Chapter 2
Abelian Categories
2.1. Additive categories
Definition 2.1.1. §
An additive category is a category with the following properties:
- i.
-
for , the set of morphisms in has an abelian group law (addition) with the property that for any diagram
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:
- An arrow from A to B, labelled f.
- An arrow from B to C, labelled g subscript (1).
- An arrow from B to C, labelled g subscript (2).
- An arrow from C to D, labelled h.
in , we have
- ii.
-
has a zero object ,
- iii.
-
admits finite coproducts.
Note that between any two objects and in an additive category , there always exists the zero morphism , the unique map factoring through . This is the identity element in the abelian group .
Examples 2.1.2. §
- a.
-
The categories and are additive categories, with the usual addition of homomorphisms.
- b.
-
The full subcategory of finitely generated -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 and by .
Lemma 2.1.3. §
Finite products exist in an additive category, and there are natural isomorphisms
for . The resulting inclusion morphisms and projection morphisms and obtained by viewing as a product and coproduct, respectively, satisfy and for , while
Proof.
We have maps by definition of the direct sum. We also have maps defined by
and the universal property of the coproduct.
We then have
and hence
Given an object and maps , we then have a map
which is unique such that
Hence satisfies the universal property of the product. □
Definition 2.1.4. §
An object in an additive category together with objects , inclusion morphisms , and projection morphisms for for which is zero if and if and for which is called a biproduct of the objects and , and we write it as .
The reader will verify the following.
Lemma 2.1.5. §
In an additive category , the biproduct is a coproduct of the via the inclusion morphisms and to the product of the via the morphisms .
The notion of a biproduct allows us to reinterpret addition in an additive category. First, note the following definitions.
Definition 2.1.6. §
Let be an object in an additive category .
- a.
-
The diagonal morphism in is the unique morphism induced by two copies of and the universal property of the product.
- b.
-
The codiagonal morphism in is the unique morphisms induced by two copies of and the universal property of the coproduct.
Definition 2.1.7. §
Let be an additive category, and let and be morphisms in . The biproduct, or direct sum, of the maps and is the morphism induced as the (morphism defined by the universal property of the) coproduct of the composite maps , the latter morphisms being inclusions.
Remark 2.1.8. §
Equivalently, the direct sum of and as in Definition 2.1.7 is induced as the product of the composite maps , 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 be two morphisms in an additive category . Then we have
Proof.
Let and respectively denote the inclusion maps and projection maps for the biproduct , and similarly for . We have
Taking the first term without loss of generality, we have
□
Definition 2.1.10. §
A functor between additive categories is called additive if for each , the map
is a group homomorphism.
Example 2.1.11. §
Let be an additive category. Then for any , the functors and may be considered as functors to , rather than . The resulting functors are additive.
Example 2.1.12. §
Let be a ring, and let be a right -module. Then the functor
given by and for is additive.
Lemma 2.1.13. §
A functor of additive categories is additive if and only if preserves biproducts, which is to say that the natural morphisms and are inverse isomorphisms for all objects in .
Proof.
Suppose first that is an additive functor. Note that and for , but by additivity of . Again by additivity of , we have
It follows that is a biproduct of and in , so in particular it is a coproduct.
On the other hand, if preserves biproducts and are morphisms in , then it is easy to see that , and Lemma 2.1.9 tells us that
□
Corollary 2.1.14. §
Let be a fully faithful functor of additive categories. Then is an additive functor.
For additive functors, we may consider a finer notion of representability.
Remark 2.1.15. §
If is an additive contravariant (resp., covariant) functor of additive categories, then we may consider it to be representable if there exists an object and a natural isomorphism (resp., ). In this case, the morphisms for will be isomorphisms of groups.
2.2. Kernels and cokernels
Remark 2.2.1. §
In an additive category, any morphism to is an epimorphism, and any morphism from is a monomorphism.
Definition 2.2.2. §
Let be an additive category, and let be a morphism in .
Example 2.2.3. §
In the category , 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 be a monomorphism, and let be the induced morphism. Since by definition of the kernel, we have , as is a monomorphism. This forces to be , since factors through . (Or, one could just apply Lemma 1.7.3.) On the other hand, suppose that has trivial kernel, and let be maps with . Then , and by universal property of the kernel, factors through , i.e., is .
The proof for cokernels is similar, or is the result on kernels in the opposite (additive) category. □
Remark 2.2.5. §
Let be an additive category that admits kernels (resp., cokernels). Then there is a functor
which takes an object in to its kernel (resp., cokernel) and a morphism in to the natural morphism between kernels (resp., cokernels).
Proposition 2.2.6. §
Let be an additive category that admits kernels and cokernels. Let be a morphism in . Then
and
Proof.
We prove the first isomorphism. Let . By Yoneda’s lemma, it suffices to show that and are naturally isomorphic. For , we have a map
that takes a morphism and composes it with the morphism given by definition of the equalizer of the maps . It is a bijection by the universal property of the equalizer.
For any and morphisms such that , note that there exists a unique morphism with . Any such that then satisfies for any such and any . On the other hand, note that the themselves satisfy the property that and are morphisms with . In other words, we have
Now, we are in an additive category, so this equals
| (2.2.1) |
By the universal property of , for any with , there is a morphism with . If , then , and this works for any with . On the other hand, itself satisfies so is such a . It follows that the set in (2.2.1) equals
By the universal property of , this is in bijection with , taking an in the set to the unique morphism to through which it factors. Clearly, the composition of these bijections is natural in , so we have the desired natural isomorphism. □
2.3. Abelian categories
Definition 2.3.1. §
An abelian category is an additive category in which
Examples 2.3.2. §
- a.
- b.
-
The full subcategory of of finitely generated -submodules is not abelian in all cases. E.g., when is commutative and non-Noetherian, we can take to be an ideal of that is not finitely generated, and so the kernel of is not in .
- 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 with having its usual topology and having the subspace topology. Then and (since is dense in , and thus every continuous map from is determined by its values on ). By Proposition 2.2.6, we have but .
Remark 2.3.3. §
Note that if is an abelian category, then so is . The roles of mono- and epimorphisms, kernels and cokernels, and images and coimages switch in and .
Proposition 2.3.4. §
The functor category from a small category to an abelian category 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 are functors, then is given by and for in . This can be used to define the addition on morphisms (i.e., natural transformations) as before.
Next, the kernel of a natural transformation is defined by and for is the kernel of the induced morphism . The cokernel is defined similarly. Note that
and similarly for images. Finally, since is abelian, the natural map is an isomorphism on objects in , hence has a natural inverse determined by the inverses of these morphisms. □
We discuss a related class of examples.
Definition 2.3.5. §
Let be a topological space and an abelian category. Consider the category with objects the open sets in and morphisms the inclusion maps between subspaces to the category .
- a.
-
A presheaf on with values in is a contrvariant functor . The category of presheaves is the functor category .
- b.
-
Suppose that admits arbitrary products. A sheaf on with values in is a presheaf such that each for is the equalizer of the diagram
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:
- 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.
- 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 is an open covering of by open subsets via the two maps on products induced by the application of to inclusion maps for each pair of open sets in . The category of sheaves is the full subcategory of the category of presheaves with objects the sheaves.
Example 2.3.6. §
Let be a topological space and an abelian category. For any , we can view as a discrete space and consider the constant sheaf
for all . If is connected, then .
Example 2.3.7. §
Let be a topological space and for . Then is the sheaf with values in of continuous functions to that for satisfies
Example 2.3.8. §
Let , and let be the presheaf such that is the abelian group of bounded continuous functions on . Then is not a sheaf, as we may consider the open covering of by intervals for , and then the function is contained in for each , so in the equalizer, but it is not in .
Terminology 2.3.9. §
In an abelian category , we will typically refer to a coproduct (when it exists) as a direct sum, and we write in place of .
2.4. Exact sequences
Suppose that and are morphisms in an abelian category with . Note that , where is an epimorphism and is a monomorphism. We have , so the fact that is an epimorphism tells us that . The universal property of the kernel then provides a morphism such that when composed with the canonical monomorphism is . Since and are both monomorphisms, which is to say have trivial kernel, we have that is a monomorphism as well.
Definition 2.4.1. §
We say that a diagram
in an abelian category is exact if and the induced monomorphism 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
in a category 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 , where runs for over the interval of integers on which is defined and for over every such but the left endpoint, if it exists.
Definition 2.4.3. §
Let be an abelian category.
- a.
-
A chain complex in is a sequence in such that for all .
- b.
-
For a chain complex and , the morphism is called the th differential in the complex .
Notation 2.4.4. §
Unless otherwise specified, the th object in a chain complex will be denoted and the th differential by . If we have multiple complexes, we will use to specify the differential on .
Notation 2.4.5. §
In an abelian category, if is a subobject of an object , we will often write to denote this and for the cokernel of the inclusion morphism . For , we will let denote the image of the composite of the inclusion with .
Definition 2.4.6. §
A sequence
in an abelian category is exact if the subdiagram is exact for each such that there are terms and in the diagram.
Definition 2.4.7. §
A long exact sequence in an abelian category is an exact sequence , where 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 be an abelian category.
- a.
-
A short exact sequence in is an exact sequence in of the form
- b.
-
A left short exact sequence in is an exact sequence in of the form
- c.
-
A right short exact sequence in is an exact sequence in of the form
Definition 2.4.10. §
Let be an abelian category.
- a.
-
We say that an epimorphism is split if there exists a morphism in with . In this case, we say that is a splitting of .
- b.
-
We say that a monomorphism is split if there exists a morphism in with . In this case, we say that is a splitting of .
- c.
-
We say that a short exact sequence
(2.4.1) in an abelian category splits if there exists an isomorphism in with and , where and are the inclusion morphisms.
Example 2.4.11. §
Any exact sequence
is split in Ab, but
is not.
Lemma 2.4.12. §
The following conditions on a short exact sequence (2.4.1) are equivalent:
- The sequence (2.4.1) splits.
- The monomorphism splits.
- The epimorphism splits.
Proof.
-
Suppose we have a splitting map . Then define by where . This is well-defined as is injective, and such an exists since
It splits as
the latter step using the fact that , which follows in turn from
- Suppose that we have a splitting map . Then define by where . To see this is well defined, note that for any .
-
Define . Its inverse is . We check
and
It remains to see that . It is easy to see this on elements of the form or and that an arbitrary element of is a sum of these.
- Set .
2.5. Exact functors
Definition 2.5.1. §
Let be an additive functor between abelian categories.
- a.
-
We say that is left exact if it preserves exact sequences of the form
which is to say that the sequence
is exact.
- b.
-
We say that is right exact if it preserves exactness of exact sequences of the form
- c.
-
We say that is exact if it is both left and right exact.
The reader will verify the following.
Lemma 2.5.2. §
An additive functor between abelian categories is exact (resp., left exact, resp., right exact) if and only if it takes short exact sequences in to short exact (resp., left short exact, resp. right short exact) sequences in . Moreover, 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 be left exact is that it preserves exactness of sequences of the form , which is to say that is exact.
Examples 2.5.4. §
- a.
- b.
-
In for a commutative ring , the functor is right exact for any -module .
- c.
-
Let be a group. For a -module , consider the abelian group
of -invariants in , with -module homomorphisms restricting to homomorphisms . The resulting functor is left exact but not in general right exact.
Recall that for an additive category , the functors (and ) may be viewed as taking values in , and clearly such functors are additive. In fact, they are also left exact.
Lemma 2.5.5. §
Let be an abelian category, and let be an object of .
Proof.
Let
be an exact sequence in . Applying , we obtain homomorphisms
of abelian groups, and we claim this sequence is exact. If , then , but is a monomorphism, so . Since is a functor, we have , and if , then . Naturality of the kernel implies that factors through a morphism . But we have canonical isomorphisms
the first as is a monomorphism, and the composite of the composite of these with the canonical morphism is . Therefore, we obtain a morphism satisfying . This proves part a, and part b is just part a with replaced by . □
Lemma 2.5.6. §
Let be an abelian category. A sequence
is exact if every sequence
is exact.
Proof.
For , we get
so we have a monomorphism . For and the natural monomorphism defined by the kernel, we have , so there exists with . We then have that factors a morphism inverse to . □
Proposition 2.5.7. §
Suppose that is an additive functor between abelian categories that admits a left (resp., right) adjoint. Then 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
is a left exact sequence in . Let be a left adjoint to . Then for any , the sequence
is left exact. Since is left adjoint to , this sequence is isomorphic to
as a sequence of abelian groups. Since this holds for all , the sequence
is exact. □
Example 2.5.8. §
The inclusion functor from the category of sheaves on a space to that of presheaves on 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 -modules for any in an arbitrary abelian category.
Theorem 2.5.9 (Freyd-Mitchell). §
If is a small abelian category, then there exists a ring and an exact, fully faithful functor -mod.
In other words, is equivalent to a full, abelian subcategory of for some ring . We can use this as follows: suppose there is a result we can prove about exact diagrams in -modules for all , 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 -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 in the (opposite category of the) full subcategory of consisting of left exact, additive functors, which is an abelian category using . While is not exact as an embedding into the category of additive functors in , it is exact as an embedding into the latter category. (Another method is to embed in the category of of cofiltered limits of the objects of , viewed as a subcategory of .) In either case, the category in question has what is called a projective generator which we can choose so that every object in is a quotient object of , and the point is that is a ring under its usual addition and composition. Then induces a functor , 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
Diagram description: The snake lemma's input diagramThe two displayed rows are exact, and the squares commute. Objects, listed by row and column:
Arrows and lines:
| (2.6.1) |
with exact rows. Then there is a homomorphism fitting into a larger commutative diagram
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:
- A dashed arrow from kernel alpha to kernel beta, without a label.
- An arrow from kernel alpha to A, without a label.
- A dashed arrow from kernel beta to kernel gamma, without a label.
- An arrow from kernel beta to B, without a label.
- An arrow from kernel gamma to C, without a label.
- A dashed curved arrow from kernel gamma to cokernel alpha, without a label.
- A dashed curved arrow from D to kernel alpha, without a label.
- An arrow from D to A, without a label.
- An arrow from A to B, without a label.
- An arrow from A to A prime, without a label.
- An arrow from B to C, without a label.
- An arrow from B to B prime, without a label.
- An arrow from C to 0 (row 3, column 6), without a label.
- An arrow from C to C prime, without a label.
- An arrow from 0 (row 5, column 2) to A prime, without a label.
- An arrow from A prime to B prime, without a label.
- An arrow from A prime to cokernel alpha, without a label.
- An arrow from B prime to C prime, without a label.
- An arrow from B prime to cokernel beta, without a label.
- An arrow from C prime to D prime, without a label.
- An arrow from C prime to cokernel gamma, without a label.
- A dashed arrow from cokernel alpha to cokernel beta, without a label.
- A dashed arrow from cokernel beta to cokernel gamma, without a label.
- A dashed curved arrow from cokernel gamma to D prime, without a label.
and the resulting eight-term sequence
is exact.
Proof.
We define as follows. For , find with . Then , so for some . Let denote the image of in . To see that this is well-defined, note that if also satisfies , then , so for some . We then have
so . But has image in , so is well-defined.
We now check that the other maps are well-defined. Since
and is injective, we have , so . Since
we have . Similarly, . Also, if , then we may lift it to , map to , and then project to . This is well-defined as any other choice of differs by some , which causes to change by the image of , which is zero. Thus induces a well-defined map
Similarly, we have a well-defined
Finally, , so , and therefore . Thus induces a map
It is clear from the above definitions that the -term sequence is a complex at all the terms but and . Let . Then is given by considering , lifting it to some , which we may take to be , and projecting to . Hence . On the other hand, if , then is given by definition by projecting to , where , hence is zero.
We now check exactness at each term. We have
so the sequence is exact at . Next, if , then there exists with .
Since and is injective, we have , or . Hence
and we have exactness at .
If , then whenever and , we have , letting denote the image of . We then have for some , so still satisfies , but . So , and we have exactness at .
If is the image of and , then there exists with . Now
so , and . Hence, we have exactness at .
If is the image of and , then there exists with . Now for some . And has image in . On the other hand, , so for some . If is the image of , then as the image of in . Thus, we have exactness at .
Finally, let be the image of . Note that . If this is zero, then for some , and is the image of the projection of to under . □
Lemma 2.6.2. §
Suppose we have a commutative diagram as in (2.6.1) and that both and are split. Then the snake map is zero.
Proof.
Let . Let split and split . By Lemma ??, we also have a splitting map
It is easy to check that . It follows that . Thus , which is the image of in , is zero. □
2.7. Complexes
Let be an abelian category. We have already defined the notion of a chain complex in 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 is a diagram
such that for all . Again, the 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 and be complexes. A morphism of complexes is a sequence of morphisms commuting with the differentials of the complexes in the sense that
for all .
Remark 2.7.4. §
We may consider the category of cochain complexes in , where morphisms of chain complexes and are morphisms for each commuting with the differentials.
The reader can easily check the following.
Proposition 2.7.5. §
Let be an abelian category. Then the category is an abelian category as well.
Definition 2.7.6. §
Let (resp., ) be a chain complex (resp., cochain complex).
- a.
-
We say that (resp., ) is bounded below if (resp., ) for all for some .
- b.
-
We say that (resp., ) is bounded above if (resp., ) for all for some .
- c.
-
We say that (resp., ) 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 .
Definition 2.7.7. §
- a.
-
We define th homology (object) of a chain complex by
for each .
- b.
-
We define the th cohomology (object) of a cochain complex by
for .
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 with and for even. Then and for even.
Lemma 2.7.10. §
Let be a morphism of complexes. Then we have natural homomorphisms
for each .
Proof.
Let and . We have for all , so
and
so
Hence, we have an induced morphism
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
in is exact if and only if each sequence
is exact in .
Theorem 2.7.13. §
Let
be an exact sequence of cochain complexes of -modules. Then we have a long exact sequence in cohomology
where is the map given by the snake lemma applied to the diagram (2.7.1) below.
Proof.
Consider the following diagram
Diagram description: Kernels and cokernels for the cohomology sequenceThe two displayed rows are exact, and the squares commute. Objects, listed by row and column:
Arrows and lines:
| (2.7.1) |
By the snake lemma, the two rows are exact, and the diagram commutes since we have and as morphisms and , respectively, by naturality of the cokernel and kernel. Since we have natural isomorphisms
we apply the snake lemma to obtain an exact sequence
and the result follows by splicing together these sequences. □
Definition 2.7.14. §
Let and be chain complexes. Let be morphisms of chain complexes.
- a.
-
A chain homotopy from to is a sequence of morphisms satisfying
for all .
- b.
-
We say that and are chain homotopic, and write , if there exists a homotopy from to .
- c.
-
If is (chain) homotopic to , then is said to be null-homotopic.
The maps defining a null-homotopy fit into a (not usually ) diagram
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:
- An arrow from ellipsis (row 1, column 1) to A superscript (i minus 1), without a label.
- An arrow from A superscript (i minus 1) to A superscript (i), labelled d superscript (i minus 1) subscript (A).
- An arrow from A superscript (i minus 1) to B superscript (i minus 1), labelled f superscript (i minus 1).
- An arrow from A superscript (i) to A superscript (i plus 1), labelled d superscript (i) subscript (A).
- An arrow from A superscript (i) to B superscript (i), labelled f superscript (i).
- An arrow from A superscript (i) to B superscript (i minus 1), labelled s superscript (i).
- An arrow from A superscript (i plus 1) to ellipsis (row 1, column 5), without a label.
- An arrow from A superscript (i plus 1) to B superscript (i plus 1), labelled f superscript (i plus 1).
- An arrow from A superscript (i plus 1) to B superscript (i), labelled s superscript (i plus 1).
- An arrow from ellipsis (row 2, column 1) to B superscript (i minus 1), without a label.
- An arrow from B superscript (i minus 1) to B superscript (i), labelled d superscript (i minus 1) subscript (B).
- An arrow from B superscript (i) to B superscript (i plus 1), labelled d superscript (i) subscript (B).
- An arrow from B superscript (i plus 1) to ellipsis (row 2, column 5), without a label.
Proposition 2.7.15. §
Assume that and are homotopic as maps . Then the maps and on homology are equal for all .
Proof.
It suffices to assume that , since the th homology functor from to is additive. So, we must show that for all , which is to say that . Since , we have
so . □
Definition 2.7.16. §
A morphism of complexes is a homotopy equivalence if there exists a morphism such that and .
Definition 2.7.17. §
Let be a complex, and let . The shift by of the complex is the complex with and for all .
Notation 2.7.18. §
Let , let be objects in an abelian category for , and let be morphisms in for . Let and , and let and denote the canonical inclusions and projections. Then we use the matrix to represent the morphism in that is the sum
Definition 2.7.19. §
Let be a morphism of complexes. The cone of is the complex with
and
Proposition 2.7.20. §
The cone of fits in a short exact sequence
and the resulting long exact sequence has th connecting homomorphism
equal to .
Proof.
The differential on preserves and agrees with the differential on , so is a subcomplex. The quotient complex has th term and differential induced by the negative of the differential on , so is canonically isomorphic to . Since , we have a natural splitting of the surjection given by the inclusion. The connecting homomorphism is then induced by the composition
which is given by the sum of and as maps to , but then has image in (since is zero on ) and therefore agrees with as a morphism to . In other words, the connecting homomorphism is canonically identified with . □
2.8. Total complexes
Definition 2.8.1. §
A double (cochain) complex in an abelian category is a complex in .
The data of a double complex consists of complexes of objects of , which is to say that each is a complex in with differentials of its own, and the are morphisms between , so commute with the differentials on these complexes. We rephrase this as follows.
Remark 2.8.2. §
A double complex in is a diagram of the form
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:
- An arrow from ellipsis (row 2, column 1) to A superscript (i minus 1,j plus 1), without a label.
- 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).
- An arrow from A superscript (i minus 1,j plus 1) to vertical ellipsis (row 1, column 2), without a label.
- 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).
- An arrow from A superscript (i,j plus 1) to vertical ellipsis (row 1, column 3), without a label.
- An arrow from A superscript (i plus 1,j plus 1) to ellipsis (row 2, column 5), without a label.
- An arrow from A superscript (i plus 1,j plus 1) to vertical ellipsis (row 1, column 4), without a label.
- An arrow from ellipsis (row 3, column 1) to A superscript (i minus 1,j), without a label.
- An arrow from A superscript (i minus 1,j) to A superscript (i,j), labelled d subscript (h) superscript (i minus 1,j).
- 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).
- An arrow from A superscript (i,j) to A superscript (i plus 1,j), labelled d subscript (h) superscript (i,j).
- An arrow from A superscript (i,j) to A superscript (i,j plus 1), labelled d subscript (v) superscript (i,j).
- An arrow from A superscript (i plus 1,j) to ellipsis (row 3, column 5), without a label.
- 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).
- An arrow from ellipsis (row 4, column 1) to A superscript (i minus 1,j minus 1), without a label.
- 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).
- 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).
- 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).
- An arrow from A superscript (i,j minus 1) to A superscript (i,j), labelled d subscript (v) superscript (i,j minus 1).
- An arrow from A superscript (i plus 1,j minus 1) to ellipsis (row 4, column 5), without a label.
- 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).
- An arrow from vertical ellipsis (row 5, column 2) to A superscript (i minus 1,j minus 1), without a label.
- An arrow from vertical ellipsis (row 5, column 3) to A superscript (i,j minus 1), without a label.
- 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.
-
each row is a complex: i.e., ,
- ii.
-
each column is a complex: i.e., , and
- iii.
-
the squares commute: i.e., .
Definition 2.8.3. §
The degree of the term of a double complex is .
Definition 2.8.4. §
Let be a double complex in an abelian category .
- a.
-
Suppose that admits coproducts. The total sum complex of is the complex
with the differential
where the differentials and are taken to be zero on the with .
- b.
-
Suppose that admits products. The total product complex of is the complex
with the differential
where the differentials and are taken to be zero on the with .
Remark 2.8.5. §
We remark that is in fact a complex (when it exists). That is,
Similarly, is a complex as well.
Definition 2.8.6. §
An th quadrant double complex, for , is a double complex such that for not in the closed th quadrant of . 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 are , , , and , respectively. The closed upper, lower, left, and right half-planes are , , , and , respectively.
Remark 2.8.8. §
If is either a first or a fourth quadrant double complex, then
and it exists for any abelian category , since there are only finitely many nonzero terms in the complex of a given degree. In this case, we simply write for either total complex.
Definition 2.8.9. §
Let be a double complex. We define the truncated quotient complexes and of by
as well as subcomplexes and by
Proposition 2.8.10. §
Suppose that is an abelian category that admits direct products (resp., direct sums). Let be an upper (or left) double half-plane complex in with exact columns (resp., exact rows) or a right (or lower) double half-plane complex with exact rows (resp., exact columns). Then (resp., ) is exact.
Proof.
By interchanging rows and columns, we may suppose that 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 has exact rows. For , consider the truncated complex , which then has exact rows as well. Then a shift by in the horizontal direction is a third quadrant double complex . If we can show to be exact (for all ), then will be exact as well, as the reader may check that
the limit taken with respect to the morphisms induced by the inclusion morphisms . 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 with exact columns, for which we must show that is exact. Since we can always shift horizontally, it suffices to check that .
We now work in the category of -modules for a ring without loss of generality. Let
with . Choose with by exactness of the th column, and set . Suppose we have defined for with
Then
which is the -coordinate of , hence . By exactness of the th column, we may then choose with
completing the induction. We then set and note that , finishing the proof. □