Lemma 78.19.6. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $j : R \to U \times _ B U$ be a pre-equivalence relation over $B$ and $g : U' \to U$ a morphism of algebraic spaces over $B$. Let $j' : R' \to U' \times _ B U'$ be the restriction of $j$ to $U'$. The map of quotient sheaves

\[ U'/R' \longrightarrow U/R \]

is injective. If $U' \to U$ is surjective as a map of sheaves, for example if $\{ g : U' \to U\} $ is an fppf covering (see Topologies on Spaces, Definition 73.7.1), then $U'/R' \to U/R$ is an isomorphism of sheaves.

**Proof.**
Suppose $\xi , \xi ' \in (U'/R')(S')$ are sections which map to the same section of $U/R$. Then we can find an fppf covering $\mathcal{S} = \{ S_ i \to S'\} $ of $S'$ such that $\xi |_{S_ i}, \xi '|_{S_ i}$ are given by $a_ i, a_ i' \in U'(S_ i)$. By Lemma 78.19.5 and the axioms of a site we may after refining $\mathcal{T}$ assume there exist morphisms $r_ i : S_ i \to R$ such that $g \circ a_ i = s \circ r_ i$, $g \circ a_ i' = t \circ r_ i$. Since by construction $R' = R \times _{U \times _ S U} (U' \times _ S U')$ we see that $(r_ i, (a_ i, a_ i')) \in R'(S_ i)$ and this shows that $a_ i$ and $a_ i'$ define the same section of $U'/R'$ over $S_ i$. By the sheaf condition this implies $\xi = \xi '$.

If $U' \to U$ is a surjective map of sheaves, then $U'/R' \to U/R$ is surjective also. Finally, if $\{ g : U' \to U\} $ is a fppf covering, then the map of sheaves $U' \to U$ is surjective, see Topologies on Spaces, Lemma 73.7.5.
$\square$

## Comments (2)

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