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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