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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?

josh
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  • 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 05:04
  • @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 05:39
  • @Nate: Okay; I thought there was something amiss, but couldn't figure out what. Thanks. – Arturo Magidin Mar 10 '12 at 06:05

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

Community
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user642796
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  • 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 06:35
  • 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? – user642796 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. – user642796 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
  • @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. – user642796 Mar 10 '12 at 13:58