Lemma 12.25.5. Let $\mathcal{A}$ be an abelian category. Let $M^\bullet $ be a complex of $\mathcal{A}$. Let

\[ a : M^\bullet [0] \longrightarrow \left(A^{0, \bullet } \to A^{1, \bullet } \to A^{2, \bullet } \to \ldots \right) \]

be a homotopy equivalence in the category of complexes of complexes of $\mathcal{A}$. Then the map $\alpha : M^\bullet \to \text{Tot}(A^{\bullet , \bullet })$ induced by $M^\bullet \to A^{0, \bullet }$ is a homotopy equivalence.

**Proof.**
The statement makes sense as a complex of complexes is the same thing as a double complex. The assumption means there is a map

\[ b : \left(A^{0, \bullet } \to A^{1, \bullet } \to A^{2, \bullet } \to \ldots \right) \longrightarrow M^\bullet [0] \]

such that $a \circ b$ and $b \circ a$ are homotopic to the identity in the category of complexes of complexes. This means that $b \circ a$ is the identity of $M^\bullet [0]$ (because there is only one term in degree $0$). Also, observe that $b$ is given by a map $b^0 : A^{0, \bullet } \to M^\bullet $ and zero in all other degrees. Thus $b$ induces a map $\beta : \text{Tot}(A^{\bullet , \bullet }) \to M^\bullet $ and $\beta \circ \alpha $ is the identity on $M^\bullet $. Finally, we have to show that the map $\alpha \circ \beta $ is homotopic to the identity. For this we choose maps of complexes $h^ n : A^{n, \bullet } \to A^{n - 1, \bullet }$ such that $a \circ b - \text{id} = d_1 \circ h + h \circ d_1$ which exist by assumption. Here $d_1 : A^{n, \bullet } \to A^{n + 1, \bullet }$ are the differentials of the complex of complexes. We will also denote $d_2$ the differentials of the complexes $A^{n, \bullet }$ for all $n$. Let $h^{n, m} : A^{n, m} \to A^{n - 1, m}$ be the components of $h^ n$. Then we can consider

\[ h' : \text{Tot}(A^{\bullet , \bullet })^ k = \bigoplus \nolimits _{n + m = k} A^{n, m} \to \bigoplus \nolimits _{n + m = k - 1} A^{n, m} = \text{Tot}(A^{\bullet , \bullet })^{k - 1} \]

given by $h^{n, m}$ on the summand $A^{n, m}$. Then we compute that the map

\[ d_{sA} \circ h' + h' \circ d_{sA} \]

restricted to the summand $A^{n, m}$ is equal to

\[ d_1^{n - 1, m} \circ h^{n, m} + (-1)^{n - 1} d_2^{n - 1, m} \circ h^{n, m} + h^{n + 1, m} \circ d_1^{n, m} + h^{n, m + 1} \circ (-1)^ nd_2^{n, m} \]

Since $h^ n$ is a map of complexes, the terms $(-1)^{n - 1} d_2^{n - 1, m} \circ h^{n, m}$ and $h^{n, m + 1} \circ (-1)^ nd_2^{n, m}$ cancel. The other two terms give $(\alpha \circ \beta )|_{A^{n, m}} - \text{id}_{A^{n, m}}$ because $a \circ b - \text{id} = d_1 \circ h + h \circ d_1$. This finishes the proof.
$\square$

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