Lemma 64.10.3. Let $S$ be a scheme. Let $U$ be a scheme over $S$. Let $j = (s, t) : R \to U \times _ S U$ be an étale equivalence relation on $U$ over $S$. If the quotient $U/R$ is an algebraic space, then $U \to U/R$ is étale and surjective. Hence $(U, R, U \to U/R)$ is a presentation of the algebraic space $U/R$.

Proof. Denote $c : U \to U/R$ the morphism in question. Let $T$ be a scheme and let $a : T \to U/R$ be a morphism. We have to show that the morphism (of schemes) $\pi : T \times _{a, U/R, c} U \to T$ is étale and surjective. The morphism $a$ corresponds to an fppf covering $\{ \varphi _ i : T_ i \to T\}$ and morphisms $a_ i : T_ i \to U$ such that $a_ i \times a_{i'} : T_ i \times _ T T_{i'} \to U \times _ S U$ factors through $R$, and such that $c \circ a_ i = a \circ \varphi _ i$. Hence

$T_ i \times _{\varphi _ i, T} T \times _{a, U/R, c} U = T_ i \times _{c \circ a_ i, U/R, c} U = T_ i \times _{a_ i, U} U \times _{c, U/R, c} U = T_ i \times _{a_ i, U, t} R.$

Since $t$ is étale and surjective we conclude that the base change of $\pi$ to $T_ i$ is surjective and étale. Since the property of being surjective and étale is local on the base in the fpqc topology (see Remark 64.4.3) we win. $\square$

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