Romyar SharifiLECTURE NOTES
READING EDITIONPDF

LECTURE NOTES / Chapter 2

Homological Algebra

Romyar Sharifi

Chapter 2 Abelian Categories

Book contents

Chapter 2
Abelian Categories

2.1. Additive categories

Definition 2.1.1.

An additive category 𝒞 is a category with the following properties:

i.

for A,B Obj(𝒞), the set of morphisms Hom𝒞(A,B) in 𝒞 has an abelian group law (addition) with the property that for any diagram

Additivity of composition. A full diagram description follows.
Diagram description: Additivity of composition

This diagram records the composable morphisms in the distributivity identity: h composed with (g subscript 1 plus g subscript 2) composed with f equals the sum of h composed with g subscript 1 composed with f and h composed with g subscript 2 composed with f.

Objects, listed by row and column:

  • Row 1, from left to right: column 1: A; column 2: B; column 3: C; column 4: D.

Arrows and lines:

  1. An arrow from A to B, labelled f.
  2. An arrow from B to C, labelled g subscript (1).
  3. An arrow from B to C, labelled g subscript (2).
  4. An arrow from C to D, labelled h.

in 𝒞, we have

h(g1 +g2)f = hg1 f +hg2 f,
ii.

𝒞 has a zero object 0,

iii.

𝒞 admits finite coproducts.

Note that between any two objects A and B in an additive category 𝒞, there always exists the zero morphism 0: A B, the unique map factoring through 0. This is the identity element in the abelian group Hom𝒞(A,B).

Examples 2.1.2.

a.

The categories 𝐀𝐛 and R-mod are additive categories, with the usual addition of homomorphisms.

b.

The full subcategory R-mod of finitely generated R-modules is an additive category.

c.

The category of topological Hausdorff abelian groups (with continuous group homomorphisms) is an additive category.

In an additive category, we denote the coproduct of two objects A1 and A2 by A1 A2.

Lemma 2.1.3.

Finite products exist in an additive category, and there are natural isomorphisms

A1 A2A1 ×A2

for A1,A2 Obj(𝒞). The resulting inclusion morphisms ιi: Ai A1 A2 and projection morphisms and pi: A1 A2 Ai obtained by viewing A1 A2 as a product and coproduct, respectively, satisfy piιi = idAi and piιj = 0 for ij, while

ι1 p1 +ι2 p2 = idA1A2.
Proof.

We have maps ιi: Ai A1 A2 by definition of the direct sum. We also have maps pi: A1 A2 Ai defined by

piιj = { idAiif i = j 0 if ij

and the universal property of the coproduct.

We then have

(ι1 p1 +ι2 p2)ιi = ιi,

and hence

ι1 p1 +ι2 p2 = idA1A2.

Given an object B Obj(𝒞) and maps gi: B Ai, we then have a map

ψ = ι1 g1 +ι2 g2: B A1 A2,

which is unique such that

piψ = gi.

Hence A1 A2 satisfies the universal property of the product.

Definition 2.1.4.

An object A in an additive category 𝒞 together with objects Ai, inclusion morphisms ιi: Ai A, and projection morphisms pi: A Ai for i {1,2} for which piιj is zero if ij and idAi if i = j and for which ι1 p1 +ι2 p2 = idA is called a biproduct of the objects A1 and A2, and we write it as A1 A2.

The reader will verify the following.

Lemma 2.1.5.

In an additive category 𝒞, the biproduct A1 A2 is a coproduct of the Ai via the inclusion morphisms ιi and to the product of the Ai via the morphisms pi.

The notion of a biproduct allows us to reinterpret addition in an additive category. First, note the following definitions.

Definition 2.1.6.

Let A be an object in an additive category 𝒞.

a.

The diagonal morphism ΔA: A AA in 𝒞 is the unique morphism induced by two copies of idA: A A and the universal property of the product.

b.

The codiagonal morphism A: AA A in 𝒞 is the unique morphisms induced by two copies of idA: A A and the universal property of the coproduct.

Definition 2.1.7.

Let 𝒞 be an additive category, and let f1: A1 B1 and f2: A2 B2 be morphisms in 𝒞. The biproduct, or direct sum, f1 f2 of the maps f1 and f2 is the morphism A1 A2 B1 B2 induced as the (morphism defined by the universal property of the) coproduct of the composite maps Ai Bi B1 B2, the latter morphisms being inclusions.

Remark 2.1.8.

Equivalently, the direct sum of f1 and f2 as in Definition 2.1.7 is induced as the product of the composite maps A1 A2 Ai Bi, the initial morphisms being projections.

Of course, we could make these definitions in an arbitrary category using products and coproducts.

Lemma 2.1.9.

Let f,g: A B be two morphisms in an additive category 𝒞. Then we have

f +g = B(f g)ΔA.
Proof.

Let ιiA and piA respectively denote the inclusion maps and projection maps for the biproduct AA, and similarly for B. We have

B(f g)ΔA = (f g)(ι1Ap1A+ι2Ap2A)Δ A = B(f g)ι1Ap1AΔ A+B(f g)ι2Ap2AΔ A.

Taking the first term without loss of generality, we have

B((f g)ι1A)(p1AΔ A) = B(ι1Bf)id A = idBf = f.

Definition 2.1.10.

A functor F : 𝒞 𝒟 between additive categories is called additive if for each A,B Obj(𝒞), the map

Hom𝒞(A,B) Hom𝒟(F (A),F (B))

is a group homomorphism.

Example 2.1.11.

Let 𝒞 be an additive category. Then for any A Obj(𝒞), the functors hA and hA may be considered as functors to 𝐀𝐛, rather than 𝐒𝐞𝐭. The resulting functors are additive.

Example 2.1.12.

Let R be a ring, and let A be a right R-module. Then the functor

tA: R-mod 𝐀𝐛

given by tA(B) = ARB and tA(f) = idAf for f : B C is additive.

Lemma 2.1.13.

A functor F : 𝒞 𝒟 of additive categories is additive if and only if F preserves biproducts, which is to say that the natural morphisms F (A1)F (A2) F (A1 A2) and F (A1 A2) F (A1)F (A2) are inverse isomorphisms for all objects A1,A2 in 𝒞.

Proof.

Suppose first that F is an additive functor. Note that F (ιipi) = idF (Ai) and F (ιipj) = F (0) for ij, but F (0) = 0 by additivity of F. Again by additivity of F, we have

F (ι1)F (p1)+F (ι2)F (p2) = F (idA1A2) = idF (A1A2).

It follows that F (A1 A2) is a biproduct of F (A1) and F (A2) in 𝒟, so in particular it is a coproduct.

On the other hand, if F preserves biproducts and f,g: A B are morphisms in 𝒞, then it is easy to see that F (f g) = F (f)F (g), and Lemma 2.1.9 tells us that

F (f +g) = F (B(f g)ΔA) = F (B) (F (f)F (g))ΔF (A) = F (f)+F (g).

Corollary 2.1.14.

Let F : 𝒞 𝒟 be a fully faithful functor of additive categories. Then F is an additive functor.

For additive functors, we may consider a finer notion of representability.

Remark 2.1.15.

If F : 𝒞 𝐀𝐛 is an additive contravariant (resp., covariant) functor of additive categories, then we may consider it to be representable if there exists an object X Obj(𝒞) and a natural isomorphism η : hX F (resp., η : hX F). In this case, the morphisms ηA for A Obj(𝒞) will be isomorphisms of groups.

2.2. Kernels and cokernels

Remark 2.2.1.

In an additive category, any morphism to 0 is an epimorphism, and any morphism from 0 is a monomorphism.

Definition 2.2.2.

Let 𝒞 be an additive category, and let f : A B be a morphism in 𝒞.

a.

The kernel kerf of f is the equalizer eq(f,0), when it exists.

b.

The cokernel cokerf is the coequalizer coeq(f,0), when it exists.

Example 2.2.3.

In the category R-mod, these definitions agree with the classical ones.

Lemma 2.2.4.

A morphism in an additive category that admits kernels is a monomorphism if and only if it has zero kernel. A morphism in an additive category that admits cokernels is an epimorphism if and only if it has zero cokernel.

Proof.

Let f : A B be a monomorphism, and let h: kerf A be the induced morphism. Since f h = 0 by definition of the kernel, we have h = 0, as f is a monomorphism. This forces kerf to be 0, since h factors through 0. (Or, one could just apply Lemma 1.7.3.) On the other hand, suppose that f has trivial kernel, and let g,h: C A be maps with f g = f h. Then f (gh) = 0, and by universal property of the kernel, gh factors through 0, i.e., is 0.

The proof for cokernels is similar, or is the result on kernels in the opposite (additive) category.

Remark 2.2.5.

Let 𝒞 be an additive category that admits kernels (resp., cokernels). Then there is a functor

ker: Mor(𝒞) 𝒞(resp., coker: Mor(𝒞) 𝒞)

which takes an object in Mor(𝒞) to its kernel (resp., cokernel) and a morphism in Mor(𝒞) to the natural morphism between kernels (resp., cokernels).

Proposition 2.2.6.

Let 𝒞 be an additive category that admits kernels and cokernels. Let f : A B be a morphism in 𝒞. Then

imfker(B cokerf)

and

coim𝑓≅coker(kerf A).
Proof.

We prove the first isomorphism. Let g: B cokerf. By Yoneda’s lemma, it suffices to show that himf and hkerg are naturally isomorphic. For C Obj(𝒞), we have a map

Hom𝒞(C,imf) {α : C Bι1 α = ι2 α}

that takes a morphism C imf and composes it with the morphism imf B given by definition of the equalizer of the maps ιi: B BAB. It is a bijection by the universal property of the equalizer.

For any D Obj(𝒞) and morphisms ϕ1,ϕ2: B D such that ϕ1 f = ϕ2 f, note that there exists a unique morphism k: BAB D with ϕi = kιi. Any α : C B such that ι1 α = ι2 α then satisfies ϕ1 α = ϕ2 α for any such ϕi: B D and any D. On the other hand, note that the ιi themselves satisfy the property that ι1 f = ι2 f and are morphisms ιi: B D with D = BAB. In other words, we have

{α : C Bι1 α = ι2 α} = {α : C Bϕ1 α = ϕ2 α if ϕ1 f = ϕ2 f for some ϕ1,ϕ2: B D (for some D Obj(𝒞))}.

Now, we are in an additive category, so this equals

{α : C Bϕ α = 0 if ϕ f = 0 for some ϕ : B D}. (2.2.1)

By the universal property of cokerf, for any ϕ : B D with ϕ f = 0, there is a morphism j: cokerf D with jg = ϕ. If gα = 0, then ϕ α = jgα = 0, and this works for any ϕ : B D with ϕ f = 0. On the other hand, g itself satisfies gf = 0 so is such a ϕ. It follows that the set in (2.2.1) equals

{α : C Bgα = 0}.

By the universal property of kerg, this is in bijection with Hom𝒞(C,kerg), taking an α in the set to the unique morphism to kerg through which it factors. Clearly, the composition of these bijections is natural in C, so we have the desired natural isomorphism.

2.3. Abelian categories

Definition 2.3.1.

An abelian category is an additive category 𝒞 in which

i.

every morphism in 𝒞 admits a kernel and a cokernel and

ii.

every morphism in 𝒞 is strict.

Examples 2.3.2.

a.

The category R-mod is abelian.

b.

The full subcategory 𝒞 of R-mod of finitely generated R-submodules is not abelian in all cases. E.g., when R is commutative and non-Noetherian, we can take I to be an ideal of R that is not finitely generated, and so the kernel of R RI is not in 𝒞.

c.

The category of topological Hausdorff abelian groups is not abelian, but it is additive and admits kernels and cokernels. For instance, consider the inclusion map ι : with having its usual topology and having the subspace topology. Then kerι = 0 and cokerι = 0 (since is dense in , and thus every continuous map from is determined by its values on ). By Proposition 2.2.6, we have imι but coimι.

Remark 2.3.3.

Note that if 𝒞 is an abelian category, then so is 𝒞op. The roles of mono- and epimorphisms, kernels and cokernels, and images and coimages switch in 𝒞 and 𝒞op.

Proposition 2.3.4.

The functor category 𝐅𝐮𝐧𝐜(𝒞,𝒟) from a small category 𝒞 to an abelian category 𝒟 is abelian.

Proof.

We sketch the proof. First, note that it is additive: we have the zero functor which sends all objects to the zero object and all morphisms to the zero (identity) morphism of the zero object, and if F,G: 𝒞 𝒟 are functors, then F G is given by (F G)(C) = F (C)G(C) and (F G)(f) = F (f)G(f) for f : A B in C. This can be used to define the addition on morphisms (i.e., natural transformations) as before.

Next, the kernel of a natural transformation η : F G is defined by (kerη)(C) = kerηC and (kerη)(f) for f : A B is the kernel of the induced morphism kerηA kerηB. The cokernel is defined similarly. Note that

(coimη)Acoker(kerη η)Acoker(kerηA ηA)coimηA,

and similarly for images. Finally, since 𝒟 is abelian, the natural map coimη imη is an isomorphism coimηA imηA on objects A in 𝒞, hence has a natural inverse determined by the inverses of these morphisms.

We discuss a related class of examples.

Definition 2.3.5.

Let X be a topological space and 𝒞 an abelian category. Consider the category 𝒰X with objects the open sets in X and morphisms the inclusion maps between subspaces to the category 𝒞.

a.

A presheaf F on X with values in 𝒞 is a contrvariant functor F : 𝒰X 𝒞. The category of presheaves is the functor category 𝐅𝐮𝐧𝐜(𝒰,𝒞).

b.

Suppose that 𝒞 admits arbitrary products. A sheaf F on X with values in 𝒞 is a presheaf F : 𝒰X 𝒞 such that each F (U) for U 𝒰X is the equalizer of the diagram

The sheaf equalizer condition. A full diagram description follows.
Diagram description: The sheaf equalizer condition

The two arrows are induced by restriction to intersections of open sets. The sheaf condition says that F(U) is the equalizer of these two arrows for the given open covering. The parallel arrows themselves are not required to be equal.

Objects, listed by row and column:

  • Row 1, from left to right: column 1: product subscript (V in script V) F(V); column 2: product subscript (V subscript (1), V subscript (2) in script V) F(V subscript (1) intersection V subscript (2)).

Arrows and lines:

  1. An arrow from product subscript (V in script V) F(V) to product subscript (V subscript (1), V subscript (2) in script V) F(V subscript (1) intersection V subscript (2)), without a label.
  2. An arrow from product subscript (V in script V) F(V) to product subscript (V subscript (1), V subscript (2) in script V) F(V subscript (1) intersection V subscript (2)), without a label.

where 𝒱 is an open covering of U by open subsets via the two maps on products induced by the application of F to inclusion maps V1 V2 Vi for each pair (V1,V2) of open sets in 𝒱. The category of sheaves is the full subcategory of the category of presheaves with objects the sheaves.

Example 2.3.6.

Let X be a topological space and 𝒞 an abelian category. For any A Obj(𝒞), we can view A as a discrete space and consider the constant sheaf

F (U) = {f : U Af continuous}

for all U 𝒰X. If U is connected, then F (U) = A.

Example 2.3.7.

Let X be a topological space and for U 𝒰X. Then 𝒪X is the sheaf with values in 𝐀𝐛 of continuous functions to that for U 𝒰X satisfies

𝒪X(U) = {continuous maps U }.

Example 2.3.8.

Let X = , and let F : 𝒰 𝐀𝐛 be the presheaf such that F (U) is the abelian group of bounded continuous functions on U. Then F is not a sheaf, as we may consider the open covering of by intervals In = (n,n) for n , and then the function f(x) = x is contained in F (In) for each n, so in the equalizer, but it is not in F ().

Terminology 2.3.9.

In an abelian category 𝒞, we will typically refer to a coproduct (when it exists) as a direct sum, and we write iIAi in place of iIAi.

2.4. Exact sequences

Suppose that f : A B and g: B C are morphisms in an abelian category 𝒞 with gf = 0. Note that f : κ λ, where λ : A imf is an epimorphism and κ : imf B is a monomorphism. We have gκ λ = 0, so the fact that λ is an epimorphism tells us that gκ = 0. The universal property of the kernel then provides a morphism α : imf kerg such that when composed with the canonical monomorphism β : kerg B is β α = κ. Since β and κ are both monomorphisms, which is to say have trivial kernel, we have that α is a monomorphism as well.

Definition 2.4.1.

We say that a diagram

A fB gC

in an abelian category 𝒞 is exact if gf = 0 and the induced monomorphism imf kerg is an isomorphism. We call such a diagram a three term exact sequence.

We generalize this notion. First, we define a sequence.

Definition 2.4.2.

A diagram of the form

An+1 dn+1AA n dnAA n1

in a category 𝒞 is a sequence, with the understanding that the diagram may terminate (i.e., be of finite length) on either, both, or neither side. We denote the data of a sequence by A = (Ai,diA), where i runs for Ai over the interval of integers on which Ai is defined and for diA over every such i but the left endpoint, if it exists.

Definition 2.4.3.

Let 𝒞 be an abelian category.

a.

A chain complex in 𝒞 is a sequence A = (Ai,diA)i in 𝒞 such that diAdi+1A = 0 for all i .

b.

For a chain complex A and i , the morphism diA: Ai Ai1 is called the ith differential in the complex A.

Notation 2.4.4.

Unless otherwise specified, the ith object in a chain complex A will be denoted Ai and the ith differential by di: Ai Ai1. If we have multiple complexes, we will use diA to specify the differential on A.

Notation 2.4.5.

In an abelian category, if A is a subobject of an object B, we will often write A B to denote this and BA for the cokernel of the inclusion morphism A B. For f : B C, we will let f(A) denote the image of the composite of the inclusion with f.

Definition 2.4.6.

A sequence

An+1 dn+1An dn1An1

in an abelian category 𝒞 is exact if the subdiagram Am+1 Am Am1 is exact for each m such that there are terms Am+1 and Am1 in the diagram.

Definition 2.4.7.

A long exact sequence in an abelian category 𝒞 is an exact sequence A = (Ai,diA), where i runs over all integers.

Definition 2.4.8.

A complex is often said to be acyclic if it is an exact sequence.

Definition 2.4.9.

Let 𝒞 be an abelian category.

a.

A short exact sequence in 𝒞 is an exact sequence in 𝒞 of the form

0 A B C 0.
b.

A left short exact sequence in 𝒞 is an exact sequence in 𝒞 of the form

0 A B C.
c.

A right short exact sequence in 𝒞 is an exact sequence in 𝒞 of the form

A B C 0.

Definition 2.4.10.

Let 𝒞 be an abelian category.

a.

We say that an epimorphism g: B C is split if there exists a morphism t : C B in 𝒞 with gt = idC. In this case, we say that t is a splitting of g.

b.

We say that a monomorphism f : A B is split if there exists a morphism s: B A in 𝒞 with sf = idA. In this case, we say that s is a splitting of f.

c.

We say that a short exact sequence

0 A fB gC 0 (2.4.1)

in an abelian category 𝒞 splits if there exists an isomorphism AC wB in 𝒞 with wιA = f and wιCg = idB, where ιA: A AC and ιC: C AC are the inclusion morphisms.

Example 2.4.11.

Any exact sequence

0 3 6 2 0

is split in Ab, but

0 2 4 2 0

is not.

Lemma 2.4.12.

The following conditions on a short exact sequence (2.4.1) are equivalent:

  1. The sequence (2.4.1) splits.
  2. The monomorphism f : A B splits.
  3. The epimorphism g: B C splits.
Proof.

  1. Suppose we have a splitting map t : C B. Then define s: B A by s(b) = a where f(a) = bt(g(b)). This is well-defined as f is injective, and such an a exists since

    g(bt(g(b))) = g(b)g(t(g(b))) = g(b)g(b) = 0.

    It splits f as

    s(f(a)) = s(b)s(t(g(b))) = s(b),

    the latter step using the fact that st = 0, which follows in turn from

    f(s(t(c))) = t(c)t(g(t(c))) = t(c)t(c) = 0.
  2. Suppose that we have a splitting map s: B A. Then define t : C B by t(c) = bf(s(b)) where g(b) = c. To see this is well defined, note that f(a)f(s(f(a))) = 0 for any a A.
  3. Define w(a,c) = f(a)+t(c). Its inverse is w(b) = (s(b),g(b)). We check

    ww(a,c) = (s(f(a)+t(c)),g(f(a)+t(c))) = (s(f(a)),g(t(c))) = (a,c)

    and

    ww(b) = w(s(b),g(b)) = f(s(b))+t(g(b)).

    It remains to see that f s+t g = idB. It is easy to see this on elements of the form f(a) or t(c) and that an arbitrary element of b is a sum of these.

  4. Set t(b) = w(0,b).

2.5. Exact functors

Definition 2.5.1.

Let F : 𝒞 𝒟 be an additive functor between abelian categories.

a.

We say that F is left exact if it preserves exact sequences of the form

0 A B C,

which is to say that the sequence

0 F (A) F (B) F (C)

is exact.

b.

We say that F is right exact if it preserves exactness of exact sequences of the form

A B C 0.
c.

We say that F is exact if it is both left and right exact.

The reader will verify the following.

Lemma 2.5.2.

An additive functor F : 𝒞 𝒟 between abelian categories is exact (resp., left exact, resp., right exact) if and only if it takes short exact sequences in 𝒞 to short exact (resp., left short exact, resp. right short exact) sequences in 𝒟. Moreover, F is exact if and only if it preserves three term exact sequences.

Remark 2.5.3.

We may extend the definition of left and right exact functors to contravariant functors. The requirement that a contravariant functor F : 𝒞 𝒟 be left exact is that it preserves exactness of sequences of the form 0 A B C, which is to say that F (C) F (B) F (A) 0 is exact.

Examples 2.5.4.

a.

If 𝒞 is abelian, then the functor F : 𝒞 𝒞 by F (A) = AA with F (f) = f f is exact.

b.

In R-mod for a commutative ring R, the functor tA is right exact for any R-module A.

c.

Let G be a group. For a [G]-module A, consider the abelian group

AG = {a A𝑔𝑎 = a for all g G}

of G-invariants in A, with [G]-module homomorphisms A B restricting to homomorphisms AG BG. The resulting functor [G]-mod 𝐀𝐛 is left exact but not in general right exact.

Recall that for an additive category 𝒞, the functors hX (and hX) may be viewed as taking values in 𝐀𝐛, and clearly such functors are additive. In fact, they are also left exact.

Lemma 2.5.5.

Let 𝒞 be an abelian category, and let X be an object of 𝒞.

a.

The functor hX: 𝒞 𝐀𝐛 is left exact.

b.

The functor hX: 𝒞op 𝐀𝐛 is left exact.

Proof.

Let

0 A fB gC 0

be an exact sequence in 𝒞. Applying hX, we obtain homomorphisms

0 Hom𝒞(X,A) hX(f)Hom𝒞(X,B) hX(g)Hom𝒞(X,C)

of abelian groups, and we claim this sequence is exact. If hX(f)(α) = 0, then f α = 0, but f is a monomorphism, so α = 0. Since hX is a functor, we have hX(g)hX(f) = 0, and if β kerhX(g), then gβ = 0. Naturality of the kernel implies that β factors through a morphism X kerg. But we have canonical isomorphisms

A coimf imf kerg,

the first as f is a monomorphism, and the composite of the composite of these with the canonical morphism kerg B is g. Therefore, we obtain a morphism α : X A satisfying f α = g. This proves part a, and part b is just part a with 𝒞 replaced by 𝒞op.

Lemma 2.5.6.

Let 𝒞 be an abelian category. A sequence

A fB gC

is exact if every sequence

Hom𝒞(X,A) hX(f)Hom𝒞(X,B) hX(g)Hom𝒞(X,C)

is exact.

Proof.

For X = A, we get

gf = hX(g)hX(f)(idA) = 0,

so we have a monomorphism s: imf kerg. For X = kerg and β : kerg B the natural monomorphism defined by the kernel, we have hX(g)(β) = gι = 0, so there exists α : kerg A with f α = β. We then have that β factors a morphism t : kerg imf inverse to s.

Proposition 2.5.7.

Suppose that G: 𝒞 𝒟 is an additive functor between abelian categories that admits a left (resp., right) adjoint. Then F is left (resp., right) exact.

Proof.

We treat the case of left exactness, the other case simply being the corresponding statement in opposite categories. Suppose that

0 A fB gC

is a left exact sequence in 𝒞. Let F be a left adjoint to G. Then for any D Obj(𝒟), the sequence

0 hF (D)(A) hF (D)(f)hF (D)(B) hF (D)(g)hF (D)(C)

is left exact. Since F is left adjoint to G, this sequence is isomorphic to

0 hD(G(A)) hD(G(f))hD(G(B)) hD(G(g))hD(G(C))

as a sequence of abelian groups. Since this holds for all D, the sequence

0 G(A) G(f)G(B) G(g)G(C)

is exact.

Example 2.5.8.

The inclusion functor from the category of sheaves on a space X to that of presheaves on X has a left adjoint called sheafification. The inclusion functor is therefore left exact, and the sheafification functor is in fact exact.

The following embedding theorem allows us to do most of the homological algebra that can be done in the category of R-modules for any R in an arbitrary abelian category.

Theorem 2.5.9 (Freyd-Mitchell).

If 𝒞 is a small abelian category, then there exists a ring R and an exact, fully faithful functor 𝒞 R-mod.

In other words, 𝒞 is equivalent to a full, abelian subcategory of R-mod for some ring R. We can use this as follows: suppose there is a result we can prove about exact diagrams in R-modules for all R, like the snake lemma. We then have the result in all abelian categories, since we can take a small full, abelian subcategory containing the objects in which we are interested and embed it into some category of left R-modules. If the result holds in that category, then by exactness of the embedding, the result will hold in the original category.

While we do not prove the Freyd-Mitchell embedding theorem, we will give a few remarks on its details (which hopefully are not so far off from the truth).

Remark 2.5.10.

To prove Theorem 2.5.9, the first step is to embed 𝒞 in the (opposite category of the) full subcategory of 𝐅𝐮𝐧𝐜(𝒞,𝐀𝐛) consisting of left exact, additive functors, which is an abelian category using h𝒞. While h𝒞 is not exact as an embedding into the category of additive functors in 𝐅𝐮𝐧𝐜(𝒞,𝐀𝐛), it is exact as an embedding into the latter category. (Another method is to embed 𝒞 in the category of 𝐏𝐫𝐨(𝒞) of cofiltered limits of the objects of 𝒞, viewed as a subcategory of 𝐅𝐮𝐧𝐜(𝒞,𝐀𝐛)op.) In either case, the category E in question has what is called a projective generator G which we can choose so that every object in 𝒞 is a quotient object of G, and the point is that RG = HomE(G,G) is a ring under its usual addition and composition. Then hG induces a functor 𝒞 RG-mod, which is the exact, fully faithful functor in question.

2.6. Standard lemmas

We begin with the following extremely useful result.

Theorem 2.6.1 (Snake lemma).

Suppose that we have a commutative diagram

The snake lemma's input diagram. A full diagram description follows.
Diagram description: The snake lemma's input diagram

The two displayed rows are exact, and the squares commute.

Objects, listed by row and column:

  • Row 1, from left to right: column 1: D; column 2: A; column 3: B; column 4: C; column 5: 0.
  • Row 2, from left to right: column 1: 0; column 2: A prime; column 3: B prime; column 4: C prime; column 5: D prime.

Arrows and lines:

  1. An arrow from D to A, labelled e.
  2. An arrow from A to B, labelled f.
  3. An arrow from A to A prime, labelled alpha.
  4. An arrow from B to C, labelled g.
  5. An arrow from B to B prime, labelled beta.
  6. An arrow from C to 0 (row 1, column 5), without a label.
  7. An arrow from C to C prime, labelled gamma.
  8. An arrow from 0 (row 2, column 1) to A prime, without a label.
  9. An arrow from A prime to B prime, labelled f prime.
  10. An arrow from B prime to C prime, labelled g prime.
  11. An arrow from C prime to D prime, labelled h prime.
(2.6.1)

with exact rows. Then there is a homomorphism ψ : kerγ cokerα fitting into a larger commutative diagram

The snake lemma's connecting path. A full diagram description follows.
Diagram description: The snake lemma's connecting path

The middle two rows are exact, and the structural squares commute. The dashed path from D through kernel alpha, kernel beta, kernel gamma, cokernel alpha, cokernel beta, cokernel gamma, and D prime is an exact eight-term sequence. The long curved arrow is the connecting map psi from kernel gamma to cokernel alpha.

Objects, listed by row and column:

  • Row 1, from left to right: column 3: kernel alpha; column 4: kernel beta; column 5: kernel gamma.
  • Row 3, from left to right: column 2: D; column 3: A; column 4: B; column 5: C; column 6: 0.
  • Row 5, from left to right: column 2: 0; column 3: A prime; column 4: B prime; column 5: C prime; column 6: D prime.
  • Row 7, from left to right: column 3: cokernel alpha; column 4: cokernel beta; column 5: cokernel gamma.

Arrows and lines:

  1. A dashed arrow from kernel alpha to kernel beta, without a label.
  2. An arrow from kernel alpha to A, without a label.
  3. A dashed arrow from kernel beta to kernel gamma, without a label.
  4. An arrow from kernel beta to B, without a label.
  5. An arrow from kernel gamma to C, without a label.
  6. A dashed curved arrow from kernel gamma to cokernel alpha, without a label.
  7. A dashed curved arrow from D to kernel alpha, without a label.
  8. An arrow from D to A, without a label.
  9. An arrow from A to B, without a label.
  10. An arrow from A to A prime, without a label.
  11. An arrow from B to C, without a label.
  12. An arrow from B to B prime, without a label.
  13. An arrow from C to 0 (row 3, column 6), without a label.
  14. An arrow from C to C prime, without a label.
  15. An arrow from 0 (row 5, column 2) to A prime, without a label.
  16. An arrow from A prime to B prime, without a label.
  17. An arrow from A prime to cokernel alpha, without a label.
  18. An arrow from B prime to C prime, without a label.
  19. An arrow from B prime to cokernel beta, without a label.
  20. An arrow from C prime to D prime, without a label.
  21. An arrow from C prime to cokernel gamma, without a label.
  22. A dashed arrow from cokernel alpha to cokernel beta, without a label.
  23. A dashed arrow from cokernel beta to cokernel gamma, without a label.
  24. A dashed curved arrow from cokernel gamma to D prime, without a label.

and the resulting eight-term sequence

D kerα kerβ kerγ cokerα cokerβ cokerγ D

is exact.

Proof.

We define ψ as follows. For c kerγ, find b B with g(b) = c. Then gβ(b) = γ(c) = 0, so β(b) = f(a) for some a A. Let ψ(c) denote the image a¯ of a in cokerα. To see that this is well-defined, note that if b2 B also satisfies g(b2) = c, then g(bb2) = 0, so b2 b = f(a) for some a A. We then have

β(b2) = β(b)+β f(a) = β(b)+fα(a),

so b2 = f(a+α(a)). But a+α(a) has image a¯ in cokerα, so ψ is well-defined.

We now check that the other maps are well-defined. Since

fα e = β f e = 0

and f is injective, we have α e = 0, so ime kerα. Since

β f(kerα) = fα(kerα) = 0,

we have f(kerα) kerβ. Similarly, g(kerβ) kerγ. Also, if a¯ cokerα, then we may lift it to a A, map to b B, and then project to b¯ cokerβ. This is well-defined as any other choice of a differs by some a A, which causes b¯ to change by the image of β(f(a)), which is zero. Thus f induces a well-defined map

f¯: cokerα cokerβ.

Similarly, we have a well-defined

g¯: cokerβ cokerγ.

Finally, img = C, so imγ img, and therefore h(imγ) = 0. Thus h induces a map

h¯: cokerγ D.

It is clear from the above definitions that the 8-term sequence is a complex at all the terms but kerγ and cokerα. Let b kerβ. Then ψ(g(b)) is given by considering β(b) = 0, lifting it to some a A, which we may take to be 0, and projecting to cokerα. Hence ψ(g(kerβ)) = 0. On the other hand, if c kerγ, then f¯(ψ(c)) is given by definition by projecting β(b) to cokerβ, where g(b) = c, hence is zero.

We now check exactness at each term. We have

kerα kerf = kerα ime = ime,

so the sequence is exact at kerα. Next, if b kerβ kerg, then there exists a A with f(a) = b.

Since fα(a) = β f(a) = 0 and f is injective, we have α(a) = 0, or a kerα. Hence

f(kerα) = ker(kerβ kerγ),

and we have exactness at kerβ.

If c kerψ, then whenever g(b) = c and f(a) = β(b), we have a¯ = 0, letting a¯ denote the image of a cokerα. We then have a = f(a) for some a A, so b2 = bf(a) still satisfies f(b2) = c, but β(b2) = 0. So b2 kerβ, and we have exactness at kerγ.

If a¯ cokerα is the image of a A and f¯(a¯) = 0, then there exists b B with β(b) = f(a). Now

γ(g(b)) = g(β(b)) = g(f(a)) = 0,

so g(b) kerγ, and ψ(g(b)) = a¯. Hence, we have exactness at cokerα.

If b¯ cokerβ is the image of b B and g¯(b¯) = 0, then there exists c C with g(b) = γ(c). Now c = g(b) for some b B. And b2 = bb has image b¯ in cokerβ. On the other hand, f(b2) = 0, so b2 = f(a) for some a A. If a¯ cokerα is the image of a, then f¯(a¯) = b¯ as the image of b2 in cokerβ. Thus, we have exactness at cokerβ.

Finally, let c¯ cokerγ be the image of c C. Note that h¯(c¯) = h(c). If this is zero, then c = g(b) for some b B, and c¯ is the image of the projection b¯ of b to cokerβ under g¯.

Lemma 2.6.2.

Suppose we have a commutative diagram as in (2.6.1) and that both g and f are split. Then the snake map ψ : kerγ cokerα is zero.

Proof.

Let c kerγ. Let t split g and s split f. By Lemma ??, we also have a splitting map

t: kerh A.

It is easy to check that tγ = β t. It follows that t(kerγ) kerβ. Thus ψ(c), which is the image of s(β(t(c))) in cokerα, is zero.

2.7. Complexes

Let 𝒞 be an abelian category. We have already defined the notion of a chain complex in 𝒞 above. In addition to chain complexes, we will also deal with cochain complexes, which are likewise defined, but with increasing superscripts replacing decreasing subscripts. This is primarily a notational convenience, but it is a very useful one. When the choice of chain or cochain complexes matters little, we will typically choose to work with cochain complexes.

Definition 2.7.1.

A cochain complex A = (Ai,dAi) is a diagram

Ai1 d Ai1Ai d AiAi+1 ,

such that dAidAi1 = 0 for all i . Again, the dAi are referred to as differentials.

Remark 2.7.2.

We refer to both chain complexes and cochain complexes as complexes, though we will usually mean the latter if we do not specify. We will often state facts simply for cochain complexes that have a direct translation to the setting of chain complexes.

Definition 2.7.3.

Let A = (Ai,dAi) and B = (Bi,dBi) be complexes. A morphism of complexes f: A B is a sequence of morphisms fi: Ai Bi commuting with the differentials of the complexes in the sense that

dBifi = fi1 d Ai

for all i .

Remark 2.7.4.

We may consider the category 𝐂𝐡(𝒞) of cochain complexes in 𝒞, where morphisms of chain complexes A = (Ai,di) and B = (Bi,di) are morphisms Ai Bi for each i commuting with the differentials.

The reader can easily check the following.

Proposition 2.7.5.

Let 𝒞 be an abelian category. Then the category 𝐂𝐡(𝒞) is an abelian category as well.

Definition 2.7.6.

Let A (resp., A) be a chain complex (resp., cochain complex).

a.

We say that A (resp., A) is bounded below if Ai = 0 (resp., Ai = 0) for all i < N for some N .

b.

We say that A (resp., A) is bounded above if Ai = 0 (resp., Ai = 0) for all i > N for some N .

c.

We say that A (resp., A) is bounded if it is both bounded below and bounded above.

In most examples that we shall explore, our chain complexes (resp., cochain complexes) will be bounded below (resp., bounded above), usually with N = 0.

Definition 2.7.7.

a.

We define ith homology (object) of a chain complex A by

Hi(A) = kerdiAimd i+1A

for each i .

b.

We define the ith cohomology (object) of a cochain complex A by

Hi(A) = kerd Aiimd Ai1

for i .

Remark 2.7.8.

For a complex of abelian groups, we speak of homology and cohomology groups, as opposed to objects.

Example 2.7.9.

Consider the chain complex A = (,dAi) with dAi = 3id 0 and dAi+1 = 02id for i even. Then Hi(A) = 2 and Hi+1(A) = 3 for i even.

Lemma 2.7.10.

Let f: A B be a morphism of complexes. Then we have natural homomorphisms

fi: Hi(A) Hi(B)

for each i .

Proof.

Let A = (Ai,dAi) and B = (Bi,dBi). We have fidAi+1 = dBi+1 fi+1 for all i, so

fi(imd Ai+1) imd Bi+1,

and

dBi(fi(kerd Ai)) = fi1(d Ai(kerd Ai)) = 0,

so

fi(kerd Ai) kerd Bi.

Hence, we have an induced morphism

fi: kerdAi imdAi+1 kerdBi imdBi+1,

as desired.

Remark 2.7.11.

Of course, the analogous result holds in cohomology as a consequence.

Remark 2.7.12.

We note that a sequence

0 AfBgC 0

in 𝐂𝐡(𝒞) is exact if and only if each sequence

0 Ai fiBi giCi 0

is exact in 𝒞.

Theorem 2.7.13.

Let

0 AfBgC 0

be an exact sequence of cochain complexes of R-modules. Then we have a long exact sequence in cohomology

Hi(A) f iHi(B) g iHi(C) iHi+1(A) ,

where i is the map given by the snake lemma applied to the diagram (2.7.1) below.

Proof.

Consider the following diagram

Kernels and cokernels for the cohomology sequence. A full diagram description follows.
Diagram description: Kernels and cokernels for the cohomology sequence

The two displayed rows are exact, and the squares commute.

Objects, listed by row and column:

  • Row 1, from left to right: column 2: cokernel d superscript (i minus 1) subscript (A); column 3: cokernel d superscript (i minus 1) subscript (B); column 4: cokernel d superscript (i minus 1) subscript (C); column 5: 0.
  • Row 2, from left to right: column 1: 0; column 2: kernel d superscript (i plus 1) subscript (A); column 3: kernel d superscript (i plus 1) subscript (B); column 4: kernel d superscript (i plus 1) subscript (C).

Arrows and lines:

  1. An arrow from cokernel d superscript (i minus 1) subscript (A) to cokernel d superscript (i minus 1) subscript (B), without a label.
  2. An arrow from cokernel d superscript (i minus 1) subscript (A) to kernel d superscript (i plus 1) subscript (A), labelled d superscript (i) subscript (A).
  3. An arrow from cokernel d superscript (i minus 1) subscript (B) to cokernel d superscript (i minus 1) subscript (C), without a label.
  4. An arrow from cokernel d superscript (i minus 1) subscript (B) to kernel d superscript (i plus 1) subscript (B), labelled d superscript (i) subscript (B).
  5. An arrow from cokernel d superscript (i minus 1) subscript (C) to 0 (row 1, column 5), without a label.
  6. An arrow from cokernel d superscript (i minus 1) subscript (C) to kernel d superscript (i plus 1) subscript (C), labelled d superscript (i) subscript (C).
  7. An arrow from 0 (row 2, column 1) to kernel d superscript (i plus 1) subscript (A), without a label.
  8. An arrow from kernel d superscript (i plus 1) subscript (A) to kernel d superscript (i plus 1) subscript (B), without a label.
  9. An arrow from kernel d superscript (i plus 1) subscript (B) to kernel d superscript (i plus 1) subscript (C), without a label.
(2.7.1)

By the snake lemma, the two rows are exact, and the diagram commutes since we have dBifi = fi+1 dAi and dCigi = gi+1 dBi as morphisms Ai Bi+1 and Bi Ci+1, respectively, by naturality of the cokernel and kernel. Since we have natural isomorphisms

Hi(A)ker(Aiimd Ai1 Ai+1) and Hi+1(A)coker(Ai kerd Ai+1),

we apply the snake lemma to obtain an exact sequence

Hi(A) f iHi(B) g iHi(C) iHi+1(A) f i+1Hi+1(B) g i+1Hi+1(C),

and the result follows by splicing together these sequences.

Definition 2.7.14.

Let A = (Ai,dAi) and B = (Bi,dBi) be chain complexes. Let f,g: A B be morphisms of chain complexes.

a.

A chain homotopy from f to g is a sequence s = (si)i of morphisms si: Ai Bi1 satisfying

figi = d Bi1 si+si+1 d Ai

for all i .

b.

We say that f and g are chain homotopic, and write f g, if there exists a homotopy from f to g.

c.

If f is (chain) homotopic to 0, then f is said to be null-homotopic.

The maps s defining a null-homotopy fit into a (not usually ) diagram

A null-homotopy. A full diagram description follows.
Diagram description: A null-homotopy

The diagonal maps are the components of a null-homotopy. This diagram is not generally commutative; the homotopy identity in the surrounding text is a sum of the two routes through a diagonal map.

Objects, listed by row and column:

  • Row 1, from left to right: column 1: ellipsis; column 2: A superscript (i minus 1); column 3: A superscript (i); column 4: A superscript (i plus 1); column 5: ellipsis.
  • Row 2, from left to right: column 1: ellipsis; column 2: B superscript (i minus 1); column 3: B superscript (i); column 4: B superscript (i plus 1); column 5: ellipsis.

Arrows and lines:

  1. An arrow from ellipsis (row 1, column 1) to A superscript (i minus 1), without a label.
  2. An arrow from A superscript (i minus 1) to A superscript (i), labelled d superscript (i minus 1) subscript (A).
  3. An arrow from A superscript (i minus 1) to B superscript (i minus 1), labelled f superscript (i minus 1).
  4. An arrow from A superscript (i) to A superscript (i plus 1), labelled d superscript (i) subscript (A).
  5. An arrow from A superscript (i) to B superscript (i), labelled f superscript (i).
  6. An arrow from A superscript (i) to B superscript (i minus 1), labelled s superscript (i).
  7. An arrow from A superscript (i plus 1) to ellipsis (row 1, column 5), without a label.
  8. An arrow from A superscript (i plus 1) to B superscript (i plus 1), labelled f superscript (i plus 1).
  9. An arrow from A superscript (i plus 1) to B superscript (i), labelled s superscript (i plus 1).
  10. An arrow from ellipsis (row 2, column 1) to B superscript (i minus 1), without a label.
  11. An arrow from B superscript (i minus 1) to B superscript (i), labelled d superscript (i minus 1) subscript (B).
  12. An arrow from B superscript (i) to B superscript (i plus 1), labelled d superscript (i) subscript (B).
  13. An arrow from B superscript (i plus 1) to ellipsis (row 2, column 5), without a label.

Proposition 2.7.15.

Assume that f and g are homotopic as maps A B. Then the maps fi and gi on homology are equal for all i .

Proof.

It suffices to assume that g = 0, since the ith homology functor from 𝐂𝐡(𝒞) to 𝒞 is additive. So, we must show that fi = 0 for all i, which is to say that fi(kerdAi) imdAi1. Since fi = dBi1 si+si+1 dAi, we have

fi(kerd Ai) = d Bi1(si(kerd Ai)) imd Bi1,

so fi = 0.

Definition 2.7.16.

A morphism of complexes f: A B is a homotopy equivalence if there exists a morphism g: B A such that gf idA and fg idB.

Definition 2.7.17.

Let A be a complex, and let j . The shift by j of the complex A is the complex A[j] with A[j]i = Ai+j and dA[j]i = (1)jdAi for all i .

Notation 2.7.18.

Let n 1, let Ai,Bi 𝒞 be objects in an abelian category 𝒞 for 1 i n, and let fi,j: Ai Bj be morphisms in 𝒞 for 1 i,j n. Let A = i=1nAi and B = i=1nBi, and let ιi: Bi B and πi: A Ai denote the canonical inclusions and projections. Then we use the matrix (fi,j) to represent the morphism f : A B in 𝒞 that is the sum

f =i=1n j=1nι jfi,jπj.

Definition 2.7.19.

Let f: A B be a morphism of complexes. The cone of f is the complex Cone(f) with

Cone(f)i = Ai+1 Bi

and

dCone(f)i = ( dAi+1 0 fi+1 dBi ) : Ai+1Bi Ai+2Bi+1.

Proposition 2.7.20.

The cone of f: A B fits in a short exact sequence

0 B Cone(f) A[1] 0,

and the resulting long exact sequence has ith connecting homomorphism

Hi(A[1]) = Hi+1(A) Hi+1(B)

equal to fi+1.

Proof.

The differential on Cone(f) preserves B and agrees with the differential on B, so B is a subcomplex. The quotient complex has ith term Ai+1 and differential induced by the negative of the differential on A, so is canonically isomorphic to A[1]. Since Cone(f)i = Ai+1 Bi, we have a natural splitting of the surjection Cone(f)i Ai+1 given by the inclusion. The connecting homomorphism is then induced by the composition

kerdAi+1 Ai+1 Cone(f)i d Cone(f)iCone(f)i+1,

which is given by the sum of dAi+1 and fi as maps to Ai+2 Bi+1, but then has image in Bi+1 (since dAi+1 is zero on dAi+1) and therefore agrees with fi+1 as a morphism to Bi+1. In other words, the connecting homomorphism is canonically identified with fi+1.

2.8. Total complexes

Definition 2.8.1.

A double (cochain) complex in an abelian category 𝒞 is a complex in 𝐂𝐡(𝒞).

The data of a double complex A⋅⋅ consists of complexes (Ai,,di,) of objects of Ch(𝒞), which is to say that each Ai,is a complex in 𝒞 with differentials of its own, and the di,are morphisms between Ai, Ai+1,, so commute with the differentials on these complexes. We rephrase this as follows.

Remark 2.8.2.

A double complex (A⋅⋅,d⋅⋅) in 𝒞 is a diagram of the form

A double complex. A full diagram description follows.
Diagram description: A double complex

Each horizontal row and each vertical column is a complex, so consecutive horizontal differentials compose to zero and consecutive vertical differentials compose to zero. The squares commute: a horizontal step followed by a vertical step equals a vertical step followed by a horizontal step. The ellipses indicate continuation in both directions.

Objects, listed by row and column:

  • Row 1, from left to right: column 2: vertical ellipsis; column 3: vertical ellipsis; column 4: vertical ellipsis.
  • Row 2, from left to right: column 1: ellipsis; column 2: A superscript (i minus 1,j plus 1); column 3: A superscript (i,j plus 1); column 4: A superscript (i plus 1,j plus 1); column 5: ellipsis.
  • Row 3, from left to right: column 1: ellipsis; column 2: A superscript (i minus 1,j); column 3: A superscript (i,j); column 4: A superscript (i plus 1,j); column 5: ellipsis.
  • Row 4, from left to right: column 1: ellipsis; column 2: A superscript (i minus 1,j minus 1); column 3: A superscript (i,j minus 1); column 4: A superscript (i plus 1,j minus 1); column 5: ellipsis.
  • Row 5, from left to right: column 2: vertical ellipsis; column 3: vertical ellipsis; column 4: vertical ellipsis.

Arrows and lines:

  1. An arrow from ellipsis (row 2, column 1) to A superscript (i minus 1,j plus 1), without a label.
  2. An arrow from A superscript (i minus 1,j plus 1) to A superscript (i,j plus 1), labelled d subscript (h) superscript (i minus 1,j plus 1).
  3. An arrow from A superscript (i minus 1,j plus 1) to vertical ellipsis (row 1, column 2), without a label.
  4. An arrow from A superscript (i,j plus 1) to A superscript (i plus 1,j plus 1), labelled d subscript (h) superscript (i,j plus 1).
  5. An arrow from A superscript (i,j plus 1) to vertical ellipsis (row 1, column 3), without a label.
  6. An arrow from A superscript (i plus 1,j plus 1) to ellipsis (row 2, column 5), without a label.
  7. An arrow from A superscript (i plus 1,j plus 1) to vertical ellipsis (row 1, column 4), without a label.
  8. An arrow from ellipsis (row 3, column 1) to A superscript (i minus 1,j), without a label.
  9. An arrow from A superscript (i minus 1,j) to A superscript (i,j), labelled d subscript (h) superscript (i minus 1,j).
  10. An arrow from A superscript (i minus 1,j) to A superscript (i minus 1,j plus 1), labelled d subscript (v) superscript (i minus 1,j).
  11. An arrow from A superscript (i,j) to A superscript (i plus 1,j), labelled d subscript (h) superscript (i,j).
  12. An arrow from A superscript (i,j) to A superscript (i,j plus 1), labelled d subscript (v) superscript (i,j).
  13. An arrow from A superscript (i plus 1,j) to ellipsis (row 3, column 5), without a label.
  14. An arrow from A superscript (i plus 1,j) to A superscript (i plus 1,j plus 1), labelled d subscript (v) superscript (i plus 1,j).
  15. An arrow from ellipsis (row 4, column 1) to A superscript (i minus 1,j minus 1), without a label.
  16. An arrow from A superscript (i minus 1,j minus 1) to A superscript (i,j minus 1), labelled d subscript (h) superscript (i minus 1,j minus 1).
  17. An arrow from A superscript (i minus 1,j minus 1) to A superscript (i minus 1,j), labelled d subscript (v) superscript (i minus 1,j minus 1).
  18. An arrow from A superscript (i,j minus 1) to A superscript (i plus 1,j minus 1), labelled d subscript (h) superscript (i,j minus 1).
  19. An arrow from A superscript (i,j minus 1) to A superscript (i,j), labelled d subscript (v) superscript (i,j minus 1).
  20. An arrow from A superscript (i plus 1,j minus 1) to ellipsis (row 4, column 5), without a label.
  21. An arrow from A superscript (i plus 1,j minus 1) to A superscript (i plus 1,j), labelled d subscript (v) superscript (i plus 1,j minus 1).
  22. An arrow from vertical ellipsis (row 5, column 2) to A superscript (i minus 1,j minus 1), without a label.
  23. An arrow from vertical ellipsis (row 5, column 3) to A superscript (i,j minus 1), without a label.
  24. An arrow from vertical ellipsis (row 5, column 4) to A superscript (i plus 1,j minus 1), without a label.

such that (for all i,j )

i.

each row is a complex: i.e., dhi+1,jdhi,j = 0,

ii.

each column is a complex: i.e., dvi,j+1 dvi,j = 0, and

iii.

the squares commute: i.e., dhi,j+1 dvi,j = dvi+1,jdhi,j.

Definition 2.8.3.

The degree of the term Ai,j of a double complex A⋅⋅ is i+j.

Definition 2.8.4.

Let A⋅⋅be a double complex in an abelian category 𝒞.

a.

Suppose that 𝒞 admits coproducts. The total sum complex Tot(A) of A⋅⋅ is the complex

Tot(A)k = iAi,ki

with the differential

dk = i(dhi,ki+(1)id vi,ki): iAi,ki iAi,k+i1

where the differentials dhi,ki and dvi,ki are taken to be zero on the Aj,kj with ji.

b.

Suppose that 𝒞 admits products. The total product complex TotΠ(A) of A⋅⋅ is the complex

TotΠ(A)k = iAi,ki

with the differential

dk = (d hi,ki+(1)id vi,ki) i: iAi,ki iAi,k+1i,

where the differentials dhi,ki and dvi,ki are taken to be zero on the Aj,kj with ji.

Remark 2.8.5.

We remark that Tot(A) is in fact a complex (when it exists). That is,

dk+1 dk = ij(dhi,k+1i+(1)id vi,k+1i)(d hj,kj+(1)jd vj,kj) =i((1)id hi,k+1id vi,ki+(1)id vi,k+1id hi1,k+1i) =i((1)id hi,k+1id vi,ki+(1)i+1d vi+1,kid hi,ki) = 0.

Similarly, TorΠ(A) is a complex as well.

Definition 2.8.6.

An nth quadrant double complex, for 1 n 4, is a double complex A⋅⋅ such that Ai,j = 0 for (i,j) not in the closed nth quadrant of 2. An upper, lower, left, or right half-plane double complex is one that is zero outside of said closed half-plane.

Remark 2.8.7.

Recall that the closed 1st, 2nd, 3rd, and 4th quadrants of 2 are 0 ×0, 0 ×0, 0 ×0, and 0 ×0, respectively. The closed upper, lower, left, and right half-planes are ×0, ×0, 0 ×, and 0 ×, respectively.

Remark 2.8.8.

If A⋅⋅ is either a first or a fourth quadrant double complex, then

Tot(A) = TotΠ(A),

and it exists for any abelian category 𝒞, since there are only finitely many nonzero terms in the complex of a given degree. In this case, we simply write Tot(A) for either total complex.

Definition 2.8.9.

Let A⋅⋅ be a double complex. We define the truncated quotient complexes (IτnA)⋅⋅ and (𝐼𝐼τnA)⋅⋅ of A⋅⋅ by

(IτnA)i,j = { Ai,j if i > n cokerdhn1,jif i = n 0 if i < n and (𝐼𝐼τnA)i,j = { Ai,j if j > n cokerdvj,n1if j = n 0 if j < n,

as well as subcomplexes (IτnA)⋅⋅ and (𝐼𝐼τnA)⋅⋅ by

(IτnA)i,j = { Ai,j if i < n kerdhn,jif i = n 0 if i > n and (𝐼𝐼τnA)i,j = { Ai,j if j < n kerdvj,nif j = n 0 if j > n.

Proposition 2.8.10.

Suppose that 𝒞 is an abelian category that admits direct products (resp., direct sums). Let A⋅⋅ be an upper (or left) double half-plane complex in 𝒞 with exact columns (resp., exact rows) or a right (or lower) double half-plane complex with exact rows (resp., exact columns). Then Tot(A) (resp., TotΠ(A)) is exact.

Proof.

By interchanging rows and columns, we may suppose that A⋅⋅ is an upper or lower half-plane complex with either exact rows or exact columns. By working in the opposite category (and changing the signs of the degrees), we may focus on the case of a lower half-plane complex. Suppose first that A⋅⋅ has exact rows. For n , consider the truncated complex (IτnA)⋅⋅, which then has exact rows as well. Then a shift by n in the horizontal direction is a third quadrant double complex Bn⋅⋅. If we can show Tot(Bn) to be exact (for all n), then Tot(A) will be exact as well, as the reader may check that

Tot(A)lim nTot(Bn),

the limit taken with respect to the morphisms induced by the inclusion morphisms Bn Bn+1. Thus, it suffices to consider the case of an exact left half-plane complex with exact rows, which is equivalent to the case of an exact lower half-plane complex A⋅⋅ with exact columns, for which we must show that TotΠ(A) is exact. Since we can always shift A⋅⋅ horizontally, it suffices to check that H0(TotΠ(A)) = 0.

We now work in the category of R-modules for a ring R without loss of generality. Let

a = (ai) TotΠ(A)0 = i=0Ai,i.

with d0(a) = 0. Choose b0 A0,1 with dv0,1(b0) = a0 by exactness of the 0th column, and set b0 = 0 A1,0. Suppose we have defined bj Ai,i1 for 0 j i with

(1)id vi,i1(b i)+dhi1,i(b i1) = ai.

Then

(1)i+1d vi+1,i1(a i+1 dhi,i1(b i)) = (1)i+1d vi+1,i1(a i+1)+(1)id hi,id vi,i1(b i) = (1)i+1d vi+1,i1(a i+1)+dhi,i(a i),

which is the (i+1,i)-coordinate of di+1(a), hence 0. By exactness of the (i+1)th column, we may then choose bi+1 Ai+1,i with

(1)i+1d vi+1,i(b i+1) = ai+1 dhi,i1(b i),

completing the induction. We then set b = (bi) TotΠ(A)1 and note that d1(b) = a, finishing the proof.

Find in the notes