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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