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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