Lemma 20.36.1. In the situation described above. Denote $\mathcal{H}^ m = H^ m(\mathcal{F}^\bullet )$ the $m$th cohomology sheaf. Let $\mathcal{B}$ be a set of open subsets of $X$. Let $d \in \mathbf{N}$. Assume

1. every open in $X$ has a covering whose members are elements of $\mathcal{B}$,

2. for every $U \in \mathcal{B}$ we have $H^ p(U, \mathcal{H}^ q) = 0$ for $p > d$ and $q < 0$1.

Then (20.36.0.1) is a quasi-isomorphism.

Proof. By Derived Categories, Lemma 13.34.4 it suffices to show that the canonical map $\mathcal{F}^\bullet \to R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n} \mathcal{F}^\bullet$ is an isomorphism. This is Lemma 20.35.9. $\square$

 It suffices if $\forall m$, $\exists p(m)$, $H^ p(U. \mathcal{H}^{m - p}) = 0$ for $p > p(m)$, see Lemma 20.35.8.

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