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I'm seeking "best possible" examples of a nonnormal quotient of a normal space, namely, a nonnormal but completely regular $T_{0}$ space $Y$ that is the quotient of a normal $T_{1}$-space $X$.

The more "concrete" and "geometric" the better!

Are there any such examples other than those constructed by the method from https://math.stackexchange.com/a/1682677/32337 (which uses ultrafilters) or https://math.stackexchange.com/a/1569463/32337 (which uses cardinality)?

Examples that avoid both cardinality and ultrafilters would be desireable.possible.

Related: Examples of a quotient of a normal topological space that is not normal?. Note that the quotient spaces in the accepted answer there are either not regular or not Hausdorff, so do not constitute a "best posssible" example.

murray
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It was shown by Hanai that a $T_1$ space is first-countable if and only if it is an open image of a metric space. See

S. Hanai, Open Mappings II, Proc. Japan Acad. 37, (1961), 233-238.

I believe the result is also often credited to Ponomarev.

Since every open surjection is quotient, to answer your question will suffice to find any first-countable, completely regular, $T_1$ space which is not normal. The Niemytzki plane works.

This particular example can be made explicit by noticing that the Niemytzki plane $N$ is locally metrisable. Thus let $\mathcal{U}$ be any open cover of $N$ by metrisable open sets and let $X=\bigsqcup_{U\in\mathcal{U}} U$ be the disjoint union of the members of $\mathcal{U}$. The obvious mappings $\pi:X\rightarrow N$ is a continuous open surjection of a metrisable space onto a non-normal space.

PatrickR
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Tyrone
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  • I don't see why a basic open neighborhood $N = {(x, 0)} \cup B_{r}((x, 0))$ of a point $(x, 0)$ on the horizontal is metrizable. In particular, for such a metric, inducing the relative topology on $N$ from the Niemytzki plane, what would be the distance between $(x, 0)$ and other points in $N$? – murray Mar 24 '22 at 19:45
  • Murray, every second-countable regular Hausdorff space is metrisable by the Urysohn Metrisation Theorem. Clearly this applies to $N$ (choose a countable neighbourhood base at each point of $B$ with rational coordinates and add to these a countable neighbourhood base for the point $(x,0)$. The resulting collection is a countable base for $N$). Of course its fairly easy to see what a compatible metric might look like... – Tyrone Mar 25 '22 at 01:11
  • Invoking the Urysohn Metrization Theorem seems to be hitting a small nail with a sledgehammer. How about a direct way? – murray Mar 25 '22 at 15:30
  • I think you saw in your other question that there is no trick. Do you still need a pointer? – Tyrone Mar 25 '22 at 16:52
  • Please seem my 2nd comment (the 3rd comment in all) to the answer https://math.stackexchange.com/a/4412135/32337. I'm just not seeing something. – murray Mar 25 '22 at 19:01
  • I anxiously await seeing a compatible metric on a basic neighborhood $N$ of $(x, 0)$ in the Niemytzki plane $\Gamma$. – murray Mar 25 '22 at 19:41