Lemma 67.45.12. 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 67.45.5. The morphism $X \to X \times _ Z Y$ is a closed immersion as $Y \to Z$ is separated, see Lemma 67.4.7. A closed immersion is finite (resp. integral), see Lemma 67.45.10. The composition of finite (resp. integral) morphisms is finite (resp. integral), see Lemma 67.45.4. Thus we win. $\square$
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