Lemma 84.35.1. Let $X$ be an algebraic space over a scheme $S$. Let $K, E \in D_\mathit{QCoh}(\mathcal{O}_ X)$. Let $a : U \to X$ be an fppf hypercovering. Assume that for all $n \geq 0$ we have

$\mathop{\mathrm{Ext}}\nolimits _{\mathcal{O}_{U_ n}}^ i(La_ n^*K, La_ n^*E) = 0 \text{ for } i < 0$

Then we have

1. $\mathop{\mathrm{Ext}}\nolimits _{\mathcal{O}_ X}^ i(K, E) = 0$ for $i < 0$, and

2. there is an exact sequence

$0 \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(K, E) \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{U_0}}(La_0^*K, La_0^*E) \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{U_1}}(La_1^*K, La_1^*E)$

Proof. Write $K_ n = La_ n^*K$ and $E_ n = La_ n^*E$. Then these are the simplicial systems of the derived category of modules (Definition 84.14.1) associated to $La^*K$ and $La^*E$ (Lemma 84.14.2) where $a : U_{\acute{e}tale}\to X_{\acute{e}tale}$ is as in Section 84.32. Let us prove (2) first. By Lemma 84.34.4 we have

$\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(K, E) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ U}(La^*K, La^*E)$

Thus the sequence looks like this:

$0 \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ U}(La^*K, La^*E) \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{U_0}}(K_0, E_0) \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{U_1}}(K_1, E_1)$

The first arrow is injective by Lemma 84.14.5. The image of this arrow is the kernel of the second by Lemma 84.14.6. This finishes the proof of (2). Part (1) follows by applying part (2) with $K[i]$ and $E$ for $i > 0$. $\square$

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