$(A_0^\bullet \to A_1^\bullet \to A_2^\bullet \to \ldots ) \longrightarrow (B_0^\bullet \to B_1^\bullet \to B_2^\bullet \to \ldots )$

be a map between two complexes of complexes of abelian groups. Set $A^{p, q} = A_ p^ q$, $B^{p, q} = B_ p^ q$ to obtain double complexes. Let $\text{Tot}_\pi (A^{\bullet , \bullet })$ and $\text{Tot}_\pi (B^{\bullet , \bullet })$ be the product total complexes associated to the double complexes. If each $A_ p^\bullet \to B_ p^\bullet$ is a quasi-isomorphism, then $\text{Tot}_\pi (A^{\bullet , \bullet }) \to \text{Tot}_\pi (B^{\bullet , \bullet })$ is a quasi-isomorphism.

Proof. Recall that $\text{Tot}_\pi (A^{\bullet , \bullet })$ in degree $n$ is given by $\prod _{p + q = n} A^{p, q} = \prod _{p + 1 = n} A^ q_ p$. Let $C_ p^\bullet$ be the cone on the map $A_ p^\bullet \to B_ p^\bullet$, see Derived Categories, Section 13.9. By the functoriality of the cone construction we obtain a complex of complexes

$C_0^\bullet \to C_1^\bullet \to C_2^\bullet \to \ldots$

Then we see $\text{Tot}_\pi (C^{\bullet , \bullet })$ in degree $n$ is given by

$\prod _{p + q = n} C^{p, q} = \prod _{p + q = n} C^ q_ p = \prod _{p + q = n} (B^ q_ p \oplus A^{q + 1}_ p) = \prod _{p + q = n} B^ q_ p \oplus \prod _{p + q = n} A^{q + 1}_ p$

We conclude that $\text{Tot}_\pi (C^{\bullet , \bullet })$ is the cone of the map $\text{Tot}_\pi (A^{\bullet , \bullet }) \to \text{Tot}_\pi (B^{\bullet , \bullet })$ (We omit the verification that the differentials agree.) Thus it suffices to show $\text{Tot}_\pi (A^{\bullet , \bullet })$ is acyclic if each $A_ p^\bullet$ is acyclic. This follows from Lemma 15.103.1. $\square$

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