Lemma 87.16.5. Let $S$ be a scheme. Let $X$ be a formal algebraic space over $S$. Let $U \to X$ be a morphism where $U$ is a separated algebraic space over $S$. Then $U \to X$ is separated.
Proof. The statement makes sense because $U \to X$ is representable by algebraic spaces (Lemma 87.11.3). Let $T$ be a scheme and $T \to X$ a morphism. We have to show that $U \times _ X T \to T$ is separated. Since $U \times _ X T \to U \times _ S T$ is a monomorphism, it suffices to show that $U \times _ S T \to T$ is separated. As this is the base change of $U \to S$ this follows. We used in the argument above: Morphisms of Spaces, Lemmas 67.4.4, 67.4.8, 67.10.3, and 67.4.11. $\square$
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