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.