
Lemma 12.22.7. Let $\mathcal{A}$ be an abelian category. Let $K^\bullet$ be a complex. Let $A^{\bullet , \bullet }$ be a double complex. Let $\alpha ^ p : K^ p \to A^{p, 0}$ be morphisms. Assume that

1. For every $n \in \mathbf{Z}$ there are only finitely many nonzero $A^{p, q}$ with $p + q = n$.

2. We have $A^{p, q} = 0$ if $q < 0$.

3. The morphisms $\alpha ^ p$ give rise to a morphism of complexes $\alpha : K^\bullet \to A^{\bullet , 0}$.

4. The complex $A^{p, \bullet }$ is exact in all degrees $q \not= 0$ and the morphism $K^ p \to A^{p, 0}$ induces an isomorphism $K^ p \to \mathop{\mathrm{Ker}}(d_2^{p, 0})$.

Then $\alpha$ induces a quasi-isomorphism

$K^\bullet \longrightarrow sA^\bullet$

of complexes. Moreover, there is a variant of this lemma involving the second variable $q$ instead of $p$.

Proof. The map is simply the map given by the morphisms $K^ n \to A^{n, 0} \to sA^ n$, which are easily seen to define a morphism of complexes. Consider the spectral sequence $({}'E_ r, {}'d_ r)_{r \geq 0}$ associated to the double complex $A^{\bullet , \bullet }$. By Lemma 12.22.6 this spectral sequence converges and the induced filtration on $H^ n(sA^\bullet )$ is finite for each $n$. By Lemma 12.22.4 and assumption (4) we have ${}'E_1^{p, q} = 0$ unless $q = 0$ and $'E_1^{p, 0} = K^ p$ with differential ${}'d_1^{p, 0}$ identified with $d_ K^ p$. Hence ${}'E_2^{p, 0} = H^ p(K^\bullet )$ and zero otherwise. This clearly implies $d_2^{p, q} = d_3^{p, q} = \ldots = 0$ for degree reasons. Hence we conclude that $H^ n(sA^\bullet ) = H^ n(K^\bullet )$. We omit the verification that this identification is given by the morphism of complexes $K^\bullet \to sA^\bullet$ introduced above. $\square$

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