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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)