Lemma 85.35.1. 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 85.14.1) associated to $La^*K$ and $La^*E$ (Lemma 85.14.2) where $a : U_{\acute{e}tale}\to X_{\acute{e}tale}$ is as in Section 85.32. Let us prove (2) first. By Lemma 85.34.4 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 85.14.5. The image of this arrow is the kernel of the second by Lemma 85.14.6. This finishes the proof of (2). Part (1) follows by applying part (2) with $K[i]$ and $E$ for $i > 0$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).