Lemma 67.8.3. Let S be a scheme. If f : X \to Y is a quasi-compact morphism of algebraic spaces over S, then the underlying map |f| : |X| \to |Y| of topological space is quasi-compact.
Proof. Let V \subset |Y| be quasi-compact open. By Properties of Spaces, Lemma 66.4.8 there is an open subspace Y' \subset Y with V = |Y'|. Then Y' is a quasi-compact algebraic space by Properties of Spaces, Lemma 66.5.2 and hence X' = Y' \times _ Y X is a quasi-compact algebraic space by Definition 67.8.2. On the other hand, X' \subset X is an open subspace (Spaces, Lemma 65.12.3) and |X'| = |f|^{-1}(|X'|) = |f|^{-1}(V) by Properties of Spaces, Lemma 66.4.3. We conclude using Properties of Spaces, Lemma 66.5.2 again that |X'| is a quasi-compact open of |X| as desired. \square
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