Lemma 29.22.1. Let f : X \to Y be a morphism of schemes. Let E \subset Y be a subset. If E is (locally) constructible in Y, then f^{-1}(E) is (locally) constructible in X.
Proof. To show that the inverse image of every constructible subset is constructible it suffices to show that the inverse image of every retrocompact open V of Y is retrocompact in X, see Topology, Lemma 5.15.3. The significance of V being retrocompact in Y is just that the open immersion V \to Y is quasi-compact. Hence the base change f^{-1}(V) = X \times _ Y V \to X is quasi-compact too, see Schemes, Lemma 26.19.3. Hence we see f^{-1}(V) is retrocompact in X. Suppose E is locally constructible in Y. Choose x \in X. Choose an affine neighbourhood V of f(x) and an affine neighbourhood U \subset X of x such that f(U) \subset V. Thus we think of f|_ U : U \to V as a morphism into V. By Properties, Lemma 28.2.1 we see that E \cap V is constructible in V. By the constructible case we see that (f|_ U)^{-1}(E \cap V) is constructible in U. Since (f|_ U)^{-1}(E \cap V) = f^{-1}(E) \cap U we win. \square
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