Lemma 100.7.10. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $W \to \mathcal{Y}$ be 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 quasi-compact, then $f$ is quasi-compact.

Proof. Assume $W \times _\mathcal {Y} \mathcal{X} \to W$ is quasi-compact. Let $\mathcal{Z} \to \mathcal{Y}$ be a morphism with $\mathcal{Z}$ a quasi-compact algebraic stack. Choose a scheme $U$ and a surjective smooth morphism $U \to W \times _\mathcal {Y} \mathcal{Z}$. Since $U \to \mathcal{Z}$ is flat, surjective, and locally of finite presentation and $\mathcal{Z}$ is quasi-compact, we can find a quasi-compact open subscheme $U' \subset U$ such that $U' \to \mathcal{Z}$ is surjective. Then $U' \times _\mathcal {Y} \mathcal{X} = U' \times _ W (W \times _\mathcal {Y} \mathcal{X})$ is quasi-compact by assumption and surjects onto $\mathcal{Z} \times _\mathcal {Y} \mathcal{X}$. Hence $\mathcal{Z} \times _\mathcal {Y} \mathcal{X}$ is quasi-compact as desired. $\square$

Comment #7435 by Anonymous on

Typo: In the second to last sentence of the proof, we should have $U' \times_{W} (W \times_{\mathcal{Y}} \mathcal{X})$ instead of $U' \times_{W} \times_{W} (W \times \mathcal{Y} \mathcal{X})$.

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