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:
we set $A[k]_ n = A_{n + k}$, and
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
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 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 define a morphism of complexes $\delta (s) : C_\bullet \to A[-1]_\bullet $.
\[ 0 \to A_\bullet \to B_\bullet \to C_\bullet \to 0 \]
\[ \pi _{n - 1} \circ d_{B, n} \circ s_ n : C_ n \to A_{n - 1} \]
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 which occur in the long exact homology sequence associated to the short exact sequence of chain complexes by Lemma 12.13.6.
\[ H_ i(\delta (s)) : H_ i(C_\bullet ) \longrightarrow H_ i(A[-1]_\bullet ) = H_{i - 1}(A_\bullet ) \]
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:
we set $A[k]^ n = A^{n + k}$, and
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
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 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 define a morphism of complexes $\delta : C^\bullet \to A[1]^\bullet $.
\[ 0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0 \]
\[ \pi ^{n + 1} \circ d_ B^ n \circ s^ n : C^ n \to A^{n + 1} \]
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 which occur in the long exact homology sequence associated to the short exact sequence of cochain complexes by Lemma 12.13.12.
\[ H^ i(\delta ) : H^ i(C^\bullet ) \longrightarrow H^ i(A[1]^\bullet ) = H^{i + 1}(A^\bullet ) \]
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
$g^ n = - h^ n$, and
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
which proves (1). We compute $(\delta ')^ n$ as follows
Since $h^ n = -g^ n$ and since $d_{A[1]}^{n - 1} = -d_ A^ n$ we conclude that (2) holds. $\square$
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