Lemma 64.16.1. Suppose given a big site $\mathit{Sch}_{fppf}$. Let $g : S \to S'$ be morphism of $\mathit{Sch}_{fppf}$. Let $j : (\mathit{Sch}/S)_{fppf} \to (\mathit{Sch}/S')_{fppf}$ be the corresponding localization functor. Let $F$ be a sheaf of sets on $(\mathit{Sch}/S)_{fppf}$. Then

for a scheme $T'$ over $S'$ we have $j_!F(T'/S') = \coprod \nolimits _{\varphi : T' \to S} F(T' \xrightarrow {\varphi } S),$

if $F$ is representable by a scheme $X \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S)_{fppf})$, then $j_!F$ is representable by $j(X)$ which is $X$ viewed as a scheme over $S'$, and

if $F$ is an algebraic space over $S$, then $j_!F$ is an algebraic space over $S'$, and if $F = U/R$ is a presentation, then $j_!F = j(U)/j(R)$ is a presentation.

Let $F'$ be a sheaf of sets on $(\mathit{Sch}/S')_{fppf}$. Then

for a scheme $T$ over $S$ we have $j^{-1}F'(T/S) = F'(T/S')$,

if $F'$ is representable by a scheme $X' \in \mathop{\mathrm{Ob}}\nolimits ((\mathit{Sch}/S')_{fppf})$, then $j^{-1}F'$ is representable, namely by $X'_ S = S \times _{S'} X'$, and

if $F'$ is an algebraic space, then $j^{-1}F'$ is an algebraic space, and if $F' = U'/R'$ is a presentation, then $j^{-1}F' = U'_ S/R'_ S$ is a presentation.

**Proof.**
The functors $j_!$, $j_*$ and $j^{-1}$ are defined in Sites, Lemma 7.25.8 where it is also shown that $j = j_{S/S'}$ is the localization of $(\mathit{Sch}/S')_{fppf}$ at the object $S/S'$. Hence all of the material on localization functors is available for $j$. The formula in (1) is Sites, Lemma 7.27.1. By definition $j_!$ is the left adjoint to restriction $j^{-1}$, hence $j_!$ is right exact. By Sites, Lemma 7.25.5 it also commutes with fibre products and equalizers. By Sites, Lemma 7.25.3 we see that $j_!h_ X = h_{j(X)}$ hence (2) holds. If $F$ is an algebraic space over $S$, then we can write $F = U/R$ (Lemma 64.9.1) and we get

\[ j_!F = j(U)/j(R) \]

because $j_!$ being right exact commutes with coequalizers, and moreover $j(R) = j(U) \times _{j_!F} j(U)$ as $j_!$ commutes with fibre products. Since the morphisms $j(s), j(t) : j(R) \to j(U)$ are simply the morphisms $s, t : R \to U$ (but viewed as morphisms of schemes over $S'$), they are still étale. Thus $(j(U), j(R), s, t)$ is an étale equivalence relation. Hence by Theorem 64.10.5 we conclude that $j_!F$ is an algebraic space.

Proof of (4), (5), and (6). The description of $j^{-1}$ is in Sites, Section 7.25. The restriction of the representable sheaf associated to $X'/S'$ is the representable sheaf associated to $X'_ S = S \times _{S'} Y'$ by Sites, Lemma 7.27.2. The restriction functor $j^{-1}$ is exact, hence $j^{-1}F' = U'_ S/R'_ S$. Again by exactness the sheaf $R'_ S$ is still an equivalence relation on $U'_ S$. Finally the two maps $R'_ S \to U'_ S$ are étale as base changes of the étale morphisms $R' \to U'$. Hence $j^{-1}F' = U'_ S/R'_ S$ is an algebraic space by Theorem 64.10.5 and we win.
$\square$

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