The Stacks project

Lemma 85.34.2. 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

\[ a^* : \mathit{QCoh}(\mathcal{O}_ X) \to \mathit{QCoh}(\mathcal{O}_ U) \]

is an equivalence fully faithful with quasi-inverse given by $a_*$. 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 85.34.1. In the proof of this lemma we have seen that $h_{-1}$ is the morphism $a_ X$ of More on Cohomology of Spaces, Section 84.7. Thus it follows from More on Cohomology of Spaces, Lemma 84.7.1 that

\[ (h_{-1})^* : \mathit{QCoh}(\mathcal{O}_ X) \longrightarrow \mathit{QCoh}(\mathcal{O}_{big}) \]

is an equivalence with quasi-inverse $h_{-1, *}$. The same holds true for the components $h_ n$ of $h$. Recall that $\mathit{QCoh}(\mathcal{O}_ U)$ and $\mathit{QCoh}(\mathcal{O}_{big, total})$ consist of cartesian modules whose components are quasi-coherent, see Lemma 85.12.10. Since the functors $h^*$ and $h_*$ of Lemma 85.7.2 agree with the functors $h_ n^*$ and $h_{n, *}$ on components we conclude that

\[ h^* : \mathit{QCoh}(\mathcal{O}_ U) \longrightarrow \mathit{QCoh}(\mathcal{O}_{big, total}) \]

is an equivalence with quasi-inverse $h_*$. Observe that $U$ is a hypercovering of $X$ in $(\textit{Spaces}/S)_{fppf}$ as defined in Section 85.21. By Lemma 85.22.1 we see that $a_{fppf}^*$ is fully faithful with quasi-inverse $a_{fppf, *}$ and with essential image the cartesian sheaves of $\mathcal{O}_{fppf, total}$-modules. Thus, by the description of $\mathit{QCoh}(\mathcal{O}_{big})$ and $\mathit{QCoh}(\mathcal{O}_{big, total})$ of Lemma 85.12.10, we get an equivalence

\[ a_{fppf}^* : \mathit{QCoh}(\mathcal{O}_{big}) \longrightarrow \mathit{QCoh}(\mathcal{O}_{big, total}) \]

with quasi-inverse given by $a_{fppf, *}$. A formal argument (chasing around the diagram) now shows that $a^*$ is fully faithful on $\mathit{QCoh}(\mathcal{O}_ X)$ and has image contained in $\mathit{QCoh}(\mathcal{O}_ U)$.

Finally, suppose that $\mathcal{G}$ is in $\mathit{QCoh}(\mathcal{O}_ U)$. Then $h^*\mathcal{G}$ is in $\mathit{QCoh}(\mathcal{O}_{big, total})$. Hence $h^*\mathcal{G} = a_{fppf}^*\mathcal{H}$ with $\mathcal{H} = a_{fppf, *}h^*\mathcal{G}$ in $\mathit{QCoh}(\mathcal{O}_{big})$ (see above). In turn we see that $\mathcal{H} = (h_{-1})^*\mathcal{F}$ with $\mathcal{F} = h_{-1, *}\mathcal{H}$ in $\mathit{QCoh}(\mathcal{O}_ X)$. Going around the diagram we deduce that $h^*\mathcal{G} \cong h^*a^*\mathcal{F}$. By fully faithfulness of $h^*$ we conclude that $a^*\mathcal{F} \cong \mathcal{G}$. Since $\mathcal{F} = h_{-1, *}a_{fppf, *}h^*\mathcal{G} = a_*h_*h^*\mathcal{G} = a_*\mathcal{G}$ we also obtain the statement that the quasi-inverse is given by $a_*$. $\square$

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