Theorem 65.10.5. 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 65.10.3 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 65.10.1 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 U. 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 65.10.1 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 65.10.2 each F_ i \to F is representable, and an open immersion. By Lemma 65.10.4 applied to (U_ i, R_ i) we see that F_ i is an algebraic space. Then by Lemma 65.10.3 we see that U_ i \to F_ i is étale and surjective. From Lemma 65.8.4 it follows that \coprod F_ i is an algebraic space. Finally, we have verified all hypotheses of Lemma 65.8.5 and it follows that F = U/R is an algebraic space. \square
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Comment #216 by David Holmes on
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