Lemma 67.17.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $U \subset X$ be an open subspace. The following are equivalent

1. for every étale morphism $\varphi : V \to X$ (of algebraic spaces) the scheme theoretic closure of $\varphi ^{-1}(U)$ in $V$ is equal to $V$,

2. there exists a scheme $V$ and a surjective étale morphism $\varphi : V \to X$ such that the scheme theoretic closure of $\varphi ^{-1}(U)$ in $V$ is equal to $V$,

Proof. Observe that if $V \to V'$ is a morphism of algebraic spaces étale over $X$, and $Z \subset V$, resp. $Z' \subset V'$ is the scheme theoretic closure of $U \times _ X V$, resp. $U \times _ X V'$ in $V$, resp. $V'$, then $Z$ maps into $Z'$. Thus if $V \to V'$ is surjective and étale then $Z = V$ implies $Z' = V'$. Next, note that an étale morphism is flat, locally of finite presentation, and locally quasi-finite (see Morphisms, Section 29.36). Thus Lemma 67.17.1 implies that if $V$ and $V'$ are schemes, then $Z' = V'$ implies $Z = V$. A formal argument using that every algebraic space has an étale covering by a scheme shows that (1) and (2) are equivalent. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).