Is the image of a constructible set between Jacobson spaces constructible if the map takes closed points to closed points? Let $X$ and $Y$ be two spectral Jacobson spaces and let $f: X \to Y$ be a spectral morphism, i.e. $f$ is continuous and the inverse image of a quasi-compact open is quasi-compact. Suppose further that $f$ takes closed points of $X$ to closed points of $Y$. Under these hypotheses, is it true that the image $f(X)$ of a constructible set $C \subset X$ is also constructible in $Y?$ If not, what are some hypotheses one can put on $X,Y$ and $f$ which makes the statement true?
 A: This is certainly not true in general. Of course in the scheme-theoretic setting where $f$ is locally of finite presentation, this is Chevalley's theorem [Tag 054K], but this is far from a purely topological theorem.
Let's produce three examples of a similar flavour. Let $Y$ be a surface over a countable field, and note that $Y$ is Jacobson by the Nullstellensatz [Tag 0478]. Enumerate all curves $C_1, C_2, \ldots$ in $Y$, choose distinct closed points $y_1,y_2,\ldots$ in $Y$ such that $y_i \not \in \bigcup_{j < i} C_j$, and write $S = \{y_1,y_2,\ldots\}$. In particular, any curve $C_j$ can only contain finitely many of the $y_i$, namely a subset of $S_{\leq j} = \{y_1,\ldots,y_j\}$. Note that $S$ is not constructible: it is dense in $Y$, but does not contain a dense open [Tag 005K].
Example 1. Let $X = Y \setminus S$, with its natural inclusion $f \colon X \hookrightarrow Y$. Then $X$ is the limit $\lim_j Y \setminus S_{\leq j}$ of spectral spaces, hence $X$ and $f$ are spectral [Tag 02AZ]. If $X^{\operatorname{cl}}$ denotes the set of closed points, note that $X^{\operatorname{cl}} = f^{-1}(Y^{\operatorname{cl}})$. If $x \in X$ is any point, then $X^{\operatorname{cl}} \cap \overline{\{x\}}$ is dense in $\overline{\{x\}}$:

*

*If $x$ is closed there is nothing to prove;

*If $x$ is a generic point of a curve $C_i$, then this is true because $S \cap C_i$ is finite;

*If $x$ is the generic point of $X$, then the closure of $X^{\operatorname{cl}}$ contains all generic points of curves, hence the generic point $x$ as well.

Thus $X$ is Jacobson [Tag 005V], and we already saw that $f$ takes $X^{\operatorname{cl}}$ to $Y^{\operatorname{cl}}$. However, the image of $f$ is $Y \setminus S$, which is not constructible because $S$ is not constructible [Tag 005H].
Example 2. Let $X = S \cup \{\eta\} \subseteq Y$ with the subspace topology, where $\eta$ is the generic point of $Y$. Since the intersection of a curve with $X$ is a finite set of closed points by the construction of $S$, we see that the strict closed subsets of $X$ are exactly the finite sets of closed points; thus $X$ is homeomorphic to the Zariski topology on any countable one-dimensional irreducible Noetherian scheme. Thus $X$ is spectral and the inclusion $f \colon X \to Y$ is spectral, and $X$ is Jacobson and $f$ takes closed points to closed points. But the image of $f$ is not constructible, because it is dense but does not contain a dense open [Tag 005K].
Example 3. By combining the two examples above, we get a space $X_1 \amalg X_2$ mapping surjectively to $Y$ with all the desired properties (spectral, Jacobson, closed points map to closed points). So demanding surjectivity of $f$ does not help.
Remark. However, the constructible topology on a spectral space is profinite (compact, Hausdorff, and totally disconnected) [Tag 0901], and spectral maps are continuous for the constructible topology [Tag 0G1J]. Thus, the image of a closed set for the constructible topology (e.g. a constructible set) is closed for the constructible topology, i.e. its complement is a (typically infinite) union of constructible sets.
For instance, the set $S$ is open for the constructible topology since each $\{y_j\} \subseteq S$ is constructible, showing that the image of the map of Example 1 is closed in the constructible topology. Likewise, $Y \setminus (S \cup \{\eta\})$ is open for the constructible topology since it is the union of the constructible sets $C_i \setminus S_{\leq i}$ for $i \in \mathbf N$, showing that the image of the map of Example 2 is closed in the constructible topology. This is probably the best you can say in general, even with additional Jacobson hypotheses.
