Lemma 101.27.16. Let f : \mathcal{X} \to \mathcal{Y} be a morphism of algebraic stacks. Let \mathcal{Z} \to \mathcal{Y} be a surjective, flat, locally finitely presented morphism of algebraic stacks. If the base change \mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z} is quasi-compact, then f is quasi-compact.
Proof. We have to show that given \mathcal{Y}' \to \mathcal{Y} with \mathcal{Y}' quasi-compact, we have \mathcal{Y}' \times _\mathcal {Y} \mathcal{X} is quasi-compact. Denote \mathcal{Z}' = \mathcal{Z} \times _\mathcal {Y} \mathcal{Y}'. Then |\mathcal{Z}'| \to |\mathcal{Y}'| is open, see Lemma 101.27.15. Hence we can find a quasi-compact open substack \mathcal{W} \subset \mathcal{Z}' mapping onto \mathcal{Y}'. Because \mathcal{Z} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Z} is quasi-compact, we know that
\mathcal{W} \times _\mathcal {Z} \mathcal{Z} \times _\mathcal {Y} \mathcal{X} = \mathcal{W} \times _\mathcal {Y} \mathcal{X}
is quasi-compact. And the map \mathcal{W} \times _\mathcal {Y} \mathcal{X} \to \mathcal{Y}' \times _\mathcal {Y} \mathcal{X} is surjective, hence we win. Some details omitted. \square
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