3
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.