Lemma 66.8.9. 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 66.4.7 because it is a section of the quasi-separated morphism $X \times _ Z Y \to X$ (a base change of $g$, see Lemma 66.4.4). The second map is quasi-compact as it is the base change of $f$, see Lemma 66.8.4. And compositions of quasi-compact morphisms are quasi-compact, see Lemma 66.8.5. $\square$
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Your email address will not be published.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.