Lemma 26.21.14. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of schemes. 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 26.21.11 because it is a section of the quasi-separated morphism $X \times _ Z Y \to X$ (a base change of $g$, see Lemma 26.21.12). The second map is quasi-compact as it is the base change of $g \circ f$, see Lemma 26.19.3. And compositions of quasi-compact morphisms are quasi-compact, see Lemma 26.19.4. $\square$

Comment #2800 by Kat Christianson on

In the second-to-last sentence, I think the map in question is a base change of $g \circ f$, not of $f$.

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