Lemma 20.11.2. Let $X$ be a ringed space. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be an open covering. There is a transformation
of functors $\textit{Mod}(\mathcal{O}_ X) \to D^{+}(\mathcal{O}_ X(U))$. In particular this provides canonical maps $\check{H}^ p(\mathcal{U}, \mathcal{F}) \to H^ p(U, \mathcal{F})$ for $\mathcal{F}$ ranging over $\textit{Mod}(\mathcal{O}_ X)$.
Proof. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. Choose an injective resolution $\mathcal{F} \to \mathcal{I}^\bullet $. Consider the double complex $\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{I}^\bullet )$ with terms $\check{\mathcal{C}}^ p(\mathcal{U}, \mathcal{I}^ q)$. There is a map of complexes
coming from the maps $\mathcal{I}^ q(U) \to \check{H}^0(\mathcal{U}, \mathcal{I}^ q)$ and a map of complexes
coming from the map $\mathcal{F} \to \mathcal{I}^0$. We can apply Homology, Lemma 12.25.4 to see that $\alpha $ is a quasi-isomorphism. Namely, Lemma 20.11.1 implies that the $q$th row of the double complex $\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{I}^\bullet )$ is a resolution of $\Gamma (U, \mathcal{I}^ q)$. Hence $\alpha $ becomes invertible in $D^{+}(\mathcal{O}_ X(U))$ and the transformation of the lemma is the composition of $\beta $ followed by the inverse of $\alpha $. We omit the verification that this is functorial. $\square$
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