Lemma 96.10.3. Let $S$ be a scheme. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. Assume $\mathcal{X}$, $\mathcal{Y}$ are representable by algebraic spaces $F$, $G$. Denote $f : F \to G$ the induced morphism of algebraic spaces, and $f_{small} : F_{\acute{e}tale}\to G_{\acute{e}tale}$ the corresponding morphism of ringed topoi. Then
\[ \xymatrix{ (\mathop{\mathit{Sh}}\nolimits (\mathcal{X}_{\acute{e}tale}), \mathcal{O}_\mathcal {X}) \ar[d]_{\pi _ F} \ar[rr]_ f & & (\mathop{\mathit{Sh}}\nolimits (\mathcal{Y}_{\acute{e}tale}), \mathcal{O}_\mathcal {Y}) \ar[d]^{\pi _ G} \\ (\mathop{\mathit{Sh}}\nolimits (F_{\acute{e}tale}), \mathcal{O}_ F) \ar[rr]^{f_{small}} & & (\mathop{\mathit{Sh}}\nolimits (G_{\acute{e}tale}), \mathcal{O}_ G) } \]
is a commutative diagram of ringed topoi.
Proof.
This is similar to Topologies, Lemma 34.4.17 (3) but there is a small snag due to the fact that $F \to G$ may not be representable by schemes. In particular we don't get a commutative diagram of ringed sites, but only a commutative diagram of ringed topoi.
Before we start the proof proper, we choose equivalences $j : \mathcal{S}_ F \to \mathcal{X}$ and $j' : \mathcal{S}_ G \to \mathcal{Y}$ which induce functors $u : F_{\acute{e}tale}\to \mathcal{X}$ and $u' : G_{\acute{e}tale}\to \mathcal{Y}$ as in the proof of Lemma 96.10.1. Because of the 2-functoriality of sheaves on categories fibred in groupoids over $\mathit{Sch}_{fppf}$ (see discussion in Section 96.3) we may assume that $\mathcal{X} = \mathcal{S}_ F$ and $\mathcal{Y} = \mathcal{S}_ G$ and that $f : \mathcal{S}_ F \to \mathcal{S}_ G$ is the functor associated to the morphism $f : F \to G$. Correspondingly we will omit $u$ and $u'$ from the notation, i.e., given an object $U \to F$ of $F_{\acute{e}tale}$ we denote $U/F$ the corresponding object of $\mathcal{X}$. Similarly for $G$.
Let $\mathcal{G}$ be a sheaf on $\mathcal{X}_{\acute{e}tale}$. To prove (2) we compute $\pi _{G, *}f_*\mathcal{G}$ and $f_{small, *}\pi _{F, *}\mathcal{G}$. To do this let $V \to G$ be an object of $G_{\acute{e}tale}$. Then
\[ \pi _{G, *}f_*\mathcal{G}(V) = f_*\mathcal{G}(V/G) = \Gamma \Big( (\mathit{Sch}/V)_{fppf} \times _{\mathcal{Y}} \mathcal{X}, \ \text{pr}^{-1}\mathcal{G}\Big) \]
see (96.5.0.1). The fibre product in the formula is
\[ (\mathit{Sch}/V)_{fppf} \times _{\mathcal{Y}} \mathcal{X} = (\mathit{Sch}/V)_{fppf} \times _{\mathcal{S}_ G} \mathcal{S}_ F = \mathcal{S}_{V \times _ G F} \]
i.e., it is the split category fibred in groupoids associated to the algebraic space $V \times _ G F$. And $\text{pr}^{-1}\mathcal{G}$ is a sheaf on $\mathcal{S}_{V \times _ G F}$ for the étale topology.
In particular, if $V \times _ G F$ is representable, i.e., if it is a scheme, then $\pi _{G, *}f_*\mathcal{G}(V) = \mathcal{G}(V \times _ G F/F)$ and also
\[ f_{small, *}\pi _{F, *}\mathcal{G}(V) = \pi _{F, *}\mathcal{G}(V \times _ G F) = \mathcal{G}(V \times _ G F/F) \]
which proves the desired equality in this special case.
In general, choose a scheme $U$ and a surjective étale morphism $U \to V \times _ G F$. Set $R = U \times _{V \times _ G F} U$. Then $U/V \times _ G F$ and $R/V \times _ G F$ are objects of the fibre product category above. Since $\text{pr}^{-1}\mathcal{G}$ is a sheaf for the étale topology on $\mathcal{S}_{V \times _ G F}$ the diagram
\[ \xymatrix{ \Gamma \Big( (\mathit{Sch}/V)_{fppf} \times _{\mathcal{Y}} \mathcal{X}, \ \text{pr}^{-1}\mathcal{G}\Big) \ar[r] & \text{pr}^{-1}\mathcal{G}(U/V \times _ G F) \ar@<1ex>[r] \ar@<-1ex>[r] & \text{pr}^{-1}\mathcal{G}(R/V \times _ G F) } \]
is an equalizer diagram. Note that $\text{pr}^{-1}\mathcal{G}(U/V \times _ G F) = \mathcal{G}(U/F)$ and $\text{pr}^{-1}\mathcal{G}(R/V \times _ G F) = \mathcal{G}(R/F)$ by the definition of pullbacks. Moreover, by the material in Properties of Spaces, Section 66.18 (especially, Properties of Spaces, Remark 66.18.4 and Lemma 66.18.8) we see that there is an equalizer diagram
\[ \xymatrix{ f_{small, *}\pi _{F, *}\mathcal{G}(V) \ar[r] & \pi _{F, *}\mathcal{G}(U/F) \ar@<1ex>[r] \ar@<-1ex>[r] & \pi _{F, *}\mathcal{G}(R/F) } \]
Since we also have $\pi _{F, *}\mathcal{G}(U/F) = \mathcal{G}(U/F)$ and $\pi _{F, *}\mathcal{G}(U/F) = \mathcal{G}(U/F)$ we obtain a canonical identification $f_{small, *}\pi _{F, *}\mathcal{G}(V) = \pi _{G, *}f_*\mathcal{G}(V)$. We omit the proof that this is compatible with restriction mappings and that it is functorial in $\mathcal{G}$.
$\square$
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