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. $\square$
Comments (0)
There are also: