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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