Lemma 66.6.6. Let $S$ be a scheme. Let $X \to Y$ be an étale morphism of algebraic spaces over $S$. If $Y$ is decent, resp. reasonable, then so is $X$.
Proof. Let $U$ be an affine scheme and $U \to X$ an étale morphism. Set $R = U \times _ X U$ and $R' = U \times _ Y U$. Note that $R \to R'$ is a monomorphism.
Let $x \in |X|$. To show that $X$ is decent, we have to show that the fibres of $|U| \to |X|$ and $|R| \to |X|$ over $x$ are finite. But if $Y$ is decent, then the fibres of $|U| \to |Y|$ and $|R'| \to |Y|$ are finite. Hence the result for “decent”.
To show that $X$ is reasonable, we have to show that the fibres of $U \to X$ are universally bounded. However, if $Y$ is reasonable, then the fibres of $U \to Y$ are universally bounded, which immediately implies the same thing for the fibres of $U \to X$. Hence the result for “reasonable”. $\square$
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