Lemma 66.17.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. If $X$ is decent (resp. is reasonable, resp. has property $(\beta )$ of Lemma 66.5.1), then $f$ is decent (resp. reasonable, resp. has property $(\beta )$).
Proof. Let $T$ be a scheme and let $T \to Y$ be a morphism. Then $T \to Y$ is representable, hence the base change $T \times _ Y X \to X$ is representable. Hence if $X$ is decent (or reasonable), then so is $T \times _ Y X$, see Lemma 66.6.5. Similarly, for property $(\beta )$, see Lemma 66.5.3. $\square$
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