Lemma 20.30.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}^\bullet$ be a bounded below complex of $\mathcal{O}_ X$-modules. There exists a quasi-isomorphism $\mathcal{F}^\bullet \to \mathcal{G}^\bullet$ where $\mathcal{G}^\bullet$ be a bounded below complex of flasque $\mathcal{O}_ X$-modules and for all $x \in X$ the map $\mathcal{F}^\bullet _ x \to \mathcal{G}^\bullet _ x$ is a homotopy equivalence in the category of complexes of $\mathcal{O}_{X, x}$-modules.

Proof. Let $\mathcal{A}$ be the category of complexes of $\mathcal{O}_ X$-modules and let $\mathcal{B}$ be the category of complexes of $\mathcal{O}_ X$-modules. Then we can apply the discussion above to the adjoint functors $f^*$ and $f_*$ between $\mathcal{A}$ and $\mathcal{B}$. Arguing exactly as in the proof of Lemma 20.30.1 we get a resolution

$0 \to \mathcal{F}^\bullet \to f_*f^*\mathcal{F}^\bullet \to f_*f^*f_*f^*\mathcal{F}^\bullet \to f_*f^*f_*f^*f_*f^*\mathcal{F}^\bullet \to \ldots$

in the abelian category $\mathcal{A}$ such that each term of each $f_*f^*\ldots f_*f^*\mathcal{F}^\bullet$ is a flasque $\mathcal{O}_ X$-module and such that for all $x \in X$ the map

$\mathcal{F}^\bullet _ x[0] \to \Big( (f_*f^*\mathcal{F}^\bullet )_ x \to (f_*f^*f_*f^*\mathcal{F}^\bullet )_ x \to (f_*f^*f_*f^*f_*f^*\mathcal{F}^\bullet )_ x \to \ldots \Big)$

is a homotopy equivalence in the category of complexes of complexes of $\mathcal{O}_{X, x}$-modules. Since a complex of complexes is the same thing as a double complex, we can consider the induced map

$\mathcal{F}^\bullet \to \mathcal{G}^\bullet = \text{Tot}( f_*f^*\mathcal{F}^\bullet \to f_*f^*f_*f^*\mathcal{F}^\bullet \to f_*f^*f_*f^*f_*f^*\mathcal{F}^\bullet \to \ldots )$

Since the complex $\mathcal{F}^\bullet$ is bounded below, the same is true for $\mathcal{G}^\bullet$ and in fact each term of $\mathcal{G}^\bullet$ is a finite direct sum of terms of the complexes $f_*f^*\ldots f_*f^*\mathcal{F}^\bullet$ and hence is flasque. The final assertion of the lemma now follows from Homology, Lemma 12.25.5. Since this in particular shows that $\mathcal{F}^\bullet \to \mathcal{G}^\bullet$ is a quasi-isomorphism, the proof is complete. $\square$

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