Lemma 67.25.5. 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 surjective, then f(X_{\text{ft-pts}}) = Y_{\text{ft-pts}}.
Proof. We have f(X_{\text{ft-pts}}) \subset Y_{\text{ft-pts}} by Lemma 67.25.4. Let y \in |Y| be a finite type point. Represent y by a morphism \mathop{\mathrm{Spec}}(k) \to Y which is locally of finite type. As f is surjective the algebraic space X_ k = \mathop{\mathrm{Spec}}(k) \times _ Y X is nonempty, therefore has a finite type point x \in |X_ k| by Lemma 67.25.3. Now X_ k \to X is a morphism which is locally of finite type as a base change of \mathop{\mathrm{Spec}}(k) \to Y (Lemma 67.23.3). Hence the image of x in X is a finite type point by Lemma 67.25.4 which maps to y by construction. \square
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