Lemma 35.8.8. Let $S$ be a scheme. Let $\tau \in \{ Zariski, \linebreak[0] {\acute{e}tale}, \linebreak[0] smooth, \linebreak[0] syntomic, \linebreak[0] fppf\} $. Each of the functors $\mathcal{F} \mapsto \mathcal{F}^ a$ of Definition 35.8.2

\[ \mathit{QCoh}(\mathcal{O}_ S) \to \mathit{QCoh}((\mathit{Sch}/S)_\tau , \mathcal{O}) \quad \text{or}\quad \mathit{QCoh}(\mathcal{O}_ S) \to \mathit{QCoh}(S_\tau , \mathcal{O}) \]

is fully faithful.

**Proof.**
(By Lemma 35.8.7 we do indeed get functors as indicated.) We may and do identify $\mathcal{O}_ S$-modules on $S$ with modules on $(S_{Zar}, \mathcal{O}_ S)$. The functor $\mathcal{F} \mapsto \mathcal{F}^ a$ on quasi-coherent modules $\mathcal{F}$ is given by pullback by a morphism $f$ of ringed sites, see Remark 35.8.6. In each case the functor $f_*$ is given by restriction along the inclusion functor $S_{Zar} \to S_\tau $ or $S_{Zar} \to (\mathit{Sch}/S)_\tau $ (see discussion of how these morphisms of sites are defined in Topologies, Section 34.11). Combining this with the description of $f^*\mathcal{F} = \mathcal{F}^ a$ we see that $f_*f^*\mathcal{F} = \mathcal{F}$ provided that $\mathcal{F}$ is quasi-coherent. Then we see that

\[ \mathop{\mathrm{Hom}}\nolimits _\mathcal {O}(\mathcal{F}^ a, \mathcal{G}^ a) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {O}(f^*\mathcal{F}, f^*\mathcal{G}) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ S}(\mathcal{F}, f_*f^*\mathcal{G}) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ S}(\mathcal{F}, \mathcal{G}) \]

as desired.
$\square$

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