Lemma 85.34.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $U$ be a simplicial algebraic space over $S$. Let $a : U \to X$ be an augmentation. If $a : U \to X$ is an fppf hypercovering of $X$, then for $\mathcal{F}$ a quasi-coherent $\mathcal{O}_ X$-module the map
is an isomorphism. Here $a : \mathop{\mathit{Sh}}\nolimits (U_{\acute{e}tale}) \to \mathop{\mathit{Sh}}\nolimits (X_{\acute{e}tale})$ is as in Section 85.32.
Proof. Consider the diagram of Lemma 85.33.1. Let $\mathcal{F}_ n = a_ n^*\mathcal{F}$ be the $n$th component of $a^*\mathcal{F}$. This is a quasi-coherent $\mathcal{O}_{U_ n}$-module. Then $\mathcal{F}_ n = Rh_{n, *}h_ n^*\mathcal{F}_ n$ by More on Cohomology of Spaces, Lemma 84.7.2. Hence $a^*\mathcal{F} = Rh_*h^*a^*\mathcal{F}$ by Lemma 85.7.3. We have
The first equality by the discussion above, the second equality because of the commutativity of the diagram in Lemma 85.25.1, the third equality by Lemma 85.22.2 as $U$ is a hypercovering of $X$ in $(\textit{Spaces}/S)_{fppf}$ and $La_{fppf}^* = a_{fppf}^*$ as $a_{fppf}$ is flat (namely $a_{fppf}^{-1}\mathcal{O}_{big} = \mathcal{O}_{big, total}$, see Remark 85.16.5), and the last equality by the already used More on Cohomology of Spaces, Lemma 84.7.2. $\square$
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