Lemma 66.45.12. 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 finite and $g$ separated then $f$ is finite.

If $g \circ f$ is integral and $g$ separated then $f$ is integral.

Lemma 66.45.12. 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 finite and $g$ separated then $f$ is finite.

If $g \circ f$ is integral and $g$ separated then $f$ is integral.

**Proof.**
Assume $g \circ f$ is finite (resp. integral) and $g$ separated. The base change $X \times _ Z Y \to Y$ is finite (resp. integral) by Lemma 66.45.5. The morphism $X \to X \times _ Z Y$ is a closed immersion as $Y \to Z$ is separated, see Lemma 66.4.7. A closed immersion is finite (resp. integral), see Lemma 66.45.10. The composition of finite (resp. integral) morphisms is finite (resp. integral), see Lemma 66.45.4. Thus we win.
$\square$

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