# Closed surjection that does not preserve regularity

Def Map $p\colon X\rightarrow Y$ is perfect if it is a closed surjection and $p^{-1}\left(\left\{y\right\}\right)$ is compact for each $y\in Y$

It is well known that perfect maps preserve regularity, i.e. if $X$ is regular, so is $Y$. What is an example of failure of this property, if we drop the compactness assumption?

Some thoughts: I believe it's enough if fibers are paracompact, instead of compact (proof goes similarly to the one I know for compact case, unless I've overlooked something silly). Fibers of closed maps from $T_1$ spaces are necessarily closed, so this rules out maps from paracompact spaces (e.g. all metric spaces, by Stone theorem), since paracompactness is weakly hereditary.

Let $X$ be a (completely) regular space which is not normal, and let $A,B \subseteq X$ be disjoint closed sets which cannot be separated by disjoint open sets. Let $Y$ be the space obtained by collapsing $A$ to a single point $a$, and $B$ to a single point $b$.
Then the quotient mapping $\pi : X \to Y$ is a continuous closed surjection, but $Y$ is not Hausdorff, let alone regular.
I think that if you modify the construction of the deleted Tychonoff plank to begin with the product of the one-point compactifications of a countably infinite discrete space $X_0$ and an uncountable discrete space $X_1$, and then remove the point $(\infty,\infty)$ (so $X = ( X_0 \cup \{ \infty \} ) \times ( X_1 \cup \{ \infty \} ) \setminus \{ (\infty,\infty) \}$) ) you obtain a completely regular non-normal space. Then $A = X_0 \times \{ \infty \}$ and $B = \{ \infty \} \times X_1$ are disjoint closed sets which cannot be separated by disjoint open sets, and if you construct $Y$ as above, then the quotient mapping $\pi : X \to Y$ has paracompact fibres. (The only non-singleton fibres are $\pi^{-1} ( a ) = A$ and $\pi^{-1} (b) = B$ which are discrete subspaces of $X$.)