Lemma 66.21.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume that $f$ is quasi-separated and locally of finite type and $Y$ quasi-separated. Let $y \in |Y|$ be a point of codimension $0$ on $Y$. The following are equivalent:

1. the set $f^{-1}(\{ y\} )$ is finite,

2. the space $|X_ k|$ is finite where $\mathop{\mathrm{Spec}}(k) \to Y$ represents $y$,

3. there exist open subspaces $X' \subset X$ and $Y' \subset Y$ with $f(X') \subset Y'$, $y \in |Y'|$, and $f^{-1}(\{ y\} ) \subset |X'|$ such that $f|_{X'} : X' \to Y'$ is finite.

Proof. Since quasi-separated algebraic spaces are decent, the equivalence of (1) and (2) follows from Lemma 66.18.10. To prove that (1) and (2) imply (3) we may and do replace $Y$ by a quasi-compact open containing $y$. Since $f^{-1}(\{ y\} )$ is finite, we can find a quasi-compact open subspace of $X' \subset X$ containing the fibre. The restriction $f|_{X'} : X' \to Y$ is quasi-compact and quasi-separated by Morphisms of Spaces, Lemma 65.8.10 (this is where we use that $Y$ is quasi-separated). Applying Lemma 66.21.1 to $f|_{X'} : X' \to Y$ we see that (3) holds. We omit the proof that (3) implies (2). $\square$

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