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)
Comment #8544 by ZL on
Comment #9128 by Stacks project on