# Proper continuous image of metrizable space

Motivated by the following post, "Gelfand duality" and the fact that "a Hausdorff continuous image of a compact metric space is metrizable", we ask:

What is a counter example of two locally compact Hausdorff spaces $X$ and $Y$ and a surjective proper continuous map $f:X\to Y$ such that $X$ is metrizable but $Y$ is not?

• Note that $f$ can be extended to the one-point compactifications, so $X$ cannot be second countable. The only locally compact, metrizable, not second countable spaces that appear in Steen-Seebach´s book are the uncountable discrete spaces. But if $X$ is discrete then $Y$ is discrete and therefore metrizable. So perhaps there are no such counterexamples. – Ramiro de la Vega Feb 2 '15 at 16:30
• @RamirodelaVega thank you for your very interesting answer. Is it easy to prove that theorem in Steen_Seebach book. I do not have that book. – Ali Taghavi Feb 2 '15 at 18:12
• @RamirodelaVega the disjoint union of uncountable discrete space and interval satisfies the same properties but is not discrete. So what is that theorem, precisely? – Ali Taghavi Feb 2 '15 at 18:20
• I was not talking about a theorem. Steen-Seebach´s book is a book of counterexamples. That´s why I said "perhaps" at the end and that´s why I posted a comment, not an answer. – Ramiro de la Vega Feb 2 '15 at 18:31
• @RamirodelaVega I am sorry about my misunderstanding. – Ali Taghavi Feb 2 '15 at 18:32

If $X$ is metrisable and $f$ is a perfect surjection onto $Y$, where perfect means continuous, closed and pre-images of singletons are compact, then $Y$ is metrisable. This is proved e.g. in Engelking's General Topology (Thm. 4.4.15), and due independently to K. Morita and S. Hanai ("Closed mappings and metric space", Proc. Japan Acad. 32 (1956), 10-14) and A.H. Stone ("Metrizability of decomposition spaces", Proc. Amer. Math. Soc 7 (1956), 690-700). The proof in Engelking is based on the Bing-Nagata-Smirnov metrisation theorem.
A proper (in the sense of inverse images of compact sets being compact) continuous map with a $k$-space as codomain is closed, if the codomain is Hausdorff (for a simple proof see this paper) so the above theorem applies, as all locally compact spaces are $k$-spaces (or compactly generated); for a proof for locally compact see this question. This means that no counterexample exists.
As an afterthought, I looked a bit more at the question whether a proper continuous image of a metric space is metrisable, without conditions. We saw above that with $Y$ being a Hausdorff $k$-space (which are necessary conditions to be metrisable anyway) the answer is yes, as we then have a perfect map. But without a condition on $Y$ we do have an example: let $X$ be the reals in the discrete topology, and $Y$ the set of reals with the topology where all subsets of $\mathbb{R} \setminus \{0\}$ are open, and a set $O$ containing $0$ is open iff $O$ is co-countable (so it's essentially the one-point Lindelöfication of the reals in the discrete topology). One checks that $Y$ is hereditarily normal (but not perfectly normal) and is not a $k$-space because the only compact subsets of $Y$ are the finite ones (and so all subsets are compactly closed, but some are not closed). The last fact also implies that $f(x) = x$ is continuous and proper, but its image is not metrisable. And $X$ is locally compact metrisable.
• I believe perfect image of metrizable is metrizable was proved independently by A.H. Stone ("Metrizability of decomposition spaces") and K. Morita - S. Hanai ("Closed mappings and metric spaces"). I don´t quite see how it follows easily from Nagata-Smirnov. Also, don´t you need $Y$ to be compactly generated (e.g. locally compact) for $f$ to be closed? – Ramiro de la Vega Feb 3 '15 at 15:44