Theorem 63.10.5. 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 $S$. Then the quotient $U/R$ is an algebraic space, and $U \to U/R$ is étale and surjective, in other words $(U, R, U \to U/R)$ is a presentation of $U/R$.

**Proof.**
By Lemma 63.10.3 it suffices to prove that $U/R$ is an algebraic space. Let $U' \to U$ be a surjective, étale morphism. Then $\{ U' \to U\} $ is in particular an fppf covering. Let $R'$ be the restriction of $R$ to $U'$, see Groupoids, Definition 39.3.3. According to Groupoids, Lemma 39.20.6 we see that $U/R \cong U'/R'$. By Lemma 63.10.1 $R'$ is an étale equivalence relation on $U'$. Thus we may replace $U$ by $U'$.

We apply the previous remark to $U' = \coprod U_ i$, where $U = \bigcup U_ i$ is an affine open covering of $S$. Hence we may and do assume that $U = \coprod U_ i$ where each $U_ i$ is an affine scheme.

Consider the restriction $R_ i$ of $R$ to $U_ i$. By Lemma 63.10.1 this is an étale equivalence relation. Set $F_ i = U_ i/R_ i$ and $F = U/R$. It is clear that $\coprod F_ i \to F$ is surjective. By Lemma 63.10.2 each $F_ i \to F$ is representable, and an open immersion. By Lemma 63.10.4 applied to $(U_ i, R_ i)$ we see that $F_ i$ is an algebraic space. Then by Lemma 63.10.3 we see that $U_ i \to F_ i$ is étale and surjective. From Lemma 63.8.4 it follows that $\coprod F_ i$ is an algebraic space. Finally, we have verified all hypotheses of Lemma 63.8.5 and it follows that $F = U/R$ is an algebraic space. $\square$

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## Comments (1)

Comment #216 by David Holmes on

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