Lemma 67.40.7. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $f : X \to Y$ be a morphism of algebraic spaces over $B$. If $X$ is universally closed over $B$ and $f$ is surjective then $Y$ is universally closed over $B$. In particular, if also $Y$ is separated and of finite type over $B$, then $Y$ is proper over $B$.
Proof. Assume $X$ is universally closed and $f$ surjective. Denote $p : X \to B$, $q : Y \to B$ the structure morphisms. Let $B' \to B$ be a morphism of algebraic spaces over $S$. The base change $f' : X_{B'} \to Y_{B'}$ is surjective (Lemma 67.5.5), and the base change $p' : X_{B'} \to B'$ is closed. If $T \subset Y_{B'}$ is closed, then $(f')^{-1}(T) \subset X_{B'}$ is closed, hence $p'((f')^{-1}(T)) = q'(T)$ is closed. So $q'$ is closed. $\square$
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Your email address will not be published.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)