I'm trying to prove that a perfect map preserves metrizability using the Nagata-Smirnov Theorem, but I got stuck, hope someone can help me solve this.
Nagata-Smirnov Metrization Theorem:
A topological space X is metrizable if and only if it is regular, Hausdorff and has a countably locally finite (i.e., σ-locally finite) basis.Definition 1:
A collection $A$ of subsets of a topological space $(X,\tau)$ is said to be locally finite if each point of $X$ has a neighborhood that intersects only finitely many elements of $A$.
Definition 2:
A collection $B$ of subsets of a topological space $(X,\tau)$ is said to be countably locally finite (or $\sigma$-locally finite) if $B$ can be written as a countable union of collections $B_n$, each of which is locally finite.
Definition 3:
$f$ is said to be a perfect function if and only if $f$ is continuous, closed, surjective and such that $f^{-1}(y)$ is compact relative to $ X$ for each $y\in{Y}$.
Problem: Let $f: X \rightarrow Y$ be a perfect function. If $X$ is metrizable then so is $Y$.
My attempt:
As $X$ is metrizable, then it is regular, Hausdorff and has a countably locally finite basis, lets say $B=(B_\alpha)_{\alpha\in{A}}$ . We already know that $f$ preserves regularity and Hausdorff, so we must show that $Y$ has a σ-locally finite basis.
Let $B'=(f({B_\alpha}^c)^c)_{\alpha\in{A}}$. We want to show $B'$ is a basis for Y:
Let $y\in{Y}$, and $U\in{Y}$ an open set. We need to find an element in $B'$ containing $y$ and contained in $U$. $f$ being perfect implies $f^{-1}(y)$ is compact, then there exist $B_1,...,B_k\in{B}$ such that $f^{-1}(y)\subset{\bigcup{B_i}}$. I say that there is a $j\in\{1,...,k\}$ such that $y\in{f(B_i^c)^c}$ (I got stuck here cause I thought I have proved it, but I didn't)
If we write $B = \bigcup _{n=1}^{\infty} B_n$ where $B_n$'s are locally finite subcollections of $B$ then we can write $$B' = \bigcup_{n=1}^{\infty}B'_n \text{ where }B'_n = \{ f(B_{\alpha}^c)^c : B_{\alpha} \in B_n \}$$ Now it remains to show that $B'_n$ is a locally finite collection.
$f^{-1}(y)$ is compact, then there exist $B_1,...,B_k\in{B}$ such that $f^{-1}(y)\subset{\bigcup{B_i}}$. Let $x_i\in{B_i}$ such that $x_i\in{f^{-1}(y)}$, then each $x_i$ has an open neighbourhood $V_i$ that intersects finitely many members of $Bn$.
Let $V = \bigcap_{i=1}^{k} f(V_i^c)^c$. Then $V$ is an open neighbourhood of $y$. (Again, I had a trouble proving this, cause I proceeded like this: If $y\not\in{\bigcap_{i=1}^{k} f(V_i^c)^c}$ then $y\not\in{f(V_j^c)^c}$ for some $j$, then $y\in{f(V_j^c)}$ $\Rightarrow$ $f^{-1}(y)\subset{f^{-1}(f(V_j^c))\subset V_j^c}$ $!$, but this is not right.)
Suppose $V\cap f(B_{\alpha}^c)^c \neq \emptyset$ for some $B_{\alpha}$ in $B_n$. This means that, $$f^{-1}(V\cap f(B_{\alpha}^c)^c)=f^{-1}(V)\cap f^{-1}(f(B_{\alpha}^c)^c) \neq \emptyset$$
Now, $$f^{-1}(V)=\bigcap_{i=1}^{m}f^{-1}(f(V_i^c)^c)=\bigcap_{i=1}^{m}f^{-1}(f(V_i^c))^c\subseteq V_i \text{ for all } i$$ (Mistake)
So, $V_i \cap B_{\alpha} \neq \emptyset $ for all $i$, meaning that there are only finitely many $B_{\alpha}$ for which $V\cap f(B_{\alpha}) \neq \emptyset$
I know, It is a mess, but I'm trying to fix it. Any suggestion is welcome.