Lemma 66.27.10. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. If $f$ is locally of finite type and a monomorphism, then $f$ is separated and locally quasi-finite.

Proof. A monomorphism is separated, see Lemma 66.10.3. By Lemma 66.27.6 it suffices to prove the lemma after performing a base change by $Z \to Y$ with $Z$ affine. Hence we may assume that $Y$ is an affine scheme. Choose an affine scheme $U$ and an étale morphism $U \to X$. Since $X \to Y$ is locally of finite type the morphism of affine schemes $U \to Y$ is of finite type. Since $X \to Y$ is a monomorphism we have $U \times _ X U = U \times _ Y U$. In particular the maps $U \times _ Y U \to U$ are étale. Let $y \in Y$. Then either $U_ y$ is empty, or $\mathop{\mathrm{Spec}}(\kappa (u)) \times _{\mathop{\mathrm{Spec}}(\kappa (y))} U_ y$ is isomorphic to the fibre of $U \times _ Y U \to U$ over $u$ for some $u \in U$ lying over $y$. This implies that the fibres of $U \to Y$ are finite discrete sets (as $U \times _ Y U \to U$ is an étale morphism of affine schemes, see Morphisms, Lemma 29.36.7). Hence $U \to Y$ is quasi-finite, see Morphisms, Lemma 29.20.6. As $U \to X$ was an arbitrary étale morphism with $U$ affine this implies that $X \to Y$ is locally quasi-finite. $\square$

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