# Do perfect maps preserve perfect normality?

A perfect map is a function $$f : X \to Y$$ between topological spaces, which is continuous, closed, surjective, and has compact fibers ($$f^{-1}(\{y\})$$ is compact for each $$y \in Y$$).

A space $$X$$ is perfectly normal, if for each closed $$V_1, V_2 \subset X$$ there exists continuous $$g : X \to \mathbb{R}$$ such that $$g^{-1}(\{i\}) = V_i$$. In particular, T1 or Hausdorff is not assumed.

I have shown that a perfect map preserves regularity, normality, and complete normality. I haven't had any luck with perfect normality. Does a perfect map preserve perfect normality?

Yes, I think. We only need closedness and continuity of $$f$$: $$Y$$ is normal (in the usual closed separation sense (or closed shrinkings), without $$T_1$$-ness assumed) by closedness of $$f$$; this is standard (let $$\{U,V\}$$ be an open cover of $$Y$$, then $$\{f^{-1}[U], f^{-1}[V]\}$$ is an open cover for $$X$$, so by normality has a closed shrinking $$\{F,G\}$$ with $$F \subseteq f^{-1}[U], G \subseteq f^{-1}[V]$$ and then closedness of the map gives us that $$\{f[F], f[G]\}$$ is a closed shrinking of $$\{U,V\}$$ as required).
Now if $$U$$ is open in $$Y$$, $$f^{-1}[U]$$ is a an $$F_\sigma$$ in $$X$$ (being open in a perfectly normal space) and so its image $$U$$ (by closedness and surjectivity of $$f$$) is also an $$F_\sigma$$. Now the usual Urysohn lemma proof applies to show $$Y$$ is perfectly normal in your sense as well.
• Nice! Not sure what you mean by the last sentence. I think Y is perfectly normal iff Y is normal and each open subset is $F_\sigma$, so we are done after showing the latter right? – kaba Sep 11 at 22:01