The Stacks project

12.14 Homotopy and the shift functor

It is an annoying feature that signs and indices have to be part of any discussion of homological algebra1.

Definition 12.14.1. Let $\mathcal{A}$ be an additive category. Let $A_\bullet $ be a chain complex with boundary maps $d_{A, n} : A_ n \to A_{n - 1}$. For any $k \in \mathbf{Z}$ we define the $k$-shifted chain complex $A[k]_\bullet $ as follows:

  1. we set $A[k]_ n = A_{n + k}$, and

  2. we set $d_{A[k], n} : A[k]_ n \to A[k]_{n - 1}$ equal to $d_{A[k], n} = (-1)^ k d_{A, n + k}$.

If $f : A_\bullet \to B_\bullet $ is a morphism of chain complexes, then we let $f[k] : A[k]_\bullet \to B[k]_\bullet $ be the morphism of chain complexes with $f[k]_ n = f_{k + n}$.

Of course this means we have functors $[k] : \text{Ch}(\mathcal{A}) \to \text{Ch}(\mathcal{A})$ which mutually commute (on the nose, without any intervening isomorphisms of functors), such that $A[k][l]_\bullet = A[k + l]_\bullet $ and with $[0] = \text{id}_{\text{Ch}(\mathcal{A})}$.

Recall that we view $\mathcal{A}$ as a full subcategory of $\text{Ch}(\mathcal{A})$, see Section 12.13. Thus for any object $A$ of $\mathcal{A}$ the notation $A[k]$ refers to the unique chain complex zero in all degrees except having $A$ in degree $-k$.

Definition 12.14.2. Let $\mathcal{A}$ be an abelian category. Let $A_\bullet $ be a chain complex with boundary maps $d_{A, n} : A_ n \to A_{n - 1}$. For any $k \in \mathbf{Z}$ we identify $H_{i + k}(A_\bullet ) \rightarrow H_ i(A[k]_\bullet )$ via the identification $A_{i + k} = A[k]_ i$.

This identification is functorial in $A_\bullet $. Note that since no signs are involved in this definition we actually get a compatible system of identifications of all the homology objects $H_{i - k}(A[k]_\bullet )$, which are further compatible with the identifications $A[k][l]_\bullet = A[k + l]_\bullet $ and with $[0] = \text{id}_{\text{Ch}(\mathcal{A})}$.

Let $\mathcal{A}$ be an additive category. Suppose that $A_\bullet $ and $B_\bullet $ are chain complexes, $a, b : A_\bullet \to B_\bullet $ are morphisms of chain complexes, and $\{ h_ i : A_ i \to B_{i + 1}\} $ is a homotopy between $a$ and $b$. Recall that this means that $a_ i - b_ i = d_{i + 1} \circ h_ i + h_{i - 1} \circ d_ i$. What if $a = b$? Then we obtain the formula $0 = d_{i + 1} \circ h_ i + h_{i - 1} \circ d_ i$, in other words, $ - d_{i + 1} \circ h_ i = h_{i - 1} \circ d_ i$. By definition above this means the collection $\{ h_ i\} $ above defines a morphism of chain complexes

\[ A_\bullet \longrightarrow B[1]_\bullet . \]

Such a thing is the same as a morphism $A[-1]_\bullet \to B_\bullet $ by our remarks above. This proves the following lemma.

Lemma 12.14.3. Let $\mathcal{A}$ be an additive category. Suppose that $A_\bullet $ and $B_\bullet $ are chain complexes. Given any morphism of chain complexes $a : A_\bullet \to B_\bullet $ there is a bijection between the set of homotopies from $a$ to $a$ and $\mathop{\mathrm{Mor}}\nolimits _{\text{Ch}(\mathcal{A})}(A_\bullet , B[1]_\bullet )$. More generally, the set of homotopies between $a$ and $b$ is either empty or a principal homogeneous space under the group $\mathop{\mathrm{Mor}}\nolimits _{\text{Ch}(\mathcal{A})}(A_\bullet , B[1]_\bullet )$.

Proof. See above. $\square$

Lemma 12.14.4. Let $\mathcal{A}$ be an abelian category. Let

\[ 0 \to A_\bullet \to B_\bullet \to C_\bullet \to 0 \]

be a short exact sequence of complexes. Suppose that $\{ s_ n : C_ n \to B_ n\} $ is a family of morphisms which split the short exact sequences $0 \to A_ n \to B_ n \to C_ n \to 0$. Let $\pi _ n : B_ n \to A_ n$ be the associated projections, see Lemma 12.5.10. Then the family of morphisms

\[ \pi _{n - 1} \circ d_{B, n} \circ s_ n : C_ n \to A_{n - 1} \]

define a morphism of complexes $\delta (s) : C_\bullet \to A[-1]_\bullet $.

Proof. Denote $i : A_\bullet \to B_\bullet $ and $q : B_\bullet \to C_\bullet $ the maps of complexes in the short exact sequence. Then $i_{n - 1} \circ \pi _{n - 1} \circ d_{B, n} \circ s_ n = d_{B, n} \circ s_ n - s_{n - 1} \circ d_{C, n}$. Hence $i_{n - 2} \circ d_{A, n - 1} \circ \pi _{n - 1} \circ d_{B, n} \circ s_ n = d_{B, n - 1} \circ (d_{B, n} \circ s_ n - s_{n - 1} \circ d_{C, n}) = - d_{B, n - 1} \circ s_{n - 1} \circ d_{C, n}$ as desired. $\square$

Lemma 12.14.5. Notation and assumptions as in Lemma 12.14.4 above. The morphism of complexes $\delta (s) : C_\bullet \to A[-1]_\bullet $ induces the maps

\[ H_ i(\delta (s)) : H_ i(C_\bullet ) \longrightarrow H_ i(A[-1]_\bullet ) = H_{i - 1}(A_\bullet ) \]

which occur in the long exact homology sequence associated to the short exact sequence of chain complexes by Lemma 12.13.6.

Proof. Omitted. $\square$

Lemma 12.14.6. Notation and assumptions as in Lemma 12.14.4 above. Suppose $\{ s'_ n : C_ n \to B_ n\} $ is a second choice of splittings. Write $s'_ n = s_ n + i_ n \circ h_ n$ for some unique morphisms $h_ n : C_ n \to A_ n$. The family of maps $\{ h_ n : C_ n \to A[-1]_{n + 1}\} $ is a homotopy between the associated morphisms $\delta (s), \delta (s') : C_\bullet \to A[-1]_\bullet $.

Proof. Omitted. $\square$

Definition 12.14.7. Let $\mathcal{A}$ be an additive category. Let $A^\bullet $ be a cochain complex with boundary maps $d_ A^ n : A^ n \to A^{n + 1}$. For any $k \in \mathbf{Z}$ we define the $k$-shifted cochain complex $A[k]^\bullet $ as follows:

  1. we set $A[k]^ n = A^{n + k}$, and

  2. we set $d_{A[k]}^ n : A[k]^ n \to A[k]^{n + 1}$ equal to $d_{A[k]}^ n = (-1)^ k d_ A^{n + k}$.

If $f : A^\bullet \to B^\bullet $ is a morphism of cochain complexes, then we let $f[k] : A[k]^\bullet \to B[k]^\bullet $ be the morphism of cochain complexes with $f[k]^ n = f^{k + n}$.

Of course this means we have functors $[k] : \text{CoCh}(\mathcal{A}) \to \text{CoCh}(\mathcal{A})$ which mutually commute (on the nose, without any intervening isomorphisms of functors) and such that $A[k][l]^\bullet = A[k + l]^\bullet $ and with $[0] = \text{id}_{\text{CoCh}(\mathcal{A})}$.

Recall that we view $\mathcal{A}$ as a full subcategory of $\text{CoCh}(\mathcal{A})$, see Section 12.13. Thus for any object $A$ of $\mathcal{A}$ the notation $A[k]$ refers to the unique cochain complex zero in all degrees except having $A$ in degree $-k$.

Definition 12.14.8. Let $\mathcal{A}$ be an abelian category. Let $A^\bullet $ be a cochain complex with boundary maps $d_ A^ n : A^ n \to A^{n + 1}$. For any $k \in \mathbf{Z}$ we identify $H^{i + k}(A^\bullet ) \longrightarrow H^ i(A[k]^\bullet )$ via the identification $A^{i + k} = A[k]^ i$.

This identification is functorial in $A^\bullet $. Note that since no signs are involved in this definition we actually get a compatible system of identifications of all the homology objects $H^{i - k}(A[k]^\bullet )$, which are further compatible with the identifications $A[k][l]^\bullet = A[k + l]^\bullet $ and with $[0] = \text{id}_{\text{CoCh}(\mathcal{A})}$.

Let $\mathcal{A}$ be an additive category. Suppose that $A^\bullet $ and $B^\bullet $ are cochain complexes, $a, b : A^\bullet \to B^\bullet $ are morphisms of cochain complexes, and $\{ h^ i : A^ i \to B^{i - 1}\} $ is a homotopy between $a$ and $b$. Recall that this means that $a^ i - b^ i = d^{i - 1} \circ h^ i + h^{i + 1} \circ d^ i$. What if $a = b$? Then we obtain the formula $0 = d^{i - 1} \circ h^ i + h^{i + 1} \circ d^ i$, in other words, $ - d^{i - 1} \circ h^ i = h^{i + 1} \circ d^ i$. By definition above this means the collection $\{ h^ i\} $ above defines a morphism of cochain complexes

\[ A^\bullet \longrightarrow B[-1]^\bullet . \]

Such a thing is the same as a morphism $A[1]^\bullet \to B^\bullet $ by our remarks above. This proves the following lemma.

Lemma 12.14.9. Let $\mathcal{A}$ be an additive category. Suppose that $A^\bullet $ and $B^\bullet $ are cochain complexes. Given any morphism of cochain complexes $a : A^\bullet \to B^\bullet $ there is a bijection between the set of homotopies from $a$ to $a$ and $\mathop{\mathrm{Mor}}\nolimits _{\text{CoCh}(\mathcal{A})}(A^\bullet , B[-1]^\bullet )$. More generally, the set of homotopies between $a$ and $b$ is either empty or a principal homogeneous space under the group $\mathop{\mathrm{Mor}}\nolimits _{\text{CoCh}(\mathcal{A})}(A^\bullet , B[-1]^\bullet )$.

Proof. See above. $\square$

Lemma 12.14.10. Let $\mathcal{A}$ be an additive category. Let

\[ 0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0 \]

be a complex (!) of complexes. Suppose that we are given splittings $B^ n = A^ n \oplus C^ n$ compatible with the maps in the displayed sequence. Let $s^ n : C^ n \to B^ n$ and $\pi ^ n : B^ n \to A^ n$ be the corresponding maps. Then the family of morphisms

\[ \pi ^{n + 1} \circ d_ B^ n \circ s^ n : C^ n \to A^{n + 1} \]

define a morphism of complexes $\delta : C^\bullet \to A[1]^\bullet $.

Proof. Denote $i : A^\bullet \to B^\bullet $ and $q : B^\bullet \to C^\bullet $ the maps of complexes in the short exact sequence. Then $i^{n + 1} \circ \pi ^{n + 1} \circ d_ B^ n \circ s^ n = d_ B^ n \circ s^ n - s^{n + 1} \circ d_ C^ n$. Hence $i^{n + 2} \circ d_ A^{n + 1} \circ \pi ^{n + 1} \circ d_ B^ n \circ s^ n = d_ B^{n + 1} \circ (d_ B^ n \circ s^ n - s^{n + 1} \circ d_ C^ n) = - d_ B^{n + 1} \circ s^{n + 1} \circ d_ C^ n$ as desired. $\square$

Lemma 12.14.11. Notation and assumptions as in Lemma 12.14.10 above. Assume in addition that $\mathcal{A}$ is abelian. The morphism of complexes $\delta : C^\bullet \to A[1]^\bullet $ induces the maps

\[ H^ i(\delta ) : H^ i(C^\bullet ) \longrightarrow H^ i(A[1]^\bullet ) = H^{i + 1}(A^\bullet ) \]

which occur in the long exact homology sequence associated to the short exact sequence of cochain complexes by Lemma 12.13.12.

Proof. Omitted. $\square$

Lemma 12.14.12. Notation and assumptions as in Lemma 12.14.10. Let $\alpha : A^\bullet \to B^\bullet $, $\beta : B^\bullet \to C^\bullet $ be the given morphisms of complexes. Suppose $(s')^ n : C^ n \to B^ n$ and $(\pi ')^ n : B^ n \to A^ n$ is a second choice of splittings. Write $(s')^ n = s^ n + \alpha ^ n \circ h^ n$ and $(\pi ')^ n = \pi ^ n + g^ n \circ \beta ^ n$ for some unique morphisms $h^ n : C^ n \to A^ n$ and $g^ n : C^ n \to A^ n$. Then

  1. $g^ n = - h^ n$, and

  2. the family of maps $\{ g^ n : C^ n \to A[1]^{n - 1}\} $ is a homotopy between $\delta , \delta ' : C^\bullet \to A[1]^\bullet $, more precisely $(\delta ')^ n = \delta ^ n + g^{n + 1} \circ d_ C^ n + d_{A[1]}^{n - 1} \circ g^ n$.

Proof. As $(s')^ n$ and $(\pi ')^ n$ are splittings we have $(\pi ')^ n \circ (s')^ n = 0$. Hence

\[ 0 = ( \pi ^ n + g^ n \circ \beta ^ n ) \circ ( s^ n + \alpha ^ n \circ h^ n ) = g^ n \circ \beta ^ n \circ s^ n + \pi ^ n \circ \alpha ^ n \circ h^ n = g^ n + h^ n \]

which proves (1). We compute $(\delta ')^ n$ as follows

\[ ( \pi ^{n + 1} + g^{n + 1} \circ \beta ^{n + 1} ) \circ d_ B^ n \circ ( s^ n + \alpha ^ n \circ h^ n ) = \delta ^ n + g^{n + 1} \circ d_ C^ n + d_ A^ n \circ h^ n \]

Since $h^ n = -g^ n$ and since $d_{A[1]}^{n - 1} = -d_ A^ n$ we conclude that (2) holds. $\square$

[1] Please let us know if you notice sign errors or if you have improvements to our conventions.

Comments (2)

Comment #5080 by Laurent Moret-Bailly on

The same warning as at the beginning of the previous section (on chains vs cochain complexes) would be appropriate here. However, I remark that 011D and 011J (on term-by-term splittings) are not exactly parallel. Any reason for this?


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0119. Beware of the difference between the letter 'O' and the digit '0'.