Let $p: X \to Y$ be a perfect map. Show that if $X$ is regular, $Y$ is regular.

Question

I'm stuck here:

Let $p: X \rightarrow Y$ be a perfect map (closed, surjective and continuous map) such that $p^{-1}(\{y\})$ is compact for each $y \in Y$.

Show that if $X$ is regular, then $Y$ is regular.

I've been struggling for several hours , but I always get nothing.

EDIT: My try. Take a $y\in Y$ and a closed space $C$ in $Y$, such that $y$ is not in $C$. I want to show that I can put $y$ and $C$ in open disjoint sets of $Y$. I consider $p^{-1}(\{y\})$ and $p^{-1}(C)$. $p^{-1}(\{y\})$ is compact, and $p^{-1}(C)$ is closed.

Answer 1

Let $y\in Y$ and $C$ be closed in $Y$ such that $y\notin C$.

Since $p$ is surjective so $p^{-1}(y)\neq\emptyset$. Since $p$ is continuous then $p^{-1}(C)$ is closed in $X$.

Also $p^{-1}(y)\cap p^{-1}(C)=\emptyset $.

Hence since $X$ is regular for each $z\in p^{-1}(y)$ we find disjoint open sets $U_z $ and $V_z$ such that $z\in U_z,p^{-1}(C)\subset V_z$ .

Then $\cup_{z\in p^{-1}(y)}\{U_z\}$ is an open cover of $p^{-1}(y)$ which is compact and hence there exists $z_1,z_2,\ldots z_n$ such that $U=\cup_{i=1}^n U_i=p^{-1}(y)$. Take those $z_1,z_2,\ldots z_n$ and consider $V_{z_1},V_{z_2},\ldots V_{z_n}$.

Consider $V=\cap_{i=1}^n V_{z_i}$

Then $p^{-1}(y)\subset U\implies y\in p(U)$ and $p^{-1}(C)\subset V\implies C\subset p(V)$ and $U\cap V=\emptyset$

EDIT: How do we obtain the open sets $W_1,W_2$ in $Y$ which separates $\{y\}$ and $C$?

Since $p^{-1}(y)\subset U$ so there exists an open set $W_1$ containing $y$ such that $p^{-1}(W_1)\subset U$,where we take $W_1=Y\setminus p(X\setminus U)$.

Similarly we obtain an open set $W_2$ containing $C$ such that $p^{-1}(W_2)\subset V$,where we take $W_2=Y\setminus p(X\setminus V)$.

Also $W_1\cap W_2 \subset U\cap V=\emptyset$