Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
Does this work as a counterexample? Take $X$ to be a union of two lines in $\mathbb{R}^2$ with the induced metric and topology; map it onto the line with the origin doubled, which is not normal and hence not metrizable. –  Arturo Magidin Mar 10 '12 at 5:04
    
@ArturoMagidin: I don't think this map is closed. If we write $X = \mathbb{R}^2 \times \{a,b\}$, then $[0,1] \times \{a\}$ is closed, but its image $[0,1] \times \{a\}$ is not closed in the quotient, since it fails to contain its limit point $(0,b)$. –  Nate Eldredge Mar 10 '12 at 5:39
    
@Nate: Okay; I thought there was something amiss, but couldn't figure out what. Thanks. –  Arturo Magidin Mar 10 '12 at 6:05

1 Answer 1

up vote 2 down vote accepted

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 $U_i \not\subseteq V$. $\Box$

share|improve this answer
    
Yes, nice example. Also as an addition: closed continuous maps do preserve paracompactness and if we add that all points have compact preimages (so we have a perfect map: closed continuous as well) then we do preserve metrisability and the example shows that even one non-compact preimages is already enough to spoil it. –  Henno Brandsma Mar 10 '12 at 6:35
    
You can simplify this a bit by letting $X=\omega\times[0,1]$ and taking $Y$ to be the quotient obtained by identifying the points $\langle n,0\rangle$ to get the hedgehog of spininess $\omega$. Or for a truly minimal skeleton, $X=\omega\times(\omega+1)$ and $Y$ is the quotient obtained by identifying the points $\langle n,\omega\rangle$. –  Brian M. Scott Mar 10 '12 at 10:07
    
@Brian: I'm confused. Hedgehogs are metrizable, so how would this give an example of a closed quotient mapping from a metrizable space onto a non-metrizable one? –  Arthur Fischer Mar 10 '12 at 10:24
    
@Brian: I think I understand. You meant for the other question, and essentially using the same proof. Yeah, your space is (quite) a bit easier to visualise. –  Arthur Fischer Mar 10 '12 at 10:35
    
@Arthur: Sorry: I shouldn’t have called it the hedgehog space. I forgot that that term is usually reserved for the metrizable version rather than quotient. –  Brian M. Scott Mar 10 '12 at 11:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.