, we have not made use of the assumption that F be right exact. However, in order to relate F to its left derived functors, we do need this assumption. For if F is right exact, we have that the sequence F P 1 F P 0 F A 0 is exact, so that H 0 (F P ) = F A. Therefore, the 0-the left derived functor of F agrees with F : L 0 F = F. Exercise 33. Check that there is indeed a natural isomorphism between these functors. Thus, for a right exact functor F, the 0-th derived functor does not give us any new information. The higher derived functors, however, will be seen to repair the lack of exactness of F. Lemma If F is exact, then all of its higher derived functors vanish, i.e. L i F = 0, all i > 0.

21 Proof. To show L i F = 0, consider a projective resolution P A, and look at the part P i+1 P i P i 1, which is exact at P i. Since F is exact, we find that F P i+1 F P i F P i 1 is again exact, and so the i-th homology of F P vanishes. Another easy consequence of the definition is the fact that the higher derived functors vanish on projectives. To see this, let A be a projective object and recall that we can form a projective resolution P A by letting P 0 = A and P i = 0 for all i > 0. Then clearly L i F (A) = 0 for all i > 0. We now consider the dual notions: Definition 3.12 (Right Derived Functors). Let F : C D be a left exact functor between abelian categories. We define a sequence of functors R i F : C D for i 0 by, for A an object of C: (R i F )(A) = H i (F I) where A I is an injective resolution of A. Again, this is functorial and does not depend on the choice of injective resolution. There is a natural isomorphism R 0 F = F, so that the 0-th right derived functor agrees with F. If A is injective, then R i F (A) = 0. Moreover, if F is exact, then all higher derived functors vanish. We end this section with a brief example about abelian groups. If M is an abelian group, then the hom functor Hom(M, ) is left exact, so by the above we may consider its right derived functors. To see what these are, consider an abelian group A and an injective resolution (4) 0 A I 0 I 1 0. Applying Hom(M, ) to this resolution gives a (non-exact) complex (5) 0 Hom(M, I 0 ) d Hom(M, I 1 ) 0. The 0-th cohomology of this complex is simply Hom(M, A). The first cohomology (i.e. the value of R i F at A) is the cokernel of d. One writes Ext(M, A) for the value of the first derived functor of Hom(M, ) at A. Thus we get an exact sequence 0 Hom(M, A) Hom(M, I 0 ) Hom(M, I 1 ) Ext(M, A) 0. This shows that the first derived functor Ext(M, ) repairs the fact that Hom(M, ) did not preserve exactness of the complex 4. Let us consider a concrete case, namely where A = M = Z/2Z. As an injective resolution of Z/2Z, we may take 0 Z/2Z Q/2Z Q/Z 0. Applying Hom(Z/2Z, ) to this sequence gives a description of the group Ext(Z/2Z, Z/2Z) as the following cokernel: Hom(Z/2Z, Q/2Z) Hom(Z/2Z, Q/Z) Ext(Z/2Z, Z/2Z) 21 0 = Z/2Z = = 2 Z/2Z Z/2Z. Thus we find that Ext(Z/2Z, Z/2Z) is isomorphic to Z/2Z.

22 22 We may also consider Hom(, A) as a contravariant left exact functor. Using projective resolutions, we may then construct the right derived functors of Hom(, A). Let us, for the moment, write Ext (, A) for the first derived functor of Hom(, A). We compute Ext (M, A) in the following example: consider again A = M = Z/2Z. As a projective resolution of M = Z/2Z, we may take 0 Z 2 Z Z/2Z 0. (Of course, other resolutions are possible, and the reader is invited to experiment with this.) Applying Hom(, Z/2Z) to this sequence gives a presentation of Ext (Z/2Z, Z/2Z) as the cokernel Hom(Z, Z/2Z) Hom(Z, Z/2Z) Ext (Z/2Z, Z/2Z) 0 = Z/2Z = = 2 Z/2Z Z/2Z. Note that Ext and Ext agree in this example; we say that Ext, considered as a functor of two variables, is balanced. Later we will see that this is holds because of general reasons. In the next section we shall also look at a more intrinsic interpretation of the groups Ext(M, A) which explains what information this group carries about the relation between M and A. A last observation for now is the following. From the fact that resolutions (both injective and projective) in the category of abelian groups may always be chosen to be trivial in degree > 1, we find that the derived functors L i F a of any functor on the category AbGrp vanish for i > 1, and similarly for the right derived functors. Exercise 34. Using projective resolutions as above, compute Ext(Z, Z/pZ) for any p > 1. Similarly for Ext(Z/pZ, Z) and Ext(Z/pZ, Z/qZ). Exercise 35. Repeat the previous exercise using injective resolutions, and verify that the results are the same. Exercise 36. Show that the derived functors are additive and hence preserve finite biproducts.

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