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 $X$.

## 29.22 Constructible sets

Constructible and locally constructible sets of schemes have been discussed in Properties, Section 28.2. In this section we prove some results concerning images and inverse images of (locally) constructible sets. The main result is Chevalley's theorem which states that the image of a locally constructible set under a morphism of finite presentation is locally constructible.

**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.
$\square$

Lemma 29.22.2. Let $f : X \to Y$ be a morphism of schemes. Assume

$f$ is quasi-compact and locally of finite presentation, and

$Y$ is quasi-compact and quasi-separated.

Then the image of every constructible subset of $X$ is constructible in $Y$.

**Proof.**
By Properties, Lemma 28.2.5 it suffices to prove this lemma in case $Y$ is affine. In this case $X$ is quasi-compact. Hence we can write $X = U_1 \cup \ldots \cup U_ n$ with each $U_ i$ affine open in $X$. If $E \subset X$ is constructible, then each $E \cap U_ i$ is constructible too, see Topology, Lemma 5.15.4. Hence, since $f(E) = \bigcup f(E \cap U_ i)$ and since finite unions of constructible sets are constructible, this reduces us to the case where $X$ is affine. In this case the result is Algebra, Theorem 10.29.10.
$\square$

Theorem 29.22.3 (Chevalley's Theorem). Let $f : X \to Y$ be a morphism of schemes. Assume $f$ is quasi-compact and locally of finite presentation. Then the image of every locally constructible subset is locally constructible.

**Proof.**
Let $E \subset X$ be locally constructible. We have to show that $f(E)$ is locally constructible too. We will show that $f(E) \cap V$ is constructible for any affine open $V \subset Y$. Thus we reduce to the case where $Y$ is affine. In this case $X$ is quasi-compact. Hence we can write $X = U_1 \cup \ldots \cup U_ n$ with each $U_ i$ affine open in $X$. If $E \subset X$ is locally constructible, then each $E \cap U_ i$ is constructible, see Properties, Lemma 28.2.1. Hence, since $f(E) = \bigcup f(E \cap U_ i)$ and since finite unions of constructible sets are constructible, this reduces us to the case where $X$ is affine. In this case the result is Algebra, Theorem 10.29.10.
$\square$

Lemma 29.22.4. Let $X$ be a scheme. Let $x \in X$. Let $E \subset X$ be a locally constructible subset. If $\{ x' \mid x' \leadsto x\} \subset E$, then $E$ contains an open neighbourhood of $x$.

**Proof.**
Assume $\{ x' \mid x' \leadsto x\} \subset E$. We may assume $X$ is affine. In this case $E$ is constructible, see Properties, Lemma 28.2.1. In particular, also the complement $E^ c$ is constructible. By Algebra, Lemma 10.29.4 we can find a morphism of affine schemes $f : Y \to X$ such that $E^ c = f(Y)$. Let $Z \subset X$ be the scheme theoretic image of $f$. By Lemma 29.6.5 and the assumption $\{ x' \mid x' \leadsto x\} \subset E$ we see that $x \not\in Z$. Hence $X \setminus Z \subset E$ is an open neighbourhood of $x$ contained in $E$.
$\square$

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