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?


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 $H_IH_{II}$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.