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?
Question about a closed mapping.
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0@Nate: Okay; I thought there was something amiss, but couldn't figure out what. Thanks. – 2012-03-10
1 Answers
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 $U_i \not\subseteq V$. $\Box$
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0@Brian: Wow! Was I confused! Contained in my rambling was a "proof" of 0=1. Yes, the "non-metrizable hedgehog" space works quite nicely in both situations. Thanks. – 2012-03-10