Lemma 29.23.6. Let $f : X \to Y$ be a morphism of schemes. Let $g : Y' \to Y$ be open and surjective such that the base change $f' : X' \to Y'$ is quasi-compact. Then $f$ is quasi-compact.

Proof. Let $V \subset Y$ be a quasi-compact open. As $g$ is open and surjective we can find a quasi-compact open $W' \subset W$ such that $g(W') = V$. By assumption $(f')^{-1}(W')$ is quasi-compact. The image of $(f')^{-1}(W')$ in $X$ is equal to $f^{-1}(V)$, see Lemma 29.9.3. Hence $f^{-1}(V)$ is quasi-compact as the image of a quasi-compact space, see Topology, Lemma 5.12.7. Thus $f$ is quasi-compact. $\square$

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