The Stacks project

Proof. This lemma collects the results shown above in a slightly different manner. First of all, by Lemma 35.10.4 we already know the output of the construction in (1), (2), or (3) ends up in $\mathit{QCoh}((\mathit{Sch}/S)_\tau , \mathcal{O})$. It remains to show in each case that the result is independent of the topology used. The key to this is that the equivalence $\mathit{QCoh}(\mathcal{O}_ S) \to \mathit{QCoh}((\mathit{Sch}/S)_\tau , \mathcal{O})$, $\mathcal{F} \mapsto \mathcal{F}^ a$ of Proposition 35.8.9 is given by the same formula independent of the choice of the topology $\tau \in \{ Zariski, {\acute{e}tale}, fppf\}$.

Proof of (1). Let $\mathcal{I} \to \mathit{QCoh}((\mathit{Sch}/S)_{fppf}, \mathcal{O})$, $i \mapsto \mathcal{F}_ i$ be a diagram. Write $\mathcal{F}_ i = \mathcal{G}_ i^ a$ so we get a diagram $\mathcal{I} \to \mathit{QCoh}(\mathcal{O}_ S)$. Then $\mathop{\mathrm{colim}}\nolimits \mathcal{F}_ i = (\mathop{\mathrm{colim}}\nolimits \mathcal{G}_ i)^ a$ in $\textit{Mod}((\mathit{Sch}/S)_\tau , \mathcal{O})$ for $\tau \in \{ Zariski, {\acute{e}tale}, fppf\}$ by Lemma 35.10.2. This proves (1).

Proof of (2). Write $\mathcal{F} = \mathcal{H}^ a$ and $\mathcal{G} = (\mathcal{I})^ a$ with $\mathcal{H}$ and $\mathcal{I}$ quasi-coherent on $S$. Then $\mathcal{F} \otimes _\mathcal {O} \mathcal{G} = (\mathcal{H} \otimes _\mathcal {O} \mathcal{I})^ a$ in $\textit{Mod}((\mathit{Sch}/S)_\tau , \mathcal{O})$ for $\tau \in \{ Zariski, {\acute{e}tale}, fppf\}$ by Lemma 35.10.2. This proves (2).

Proof of (3). Let $\mathcal{F}$ and $\mathcal{G}$ be in $\mathit{QCoh}((\mathit{Sch}/S)_{fppf}, \mathcal{O})$. Write $\mathcal{F} = \mathcal{H}^ a$ with $\mathcal{H}$ quasi-coherent on $S$. By Lemma 35.8.10 we have

This explains the parenthetical statement of part (3). Now, if these equivalent conditions hold, then $\mathcal{H}$ is finite locally free. The construction of $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{F}, \mathcal{G})$ in Modules on Sites, Section 18.27 depends only on $\mathcal{F}$ and $\mathcal{G}$ as presheaves of modules (only whether the output $\mathop{\mathcal{H}\! \mathit{om}}\nolimits$ is a sheaf depends on whether $\mathcal{F}$ and $\mathcal{G}$ are sheaves). $\square$