Lemma 12.25.4 (0133)—The Stacks project

Lemma 12.25.4. 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

For every $n \in \mathbf{Z}$ there are only finitely many nonzero $A^{p, q}$ with $p + q = n$.
We have $A^{p, q} = 0$ if $q < 0$.
The morphisms $\alpha ^ p$ give rise to a morphism of complexes $\alpha : K^\bullet \to A^{\bullet , 0}$.
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

\[ K^\bullet \longrightarrow \text{Tot}(A^{\bullet , \bullet }) \]

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 $K^ n \to A^{n, 0} \to \text{Tot}^ n(A^{\bullet , \bullet })$, 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 12.25.3 this spectral sequence converges and the induced filtration on $H^ n(\text{Tot}(A^{\bullet , \bullet }))$ is finite for each $n$. By Lemma 12.25.1 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 $H^ n(\text{Tot}(A^{\bullet , \bullet })) = H^ n(K^\bullet )$. We omit the verification that this identification is given by the morphism of complexes $K^\bullet \to \text{Tot}(A^{\bullet , \bullet })$ introduced above. $\square$

Comments (1)

Comment #9490 by Elías Guisado on July 04, 2024 at 03:01

Typo: In the proof, third-to-last sentence, I think it should be ${}'d_2^{p,q}={}'d_3^{p,q}=\dots=0$ .

I don't understand “hence we conclude that $H^n(\operatorname{Tot}(A^{\bullet,\bullet}))=H^n(K^{\bullet})$ .” I asked about this and expressed my concerns in MOs; by the time of publication of this comment no one has answered yet.

In Kashiwara, Schapira, Categories and Sheaves, a stronger result (Theorem 12.5.4) is proven without resorting to spectral sequences. It says that if $f:X\to Y$ is a morphism of double complexes that are diagonally bounded (i.e., with a finite amount of nonzero terms in each diagonal $p+q=n$ ) and such that $H_{II}(H_I(X))\to H_{II}(H_I(Y))$ is an isomorphism, then $\operatorname{Tot}f:\operatorname{Tot}X\to\operatorname{Tot}Y$ is a quasi-isomorphism.

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1 comment(s) on Section 12.25: Spectral sequences: double complexes

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