Lemma 66.27.9. Let $S$ be a scheme. Let $f : X \to Y$ be a finite type morphism of algebraic spaces over $S$. Let $y \in |Y|$. There are at most finitely many points of $|X|$ lying over $y$ at which $f$ is quasi-finite.

Proof. Choose a field $k$ and a morphism $\mathop{\mathrm{Spec}}(k) \to Y$ in the equivalence class determined by $y$. The fibre $X_ k = \mathop{\mathrm{Spec}}(k) \times _ Y X$ is an algebraic space of finite type over a field, in particular quasi-compact. The map $|X_ k| \to |X|$ surjects onto the fibre of $|X| \to |Y|$ over $y$ (Properties of Spaces, Lemma 65.4.3). Moreover, the set of points where $X_ k \to \mathop{\mathrm{Spec}}(k)$ is quasi-finite maps onto the set of points lying over $y$ where $f$ is quasi-finite by Lemma 66.27.2. Choose an affine scheme $U$ and a surjective étale morphism $U \to X_ k$ (Properties of Spaces, Lemma 65.6.3). Then $U \to \mathop{\mathrm{Spec}}(k)$ is a morphism of finite type and there are at most a finite number of points where this morphism is quasi-finite, see Morphisms, Lemma 29.20.14. Since $X_ k \to \mathop{\mathrm{Spec}}(k)$ is quasi-finite at a point $x'$ if and only if it is the image of a point of $U$ where $U \to \mathop{\mathrm{Spec}}(k)$ is quasi-finite, we conclude. $\square$

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