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.
$\square$

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