The Stacks project

Lemma 65.8.9. Let $S$ be a scheme. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of algebraic spaces over $S$. If $g \circ f$ is quasi-compact and $g$ is quasi-separated then $f$ is quasi-compact.

Proof. This is true because $f$ equals the composition $(1, f) : X \to X \times _ Z Y \to Y$. The first map is quasi-compact by Lemma 65.4.7 because it is a section of the quasi-separated morphism $X \times _ Z Y \to X$ (a base change of $g$, see Lemma 65.4.4). The second map is quasi-compact as it is the base change of $f$, see Lemma 65.8.4. And compositions of quasi-compact morphisms are quasi-compact, see Lemma 65.8.5. $\square$


Comments (0)


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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 03KS. Beware of the difference between the letter 'O' and the digit '0'.