1. Abelian categories

Most of homological algebra can be carried out in the setting of abeliancategories, a class of categories which includes on the one hand all categoriesof modules and on the other hand categories of (abelian) sheaves. In thissection we give a quick introduction to abelian categories. There are threelayers of structure, and we discuss each in succession.

1.1. 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 carrysome additional structure. For example, consider the category Pos of posetsand 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. Sim-ilarly, in the category Rel of sets and relations we may define an orderingon 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 abeliangroups, the homset AbGrp(A,B) is more than a mere set: it is itself anabelian group under the operation of pointwise addition. More generally, inany category of modules we have that R−Mod(A,B) is an abelian group.

The phenomenon that categories may have homsets with additional struc-ture goes by the name of enrichment : we say that the category Rel is en-riched in posets, or that the category of R-modules is enriched in abeliangroups. We will not go into the theory of enriched categories in general (astandard reference is Max Kelly’s book “Enriched Category Theory”) butonly concentrate on the abelian group case.

Definition 1.1. A category C is called pre-additive (equivalently: enrichedin 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

C(B,C)× C(A,B) −→ C(A,C)

are additive in each variable.

Explicitly, the condition on composition means that given a diagram

Af// B

g//

h// C

k // D

we have(g + h)f = gf + hf ; k(g + h) = kg + kh.

Note that pre-additivity is really an extra piece of structure on the cat-egory: in particular, a category may be pre-additive in different ways. Tosee 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 haveZ/4Z 6∼= Z/2Z ⊕ Z/2Z there are two different abelian group structures onHom(A,B).Examples 1.2.

(1) As mentioned earlier, any category of R-modules is pre-additivewhen we define a pointwise group structure on the homsets. Inparticular the category of abelian groups is pre-additive.

1

2

(2) Recall that a one-object category is the same thing as a monoid. Nowa one-object pre-additive category is the same thing as a monoidwhich also has the structure of an abelian group, in such a waythat the multiplication (which is the composition in the one-objectcategory) preserves addition. Such a structure is commonly calleda ring. So, if you like, you can regard a pre-additive category as a“multi-object” ring.

(3) If C is pre-additive then so is the opposite category Cop.(4) If C is pre-additive then so is any functor category CD.

Exercise 1. Work out the last example.

Exercise 2. Show that pre-additive structure on a category C correspondsto a lift of the Hom-functor of C, as in

AbGrp

U

��

Cop × C

99ss

ss

s

C(−,−)// Set

where U is the forgetful functor.

Definition 1.3 (Additive functor). Let F : C −→ D be a functor betweenpre-additive categories. We say F is additive whenever for each pair ofobjects A,B of C the induced function

C(A,B) −→ D(FA,FB)

is an additive map.

The apparent terminological mismatch (pre-additive vs. additive) will becleared up in the next section.Examples 1.4.

(1) The forgetful functor R−Mod −→ AbGrp is additive.(2) When C and D are both pre-additive categories, then we may con-

sider the (full) subcategory of CD on those objects which are additivefunctors C −→ D. This is again a pre-additive category.

(3) Let R be a ring, considered as one-object pre-additive category. Bythe previous example the category of additive functors from R tothe category of abelian groups is again a pre-additive category.

(4) For any pre-additive category with terminal object, the global sec-tions functor factors additively through the category of abelian groups.

Exercise 3. Show that global sections is an additive functor.

Exercise 4. Prove that the category of additive functors from R to AbGrpis equivalent (via an additive functor!) to the category of R-modules.

Exercise 5. Let A be an object of a pre-additive category C. Show thatEnd(A) = C(A,A) is a ring.

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1.2. Additive categories. We now add some extra structure to our pre-additive categories:

Definition 1.5 (Additive category). A pre-additive category C is additiveif it has finite products and a zero object.

Most examples of pre-additive categories given before are in fact additive:for any ring R, the category of R-modules is additive; if C is additive thenso is CD for any D, and so is the subcategory on the additive functors D−→ C if D is also (pre-)additive.

The correct notion of functor between additive categories is the same asfor pre-additive categories. The reason is that, as we shall see below, anyadditive functor automatically preserves finite products and zero objects.

We now show that the notion of additive category is, in spite of theapparent non-duality of the definition, is self-dual.

For that, we show that any additive category not only has products butalso has finite coproducts. In fact:

Lemma 1.6. If C is additive, then C has biproducts.

Proof. Given two objects A,B write A⊕B for the product of A and B. Weshall show that A ⊕ B is (the vertex of) a coproduct of A and B. Definecoproduct inclusions by

〈1A, 0〉 : A −→ A⊕B; 〈0, 1B〉 : B −→ A⊕B.Given morphisms f : A −→ C and g : B −→ C define [f, g] : A ⊕ B −→ C asfπA + gπB. Remaining details are left as exercise. �

Exercise 6. Complete the proof.

We now obtain:

Corollary 1.7. If C is additive then so is Cop.

Note that we used that additive structure on homsets (and the zero maps)to define the coproduct structure. Thus, the additive structure on C andthe finite products are related! To make this clearer, consider two parallelmaps f, g : A −→ B. On the one hand, since our category is pre-additive, wemay use the abelian group structure on C(A,B) to form f + g. But on theother hand we can consider

Aδ−−→ A⊕A [f, g]−−−−→ B.

Here, δ is the diagonal map. From the construction of [f, g] it is clear thatf + g = [f, g]δ. Thus, the addition on C(A,B) can be recovered from thebiproduct structure. Moreover, the map A −→ 0 −→ B is indeed the unitelement of C(A,B) (check!).

This means that in an additive category the pre-additive structure (i.e.the abelian group structures on C(A,B) is uniquely determined. Conse-quently, unlike for pre-additive categories, it is impossible for a category tobe additive in more than one way.

Exercise 7. Let F : C −→ D be an additive functor between additive cate-gories. Show that F preserves finite products and the zero object.

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1.3. Abelian categories. The final step to make is to add to the notionof additive category some exactness properties:

Definition 1.8 (Abelian Category). An additive category C is called abelianif:

(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 onlyif Cop is.

The conditions on monomorphisms and epimorphisms imply that theseare regular; in particular, an abelian category is balanced, in the sense thatany 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 ringR.

(2) If C is abelian then so is CD 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 isabelian is a consequence of the fact that this category is of the formAbGrpR 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 1.10. In an abelian category, every morphism factors as an epi-morphism followed by a monomorphism. Moreover, such factorization isessentially unique.

Proof. Consider a morphism f : A −→ B. First form the kernel Ker(f) i−→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 factorizationf = me where m : Coker(Ker(f)) −→ B. Usually, one writes Im(f) forCoker(Ker(f)). It is now straightforward to check that m is a monomor-phism. Moreover, given any other factorization f = m′e′ with m′ monic, e′

epi, we get a diagram

Im(f)m

""FFFFFFFF

���������

A

e<<yyyyyyyyy

e′##GGGGGGGGG B

Dm′

;;wwwwwwwww

where the dotted arrow arises from the universal property of the cokernel.Using the universal property of e′ one obtains an arrow in the oppositedirection and it is readily seen that these are inverse. �

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Exercise 8. Complete the details.

Let C be an additive category. Then a subcategory D is called an additivesubcategory of C if D is itself additive and the inclusion functor preservesthe additive structure.

For example, the full subcategory of R−Mod on the projective R-modulesis an additive subcategory, as is the full subcategory on the free R-modules.

Exercise 9. If S is a subring of R, show that S −Mod may be regarded asan additive subcategory of R−Mod.

Similarly, a subcategory of an abelian category is called an abelian sub-category if it is itself abelian and when the inclusion functor preserves allstructure. For example, the category of finite abelian groups is an abeliansubcategory of the category of all abelian groups.

Exercise 10. Let R be a ring, and consider the category of finitely generatedR-modules. Show that this is an additive subcategory of R − Mod. Notethat in general it will not be abelian since it need not contain all kernels(a submodule of a finitely generated module need not be finitely generatedagain).

You might expect a corresponding definition of abelian functor (say asand additive functor which preserves kernels and cokernels) but that’s tooquick: first, while almost all functors that naturally arise between abeliancategories are additive, many of them are not exact (i.e. preserve the kernelsand cokernels). Second, from the point of view of homological algebra theexact functors are not the most interesting ones: rather, the focus is onunderstanding how far certain additive functors are from being exact. Wethus make the following definitions:

Definition 1.11 (Exact Functor). Let F : C −→ D be an additive functorbetween abelian categories. Then F is called

(i) left exact if F preserves kernels(ii) right exact if F preserves cokernels

(iii) exact if it is both left- and right exact.

Many interesting functors are either left- or right exact, but not both. Asan important example, consider the functor

HomR(A,−) : R−Mod −→ AbGrp,

for a fixed R-module A. Recall that on arrows Bf−−→ C, this functor

is defined to act by postcomposition, i.e. HomR(A, f) : HomR(A,B) −→HomR(A,C) sends k : A −→ B to fk. We shall often write f∗ for this map.It is easy to see that this functor takes values in abelian groups, becausepostcomposition preserves the abelian group structure. For the same reason,HomR(A,−) is additive: the map

R−Mod(B,C) −→ AbGrp(HomR(A,B), HomR(A,C))

is a homomorphism, since for f, g : B −→ C we have

(f + g)∗(k) = (f + g)k = fk + gk = f∗k + g∗k

so that (f + g)∗ = f∗ + g∗.

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To see that HomR(A,−) is left exact, consider a kernel Ker(f) m−−→ Mf−−→ N . We wish to show that HomR(A,M) m∗−−−→ HomR(A,Ker(f)) is the

kernel of HomR(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 theform m∗(q).

However, HomR(A,−) is typically not right exact. Indeed, consider anepimorphism M

e−−→ N . To say that e∗ : HomR(A,M) −→ HomR(A,N) issurjective is to say that every f : A −→ N is in the image of e∗, i.e. is of theform f = ef for some f : A −→M . See the diagram

M

e����

Af//

f>>}

}}

}N.

Thus, we see:

Proposition 1.12. For a fixed R-module A, the (covariant) hom-functorHomR(A,−) : R−Mod −→ AbGrp is left exact. Moreover, it is exact if andonly if A is projective.

We may dualize all of the above, and consider, for a fixed R-module B,the (contravariant!) functor

HomR(−, B) : (R−Mod)op −→ AbGrp.

Using the fact that kernels in Cop are cokernels in C, one finds thatHomR(−, B)preserves kernels but in general not cokernels. Therefore:

Proposition 1.13. For a fixed R-module B, the contravariant hom-functorHomR(−, B) : (R −Mod)op −→ AbGrp is left exact. Moreover, it is exact ifand only if B is injective.

Exercise 11. Consider the forgetful functor R − Mod −→ AbGrp. Is thisfunctor left exact? Right exact?

Exercise 12. Suppose F : C −→ D is an additive functor which has a leftadjoint. Prove that F is left exact. Formulate and derive a dual statement.

1.4. The Freyd-Mitchell embedding theorem. Categories of modulesare not only the motivating example of abelian categories, but they arealso a universal example, in the sense that for most questions about abeliancategories can be settled in categories of modules.

More concretely, there are several related embedding theorems (for thereader interested in more information on this, see Peter Freyd’s book onAbelian Categories, freely available online). We state one useful versionhere:

Theorem 1.14. Let C be a (small) abelian category. Then there exists aring R and a full embedding C −→ R −Mod. This embedding preserves allabelian structure (i.e. is additive and preserves kernels and cokernels).

So, every small abelian category is a full abelian subcategory of a categoryof modules. The power of this result lies in the fact that we can now reduce

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problems about abelian categories in general to problems about categoriesof modules. More precisely, in order to show that a given statement holds inan arbitrary abelian category it suffices to show that it holds in categoriesof modules. As an illustration, consider Lemma 1.10: we gave a generalarrow-theoretic proof of this, but in a category of modules there is a moreintuitive proof using elements and functions (by using the usual surjective-injective factorization of a homomorphism). The embedding theorem nowsays that it would have been sufficient to prove the lemma only for categoriesof modules.

There are a few things to keep in mind: first, one cannot use the theo-rem to generalize every possible statement about categories of modules toarbitrary abelian categories. For example, consider the statement “Abeliancategories are complete”. Certainly categories of modules are complete, butnot every abelian category is (for example, the category of finite abeliangroups is not). The problem here is that the statement is a statement aboutarbitrarily large collections of objects and arrows, while the embedding the-orem only applies to small abelian categories. By contrast, a statement like“In an abelian category the pullback of an epimorphism is again an epimor-phism” works fine, since this involves only finitary data about objects andarrows in the category.

In short, the embedding theorem works if you have a statement whichis really a statement about small subcategories of abelian categories. Notethat given a small set of objects and arrows of a given abelian category wemay always consider the abelian subcategory generated by these objects andarrows, and that this subcategory is again small.

Exercise 13. Give examples of statements which are true in some abeliancategories but not in all categories of modules.

Exercise 14. Give an example of an abelian category which does not haveenough projectives. Why does the embedding theorem not apply?

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2. Chain complexes and homology

In the previous section we saw that the hom functor Hom(A,−) is ingeneral not exact, but only left exact: if the object A is projective, then allthe necessary liftings exist and Hom(A,−) is exact; thus we may regard thefailure of exactness as a result of obstructions to forming certain liftings.We would like to have a way to measure how far Hom(A,−) is from beingexact or, equivalently, how far A is from being projective. More generally,given a left exact functor F , we would like to measure and describe whatthe obstructions are to F being exact. Homological algebra achieves this byassociating to F a sequence of functors called derived functors. These derivedfunctors are defined using resolutions, which in turn are a special kind ofchain complexes. Roughly speaking, the purpose of chain complexes is tocapture and organize the relevant information about the objects/functors inquestion.

In this section we introduce the key notions of chain complexes and ho-mology, leaving resolutions and derived functors for the next section.

We discuss matters on the level of abelian categories, but the reader maypretend that we’re working in a category of modules. In fact, we will oftendo so ourselves with no loss of generality because of the Freyd-Mitchellembedding theorem.

2.1. Chain complexes. We begin by defining the notion of a chain com-plex, and of a chain map between chain complexes. Let C be an abeliancategory. Then a Z-graded object in C is a family (Cn)n∈Z of objects in-dexed by the integers. Clearly, a Z-graded object is the same thing as adiagram in C of type Z, regarded as a discrete category.

Given two Z-graded objects C = (Cn), D = (Dn), a graded map of degreep from C to D is a Z-indexed family of morphisms fn : Cn −→ Dn+p. Ofparticular importance is the case p = 0; the category of Z-graded objectsand graded maps of degree 0 is simply the functor category CZ.

A differential on a graded object C = (Cn) is a degree −1 map d from Cto itself. Explicitly, this is a family of maps dn : Cn −→ Cn−1 indexed by theintegers. Thus, a picture of a graded object C equipped with a differentiald looks like:

· · · // Cn+1dn+1

// Cndn // Cn−1

dn−1// · · ·.

Definition 2.1 (Chain Complex). A chain complex in C is a Z-graded objectC = (Cn) equipped with a differential d = (dn) subject to the requirementthat d2 = 0 (meaning: dndn+1 = 0 for all n ∈ Z).

We usually write (C, d) for chain complexes; it is also customary to sup-press subscripts in d = dn as much as possible.

The condition d2 = 0 ensures that we have an inclusion Im(dn+1) ⊆Ker(dn). (Strictly speaking, one has to express this more carefully if workingin a general abelian category, by saying that the inclusion of Im(dn+1) intoCn factors through Ker(dn).) Thus we may form the quotient

Hn(C) = Ker(dn)/Im(dn+1).

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This is called the n-th homology object of C. In a category of modules, this isthus a quotient module; in a general abelian category it may be constructedas the cokernel of the inclusion Im(dn+1) −→ Ker(dn).

Since we have an object Hn(C) for each n ∈ Z, H(C) = (Hn(C))n∈Z isitself a Z-graded object of C.

The following notation and terminology is standard, especially in cate-gories of modules: Zn = Zn(C) is the object Ker(dn) ⊆ Cn, and is calledthe object of n-cycles of C; Bn = Bn(C) is the object Im(dn+1) ⊆ Cn, andis called the object of n-boundaries of C; in concrete cases we call elementsof Zn n-cycles and elements of Bn n-boundaries. If two elements of Zn differby a boundary (i.e. if a − b ∈ Bn for a, b ∈ Zn) then we say that a and bare homologous. Finally, the elements of Hn(C) = Zn(C)/Bn(C) are calledhomology classes of C.

Examples of chain complexes will be given below. For now, we note thefollowing: given any composable pair of maps A

p−−→ Bq−−→ C, we say that

this diagram is exact at B if Ker(q) = Im(p). In a chain complex C, weonly have Im(dn+1) ⊆ Ker(dn), so C is not necessarily exact at Cn. Thehomology object Hn(C) measures how far C is from being exact at Cn. Thisexplains the following definition and terminology:

Definition 2.2 (Exact Complex, Acyclic Complex). A chain complex Cnis exact if it is exact at each Cn. It is called acyclic if Hn(C) = 0 for all n.

Clearly a chain complex is exact if and only if it is acyclic. The termacyclic stems from the fact that if the homology Hn(C) (cycles moduloboundaries) 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 thedifferentials. Explicitly, for each n ∈ Z, the following diagram is required tocommute:

Cnfn

//

d��

Dn

d′

��

Cn−1fn−1

// Dn−1.

Morphisms of chain complexes are usually simply called chain maps. Itis clear that such maps compose and that the degreewise identity is a chainmap. Thus we obtain a category Ch(C) of chain complexes in C and chainmaps between them.

Exercise 15. Show that the category Ch(C) is a full subcategory of thediagram category CZ

opwhere 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 = (Zn) of cycles can be made into a subcomplex

of C;

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(ii) the graded object B = (Bn) of boundaries can be made into a sub-complex of C;

(iii) the graded object H = (Hn) can be made into a complex, but hastrivial differential.

Exercise 17. Let (C, d) be a chain complex of R-modules. Suppose cn areelements of R, and define a new differential by letting d′n = cndn. Checkthat (C, d′) is again a chain complex. Now suppose that every cn ∈ {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 dif-ferentials ensures the following:

Exercise 18. Any chain map f : C −→ D sends cycles to cycles and bound-aries to boundaries, and induces morphisms in homology Hn(f) : Hn(C)−→ Hn(D).

As a consequence of the preceding exercise, we find that the assignmentC 7→ Hn(C) is a functor Ch(C) −→ C for each n ∈ Z. This is useful forvarious reasons; for example, one now knows that if two chain complexesare isomorphic, then they have isomorphic homology.

Example 2.4 (Simplicial homology). The following example of a chaincomplex stems from topology. Consider the CW-complex X which lookslike

•

@@@@@@@ •

•That is, X is obtained by glueing three copies of the unit interval into atriangle. 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:

af

h >>>>>>>> b

g

c

We now build a chain complex S(X) from X. We define S(X)0 to be the freeabelian group on the generating set {a, b, c} and we let S(X)1 be the freeabelian group on the generating set {f, g, h}. We let S(X) be 0 in all otherdegrees. (We say that the chain complex S(X) is concentrated in degrees0, 1). Writing 〈a, b, c〉 for the free abelian group on {a, b, c} and so on, wethus obtain

· · · // 0 // 〈f, g, h〉 d // 〈a, b, c〉 // 0

· · · // 0 // S(X)1d // S(X)0

// 0

We must now define a differential d : S(X)1 −→ S(X)0; it suffices tospecify 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

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the combinatorial information about the space X; the differentials encodethe way the 1-cells are glued together.

Now we may compute the homology groups of S(X); in degree 0, wefind (check) that the generator a is not in the image of d, and thus d isnot surjective. However, d(f) and d(g) are linearly independent so we maydescribe the image of d as the subgroup of 〈a, b, c〉 generated by (b− a) and(c − b). Therefore, the group H0(S(X)) = 〈a, b, c〉/Im(d) is isomorphic toZ.

In degree 1, we find that Ker(d) contains a non-zero element, namelyf + g−h. It is easily seen that any other element of Ker(d) is a multiple ofthis element, so that H1(S(X)) = Ker(d) is the free abelian group on onegenerator f + g − h. For those acquainted with singular homology groupsthis will not come as a surprise, since H1(S1) = Z.

Exercise 19. Consider this time, instead of the open triangle in the exampleabove, 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 exampleabove to this situation. What are the 1-boundaries? What is the homology?Do the same for the solid tetrahedron and the hollow tetrahedron (union offour triangles).

In practice, one often deals with complexes which are zero in all negativedegrees (we say that such a complex is concentrated in positive degrees, orthat it is a non-negative complex). To indicate this, we write

· · · −→ Cn −→ Cn−1 −→ · · · −→ C1 −→ C0 −→ 0

Similarly, a complex which does not have nonzero terms above dimension nwill be written

0 −→ Cn −→ Cn−1 −→ · · · .

We conclude this section with some structural properties of the construc-tion of Ch(C) from C. Most details will be left as exercises.

First, the category Ch(C) inherits all limits and colimits which happen toexist in C (and since C was assumed to be abelian, this includes biproducts,zero object, kernels and cokernels). This is proved as follows: the categoryCh(C) is a full subcategory of a category of diagrams in C. We know that thiscategory of diagrams inherits all these limits and colimits, since they maybe 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 7→ Ch(C) is a functor from the cate-gory of abelian categories and additive functors to itself.

By the previous exercise, we only need to verify functoriality; we leave itas an exercise to the reader to check that an additive functor F : C −→ Dinduces an additive functor Ch(F ) : Ch(C) −→ Ch(D).

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2.2. Cohomology. We shall now briefly indicate how the above conceptsdualize.

First of all, a cochain complex in C is a diagram

· · · // Cn−1 dn−1// Cn

dn// Cn+1 d

n+1// · · ·

such that d2 = 0. The elements of the objects Cn (again, if we are ina category of modules) are referred to as n-cochains, those of Ker(dn) asn-cocycles, and those of Im(dn−1) as n-coboundaries.

Moreover, we define the n-th cohomology object of C to be the quotient

Hn(C) = Ker(dn)/Im(dn−1).

Again we usually denote cochain complexes by (C, d), or by (Cn, dn). Amorphism of cochain complexes f : (C, d) −→ (D, d′) (usually called a cochainmap) is a graded map fn : Cn −→ Dn commuting with differentials. Anymorphism of cochain complexes sends cocycles to cocycles and coboundariesto coboundaries, and so induces a morphism Hn(f) : Hn(C) −→ Hn(D) incohomology. We denote by CoCh(C) the category of cochain complexes in Cand cochain maps. Each Hn(−) is a functor CoCh(C) −→ C.

The following proposition is easily obtained by dualizing results in theprevious section:

Proposition 2.6. The category CoCh(C) is abelian. The assignment C 7→CoCh(C) is a functor from the category of abelian categories and additivefunctors to itself.

Exercise 21. Prove the proposition by indicating how to dualize the resultsabout 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 objectof C. Verify that we have a cochain complex whose degree n object isHom(Cn, B), and whose codifferentials are given by composition with thedifferentials of C. Also show that this process is contravariantly functorialin C, so gives rise to a functor Hom(−, B) : (Ch(C))op −→ CoCh(C). Alsoshow that the process is covariantly functorial in B.

2.3. Homotopy. We will now turn to the question of when two chain com-plexes have the same (isomorphic) homology, as well as the question of whentwo chain maps induce the same map in homology. Because of the functo-riality results of the first section, we know that isomorphic chain complexeshave isomorphic homology; however, the converse does not hold as witnessedby the example after this definition.

Definition 2.7 (Quasi-isomorphism). A morphism f : C −→ D of chaincomplexes is a quasi-isomorphism if Hn(f) is an isomorphism for all n.

13

Here is an example of a quasi-isomorphism (which is not an isomorphism).Consider

(1) 0 // Z 4 //

f2

��

Z d //

f1��

Z/2Z //

f0��

0

0 // 0 // Z/4Z d′ // Z/2Z // 0

The differential d is the quotient n 7→ n(mod2), as is d′. The first homologyof the top complex is 2Z/4Z ∼= Z/2Z. All other homology groups of thetop complex are trivial. The second complex also has homology Z/2Z indegree one and has trivial homology in all other degrees. The chain map fis defined by f1(n) = n(mod2), and f2 = 1. It is now easily verified that f1

induces an isomorphism in homology. Clearly, f is itself not an isomorphismof chain complexes.

In general it is not easy to determine whether a given map is a quasi-isomorphism. Also, the question whether Hn(f) = Hn(g) (or, equivalently,when f − g induces the zero map in homology) is a hard one. Therefore weshall now investigate a stronger notion which is both better behaved andmore managable in practice.

Definition 2.8. Let f, g : (C, d) −→ (D, d′) be chain maps. A homotopyfrom f to g is a degree 1 map αn : Cn −→ Dn+1 such that

fn − gn = d′n+1αn + αn−1dn.

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 functionsof topological spaces. Diagrammatically, a homotopy α from f to g may bepictured as

· · · d // Cn+1

αn+1yyy

||yyyy f

��

g

��

d // Cnd //

αnyyyy

||yyyy f

��

g

��

Cn−1d //

αn−1yyyy

||yyyy f

��

g

��

· · ·

· · · d′ // Dn+1d′ // Dn

d′ // Dn−1d′ // · · ·

Example 2.9 (Simplicial Homotopy). Let us give an example of homotopywhich will make clear how this notion is related to the notion of homotopybetween continuous maps of spaces. Consider the following two spaces: thefirst 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 edgeswe label f, g, h and whose 2-cell is labelled γ. To these spaces we associatechain complexes S(I) and S(Y ), respectively. S(Y ) is the following complex:

· · · // 0 // 〈γ〉 d2 // 〈f, g, h〉 d1 // 〈a, b, c〉 // 0

· · · // 0 // S(Y )2d2 // S(Y )1

d1 // S(Y )0// 0

14

Here, the map d1 is defined as in example 2.4, while d2(γ) = g−h+ f . Thecomplex S(I) is:

· · · // 0 // 〈i〉 d // 〈0, 1〉 // 0

· · · // 0 // S(I)1d // 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 fromφ to ψ.

In degree 0, define α0(0) = 0, and α0(1) = −g. Then certainly we have

dα0(0) = 0 = a− a = φ0(0)− ψ0(0),

and alsodα0(1) = d(−g) = −dg = b− c = φ0(1)− ψ0(1).

In degree 1, let α1(i) = γ. Then

dα1(i) + α0d(i) = dγ + α0(1− 0) = (g− h+ f)− g = f − h = φ1(i)− ψ1(i).

Thus, α is indeed a homotopy from α to β.

Thus, the situation is very similar to that in algebraic topology, whereone has two relations on continuous functions: the first is that of beinghomotopic (this is the stronger relation) and the second is that of induc-ing the same homomorphism in singular homology. In fact, the ambitiousreader may wish to verify how the passage from topological spaces to chaincomplexes sends homotopic maps of spaces to homotopic chain maps.

The following exercises develop a few elementary facts about the homo-topy relation:

Exercise 23. Let F : C −→ D be an additive functor and let f ' g in C.Then Ff ' Fg in D.

Exercise 24. The relation ' is an equivalence relation on the set of chainmaps from C to D. Moreover, the homotopy relation is compatible withcomposition 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 thecategory of chain complexes. Therefore, we may consider the quotient cat-egory Ch(C)/', usually called the homotopy category. Its objects are chaincomplexes, but morphisms are now homotopy classes of chain maps. Thehomotopy category is additive. Note that there is a canonical additive quo-tient functor Ch(C) −→ Ch(C)/ '.

Exercise 25. Show that the quotient functor QC : 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 functorF̂ : Ch(C)/ '−→ D such that F̂QC = F . In particular, conclude that QC isnatural in C.

15

Returning to the original question about when two chain maps inducethe same map in homology, we now give the main observation about chainhomotopy:

Lemma 2.10. If f ' g then Hn(f) = Hn(g).

Proof. We reason with modules (the reader may wish to construct an element-free proof). Of course it is sufficient to show that Hn(f−g) = 0, and that inturn is established by showing that fc−gc is a boundary for all n-cycles c. Soconsider c ∈ Zn(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 homo-topy. 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:

(2) 0 // Z 2 // Z // Z/2Z // 0

This complex (call it C) is exact, so has trivial homology. Therefore anytwo maps from C to itself will induce the same map in homology. Considerthe two maps 1, 0 : C −→ C, the identity and the zero map. We claim thatthere cannot exist a homotopy between them. Indeed, a homotopy α wouldin particular give a map α0 : Z/2Z −→ Z which would be a section of thequotient 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 (where0 : C −→ D is the zero map). Then we may say that a chain complex C iscontractible if the identity map 1 : C −→ C is null-homotopic. The homotopyfrom 1 to 0 is then called a contracting homotopy for C. Clearly, if C iscontractible, then Hn(C) = 0 for all n.

Exercise 26. Verify this.

Lastly, We say that a chain map f is a homotopy equivalence if it has ahomotopy inverse, i.e. a map g in the opposite direction for which fg ' 1and gf ' 1. We may define two chain complexes C,D to be homotopy-equivalent if there exists a homotopy equivalence between them.

Note that the above example 2 of an acyclic complex is in fact an exampleof a complex for which the map C −→ 0 is a quasi-isomorphism but nota 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 getssent by QC to an isomorphism in the homotopy category. Also show that twocomplexes are homotopy equivalent if and only if they become isomorphicin the homotopy category.

Exercise 28. A chain complex C is called split if there is a degree +1 maps from C to itself such that dsd = d. Show that any contractible complex issplit. Show that any complex of vector spaces is split. Also give an exampleof an exact complex which is not split.

16

3. Derived functors

We now turn to the key construction in homological algebra, namely thatof the derived functors of an additive functor. The crucial idea is that of aresolution of an object, which is a special kind of complex. The first sectiondescribes resolutions in general, as well as a method for constructing them.Then in the second section we prove the technically important comparisontheorem, which tells us that, up to homotopy, taking resolutions is a func-torial process. With that in place, we can define derived functors and studysome of their properties.

3.1. Resolutions. Resolutions are a way of replacing, or approximating, anobject in an abelian category by a complex of better-behaved objects. Thisidea 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, thisconsists 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. Theseassemble 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, orresolution, of A by free objects.

Definition 3.1 (Resolution, Projective Resolution). Let A be an object inC. A (left) resolution of A is an exact chain complex of the form

· · · −→ Pn −→ Pn−1 −→ · · · −→ P1 −→ P0ε−→ A −→ 0.

We write P ε−→ A for such a resolution; the map ε is called the augmentationmap. A resolution P ε−→ A of A is called projective if each Pi is a projectiveobject.

Clearly, the sequence (3) is a projective (even free) resolution of theabelian group A. Thus, we have shown that every abelian group admitsa projective resolution.

There are a few things to be noted about the definition. First of all, onemay consider the complex P of projectives in its own right, i.e. as

· · · −→ Pn −→ Pn−1 −→ · · · −→ P1 −→ P0 −→ 0.

Of course, this complex P is no longer exact at the zero term, but it is stillexact for every n > 0. However, we have H0(P ) ∼= A, and therefore to saythat P ε−→ A is a projective resolution of A is the same as saying that P is acomplex of projectives which is exact for all n > 0 and for which H0(P ) ∼= A.

Exercise 29. Check this.

Exercise 30. Yet another way of defining a projective resolution of A is thefollowing: a projective resolution of A is a complex P of projective objects,together with a homotopy equivalence P −→ A, where A is the complexwhose 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 canbe covered by a projective object P0. The following lemma tells us that if

17

every object can be covered by a projective, then we can construct projectiveresolutions.

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 Pε−→ A inductively.

To define P0ε−→ A choose any projective cover of A. Clearly this will make

P0ε−→ A −→ 0 exact at A.

Next, assume we have constructed

Pn−1dn−1−−−−→ Pn−2 −→ · · · −→ P1

d1−−→ P0ε−→ A,

and that we wish to define Pn. First note that if we took the kernelKer(dn−1) −→ Pn−1

dn−1−−−−→ Pn−2, we would have exactness at Pn−1. How-ever we cannot let Pn = Ker(dn−1) because it need not be projective. Thuswe cover it by a projective and take that to be Pn:

Pn

en %% %%JJJJJJJJJJdn:=inen // Pn−1

dn−1// Pn−2

Ker(dn−1)in

99rrrrrrrrrr

Since en is epi, we have that Im(dn) = Im(inen) = Im(in) = Ker(dn−1) sothe complex is exact at Pn−1 as needed. �

Note that the example of abelian groups at the beginning of the sectionmay be seen as arising in this way: the process halts after one step, aftertaking the kernel of the projective cover F (X) −→ A. This is special toabelian groups, because of the fact that a subgroup of a free abelian group isagain free. In some cases, even shorter resolutions may be formed: considera field R and an R-module A (i.e. a vector space over R). Then A is itselfprojective, being free, and thus a projective resolution of A would be

0 −→ P0 = A1−−→ A −→ 0.

The projective dimension of a ring R is the (supremum over all modulesof the) minimum number n needed to form a projective resolution

0 −→ Pn −→ · · ·P1 −→ P0 −→ A −→ 0

of an R-module A. Thus, fields have projective dimension 0, while Z hasprojective dimension 1. Other rings however, need not have finite projectivedimension.

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 thefollowing is a projective resolution of Z/4Z:

· · · ·4−−→ R·2−−→ R

·4−−→ R −→ Z/4Z −→ 0.

Such a resolution is called periodic.

We now briefly discuss the dual notion of injective resolution.

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 −→ Aη−−→ I0 −→ I1 −→ · · · −→ In−1 −→ In −→ · · · .

We writeAη−−→ I for such a resolution; the map η is called the co-augmentation

map. A resolution Aη−−→ I of A is called injective if each Ii is an injective

object.

Note that this is equivalent to having a complex I which is exact at alln > 0 and for which H0(I) ∼= A.

Proposition 3.2 dualizes:

Proposition 3.4. If C is an abelian category with enough injectives, thenevery object admits an injective resolution.

As an example of an injective resolution, consider an abelian group A. Weknow that we may embed this group in an injective (=divisible) group I0, sayby taking I0 =

∏a∈A Q/Z. Then let I1 be the cokernel of this embedding;

because quotients of divisible groups are again divisible, I1 is injective, sowe obtain a resolution

0 −→ A −→ I0 −→ I1 −→ 0.

The injective dimension of a ring R is defined in a way dual to that of pro-jective dimension. The above argument shows that the injective dimensionof Z is 1. The injective dimension of a field is 0.

3.2. The comparison theorem. Suppose that we are in an abelian cat-egory with enough projectives. We know now that every object admits aprojective resolution. However, the way we constructed this resolution wasby choosing projective covers at each degree, so there is in general no reasonwhy this resolution is in any way canonical, or unique. In this section weprove a key technical result which allows us to express in what way differentprojective resolutions are related. As a consequence of the result, we obtainthe important fact that projective resolutions are unique up to homotopyequivalence.

Before we prove the comparison theorem, we need a lemma which willfacilitate the reasoning:

Lemma 3.5. Let P be a projective object, let C be an exact chain complexand let f : P −→ Cn be a map. Then if dnf = 0, there exists a map f̃ : P−→ Cn+1 such that dn+1f̃ = f .

Proof. Factor dn+1 as Cn+1 � Im(dn+1) = Zn(C) −→ Cn. Because of theuniversal property of the kernel of dn, the assumption that dnf = 0 impliesthat f factors through Zn(C). But now we may use the fact that P isprojective to obtain the desired factorization:

Pf̃

zzuu

uu

u

�����

f

##GGGGGGGGG

Cn+1// // Zn(C) // Cn

�

19

Theorem 3.6 (Comparison Theorem). Let A,B be objects, let P ε−→ A and

Qε′−−→ 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:

· · · // Pn

fn

�����

d // Pn−1//

fn−1

�����

· · · // P1//

f1����� P0

ε // //

f0����� A

f ′

��

// 0

· · · // Qnd′ // Qn−1

// · · · // Q1// Q0

ε′ // // B // 0

Proof. A chain map f is constructed inductively: for f0, use the fact that P0

is projective and ε′ epi. Inductively, assume that we have already constructedf0, . . . , fn−1 and that we wish to construct fn : Pn −→ Qn. Observe thatthe composite d′fn−1d : Pn −→ Qn−2 is zero. Therefore lemma 3.5 appliesto fn−1d : Pn −→ Qn−1, and we take fn to be a lift of fn−1d.

We now need to show that if f, g : P −→ Q both are liftings of f ′, thenf ' g. Equivalently, we may show that the difference h = f − g is null-homotopic. To this end, we construct a homotopy α, again by induction.

For the base case, we need α0 : P0 −→ Q1 such that d′α0 = h0 = f0 − g0.Note that the map h0 satisfies ε′h0 = ε′(f0−g0) = ε′f0−ε′g0 = f ′ε−f ′ε = 0.Therefore, lemma 3.5 applies and we may let α0 be a lift of h0.

Inductively, assume that we have already constructed α0, . . . , αn−1 suchthat hn−1 = αn−2d

′+dαn−1 and that we wish to construct αn : Pn −→ Qn+1

such that hn = d′αn + αn−1d, i.e. d′αn = hn − αn−1d. That means that thedesired αn is a lift of the map hn − αn−1d, and we wish to use lemma 3.5once more to define this lift. So we compute

d′(hn − αn−1d) = d′hn − d′αn−1d

= d′hn − (hn−1 − αn−2d)d by IH= d′hn − hn−1d+ αn−2dd

= d′hn − hn−1d since dd = 0= 0 since d′h = hd

�

Corollary 3.7. Any two projective resolutions of A are homotopy equiva-lent.

Proof. If P −→ A and Q −→ A are both projective resolutions of A, then wemay apply the comparison theorem to the identity on A in two ways, thefirst resulting in a chain map f : P −→ Q and the other in g : Q −→ P . Thecomposite gf : P −→ P need not be the identity, but by the second part ofthe comparison theorem is homotopy equivalent to the identity. Similarly,fg ' 1Q. �

In particular, any two projective resolutions of A have the same homology.

Exercise 32. Illustrate the construction of the comparison map for thecase where f : A −→ B is the abelian group map Z/2Z ·2−−→ Z/4Z using freeresolutions of A,B as in the example at the beginning of the section.

20

As expected, everything dualizes. We formulate the comparison theoremand its corollary for injective resolutions.

Theorem 3.8 (Comparison Theorem). Let A,B be objects, let Aη−−→ I and

Bη′−−→ 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.

3.3. Derived functors. We now have everything set up to introduce thecentral concept of left- and right derived functors.

Definition 3.10 (Left Derived Functors). Let F : C −→ D be a right exactfunctor between abelian categories. We define a sequence of functors LiF : C−→ D for i ≥ 0 by, for A an object of C:

(LiF )(A) = Hi(FP )

where P −→ A is a projective resolution of A.

Thus to compute LiF (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:

· · · −→ FPn+1 −→ FPn −→ FPn−1 −→ · · · −→ · · ·FP1 −→ FP0 −→ 0.

Note that because F is not necessarily exact, the complex FP willnot be acyclic in general.

(3) Take the i-th homology of FP .Note that the corollary to the comparison theorem guarantees that, up

to isomorphism, LiF (A) does not depend on a choice of resolution P . Next,the comparison theorem also gives for free that LiF is functorial. Moreover,since homology functors are additive, each of the derived functors is alsoadditive.

In the definition of LiF , we have not made use of the assumption that Fbe right exact. However, in order to relate F to its left derived functors, wedo need this assumption. For if F is right exact, we have that the sequence

FP1 −→ FP0 −→ FA −→ 0

is exact, so that H0(FP ) ∼= FA. Therefore, the 0-the left derived functor ofF agrees with F :

L0F ∼= F.

Exercise 33. Check that there is indeed a natural isomorphism betweenthese functors.

Thus, for a right exact functor F , the 0-th derived functor does not giveus any new information. The higher derived functors, however, will be seento repair the lack of exactness of F .

Lemma 3.11. If F is exact, then all of its higher derived functors vanish,i.e. LiF = 0, all i > 0.

21

Proof. To show LiF = 0, consider a projective resolution P −→ A, and lookat the part Pi+1 −→ Pi −→ Pi−1, which is exact at Pi. Since F is exact, wefind that FPi+1 −→ FPi −→ FPi−1 is again exact, and so the i-th homologyof FP vanishes. �

Another easy consequence of the definition is the fact that the higherderived functors vanish on projectives. To see this, let A be a projectiveobject and recall that we can form a projective resolution P −→ A by lettingP0 = A and Pi = 0 for all i > 0. Then clearly LiF (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 exactfunctor between abelian categories. We define a sequence of functors RiF : C−→ D for i ≥ 0 by, for A an object of C:

(RiF )(A) = H i(FI)

where A −→ I is an injective resolution of A.

Again, this is functorial and does not depend on the choice of injectiveresolution. There is a natural isomorphism R0F ∼= F , so that the 0-th rightderived functor agrees with F . If A is injective, then RiF (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 isan abelian group, then the hom functor Hom(M,−) is left exact, so by theabove we may consider its right derived functors. To see what these are,consider an abelian group A and an injective resolution

(4) 0 −→ A −→ I0 −→ I1 −→ 0.

Applying Hom(M,−) to this resolution gives a (non-exact) complex

(5) 0 −→ Hom(M, I0) d−−→ Hom(M, I1) −→ 0.

The 0-th cohomology of this complex is simply Hom(M,A). The first co-homology (i.e. the value of RiF at A) is the cokernel of d. One writesExt(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, I0) −→ Hom(M, I1) −→ Ext(M,A) −→ 0.

This shows that the first derived functor Ext(M,−) repairs the fact thatHom(M,−) did not preserve exactness of the complex 4.

Let us consider a concrete case, namely where A = M = Z/2Z. As aninjective 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 groupExt(Z/2Z,Z/2Z) as the following cokernel:

Hom(Z/2Z,Q/2Z) //

∼=��

Hom(Z/2Z,Q/Z) // //

∼=��

Ext(Z/2Z,Z/2Z) //

∼=��

0

Z/2Z ·2 // Z/2Z // Z/2Z.

Thus we find that Ext(Z/2Z,Z/2Z) is isomorphic to Z/2Z.

22

We may also consider Hom(−, A) as a contravariant left exact functor.Using projective resolutions, we may then construct the right derived func-tors of Hom(−, A). Let us, for the moment, write Ext′(−, A) for the firstderived functor of Hom(−, A). We compute Ext′(M,A) in the followingexample: consider again A = M = Z/2Z. As a projective resolution ofM = Z/2Z, we may take

0 −→ Z ·2−−→ Z −→ Z/2Z −→ 0.

(Of course, other resolutions are possible, and the reader is invited to ex-periment with this.) Applying Hom(−,Z/2Z) to this sequence gives a pre-sentation 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, consideredas a functor of two variables, is balanced. Later we will see that this is holdsbecause of general reasons. In the next section we shall also look at amore intrinsic interpretation of the groups Ext(M,A) which explains whatinformation 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 alwaysbe chosen to be trivial in degree > 1, we find that the derived functors LiFa of any functor on the category AbGrp vanish for i > 1, and similarly forthe 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, andverify that the results are the same.

Exercise 36. Show that the derived functors are additive and hence pre-serve finite biproducts.

23

4. Extensions

The previous section introduced in particular the derived functors of thehom-functors, which were called Ext-functors. We now turn to a more con-crete interpretation of the groups Ext(A,B) in order to understand whatthey tell us about A,B.

4.1. Short exact sequences. We concentrate on a certain type of dia-grams, namely those of the form

ξ : 0 −→ Bm−−→ X

e−−→ A −→ 0.

Diagrams of this form which are exact are called short exact sequences (incontrast with the long exact sequences which we will encounter later). Notethat for ξ to be exact it is necessary and sufficient that m is the kernel of eand e is the cokernel of m (and in particular, m is monic and e is epi).

We may organize short exact sequences in a category S(C), whose objectsare sequences ξ as above, and whose maps are commutative diagrams

0 // B

φ′

��

// X //

φ��

A //

φ′′

��

0

0 // B′ // X ′ // A′ // 0.

Given two objects A,B, one may ask if it is possible to form a short exactsequence of the form ξ, thus relating A and B via a third object X. Thisproblem is known as the extension problem. More precisely, it asks for adescription of all possible short exact sequences of the form ξ for fixed A,B.Such a sequence is also called an extension of A by B. Note that to givesuch an extension, one must not only specify the object X, but also the twomaps m, e.

Of course, we have to decide when we call two extensions “the same”.This is given by the following definition:

Definition 4.1. Let A,B be objects and ξ, ξ′ be two extensions of A by B.Then ξ and ξ′ are called equivalent if there is an isomorphism φ : X −→ X ′

which makes the following diagram commute:

(6) ξ : 0 // B

=

��

// X //

φ

��

A //

=

��

0

ξ′ : 0 // B // X ′ // A // 0.

We write Ext(A,B) for the set1 of equivalence classes of extensions of A byB.

In what follows we shall be mainly interested in extensions up to equiva-lence, but simply speak of extensions in order to avoid lengthy phrases.

The following lemma tells us that in order to test whether two extensionsare equivalent, it suffices to find a morphism φ which makes the diagram (6)

1Strictly speaking, we don’t know a priori whether this is a set or a proper class, butwe shall not worry about this now.

24

commute. In other words, if φ as above makes the diagram commute, thenit is automatically an isomorphism.

Lemma 4.2. Consider a commutative diagram with exact rows:

0 // B

φ′

��

// X //

φ��

A //

φ′′

��

0

0 // B′ // X ′ // A′ // 0

in which both φ′ and φ′′ are isomorphisms. Then φ is also an isomorphism.

Proof. We first show that φ is monic, i.e. that Ker(φ) = 0. To this end itsuffices to show that Ker(φ) i−→ X is the zero map. In doing so, we willmake use of the following elementary fact: if a composite gf = 0 and g ismonic, then f = 0.

First, we show that ei : Ker(φ) −→ A is the zero map. Indeed, φ′ei′ =e′φi = 0, so since φ′′ is iso, hence monic, ei = 0. Therefore i factors throughthe kernel of i, say via j : Ker(φ) −→ B, with mj = i.

Next, observe that 0 = φi = φmj = m′φ′j. But m′ is monic and so is φ′,and thus j = 0. But then it follows that i = mj = m0 = 0 and we’re done.

The proof that φ is surjective is dual and left to the reader. �

Exercise 37. Give a direct proof that the map φ is surjective.

We next consider some examples of extensions.

Example 4.3. For concreteness’ sake we work in the category of abeliangroups.

(1) Consider the case A = Z/2Z and B = Z. The following is an exten-sion of A by B:

ξ1 : 0 −→ Z ·2−−→ Z e−−→ Z/2Z −→ 0

where e is the canonical map x 7→ x(mod2). There is another exten-sion, namely

ξ2 : 0 −→ Z −→ Z/2Z⊕ Z −→ Z/2Z −→ 0

where the maps are the evident inclusion and projection.(2) This time let A = Z/pZ with p prime and B = Z. We now have the

following extensions:

ξi : 0 −→ Z ·p−−→ Z ei−−→ Z/pZ −→ 0

with ei(x) = ix(modp), for i = 1 . . . , p− 1, and

ξp : 0 −→ Z −→ Z⊕ Z/pZ −→ Z/pZ −→ 0.

(3) Now let A = Z/2Z and B = Z/4Z. We have extensions

ξ1 : 0 −→ Z/4Z ·2−−→ Z/8Z −→ Z/2Z −→ 0

and

ξ2 : 0 −→ Z/4Z −→ Z/4Z⊕ Z/2Z −→ Z/2Z −→ 0.

25

It is possible to prove that the extensions given in each of the aboveexamples are in a sense exhaustive, i.e. that up to equivalence these are theonly possible extensions. This will in fact follow from later results, when werelate the set of extensions to the group R1(Hom(A,−))(B).

As witnessed by the examples, the set Ext(A,B) always contains at leastone element, namely

0 −→ BιB−−→ B ⊕A πA−−−→ A −→ 0.

Extensions of this form are called split, because the epimorphism πA admits asection, namely the inclusion ιA : A −→ B⊕A. Similarly, the monomorphismm admits a retraction, namely the projection πB : B ⊕A −→ B.

The following characterization of split short exact sequences follows fromthe characterization of biproducts.

Lemma 4.4. For a short exact sequence

ξ : 0 −→ Bm−−→ X

e−−→ A −→ 0

the following are equivalent:(i) e is a split epimorphism (i.e. there exists s : A −→ X with es = 1)(ii) m is a split monomorphism (i.e. there exists k : X −→ B with km = 1)

(iii) ξ is split

Exercise 38. Prove the lemma.

The split extension is considered to be the trivial solution to the extensionproblem. Also note the following lemma:

Lemma 4.5. If either A is projective or B is injective, then every extensionof A by B is split, and hence Ext(A,B) = 0.

We write Ext(A,B) = 0 to indicate that this set doesn’t contain non-trivial (i.e. non-split) extensions.

4.2. Extensions as a functor. Our next task is to show that the construc-tion of the set Ext(A,B) is in fact functorial in A and B. This will involvesome techniques for manipulating extensions which will be useful later onas well.

We begin by proving a technical result:

Lemma 4.6. Consider a commutative diagram of the following form:

(7) 0 // B

k��

m // Xe //

φ��

A //

=

��

0

0 // B′m′ // X ′

e′ // A // 0

where the top row is exact. Then the bottom row is exact if, and only if, theleft-hand square is a pushout.

Proof. Assume first that the square is a pushout. By general properties ofpushouts in abelian categories, we know that the map m′ must be a monicagain, and hence to show that the bottom row is exact it suffices to showthat e′ is the cokernel of m′. Consider any h : X ′ −→ Z for which hm′ = 0.Then also 0 = hm′k = hφm, and therefore the map hφ factors through e,

26

since e is the cokernel of m. Thus we get a map h′ : A −→ Z with h′e = hφ,and hence also h′e′φ = hφ. Moreover, both composites hm′ = 0 = h′e′m′

are equal, so by the universal property of the pushout h′e′ = h, and we haveshown that h factors through e′. It is left as a routine exercise to verify thath′ is unique with this property.

Next, assume that the bottom row is exact. Consider the pushout of mand k:

Bm //

k��

X

l��

φ

��000000000000000

B′n //

m′((PPPPPPPPPPPPPPP Y

γA

A

AA

X ′

where the dotted arrow arises by the universal property of the pushout.We wish to show that this map γ is an isomorphism. Using the pushoutproperty again, construct f : Y −→ A with fn = 0, lf = e. Then considerthe diagram

B′n // Y

γ

��

f// A

B′m′ // X ′

e′// A,

in which the left hand square commutes by construction of γ and where theright hand square commutes because e′γ and f have the same compositeswith l and n.

Thus by Lemma 4.2, the map γ is an isomorphism. �

There is an evident dual result:

Lemma 4.7. Consider a commutative diagram

(8) 0 // B

=

��

m // Xe //

φ��

A //

f��

0

0 // Bm′ // X ′

e′ // A′ // 0

where the bottom row is exact. Then the top row is exact if, and only if, theright-hand square is a pullback.

Exercise 39. Prove this directly.

Lemmata 7 and 8 are useful in many situations. For now, we deduce thefollowing functoriality result:

Proposition 4.8. The assignment (A,B) 7→ Ext(A,B) is contravariantlyfunctorial in A and covariantly functorial in B. In fact, Ext(−,−) is abifunctor AbGrpop × AbGrp −→ Set.

Later we shall see that this functor in fact takes values in abelian groups.

27

Proof. We show how a morphism k : B −→ B′ induces a function k∗ :Ext(A,B) −→ Ext(A,B′), leaving the functoriality in A as an exercise. Todefine k∗ we use Lemma 7: if ξ is an element of Ext(A,B), then form thepushout along k to obtain

ξ : 0 // B

k��

m // Xe //

φ

��

A //

=

��

0

g∗ξ : 0 // B′m′ // X ′

e′ // A // 0

Elementary properties of pushouts now guarantee that this assignment isindeed functorial. �

Exercise 40. Verify that k∗ as defined is indeed a functor.

Exercise 41. Show directly how to define, given l : A′ −→ A a functionl∗ : Ext(A,B) −→ Ext(A′, B). Finally, show that these two operations arecompatible, in that l∗k∗ ∼= k∗l

∗.

Exercise 42. Show that if ξ is a split extension, then so are l∗ξ and k∗ξ.

4.3. Relating Ext and Ext. Our main aim is now to show that the firstright derived functor of Hom(A,−) is isomorphic to the functor Ext(A,−),and similarly in the other variable. As it stands, the derived functor takesvalues in abelian groups, while Ext(A,−) is a Set-valued functor. However,when we show that the two are isomorphic as Set-valued functors, it willfollow that Ext(A,B) carries an abelian group structure, and that Ext maybe viewed as an AbGrp-valued functor as well.

We recall that in order to find Ext(A,B), one fixes a projective presen-tation of A:

0 −→ Ri−→ F

q−−→ A −→ 0(one may take F to be a free group on any set of generators for A, and R isthen the subgroup on the relations). Then apply Hom(−, B) to the map i;this gives a complex

0 −→ Hom(F,B) i∗−−→ Hom(R,B) −→ 0,

and Ext(A,B) is the cokernel of this map i∗. Explicitly, an element ofExt(A,B) is an equivalence class of maps k : R −→ B. Two such mapsk, k′ : R −→ B are equivalent if there exists a map h : F −→ B such thatk − k′ = hi.

We shall define two functions

Φ = ΦA,B : Ext(A,B) −→ Ext(A,B) Ψ = ΨA,B : Ext(A,B) −→ Ext(A,B).

First, suppose we are given an extension ξ. Consider the diagram:

0 // R

k�����

i // Fq//

φ

����� A //

=

��

0

ξ : 0 // Bm // X

e // A // 0

where the top row is the projective presentation of A and the bottom row isthe given extension. Since F is projective and e epi, we may choose a lift φ

28

as in the diagram. Then φ will satisfy eφi = 0, so the composite eφ factorsuniquely through the kernel B, say via k. We let

Φ(ξ) = [k],

the equivalence class of k in Ext(A,B).Of course we need to show that a different choice of lift φ′ would give

the same element of Ext(A,B). So let φ′ another map with eφ′ = q. Thenφ′ factors through B, say via k′ : R −→ B. Note first that φ − φ′ factorsthrough B since e(φ− φ′) = eφ− eφ′ = q− q = 0. So φ− φ′ = mh for someh : F −→ B. Now calculate

m(k − k′) = mk −mk′

= φi− φ′i= (φ− φ′)i= mhi

Since m is monic, this implies k− k′ = hi, which shows that k and k′ are inthe same cohomology class of Ext(A,B).

In the other direction, suppose we are given an element [k] ∈ Ext(A,B),represented by k : R −→ B. We use Lemma 7 to construct an extension ξ ofA by B as in

0 // R

k��

i // Fe //

φ

��

A //

=

��

0

ξ : 0 // Bm // X

e // A // 0

We now putΨ([k]) = ξ.

Again, we need to prove that this does not depend on the choice of repre-sentative k.To this end, it suffices to show that given any h : F −→ B, themorphism k′ = k + hi induces the same extension as k. But the diagram

0 // R

k′=k+hi��

i // Fe //

φ+mh

��

A //

=

��

0

0 // Bm // X

e // A // 0

commutes (check!) and both rows are exact. Thus by Lemma 7 again, theleft hand square is a pushout. This shows that k and k′ induce the sameextension (up to equivalence of extensions).

Exercise 43. Show that the two operations Φ and Ψ are both natural in Aand B.

Exercise 44. Show that the two operations Φ and Ψ are mutually inverse.

We may thus conclude:

Theorem 4.9. There is a bijection, natural in A and B, between the setsExt(A,B) and Ext(A,B).

29

This theorem gives an interpretation of the groups Ext(A,B), purely interms of A and B. In particular, the existence of non-trivial extensions ofA by B witness the non-projectivity of A (resp. the non-injectivity of B).

As stated before, the above bijection induces an abelian group structureon the set of extensions Ext(A,B), which we now describe explicitly.

First, the neutral element in Ext(A,B) is the split extension of A by B.One easily verifies that

0 // R

0��

i // Fe //

φ

��

A //

=

��

0

ξ : 0 // B // B ⊕A // A // 0

is commutative, and hence that Φ sends the zero element of Ext(A,B) tothe split extension of A by B.

To describe the addition in Ext(A,B) consider two extensions ξ and ξ′.To form their sum, consider the extension

ξ ⊕ ξ′ : 0 −→ B ⊕B m⊕m′−−−−−→ X ⊕X ′ e⊕ e′−−−−→ A⊕A −→ 0,

the result of taking the direct sum of ξ and ξ′. Now define

ξ + ξ′ = 〈1A, 1A〉∗[1B, 1B]∗(ξ ⊕ ξ′)where 〈1A, 1A〉 is the diagonal A −→ A⊕A and [1B, 1B] : B⊕B −→ B is thecodiagonal. This gives an extension of A by B. We show that under thisdefinition, the map Φ preserves addition.

Consider two maps k, k′ : R −→ B. Their sum, k+k′ induces an extensionΦ([k + k′]) of A by B, which may be constructed as follows:

0 // R

〈k,k′〉��

i // Fe //

ψ

��

A //

=

��

0

µ : 0 // B ⊕B n //

[1B ,1B ]

��

Yf//

χ

��

A

=

��

// 0

Φ([k + k′]) = [1b, 1b]∗(µ) 0 // B // X ′′ // A // 0

Here, the two left hand squares are pushouts, and the bottom sequence isΦ([k + k′]). The sequence in the middle fits into a commutative diagram

µ = 〈1A, 1A〉∗(ξ + ξ′) 0 // B ⊕B n //

=

��

Yf

//

��

A

〈1A,1A〉��

// 0

ξ + ξ′ : 0 // B ⊕B m+m′// X ⊕X ′

e⊕e′// A⊕A // 0

where the middle arrow is induced by the pushout property. Since thediagram commutes and has exact rows, the right-hand square is a pullback.This shows that the extension Φ([k + k′]) is the same as the sum of Φ([k])and Φ([k′]) as defined before.

Exercise 45. Check directly that the addition defined on Ext(A,B) makesit into an abelian group.

30

Exercise 46. Check directly that the abelian group structure on Ext(A,B)is preserved by the pullback and pushout operations, so that Ext(−,−) isin fact a bifunctor landing in the category AbGrp.

Exercise 47. Construct the group multiplication table of Ext(Z/4Z,Z).

31

5. The long exact homology sequence

We return to the general study of derived functors. We have seen that,for any right exact functor F : C −→ D, there exists a sequence of leftderived functors L0(F ), L1(F ), L2(F ), . . ., and we have seen that L0(F ) ∼= F .However, we do not know yet how the higher derived functors are related toF , and to each other. We aim to show that for any short exact sequence 0−→ A −→ B −→ C −→ 0 in C there exists an exact sequence of the form

· · · −→ Ln+1F (C) −→ LnF (A) −→ LnF (B) −→ LnF (C) −→ Ln−1F (A) −→ · · ·

This sequence tells us how the derived functors are related. Moreover, itmakes precise how the derived functors repair the fact that F is not leftexact.

5.1. Connecting homomorphisms. The key ingredient for building thelong exact sequence are the so-called connecting homomorphisms Ln+1F (C)−→ LnF (A). The existence of these maps is a formal consequence of thefamous Snake Lemma, to which we first turn attention.

Lemma 5.1 (Snake Lemma). Consider a commutative diagram of the form

Am //

p

��

Be //

q

��

C //

r

��

0

0 // A′m′ // B′

e′ // C ′

in which the rows are exact. Then there is a connecting homomorphismδ : Ker(r) −→ Coker(p) fitting into an exact sequence(9)Ker(p) −→ Ker(q) −→ Ker(r) δ−−→ Coker(p) −→ Coker(q) −→ Coker(r).

Proof. We shall prove this element-theoretically, i.e. in a category of mod-ules. The definition of the map δ is best understood when considering theexpanded diagram

Ker(p) //

��

��

Ker(q) //

��

��

Ker(r)��

��

Am //

p

��

Be // //

q

��

C //

r

��

0

0 // A′ //m′ //

����

B′e′ //

����

C ′

����

Coker(p) // Coker(q) // Coker(r).

To define δ : Ker(r) −→ Coker(p), take an element c ∈ Ker(r). Because eis surjective, we may pick an element b ∈ B with e(b) = c. Then apply q tothis element, as to obtain q(b) ∈ B′. This element is in the kernel of e′, sincee′q(b) = re(b) = r(c) = 0. Since the bottom row is exact, q(b) ∈ Im(m′),i.e. there is a unique a′ ∈ A′ with m′(a′) = q(b). Now let δ(c) = [a′], theclass of a′ in the quotient Coker(p).

32

We need to verify that this is well-defined, in that it is independent ofthe choice of lift e(b) = c. So suppose that another lift b0 with e(b0) = cwas chosen, giving rise to an element a′0 ∈ A′ with m′(a′0) = q(b0). We needto prove that the difference a′ − a′0 is in the image of p, so that [a′] = [a′0].Note that since e(b) = e(b0), we have that b−b0 ∈ Ker(e) = Im(m), so thatthere is an element a ∈ A with m(a) = b− b0. But then m′p(a) = qm(a) =q(b−b0) = m′(a−a0), and because m′ is monic it follows that p(a) = a−a0,showing that a− a0 is in the image of p. �

Exercise 48. Complete the proof by showing that the sequence (9) is indeedexact.

We wish to use the snake lemma to form a long exact sequence in ho-mology; among other things, this will provide us with a way of relating thehomology groups in different dimensions. However, in order to do so, weneed to re-express homology groups as kernels and cokernels, because this ishow the snake lemma is formulated. This is done in the following auxiliarylemma, the proof of which is left as an instructive exercise:

Lemma 5.2. Let C be a chain complex. Then there are maps γn : Coker(dn+1)−→ Ker(dn−1) with the property that Ker(γn) ∼= Hn(C) and Coker(γn) ∼=Hn−1(C). Moreover, these maps are natural in C.

Exercise 49. Construct these maps (use the universal properties of coker-nels and kernels) and prove the lemma.

We recall that the notion of a short exact sequence makes sense in anyabelian category. In particular, we may consider short exact sequences ofchain complexes. Since kernels and cokernels are computed “degreewise”,to say that a diagram 0 −→ A

m−−→ Be−−→ C −→ 0 of chain complexes is short

exact is to say that each 0 −→ Anmn−−−→ Bn

en−−→ Cn −→ 0 is short exact.

Proposition 5.3 (Long Exact Homology Sequence). Consider a short exactsequence ξ : 0 −→ A −→ B −→ C −→ 0 in Ch(C). Then there exist mapsδn : Hn(C) −→ Hn−1(A) fitting in a long exact sequence

· · · −→ Hn(A) −→ Hn(B) −→ Hn(C) δn−−→ Hn−1(A) −→ Hn−1B −→ Hn−1(C) −→ · · · .

Moreover, the maps δn are natural in ξ.

Proof. Consider the following diagram:

Coker(dAn+1) //

γAn

��

Coker(dBn+1) //

γBn

��

Coker(dCn+1) //

γCn

��

0

0 // Ker(dAn−1) // Ker(dBn−1) // Ker(dCn−1).

Here, the differentials of A,B,C have been written dA, dB and dC , respec-tively. The vertical maps are the ones arising as in Lemma 5.2. Now applythe Snake Lemma. �

33

We have not proved the naturality statement. Explicitly, this means thatfor a commutative diagram

0 // A //

��

B //

��

C

��

// 0

0 // A′ // B′ // C ′ // 0

with exact rows, the resulting diagram

Hn(C)δn //

��

Hn−1(A)

��

Hn(C ′)δn // Hn−1(A′)

also commutes.

Exercise 50. Verify this (you may want to verify a similar statement forthe Snake Lemma).

Exercise 51. Formulate and prove a dual version, showing that there existsa long exact cohomology sequence.

5.2. The long exact sequence for derived functors. We now considera right exact functor F : C −→ D and wish to show that for each short exactsequence 0 −→ A −→ B −→ C −→ 0 in C there are connecting homomorphismsδn : LnF (C) −→ Ln−1F (A), resulting in a long exact sequence. In order toexploit the long exact homology sequence, we need to have a short exactsequence of complexes, and these complexes will be projective resolutions ofA,B and C, respectively. However, if we take arbitrary projective resolu-tions P,Q and R of A,B and C, respectively, there is no reason why 0 −→ P−→ Q −→ R −→ 0 would be exact. Thus we need a little care in showing thatwe can choose P,Q and R in the right way. The Horseshoe Lemma belowtells us that given fixed P and R, we can always choose Q in such a waythat 0 −→ P −→ Q −→ R −→ 0 is short exact.

Lemma 5.4 (Horseshoe Lemma). Let 0 −→ A −→ B −→ C −→ 0 be a shortexact sequence, and let P ε−→ A and R

ρ−−→ C be projective resolutions.Then there exists a projective resolution Q

χ−−→ B making the complex 0−→ P −→ Q −→ R −→ 0 exact.

Proof. Since we are aiming to define in particular for each n a short exactsequence 0 −→ Pn −→ Qn −→ Rn −→ 0 of projectives and since such a sequencewill always be split exact, we may just as well let Qn = Pn ⊕ Rn for all n.We must define the differentials of Q. Consider

0 // A //m // B

e // // C // 0

0 // P0

ε

OO

// P0 ⊕R0

χ

OO

// R0

ρ

OOρ̃

ddII

II

I// 0

The map χ is defined to be the copairing of mε and a lift ρ̃ : R0 −→ B of ρalong e (which exists since R0 is projective). It is easily checked that χ isan epimorphism.

34

For the next stage, consider

0 // A //m // B

e // // C // 0

0 // P0

ε

OO

// P0 ⊕R0

χ

OO

// R0

ρ

OO

// 0

0 // Ker(ε)

OO

// Ker(χ)

OO

// Ker(ρ)

OO

// 0.

The bottom row is again exact (this can be checked directly or seen as aninstance of the so-called 9-Lemma). The objects Ker(ε),Ker(χ) and Ker(ρ)need not be projective but we have covers P1 −→ Ker(ε) and R1 −→ Ker(ρ).Using the same idea as in the previous step we define an epimorphism P1⊕R1

−→ Ker(χ), making the diagram commute:

0 // A //m // B

e // // C // 0

0 // P0

ε

OO

// P0 ⊕R0

χ

OO

// R0

ρ

OO

// 0

0 // Ker(ε) //

OO

Ker(χ)

OO

// Ker(ρ)

OO

// 0

0 // P1

OOOO

// P1 ⊕R1

OO���

// R1//

OOOO

0

We now proceed inductively. �

Now we have a split short exact sequence 0 −→ P −→ Q −→ R −→ 0 ofchain complexes, and we may apply the functor F to this diagram, giving

0 −→ FP −→ FQ −→ FR −→ 0

which is again (split) exact. Thus we may apply the construction of thelong homology sequence, and this gives the desired long exact sequence ofleft derived functors:

· · · −→ Ln+1F (C)δn+1−−−−→ LnF (A) −→ LnF (B) −→ LnF (C) δn−−→ Ln−1F (A) −→ · · ·

In particular we find at low degree:

· · · −→ L1F (C) δ1−−→ F (A) −→ F (B) −→ F (C) −→ 0

Exercise 52. Show that F is exact if and only if L1F = 0.

Exercise 53. Formulate and derive a dual statement exhibiting a long exactsequence of right derived functors.

Exercise 54. Let 0 −→M −→ P −→ A −→ 0 be a short exact sequence withP projective, and let F be right exact. Show that for n > 1, there is anisomorphism LnF (A) ∼= Ln−1F (M).

35

From the naturality of the long exact homology sequence it follows thatthe long exact sequence of derived functors is also natural. We leave it tothe reader to formulate what this means. (A rigorous proof is a bit of work!)

We end the discussion with an example. Consider an object A, giving aleft exact functor Hom(A,−). To say that this functor is left exact meansthat for each short exact sequence 0 −→ C ′ −→ C −→ C ′′ −→ 0, the inducedsequence

(10) 0 −→ Hom(A,C ′) −→ Hom(A,C) −→ Hom(A,C ′′)

is exact. The functor Hom(A,−) has right derived functors, which we shallwrite Extn(A,−). these fit into a long exact cohomology sequence extendingthe sequence (10) to the right:

0 // Hom(A,C ′) // Hom(A,C) // Hom(A,C ′′)δ0

ssggggggggggggggggggggggg

Ext1(A,C ′) // Ext1(A,C) // Ext1(A,C ′′)δ1

ssggggggggggggggggggggggg

Ext2(A,C ′) // · · ·

This sequence is called the long exact Ext-sequence (in the second variable).In the case of abelian groups, we have seen that this sequence trivializes afterdegree 1, but over general rings this will not be the case.

Exercise 55. Show that for abelian groups, the functor Ext1(A,−) pre-serves epimorphisms. (Hint: use the long exact sequence.)

Exercise 56. Construct a long exact Ext-sequence in the first variable.

36

6. Tensor products and torsion

In the previous section we discussed one of the leading examples of derivedfunctors, namely the Ext-functors. This section deals with the other leadingexample, the Tor-functors. In a certain sense these all other examples canbe reduced to these two.

The Tor-functors are the left-derived functors of functors which ariseby forming tensor products. We shall first describe the general categoricalsetting in which this makes sense; then we interpret what this means incategories of modules; finally, we introduce the notion of torsion and explainhow the derived functors can be understood in terms of torsion.

6.1. Monoidal categories. A monoid may be regarded as a set equippedwith a unital and associative multiplication on it. It is natural to ask whetherwe may replace the word “set” by “category”. This leads to the considerationof categories equipped with a “multiplication” operation. In most caseshowever, it turns out that such operations are not strictly associative orunital, but only up to canonical isomorphism. We arrive at the followingdefinition.

Definition 6.1 (Monoidal Category). A monoidal category is a structure(C,⊗, I, α, λ, ρ), where

• C is a category.• C×C

⊗−−→ C is a bifunctor, written (C,D) 7→ C ⊗D, (f, g) 7→ f ⊗ g.We call C ⊗D the tensor product of C and D.• I is an object of C called the unit.• α is a natural isomorphism α = αA,B,C : A⊗(B⊗C) −→ (A⊗B)⊗C,

called the associativity isomorphism.• λ is a natural isomorphism λ = λA : I ⊗A −→ A.• ρ is a natural isomorphism ρ = ρA : A⊗ I −→ A.

These isomorphisms are required to satisfy the following coherence axioms:the diagrams

A⊗ (B ⊗ (C ⊗D))

αA,B,C⊗D

��

A⊗αB,C,D// A⊗ ((B ⊗ C)⊗D)

αA,B⊗C,D

��

(A⊗B)⊗ (C ⊗D)

αA⊗B,C,D**TTTTTTTTTTTTTTTT

(A⊗ (B ⊗ C))⊗D

αA,B,C⊗Dttjjjjjjjjjjjjjjjj

((A⊗B)⊗ C)⊗D

A⊗ (I ⊗B)αA,I,B

//

A⊗λB &&NNNNNNNNNNN(A⊗ I)⊗B

ρA⊗Bxxppppppppppp

A⊗B

I ⊗ IλI //

ρI

// I

should commute.

37

In the special case where all of α, λ, ρ are equalities, we say that (C,⊗, I)is a strict monoidal category.

The three coherence diagrams ensure that different ways of reassociatingand reducing expressions involving the tensor product and the unit lead tothe same result. In fact, the so-called coherence theorem for monoidal cate-gories states that all such diagrams (i.e. built from objects of C,⊗, I, α, λ, ρ)commute. A consequence is that every monoidal category is equivalent to astrict one, and that this equivalence respects the monoidal structure. (Formore on these matters, see MacLane’s Categories for the Working Mathe-matician.)Examples 6.2.

(1) Whenever C is a category with finite products, we may let ⊗ = ×and I = 1. The universal property of the product and terminalobject then result in suitable coherence isomorphisms α, λ, ρ. Thusthe notion of a tensor product may be thought of as a generalizationof that of a cartesian product. However, it is important to keep inmind that products are special in several regards: first, being limits,products have a universal property, whereas tensor products maynot have any universal property; second, unlike products, a tensorproduct may not have projections and diagonals.

(2) Similarly, finite coproducts induce monoidal structure on C.(3) If C is any category, then we may consider the category End(C)

whose objects are endofunctors C −→ C and whose morphisms arenatural transformations. Composition of endofunctors gives a strictmonoidal structure on End(C).

(4) Consider the braid category, whose objects are finite non-empty or-dinals and whose maps are braids. The operation n⊗m = n+m ex-tends to a bifunctor. With tensor unit 0, this gives a strict monoidalstructure.

(5) If R is a commutative ring, then the usual tensor product of modulesmakes R − Mod into a monoidal category. This example will bestudied in detail later on.

Exercise 57. Verify the details in the first (or in the second) example.

Exercise 58. Show in detail that End(C) is strict monoidal under compo-sition.

From the examples it should be clear that a category C may carry manydifferent monoidal structures at the same time. For example, the categorySet has products and coproducts, each of which may be regarded as a tensorproduct. Or the category R − Mod: it has biproducts, but also the usualtensor product.

One important feature of monoidal structure on a category is that itsupports a notion of internal monoid, again generalizing the notion of amonoid in the category of sets.

Definition 6.3 (Monoid). Let (C,⊗, I, α, λ, ρ) be a monoidal category. Amonoid in C is a structure (M,η, µ), where η : I −→ M , µ : M ⊗M −→ Msuch that the following diagrams commute:

38

M ⊗ (M ⊗M) α //

M⊗µ��

(M ⊗M)⊗Mµ⊗M

// M ⊗Mµ

��

M ⊗Mµ

// M

M ⊗ IM⊗η

//

ρM##GGGGGGGGG M ⊗M

µzzvvvvvvvvv

M

M I ⊗M

λM

;;wwwwwwwww

η⊗M// M ⊗M

µddHHHHHHHHH

Examples 6.4.(1) In the category of sets with cartesian product as monoidal structure,

a monoid is just a monoid in the usual sense.(2) A monoid in the category of vector spaces (over a field k) is usually

called a (unital and associative) k-algebra.(3) A monoid in the category of abelian groups is just a ring.(4) A monoid in the category End(C) is the same thing as a monad on

C.

Exercise 59. Verify the details in the last example.

Exercise 60. Define a suitable notion of homomorphism of monoids, andshow that this gives a category Mon(C) of monoids in C.

Exercise 61. Also define the notion of a comonoid in C, and define acategory of comonoids in C.

Another noticable difference between cartesian products and general ten-sor products is that the latter need not be symmetrical. This is easily seenfrom the example End(C) above: the composite FG need not be equal toGF (not even isomorphic). This motivates the following definition:

Definition 6.5. A symmetric monoidal category is a monoidal category(C,⊗, I, α, λ, ρ) equipped with a natural isomorphism σ = σA,B : A ⊗ B−→ B ⊗A. The following coherence diagrams are required to commute:

A⊗ (B ⊗ C)A⊗σB,C

vvnnnnnnnnnnnnαA,B,C

((QQQQQQQQQQQQ

A⊗ (C ⊗B)

αA,C,B

��

(A⊗B)⊗ CσA⊗B,C

��

(A⊗ C)⊗B

σA,C⊗B ((QQQQQQQQQQQQC ⊗ (A⊗B)

αC,A,Bvvnnnnnnnnnnnn

(C ⊗A)⊗B

A⊗B 1 //

σA,B%%KKKKKKKKK A⊗B

B ⊗AσB,A

99sssssssss

39

The reader may verify that cartesian products and coproducts satisfy thissymmetry condition.

From the definition of monoidal category it follows that for a fixed objectB, there is a functor

−⊗B : C −→ C.

What properties do these functors have? In the category of sets, functorsof the form −×B have a right adjoint, given by exponentiation: there is anatural bijection

Set(X,Y B) ∼= Set(X ×B, Y ).A monoidal category is called closed if each functor − ⊗ B has a right

adjoint. This adjoint is usually written Hom(B,−), because the intuitionis that of an internal Hom-object. If C is not symmetric, one distinguishedbetween right-closedness (all −⊗B have right adjoints) and left-closedness(all A ⊗ − have right adjoints) but we will be mainly concerned with thesymmetric case. If C is closed, then the functors −⊗B preserve all existingcolimits.

6.2. Tensor products of modules. We now focus on the category R−Modand describe the monoidal structure. Assume R to be commutative, so thatwe may identify left- and right modules.

Definition 6.6. Let M,N, V be modules, and let f : M × N −→ V be afunction. Then f is said to be bilinear if the following equations hold:

f(m, rn) = f(mr, n)f(m+m′, n) = f(m,n) + f(m′, n)f(m,n+ n′) = f(m,n) + f(m,n′).

Such a map f is a homomorphism of R-modules in each variable sepa-rately, but is generally not a homomorphism M ⊕ N −→ V . The tensorproduct M ⊗ N (also written M ⊗R N if confusion about the base ring ispossible) of M and N is a representing object for such bilinear maps, in thefollowing sense: there is a map M ×N γ−−→M ⊗N which has the universalproperty that any bilinear map f : M ×N −→ V factors uniquely as f̂γ fora homomorphism f̂ : M ⊗N −→ V .

M ×Nγ//

f&&LLLLLLLLLLL M ⊗N

f̂�����

V

The tensor product M⊗N is constructed as follows. The underlying abeliangroup of M ⊗ N is generated by formal symbols {m ⊗ n|m ∈ M,n ∈ N},and subject to relations

m⊗ rn = mr ⊗ n(m+m′)⊗ n = m⊗ n+m′ ⊗ nm⊗ (n+ n′) = m⊗ n+m⊗ n′.

Next, define an action of R on M ⊗N via

r(m⊗ n) = m⊗ rn.

40

It is now a straightforward exercise to verify that this is indeed an R-module,and that the map γ(m,n) = m⊗ n has the required universal property.

Exercise 62. Show that the assignment (M,N) 7→ M ⊗ N extends to abifunctor on the category of R-modules. Then show that R − Mod is asymmetric monoidal category with R as tensor unit.

We give a few examples of calculations with tensor products; more exam-ples will appear in the next section.Examples 6.7.

(1) If M = F (X) is the free module on X, then F (X)⊗N =∐x∈X N .

This may be calculated directly, or by using the fact (see below) thatthe tensor product commutes with colimits.

(2) Over the ring Z, we have Z/pZ ⊗ B = B/pB, where pB is thesubgroup consisting of the elements of the form {pb|b ∈ B}. Thus,tensoring with Z/pZ has the effect of “killing off” elements in Bwhich are divisible by p.

(3) From the previous example it follows that Z/pZ ⊗ Q = 0, and thatZ/pZ⊗ Z/qZ = 0 if p, q are relatively prime.

(4) The tensor product depends on the ground ring! For example, overZ we have Z/2Z ⊗ Z/4Z = Z/2Z, but over the ring Z/2Z we haveZ/2Z⊗Z/2Z Z/4Z = Z/4Z (since the ground ring is the tensor unit).

(5) It may happen that A 6= 0 but that A⊗A = 0. (Hint: look at Q/Z.)

The first two of these examples both follow from the fact that the tensorproduct commutes with all colimits; to obtain the second example from thisfact, consider the exact sequence

0 −→ Z p−−→ Z −→ Z/pZ −→ 0

and apply the right exact functor −⊗B, obtaining

Z⊗B −→ Z⊗B −→ Z/pZ⊗B −→ 0,

which is isomorphic to

Bp−−→ B −→ B/pB −→ 0.

We will now show that R − Mod is in fact monoidal closed. Fixing amodule M , we need a right adjoint to −⊗M . We claim that the functor

Hom(M,−) : R−Mod −→ R−Mod; B 7→ Hom(M,B)

is such a right adjoint. Here, we regard Hom(M,A) as an R-module via thepointwise action

(rf)(m) = f(rm),for f : M −→ B, r ∈ R,m ∈M . To verify the adjunction we have to exhibita natural bijection

(11) Φ : R−Mod(A⊗M,B)∼=−−→ R−Mod(A,Hom(M,B)).

We define Φ(h), for h : A⊗M −→ B, via

Φ(h)(a) = h̃ : M −→ B; h̃(m) = h(a,m).

It is easily shown that this is natural in A,B and bijective.

41

Exercise 63. Prove the remaining details.

For obvious reasons, the adjunction − ⊗M a Hom(M,−) is referred toas a Hom-tensor adjunction. In fact, there exist many adjunctions of thisform. While the above adjunction arose by considering hom and tensor tobe endofunctors of the category of modules, we may also regard adjunctionsbetween different categories of modules. For example, the functor − ⊗ Rmay also be viewed as a functor

−⊗R : R−Mod −→ AbGrp.

Moreover, the hom-functor Hom(R,−) may be regarded as a functor

Hom(R,−) : AbGrp −→ R−Mod,

where Hom(R,A) is the abelian group of additive maps R −→ A, equippedwith the pointwise R-action. We then get an adjunction

R−Mod−⊗R⊥--AbGrp.

Hom(R,−)mm

This is precisely the adjunction we found earlier, where instead we regarded−⊗R as the forgetful functor.

Exercise 64. Let φ : R −→ S be a ring homomorphism (you may assumethat the rings are commutative), let M be an R-module and N be an S-module.

(i) Define an R-module structure on the underlying abelian group of N ,by letting R act as rn = φ(r)n. Show that this is well-defined andgives rise to a functor φ∗ : S − Mod −→ R − Mod, sometimes calledrestriction of scalars.

(ii) Recall that S is canonically an R-module via φ. Therefore we can formM⊗RS, which is again an R-module. Regard M⊗RS as an S-modulevia s(m ⊗ n) = m ⊗ sn. Show that this gives a functor φ! : R −Mod−→ S −Mod, called extension of scalars.

(iii) Show that φ! is the left adjoint to φ∗.The above situation is referred to as change-of-rings.

Exercise 65. By an R-algebra we mean a ring S together with a ringhomomorphism φ : R −→ S. These form a category, which is equivalent tothe co-slice category R/Ring.

Show that an R-algebra is the same thing as a monoid object in themonoidal category R−Mod. Show that in fact this gives an equivalence ofcategories

Mon(R−Mod) ' R/Ring.

6.3. Torsion. Before we try to understand the derived functors of the form−⊗B we rehearse some notions from group- and module theory.

Definition 6.8. Let R be a ring without zero divisors, M an R-module andm ∈ M . The element m is a torsion element if there is an element r ∈ R,r 6= 0, with rm = 0. We write T (M) for the set of torsion elements of M .

42

It is a straightforward exercise that T (M) is in fact a submodule of M .If T (M) = 0, we say that M is a torsion-free module; if T (M) = M , thenM is said to be a torsion module.

In the case of abelian groups, it is easy to see that free groups are torsion-free. Moreover, any finite group must be torsion. Also, it is easily provedthat the direct sum of two torsion groups is again torsion.

At this point it is useful to recall the structure theorem for finitely gen-erated abelian groups, which states that if A is finitely generated, then itdecomposes as a direct sum of a free group and a finite direct sum of finitecyclic groups:

A ∼= Zm ⊕ Z/p1Z⊕ · · · ⊕ Z/pkZ.Thus, A splits up into a torsion-free part Zm and a torsion part. The torsionpart is precisely the torsion subgroup T (A).

Over nice rings (principal ideal domains) there is a corresponding theoremfor modules, which states that any finitely generated module M admits adecomposition M ∼= F ⊕ T (M).

Here are some more examples:Examples 6.9.

(1) The abelian groups Z,Q,R are torsion-free.(2) In the abelian group Q/Z, every element is torsion.(3) The abelian group R/Z is not torsion-free; its torsion subgroup is

Q/Z.

Lemma 6.10. If A is a torsion group then A⊗Q = 0.

Proof. If pa = 0 for some p then a⊗ 1 = a⊗ pp = ap⊗ 1

p = 0. �

This makes precise the idea that tensoring with the rationals kills off alltorsion elements. The reader should verify that tensoring with Q/Z has thesame effect.

There is another useful exact sequence:

0 −→ pB −→ B −→ B ⊗ Z/pZ −→ 0.

One can verify this directly, but it will also follow from considering derivedfunctors as we shall see next.

6.4. Tor-groups. Since − ⊗ B is right exact it has left derived functorsLn(− ⊗ B). To compute the value at an object A, we choose a projectiveresolution of A:

· · ·Pn −→ Pn−1 −→ · · · −→ P1 −→ P0ε−→ A −→ 0

and apply the functor −⊗B to the complex P , to get

· · ·Pn ⊗B −→ Pn−1 ⊗B −→ · · · −→ P1 ⊗B −→ P0 ⊗B −→ 0.

Then Ln(−⊗B)(A) is by definition the n-th homology of this complex.The objects Ln(− ⊗ B)(A) are usually denoted Torn(A,B), and are re-

ferred to as tor-groups.We will first do an easy calculation in abelian groups which shall give us

an idea of what these groups tell us. Take A = Z/pZ, and consider thestandard projective resolution

0 −→ Z p−−→ Z −→ Z/2Z −→ 0.

43

Applying −⊗B we get:

0 // Z⊗Bp⊗B

//

∼=��

Z⊗B //

∼=��

0

Zψ

// Z

where it is easily verified that the induced map is indeed multiplication byp again. Therefore we have

Tor0(Z/pZ, B) = Coker(ψ) = B/pB; Tor1(Z/pZ, B) = Ker(ψ) = pB.

The group pB is by definition the subgroup of p on the elements which arep-torsion, i.e. those b for which pb = 0.

From the fact that the 0-th derived functor of F is isomorphic to F , weget

Z/pZ⊗B = B/pB.

From this example we also see that the group Tor1(Z/pZ, B) gives us afragment of the torsion subgroup of B, namely those elements which arep-torsion. Our next aim is to show that the entire torsion subgroup T (B)can be obtained by replacing the group Z/pZ with Q/Z.

In order to do this, we mention without proof a fact about derived func-tors. For this, recall that a poset is called directed if any two elements havean upper bound.

Lemma 6.11. Let P be a directed poset and let G : P −→ C be a diagram inan abelian category C. Then if F : C −→ D is right exact, its derived functorspreserve the colimit of G, in the sense that

LnF (ColimG) ∼= Colim(LnFG).

The directed diagram we are interested in is the following: the objectsare the abelian groups Z/pZ, and there is a unique arrow Z/pZ n−−→ Z/npZ.This diagram is clearly directed, and we have

ColimZp ∼= Q/Z.

(Of course, this suppresses some notation.)We now derive:

Proposition 6.12. For any abelian group B, Tor1(Q/Z, B) ∼= T (B).

Proof. We have

Tor1(Q/Z, B) ∼= Tor1(ColimZ/pZ, B) ∼= ColimTor1(Z/pZ, B) ∼= ColimpB.

The latter colimit is the diagram of inclusions of all p-torsion subgroups ofB, and its union (colimit) is T (B). �

In particular we find that if B is torsion-free, then Tor1(Q/Z, B) = 0.We can strenghthen this result:

Lemma 6.13. If B is torsion-free, then Tor1(A,B) = 0 for all A.

44

Proof. We invoke a result from group theory which says that every abeliangroup is the directed colimit of its finitely generated subgroups. So letA ∼= ColimAi where the Ai are finitely generated. Then

Tor1(A,B) ∼= ColimTor1(Ai, B)

since Tor1 commutes with directed colimits. Thus it suffices to show theresult for finitely generated abelian groups A. For that case, write

A ∼= Zm ⊕ Z/p1Z⊕ · · · ⊕ Z/pkZ.Then

Tor1(A,B) ∼= Tor1(Zm, B)⊕ Tor1(Zp1Z, B)⊕ · · · ⊕ Tor1(Z/pkZ, B).

Since B has no torsion, all the summands Tor1(Z/piZ, B) vanish. Since Zmis projective, the summand Tor1(Zm, B) vanishes as well. �

We conclude:

Theorem 6.14. For an abelian group B, the following are equivalent:(i) B is torsion-free(ii) Tor1(A,B) = 0 for all A

(iii) Tor1(B,A) = 0 for all A(iv) B is flat, i.e. the functor −⊗B is exact.

Proof. We have shown (i)⇒ (ii). The implication (ii)⇔ (iii) is immediatefrom the fact that the tensor product is symmetric. Moreover, (iii) ⇔ (iv)follows from general facts about derived functors. In particular, in this caseTor1(Q/Z, B) = 0, so B is torsion-free. �

Exercise 66. Construct a long exact Tor-sequence for abelian groups.

6.5. Balancing Tor and Ext. In the previous section we defined Tor(A,B)as the derived functors of − ⊗ B, using a projective resolution of A. Bysymmetry, we could also have defined the derived functors of A ⊗ − usinga projective resolution of B. Similar considerations hold for the functorsExt(−,−). We shall now show that both methods lead to the same result:this is often expressed by saying that Tor(−,−) and Ext(−,−) are balanced.We shall prove the result only for Tor, in the case of abelian groups.

First, we introduce some useful terminology and notation:

Definition 6.15. Given a commutative square Σ:

Ae //

f��

B

g

��

A′e′ // B′

we let

Ker(Σ) = Ker(ge)/(Ker(e)+Ker(f)); Im(Σ) = (Im(g)∩Im(e′))/Im(ge).

We note that in the following special cases

0 //

��

Σ

B��

��

A // //

����

B

��

A′ // // B′ A′ // 0

45

we have Im(Σ) ∼= B and Ker(Θ) ∼= A′.The main point of the definition is the following lemma, which shows that

morphisms of short exact sequences may be viewed as short exact sequencesof squares:

Lemma 6.16. Let

0 // Am //

Σf��

B

Θ

e //

g

��

C

h��

// 0

0 // A′m′ // B′

e′ // C ′ // 0

be a morphism of short exact sequences. Then Ker(Θ) ∼= Im(Σ).

Proof. We show that there is an isomorphism g : Ker(Θ) −→ Im(Σ) inducedby g. Given b ∈ Ker(he) we clearly have g(b) ∈ Ker(e′) = Im(m′), so wehave a function g : Ker(he) −→ Im(g) ∩ Im(m′). To show that this is well-defined when we divide by Ker(g) + Ker(e), note first that if x ∈ Ker(g)then g(x) = 0. Moreover, if x ∈ Ker(e) = Im(m) then x = m(a) for somea, and hence g(x) = gm(a) ∈ Im(gm), so vanishes in Im(Σ).

Next, we show that g is injective. If g(x) ∈ Im(gm) then we have g(x) =gm(a) for some a ∈ A. Then x −m(a) ∈ Ker(g), while m(a) ∈ Im(m) =Ker(e). Thus x ∈ Ker(g) +Ker(e).

Finally, g is surjective: given y ∈ Im(g) ∩ Im(m′), write y = g(x) forsome x ∈ B. Then he(x) = e′g(x) = e′(y) = 0, since y ∈ Im(m′) = Ker(e′).Thus x ∈ Ker(he). �

With this in place, we can prove the desired result about the functorTor1(−,−). Fix two abelian groups A,B, and fix projective resolutions

0 −→ Rm−−→ F

e−−→ A −→ 0

of A, and

0 −→ Sn−−→ G

f−−→ B −→ 0

of B. Now we form the “tensor product” of these resolutions:

0 //

��

Σ5

Ker(m⊗B)��

��

0

��

R⊗ S //

��

Σ3

R⊗G

��

Σ4

// // R⊗Bm⊗B��

R

��

0 //

��

Σ1

F ⊗ S

Σ2

//

����

F ⊗G

��

// F ⊗B

��

F

��

Ker(A⊗ n) // // A⊗ SA⊗n

// A⊗G // A⊗B A

��

0 // S // G // B // 0

Observe that Im(Σ1) ∼= Ker(A ⊗ n) = Tor(A,B) where the latter is com-puted using the projective resolution of B, while Im(Σ5) ∼= Ker(m⊗B) =

46

Tor(A,B) where the latter is computed using a projective resolution of A.To see that these are the same, we use the lemma repeatedly:

Im(Σ1) ∼= Ker(Σ2) ∼= Im(Σ3) ∼= Ker(Σ4) ∼= Im(Σ5).

This shows that both ways of computing Tor(A,B) yield the same result.

Exercise 67. Mimick the above proof for the functor Ext(−,−).