Lemma 12.26.4. Let $M^\bullet$ be a complex of abelian groups. Let

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

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

Proof. Abbreviating $T^\bullet = \text{Tot}_\pi (A^{\bullet , \bullet })$ we define

$T^ n = \prod \nolimits _{p + q = n} A^{p, q} = \prod \nolimits _{p + q = n} A_{-p}^ q \quad \text{with}\quad \text{d}_{T^\bullet }^ n = \prod \nolimits _{n = p + q} (f_{-p}^ q + (-1)^ pd_{A_{-p}^\bullet }^ q)$

where $f_ p^\bullet : A_ p^\bullet \to A_{p - 1}^\bullet$ are the maps of complexes in the lemma. We will show that $T^\bullet$ is acyclic when $M^\bullet$ is the zero complex. This will suffice by the following trick. Set $B_ n^\bullet = A_{n + 1}^\bullet$ and $B_0^\bullet = M^\bullet$. Then we have an exact sequence

$\ldots \to B_2^\bullet \to B_1^\bullet \to B_0^\bullet \to 0 \to 0$

as in the lemma. Let $S^\bullet = \text{Tot}_\pi (B^{\bullet , \bullet })$. Then there is an obvious short exact sequence of complexes

$0 \to M^\bullet \to S^\bullet \to T^\bullet [1] \to 0$

and we conclude by the long exact cohomology sequence. Some details omitted.

Assume $M^\bullet = 0$. We will show $H^0(T^\bullet ) = 0$. The same argument works for other degrees. Let $x =(x_ n) \in \mathop{\mathrm{Ker}}(d_{T^\bullet })$ map to $\xi \in H^0(T^\bullet )$ with $x_ n \in A^{-n, n} = A_ n^ n$. Since $M^0 = 0$ we find that $x_0 = f_1^0(y_0)$ for some $y_0 \in A_1^0$. Then $x_1 - d^0_{A_1^\bullet }(y_0) = f_2^1(y_1)$ because it is mapped to zero by $f_1^1$ as $x$ is a cocycle. for some $y_1 \in A_2^1$. Continuing, using induction, we find $y = (y_ i) \in T^{-1}$ with $d_{T^\bullet }(y) = x$ as desired. $\square$

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