Lemma 67.6.5. Let $S$ be a scheme. Let $X$, $Y$ be algebraic spaces over $S$. Let $X \to Y$ be a representable morphism. If $Y$ is decent (resp. reasonable), then so is $X$.
Proof. Translation of Lemma 67.5.3. $\square$
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Your email address will not be published.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.