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The following question arose while studying chapter 12.5 of the book Categories and Sheaves by Kashiwara and Schapira, abbreviated in the following by [KS].

Let $X$ be a double complex (with cohomological convention, i.e. with differentials of degree $1$) in an abelian category such that for any $n\in\mathbb Z$, the set $\{(k,l)\in\mathbb Z\times\mathbb Z~|~ k+l=n~\text{and}~X^{k,l}\neq 0\}$ is finite. Denote by $\operatorname{tot}X$ the associated total complex (defined by $(\operatorname{tot}X)^{n}=\oplus_{k+l=n} X^{k,l}$ and with the usual differential).

Define $H_I(X)$ via $H_I(X)^{k,l}=H^k(X^{\cdot,l})$ with differential $d_{II}$ (in the second coordinate) induced by $d_{II}$ on $X$ and with the zero differential in the first coordinate. Define $H_{II}$ in a dual manner. Then we obtain a double complex $H_{II}H_{I}(X)$ with zero differentials in each coordinate. One then obtains ([KS], Theorem 12.5.4):

Let $f\colon X\to Y$ be a morphism of double complexes (both satisfying the finiteness constraint above) and suppose that $H_{II}H_I(X)\to H_{II}H_I(Y)$ is an isomorphism. Then $\operatorname{tot} X\to \operatorname{tot} Y$ is a quasi-isomorphism.

I understand this theorem as well as the proof. What I don't understand is the following corollary ([KS], Corollary 12.5.5 (i)):

Assume $X$ has exact rows (i.e. $X^{k,\cdot}$ is exact for each $k$). Then $\operatorname{tot}X$ is quasi-isomorphic to zero.

Replacing "rows" by "columns", the statement is clear. But I don't see how this can be true for exact rows, since in the above theorem we first take homology along the columns which does not necessarily vanish, and I don't think that the complex $$\cdots\to H^{k}(X^{\cdot,l-1})\to H^{k}(X^{\cdot,l})\to H^{k}(X^{\cdot,l+1})\to \cdots$$ is necessarily exact, given that $$\cdots\to X^{k,l-1}\to X^{k,l}\to X^{k,l+1}\to\cdots$$ is exact.

So is this statement even true or is it a typo?

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Up to isomorphism, $\operatorname{tot} X$ doesn't change if you switch rows and columns. so whatever is true for columns is true for rows.

Alternatively, Theorem 12.5.4 is true, with the same proof, if you replace $H_{II}H_I$ with $H_IH_{II}$.

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  • $\begingroup$ Ah, I thought about that and tried to prove it but got confused about the signs and then I had doubts if it is even true. Thank you, I will have a look at it again. $\endgroup$ – asdq May 28 at 9:30
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    $\begingroup$ @asdq For an explanation of why the total complex doesn't, up to isomorphism, depend on the sign convention you use, you might want to look at this question: math.stackexchange.com/q/3147787/88262 $\endgroup$ – Jeremy Rickard May 28 at 9:47
  • $\begingroup$ Thank you, this is very helpful! $\endgroup$ – asdq May 28 at 10:40

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