Chevalley's theorem says that if $f \colon X \to Y$ is a morphism of finite presentation of schemes and $C \subset X$ is constructible, then $f(C)$ is constructible. Here, constructible means $C$ is contained in the algebra of sets generated by retrocompact opens. I'm interested, for no particular reason, in whether this remains true under weaker assumptions:
- What if the morphism is merely finite type and not finite presentation?
- What if, instead of taking the usual definition of constructible, we just take finite unions of locally closed sets?
In other words, can one produce a counterexample to the following claim?
Let $f \colon X \to Y$ be a morphism of finite type and $C \subset X$ a finite union of locally closed sets. Then $f(C)$ is also a finite union of locally closed sets.
I expected this to be easy, and maybe it is, but I can't find any natural-seeming counterexample.