Proposition 36.9.5. In Situation 36.9.1. For every object E of D_\mathit{QCoh}(\mathcal{O}_ X) the map (36.9.4.1) is an isomorphism.
Proof. By Lemma 36.3.5 we may assume that E is given by a complex of quasi-coherent sheaves \mathcal{F}^\bullet . Let M^\bullet = \Gamma (X, \mathcal{F}^\bullet ) be the corresponding complex of A-modules. By Lemmas 36.9.3 and 36.9.4 we have quasi-isomorphisms
\mathop{\mathrm{colim}}\nolimits _ e \mathop{\mathrm{Hom}}\nolimits ^\bullet (I^\bullet (f_1^ e, \ldots , f_ r^ e), M^\bullet ) \longrightarrow \text{Tot}(\check{\mathcal{C}}_{alt}^\bullet (\mathcal{U}, \mathcal{F}^\bullet )) \longrightarrow R\Gamma (U, \mathcal{F}^\bullet )
By More on Algebra, Lemma 15.73.2 and Equation (15.73.0.2) taking H^0 of the complex \mathop{\mathrm{Hom}}\nolimits ^\bullet (I^\bullet (f_1^ e, \ldots , f_ r^ e), M^\bullet ) computes \mathop{\mathrm{Hom}}\nolimits in D(A). Thus taking H^0 on both sides we obtain
\mathop{\mathrm{colim}}\nolimits _ e \mathop{\mathrm{Hom}}\nolimits _{D(A)}(I^\bullet (f_1^ e, \ldots , f_ r^ e), M^\bullet ) = H^0(U, E)
Since \mathop{\mathrm{Hom}}\nolimits _{D(A)}(I^\bullet (f_1^ e, \ldots , f_ r^ e), M^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ X)}(I_ e, E) by Lemma 36.3.5 the lemma follows. \square
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