The Stacks project

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

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

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

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 $a_*$. $\square$