Lemma 12.26.2. 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 such that for all $p \in \mathbf{Z}$ the complexes

$\ldots \to \mathop{\mathrm{Ker}}(d_{A_2^\bullet }^ p) \to \mathop{\mathrm{Ker}}(d_{A_1^\bullet }^ p) \to \mathop{\mathrm{Ker}}(d_{A_0^\bullet }^ p) \to \mathop{\mathrm{Ker}}(d_{M^\bullet }^ p) \to 0$

are exact as well. Set $A^{p, q} = A_{-p}^ q$ to obtain a double complex. Then $\text{Tot}(A^{\bullet , \bullet }) \to M^\bullet$ induced by $A_0^\bullet \to M^\bullet$ is a quasi-isomorphism.

Proof. Using the short exact sequences $0 \to \mathop{\mathrm{Ker}}(d^ p_{A_ n^\bullet }) \to A_ n^ p \to \mathop{\mathrm{Im}}(d^ p_{A_ n^\bullet }) \to 0$ and the assumptions we see that

$\ldots \to \mathop{\mathrm{Im}}(d_{A_2^\bullet }^ p) \to \mathop{\mathrm{Im}}(d_{A_1^\bullet }^ p) \to \mathop{\mathrm{Im}}(d_{A_0^\bullet }^ p) \to \mathop{\mathrm{Im}}(d_{M^\bullet }^ p) \to 0$

is exact for all $p \in \mathbf{Z}$. Repeating with the exact sequences $0 \to \mathop{\mathrm{Im}}(d^{p - 1}_{A_ n^\bullet }) \to \mathop{\mathrm{Ker}}(d^ p_{A_ n^\bullet }) \to H^ p(A_ n^\bullet ) \to 0$ we find that

$\ldots \to H^ p(A_2^\bullet ) \to H^ p(A_1^\bullet ) \to H^ p(A_0^\bullet ) \to H^ p(M^\bullet ) \to 0$

is exact for all $p \in \mathbf{Z}$.

Write $T^\bullet = \text{Tot}(A^{\bullet , \bullet })$. We will show that $H^0(T^\bullet ) \to H^0(M^\bullet )$ is an isomorphism. The same argument works for other degrees. Let $x \in \mathop{\mathrm{Ker}}(\text{d}_{T^\bullet }^0)$ represent an element $\xi \in H^0(T^\bullet )$. Write $x = \sum _{i = n, \ldots , 0} x_ i$ with $x_ i \in A_ i^ i$. Assume $n > 0$. Then $x_ n$ is in the kernel of $d_{A_ n^\bullet }^ n$ and maps to zero in $H^ n(A_{n - 1}^\bullet )$ because it maps to an element which is the boundary of $x_{n - 1}$ up to sign. By the first paragraph of the proof, we find that $x_ n \bmod \mathop{\mathrm{Im}}(d^{n - 1}_{A_ n^\bullet })$ is in the image of $H^ n(A_{n + 1}^\bullet ) \to H^ n(A_ n^\bullet )$. Thus we can modify $x$ by a boundary and reach the situation where $x_ n$ is a boundary. Modifying $x$ once more we see that we may assume $x_ n = 0$. By induction we see that every cohomology class $\xi$ is represented by a cocycle $x = x_0$. Finally, the condition on exactness of kernels tells us two such cocycles $x_0$ and $x_0'$ are cohomologous if and only if their image in $H^0(M^\bullet )$ are the same. $\square$

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