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Let $(X^{n,m}, d'^{n,m}, d''^{n,m})$ be a double complex. In Kashiwara, Schapira Categories and Sheaves as a corollary of

Theorem. Let $f: X \rightarrow Y$ be a morphism of bounded double complexes such that $f$ induces an isomorphism $f: H_{II}(H_{I}(X)) \rightarrow H_{II}(H_{I}(Y))$, then $f$ induces an isomorphism $\text{tot}(X) \cong \text{tot}(Y)$

it is proved that:

  1. If the rows of $X^{\bullet,\bullet}$ are exact then $\text{tot}(X^{\bullet, \bullet})$ is quasi isomorphic to $0$
  2. If the rows of $X^{j,\bullet}$ are exact except for $j = p$ then $\text{tot}(X^{\bullet,\bullet})$ is quasi isomorphic to $X^{p, \bullet}[-p]$

To prove 2 they say to apply the theorem to the morphisms $\sigma_{I}^{\geq p}X \rightarrow X$ and $\sigma_{I}^{\leq p} \sigma_{I}^{\geq p} X \rightarrow \sigma_{I}^{\geq p}X$, where $\sigma_{I}^{\geq p}$ is the stupid truncation which leaves the rows greater or equal than p. I don't understand why the hypotheses of the theorem are satisfied. Any help?

($H_I$ is the vertical cohomology and $H_{II}$ is the horizontal cohomology)

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After taking the cohomology of the rows, only one row survives. Then taking cohomology of the columns has no effect. Thus, in this case $H_{II}(H_I(X))$ is isomorphic to the cohomology of this row. But the fact that all the rows except the $p$-th one are exact is true also for the stupid truncation, because we only deleted rows which where exact already. So the same is true after the truncation, namely $H_{II}(H_I(\sigma^{\le p}X))$ is also the same as the cohomology of the p-th row. The map between them induce the identity on the cohomology of this row for obvious reasons, so the lemma applies.

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  • $\begingroup$ Maybe I am misunderstanding $H_I$ and $H_{II}$: $H_I(X)$ means taking cohomology of the vertical arrows and $H_{II}$ of the horizontal arrows, am I right? If it is so, how can you say that $H_{II}(H_I(X))$ is the cohomology of the only non-exact row? $\endgroup$ Mar 13, 2018 at 21:11
  • $\begingroup$ Oh, I see the confusion. For the proof of this statement you need to use the lemma for first cohomology of the rows and then columns. The lemma holds of course for both options by symmetry, but the one you need to use is this. $\endgroup$
    – S. carmeli
    Mar 13, 2018 at 21:28
  • $\begingroup$ Ok, perfect, I thought it was this the problem. A last question: why can't I directly use the map from the double stupid truncation to $X$? $\endgroup$ Mar 13, 2018 at 21:33
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    $\begingroup$ The row itself don't map to the double complex, the vertical differential don't match. Actually, you have a map from the complex obtained by deleting lower rows to $X$, and then a map from this smaller complex to the p-th row, so we actually have a "roof", not a map in any direction between them. $\endgroup$
    – S. carmeli
    Mar 13, 2018 at 21:39

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