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.

  • $\begingroup$ 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? $\endgroup$ – kaba Sep 11 at 22:01
  • $\begingroup$ Oh right, you mean showing exactly that equivalence:) $\endgroup$ – kaba Sep 11 at 22:09
  • $\begingroup$ @kaba I'm using that equivalence. I think it still works in the absence of other separation axioms I mean. $\endgroup$ – Henno Brandsma Sep 11 at 22:10

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