Lemma 101.17.6. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $W \to \mathcal{Y}$ be a surjective, flat, and locally of finite presentation where $W$ is an algebraic space. If the base change $W \times _\mathcal {Y} \mathcal{X} \to W$ is locally of finite type, then $f$ is locally of finite type.

Proof. Choose an algebraic space $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Choose an algebraic space $U$ and a surjective smooth morphism $U \to V \times _\mathcal {Y} \mathcal{X}$. We have to show that $U \to V$ is locally of finite presentation. Now we base change everything by $W \to \mathcal{Y}$: Set $U' = W \times _\mathcal {Y} U$, $V' = W \times _\mathcal {Y} V$, $\mathcal{X}' = W \times _\mathcal {Y} \mathcal{X}$, and $\mathcal{Y}' = W \times _\mathcal {Y} \mathcal{Y} = W$. Then it is still true that $U' \to V' \times _{\mathcal{Y}'} \mathcal{X}'$ is smooth by base change. Hence by our definition of locally finite type morphisms of algebraic stacks and the assumption that $\mathcal{X}' \to \mathcal{Y}'$ is locally of finite type, we see that $U' \to V'$ is locally of finite type. Then, since $V' \to V$ is surjective, flat, and locally of finite presentation as a base change of $W \to \mathcal{Y}$ we see that $U \to V$ is locally of finite type by Descent on Spaces, Lemma 74.11.9 and we win. $\square$

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