Lemma 67.10.7 (042R)—The Stacks project

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Table of contents
Part 4: Algebraic Spaces
Chapter 67: Morphisms of Algebraic Spaces
Section 67.10: Monomorphisms
Lemma 67.10.7 (cite)

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Lemma 67.10.7. An immersion of algebraic spaces is a monomorphism. In particular, any immersion is separated.

Proof. Let $f : X \to Y$ be an immersion of algebraic spaces. For any morphism $Z \to Y$ with $Z$ representable the base change $Z \times _ Y X \to Z$ is an immersion of schemes, hence a monomorphism, see Schemes, Lemma 26.23.8. Hence $f$ is representable, and a monomorphism. $\square$

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