TITLE: Question about a closed mapping.
QUESTION [5 upvotes]: We just learned about quotient mappings and various properties of the quotient topology. I'm curious about metrizability under these mappings. Namely, if $f: X \rightarrow Y$ is a closed continuous surjection and $X$ is metrizable, does it follow that $Y$ is metrizable?

REPLY [2 votes]: I think my answer to this question provides a counterexample.  Basically, the space $Y$ described there fails to be first countable.
Fact: $*$ does not have a countable base in $Y$.
proof:  If $\{ U_i : i \in \mathbb{N} \}$ is a family of open neighbourhoods of $*$ in $Y$, then without loss of generality we may assume that each is of the form:
$$ U_i = \bigcup_{n \in \mathbb{N}} ( (n - \varepsilon_{i,n} , n + \varepsilon_{i,n} ) \setminus \{ n \} ) \cup \{ * \}.$$
We may also assume that $\varepsilon_{i,n} \leq \frac{1}{2}$ for all $i,n$.
For each $i \in \mathbb{N}$ define $\delta_i = \min \{ \frac{\epsilon_{i,n}}{2} : n \leq i \}$.  Now define $$V = \bigcup_{n \in \mathbb{N}} ( ( n - \delta_n , n + \delta_n ) \setminus \{ n \} ) \cup \{ * \}.$$
Given $i \in \mathbb{N}$, since $\delta_i < \varepsilon_{i,i}$, it follows that $U_i \not\subseteq V$.  $\Box$