Lemma 12.26.1. Let $M^\bullet$ be a complex of abelian groups. Let

$0 \to M^\bullet \to A_0^\bullet \to A_1^\bullet \to A_2^\bullet \to \ldots$

be an exact complex of complexes of abelian groups. Set $A^{p, q} = A_ p^ q$ to obtain a double complex. Then the map $M^\bullet \to \text{Tot}(A^{\bullet , \bullet })$ induced by $M^\bullet \to A_0^\bullet$ is a quasi-isomorphism.

Proof. If there exists a $t \in \mathbf{Z}$ such that $A_0^ q = 0$ for $q < t$, then this follows immediately from Lemma 12.25.4 (with $p$ and $q$ swapped as in the final statement of that lemma). OK, but for every $t \in \mathbf{Z}$ we have a complex

$0 \to \sigma _{\geq t}M^\bullet \to \sigma _{\geq t}A_0^\bullet \to \sigma _{\geq t}A_1^\bullet \to \sigma _{\geq t}A_2^\bullet \to \ldots$

of stupid truncations. Denote $A(t)^{\bullet , \bullet }$ the corresponding double complex. Every element $\xi$ of $H^ n(\text{Tot}(A^{\bullet , \bullet }))$ is the image of an element of $H^ n(\text{Tot}(A(t)^{\bullet , \bullet }))$ for some $t$ (look at explicit representatives of cohomology classes). Hence $\xi$ is in the image of $H^ n(\sigma _{\geq t}M^\bullet )$. Thus the map $H^ n(M^\bullet ) \to H^ n(\text{Tot}(A^{\bullet , \bullet }))$ is surjective. It is injective because for all $t$ the map $H^ n(\sigma _{\geq t}M^\bullet ) \to H^ n(\text{Tot}(A(t)^{\bullet , \bullet }))$ is injective and similar arguments. $\square$

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).