Lemma 67.27.9. 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 66.4.3). 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 67.27.2. Choose an affine scheme U and a surjective étale morphism U \to X_ k (Properties of Spaces, Lemma 66.6.3). 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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