I was reading Chapter 2 of ''Algebraic Topology for Data Scientists'' by Michel S. Postol. I had before a beginner's knowledge on Algebraic Topology and therefore wanted to at least revisit the pages of these introductory chapters. In Chapter 2, Postol revisits the notions of Separation Axioms in topological spaces. This has been my first time as an undergraduate looking at the concept of regular spaces and $T_3$ spaces. In theorem 2.5.4. he states that quotients of $T_3$ spaces need not be regular. Chapter 2 is written as a revisiting of concepts, and do not always provide proofs and examples as they are assumed to be already known by the reader. I would like to ask for an example to prove exactly that statement. I have already thought that I would not be able to find one with folded polygons as they are locally Euclidean (I am assuming that locally Euclidean spaces are regular, which seems to be quite easily provable through the coordinates maps bringing the property up to the space from $\mathbb{R}^n$), so maybe I should think of more complex examples.
May someone could bring up commonly known examples in the culture of topology?
Thanks in advanced for your time.
Any $T_3$ (or even just $T_1$) non-discrete space has a non-regular quotient.
To see this, if $X$ is $T_1$ and not discrete, then let $A\subset X$ be non-closed, and for $x,y\in X$, let $x\sim y$ if and only if either $x,y\in A$ or $x=y$. Then $X/\sim$ is $T_0$ (all singletons of $X/\sim$ are closed except for one, so we never have two topologically indistinguishable points, since such points would reside in each other's closure), and yet is not $T_1$, since the image of $A$ in $X/\sim$ corresponds to a non-closed singleton.
A fortiori, $X/\sim$ is not $T_3$, and since $T_0$ regular spaces are $T_3$, we see that $X/\sim$ is not regular either.
Lest you think that the quotients that fail to be regular are weird, non-$T_1$ spaces, note also that a $T_1$ quotient of a $T_3$ space can fail to be $T_2$ (hence certainly not $T_3$, and by the same reasoning above, not regular either).
For example, consider the line with the doubled origin, viewed as a quotient of two disjoint lines.
Finally, even if a quotient of a $T_3$ space happens to be $T_2$, it can still fail to be $T_3$ (and therefore also fail regularity).
For an example of this (there might be simpler ones but this is what came to my mind), let $X=([0,\omega_1]\times [0,\omega])\backslash \{(\omega_1,\omega)\}$, where the product is given the product topology. This is the deleted Tychonoff plank. Let $A=\{\omega_1\}\times [0,\omega)$, and define $\sim$ as before. Then since $X$ is $T_3$ (though not $T_4$) and $A$ is closed, it easily follows that $X/\sim$ is $T_2$.
But $X/\sim$ cannot be regular, since the image of $[0,\omega_1)\times \{\omega\}$ is closed but cannot be separated from the image of $A$, which is a singleton.
(More generally, we can do this trick with any $T_3$ space that fails to be $T_4$, by finding two disjoint closed sets that cannot be separated by disjoint open sets, and collapsing one of them to a point.)
Remark.
I missed earlier that you said this:
This need not be the case. For example, the line with the doubled origin mentioned above is locally Euclidean, yet not even Hausdorff.