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

$\mathcal{F} \to Ra_*(a^*\mathcal{F})$

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

\begin{align*} Ra_*(a^*\mathcal{F}) & = Ra_*Rh_*h^*a^*\mathcal{F} \\ & = Rh_{-1, *}Ra_{fppf, *}a_{fppf}^*(h_{-1})^*\mathcal{F} \\ & = Rh_{-1, *}(h_{-1})^*\mathcal{F} \\ & = \mathcal{F} \end{align*}

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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