Lemma 12.13.1. Let $\mathcal{A}$ be an additive category. Let $f, g : B_\bullet \to C_\bullet $ be morphisms of chain complexes. Suppose given morphisms of chain complexes $a : A_\bullet \to B_\bullet $, and $c : C_\bullet \to D_\bullet $. If $\{ h_ i : B_ i \to C_{i + 1}\} $ defines a homotopy between $f$ and $g$, then $\{ c_{i + 1} \circ h_ i \circ a_ i\} $ defines a homotopy between $c \circ f \circ a$ and $c \circ g \circ a$.

## 12.13 Complexes

Of course the notions of a chain complex and a cochain complex are dual and you only have to read one of the two parts of this section. So pick the one you like. (Actually, this doesn't quite work right since the conventions on numbering things are not adapted to an easy transition between chain and cochain complexes.)

A *chain complex $A_\bullet $* in an additive category $\mathcal{A}$ is a complex

of $\mathcal{A}$. In other words, we are given an object $A_ i$ of $\mathcal{A}$ for all $i \in \mathbf{Z}$ and for all $i \in \mathbf{Z}$ a morphism $d_ i : A_ i \to A_{i - 1}$ such that $d_{i - 1} \circ d_ i = 0$ for all $i$. A *morphism of chain complexes $f : A_\bullet \to B_\bullet $* is given by a family of morphisms $f_ i : A_ i \to B_ i$ such that all the diagrams

commute. The *category of chain complexes of $\mathcal{A}$* is denoted $\text{Ch}(\mathcal{A})$. The full subcategory consisting of objects of the form

is denoted $\text{Ch}_{\geq 0}(\mathcal{A})$. In other words, a chain complex $A_\bullet $ belongs to $\text{Ch}_{\geq 0}(\mathcal{A})$ if and only if $A_ i = 0$ for all $i < 0$.

Given an additive category $\mathcal{A}$ we identify $\mathcal{A}$ with the full subcategory of $\text{Ch}(\mathcal{A})$ consisting of chain complexes zero except in degree $0$ by the functor

By abuse of notation we often denote the object on the right hand side simply $A$. If we want to stress that we are viewing $A$ as a chain complex we may sometimes use the notation $A[0]$, see Section 12.14.

A *homotopy $h$* between a pair of morphisms of chain complexes $f, g : A_\bullet \to B_\bullet $ is a collection of morphisms $h_ i : A_ i \to B_{i + 1}$ such that we have

for all $i$. Two morphisms $f, g : A_\bullet \to B_\bullet $ are said to be *homotopic* if a homotopy between $f$ and $g$ exists. Clearly, the notions of chain complex, morphism of chain complexes, and homotopies between morphisms of chain complexes make sense even in a preadditive category.

**Proof.**
Omitted.
$\square$

In particular this means that it makes sense to define the category of chain complexes with maps up to homotopy. We'll return to this later.

Definition 12.13.2. Let $\mathcal{A}$ be an additive category. We say a morphism $a : A_\bullet \to B_\bullet $ is a *homotopy equivalence* if there exists a morphism $b : B_\bullet \to A_\bullet $ such that there exists a homotopy between $a \circ b$ and $\text{id}_ A$ and there exists a homotopy between $b \circ a$ and $\text{id}_ B$. If there exists such a morphism between $A_\bullet $ and $B_\bullet $, then we say that $A_\bullet $ and $B_\bullet $ are *homotopy equivalent*.

In other words, two complexes are homotopy equivalent if they become isomorphic in the category of complexes up to homotopy.

Lemma 12.13.3. Let $\mathcal{A}$ be an abelian category.

The category of chain complexes in $\mathcal{A}$ is abelian.

A morphism of complexes $f : A_\bullet \to B_\bullet $ is injective if and only if each $f_ n : A_ n \to B_ n$ is injective.

A morphism of complexes $f : A_\bullet \to B_\bullet $ is surjective if and only if each $f_ n : A_ n \to B_ n$ is surjective.

A sequence of chain complexes

\[ A_\bullet \xrightarrow {f} B_\bullet \xrightarrow {g} C_\bullet \]is exact at $B_\bullet $ if and only if each sequence

\[ A_ i \xrightarrow {f_ i} B_ i \xrightarrow {g_ i} C_ i \]is exact at $B_ i$.

**Proof.**
Omitted.
$\square$

For any $i \in \mathbf{Z}$ the $i$th *homology group* of a chain complex $A_\bullet $ in an abelian category is defined by the following formula

If $f : A_\bullet \to B_\bullet $ is a morphism of chain complexes of $\mathcal{A}$ then we get an induced morphism $H_ i(f) : H_ i(A_\bullet ) \to H_ i(B_\bullet )$ because clearly $f_ i(\mathop{\mathrm{Ker}}(d_ i : A_ i \to A_{i - 1})) \subset \mathop{\mathrm{Ker}}(d_ i : B_ i \to B_{i - 1})$, and similarly for $\mathop{\mathrm{Im}}(d_{i + 1})$. Thus we obtain a functor

Definition 12.13.4. Let $\mathcal{A}$ be an abelian category.

A morphism of chain complexes $f : A_\bullet \to B_\bullet $ is called a

*quasi-isomorphism*if the induced map $H_ i(f) : H_ i(A_\bullet ) \to H_ i(B_\bullet )$ is an isomorphism for all $i \in \mathbf{Z}$.A chain complex $A_\bullet $ is called

*acyclic*if all of its homology objects $H_ i(A_\bullet )$ are zero.

Lemma 12.13.5. Let $\mathcal{A}$ be an abelian category.

If the maps $f, g : A_\bullet \to B_\bullet $ are homotopic, then the induced maps $H_ i(f)$ and $H_ i(g)$ are equal.

If the map $f : A_\bullet \to B_\bullet $ is a homotopy equivalence, then $f$ is a quasi-isomorphism.

**Proof.**
Omitted.
$\square$

Lemma 12.13.6. Let $\mathcal{A}$ be an abelian category. Suppose that

is a short exact sequence of chain complexes of $\mathcal{A}$. Then there is a canonical long exact homology sequence

**Proof.**
Omitted. The maps come from the Snake Lemma 12.5.17 applied to the diagrams

A *cochain complex $A^\bullet $* in an additive category $\mathcal{A}$ is a complex

of $\mathcal{A}$. In other words, we are given an object $A^ i$ of $\mathcal{A}$ for all $i \in \mathbf{Z}$ and for all $i \in \mathbf{Z}$ a morphism $d^ i : A^ i \to A^{i + 1}$ such that $d^{i + 1} \circ d^ i = 0$ for all $i$. A *morphism of cochain complexes $f : A^\bullet \to B^\bullet $* is given by a family of morphisms $f^ i : A^ i \to B^ i$ such that all the diagrams

commute. The *category of cochain complexes of $\mathcal{A}$* is denoted $\text{CoCh}(\mathcal{A})$. The full subcategory consisting of objects of the form

is denoted $\text{CoCh}_{\geq 0}(\mathcal{A})$. In other words, a cochain complex $A^\bullet $ belongs to the subcategory $\text{CoCh}_{\geq 0}(\mathcal{A})$ if and only if $A^ i = 0$ for all $i < 0$.

Given an additive category $\mathcal{A}$ we identify $\mathcal{A}$ with the full subcategory of $\text{CoCh}(\mathcal{A})$ consisting of cochain complexes zero except in degree $0$ by the functor

By abuse of notation we often denote the object on the right hand side simply $A$. If we want to stress that we are viewing $A$ as a cochain complex we may sometimes use the notation $A[0]$, see Section 12.14.

A *homotopy $h$* between a pair of morphisms of cochain complexes $f, g : A^\bullet \to B^\bullet $ is a collection of morphisms $h^ i : A^ i \to B^{i - 1}$ such that we have

for all $i$. Two morphisms $f, g : A^\bullet \to B^\bullet $ are said to be *homotopic* if a homotopy between $f$ and $g$ exists. Clearly, the notions of cochain complex, morphism of cochain complexes, and homotopies between morphisms of cochain complexes make sense even in a preadditive category.

Lemma 12.13.7. Let $\mathcal{A}$ be an additive category. Let $f, g : B^\bullet \to C^\bullet $ be morphisms of cochain complexes. Suppose given morphisms of cochain complexes $a : A^\bullet \to B^\bullet $, and $c : C^\bullet \to D^\bullet $. If $\{ h^ i : B^ i \to C^{i - 1}\} $ defines a homotopy between $f$ and $g$, then $\{ c^{i - 1} \circ h^ i \circ a^ i\} $ defines a homotopy between $c \circ f \circ a$ and $c \circ g \circ a$.

**Proof.**
Omitted.
$\square$

In particular this means that it makes sense to define the category of cochain complexes with maps up to homotopy. We'll return to this later.

Definition 12.13.8. Let $\mathcal{A}$ be an additive category. We say a morphism $a : A^\bullet \to B^\bullet $ is a *homotopy equivalence* if there exists a morphism $b : B^\bullet \to A^\bullet $ such that there exists a homotopy between $a \circ b$ and $\text{id}_ A$ and there exists a homotopy between $b \circ a$ and $\text{id}_ B$. If there exists such a morphism between $A^\bullet $ and $B^\bullet $, then we say that $A^\bullet $ and $B^\bullet $ are *homotopy equivalent*.

In other words, two complexes are homotopy equivalent if they become isomorphic in the category of complexes up to homotopy.

Lemma 12.13.9. Let $\mathcal{A}$ be an abelian category.

The category of cochain complexes in $\mathcal{A}$ is abelian.

A morphism of cochain complexes $f : A^\bullet \to B^\bullet $ is injective if and only if each $f^ n : A^ n \to B^ n$ is injective.

A morphism of cochain complexes $f : A^\bullet \to B^\bullet $ is surjective if and only if each $f^ n : A^ n \to B^ n$ is surjective.

A sequence of cochain complexes

\[ A^\bullet \xrightarrow {f} B^\bullet \xrightarrow {g} C^\bullet \]is exact at $B^\bullet $ if and only if each sequence

\[ A^ i \xrightarrow {f^ i} B^ i \xrightarrow {g^ i} C^ i \]is exact at $B^ i$.

**Proof.**
Omitted.
$\square$

For any $i \in \mathbf{Z}$ the $i$th *cohomology group* of a cochain complex $A^\bullet $ is defined by the following formula

If $f : A^\bullet \to B^\bullet $ is a morphism of cochain complexes of $\mathcal{A}$ then we get an induced morphism $H^ i(f) : H^ i(A^\bullet ) \to H^ i(B^\bullet )$ because clearly $f^ i(\mathop{\mathrm{Ker}}(d^ i : A^ i \to A^{i + 1})) \subset \mathop{\mathrm{Ker}}(d^ i : B^ i \to B^{i + 1})$, and similarly for $\mathop{\mathrm{Im}}(d^{i - 1})$. Thus we obtain a functor

Definition 12.13.10. Let $\mathcal{A}$ be an abelian category.

A morphism of cochain complexes $f : A^\bullet \to B^\bullet $ of $\mathcal{A}$ is called a

*quasi-isomorphism*if the induced maps $H^ i(f) : H^ i(A^\bullet ) \to H^ i(B^\bullet )$ is an isomorphism for all $i \in \mathbf{Z}$.A cochain complex $A^\bullet $ is called

*acyclic*if all of its cohomology objects $H^ i(A^\bullet )$ are zero.

Lemma 12.13.11. Let $\mathcal{A}$ be an abelian category.

If the maps $f, g : A^\bullet \to B^\bullet $ are homotopic, then the induced maps $H^ i(f)$ and $H^ i(g)$ are equal.

If $f : A^\bullet \to B^\bullet $ is a homotopy equivalence, then $f$ is a quasi-isomorphism.

**Proof.**
Omitted.
$\square$

Lemma 12.13.12. Let $\mathcal{A}$ be an abelian category. Suppose that

is a short exact sequence of cochain complexes of $\mathcal{A}$. Then there is a long exact cohomology sequence

The construction produces long exact cohomology sequences which are functorial in the short exact sequence and compatible with shifts.

**Proof.**
For the horizontal maps $H^ i(A^\bullet ) \to H^ i(B^\bullet )$ and $H^ i(B^\bullet ) \to H^ i(C^\bullet )$ we use the fact that $H^ i$ is a functor, see above. For the “boundary map” $H^ i(C^\bullet ) \to H^{i + 1}(A^\bullet )$ we use the map $\delta $ of the Snake Lemma 12.5.17 applied to the diagram

This works as the kernel of the right vertical map is equal to $H^ i(C^\bullet )$ and the cokernel of the left vertical map is $H^{i + 1}(A^\bullet )$. We omit the verification that we obtain a long exact sequence and we omit the verification of the properties mentioned at the end of the statement of the lemma. $\square$

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