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)



Double complexes with exact rows
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. 


