Lemma 65.4.10. Let $S$ be a scheme. Let $X$ be an algebraic spaces over $S$. Let $U$ be a scheme and let $f : U \to X$ be an étale morphism. Let $X' \subset X$ be the open subspace corresponding to the open $|f|(|U|) \subset |X|$ via Lemma 65.4.8. Then $f$ factors through a surjective étale morphism $f' : U \to X'$. Moreover, if $R = U \times _ X U$, then $R = U \times _{X'} U$ and $X'$ has the presentation $X' = U/R$.

Proof. The existence of the factorization follows from Lemma 65.4.9. The morphism $f'$ is surjective according to Lemma 65.4.4. To see $f'$ is étale, suppose that $T \to X'$ is a morphism where $T$ is a scheme. Then $T \times _ X U = T \times _{X'} U$ as $X' \to X$ is a monomorphism of sheaves. Thus the projection $T \times _{X'} U \to T$ is étale as we assumed $f$ étale. We have $U \times _ X U = U \times _{X'} U$ as $X' \to X$ is a monomorphism. Then $X' = U/R$ follows from Spaces, Lemma 64.9.1. $\square$

Comment #2965 by Yu-Liang Huang on

A typo in the third row of the proof, $X"$ should be $X'$.

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