According to Wikipedia Quotient topology is ill-behaved with respect to Separation Axioms,locally compactness and simply connectedness.

I have examples to support this argument for locally compactness [$\mathbb{R}/\mathbb{N}$ will do the job] and for simply connectedness [identifying $0$ and $1$ in $[0,1]$ gives a circle which is not simply connected] but cannot think of any suitable example for separation axioms. Need help!

  • $\begingroup$ You first example is also not first countable, or second countable, e.g. $\endgroup$ – Henno Brandsma Mar 31 '18 at 5:30

Let $X$ be $\mathbb{R}$ in the usual topology, define an equivalence relation on $X$ by $x \sim y$ iff $x-y \in \mathbb{Q}$, let $Y$ be the set of classes, and let $q$ be the standard map that sends a point to its class. Then $Y$ in the quotient topology induced by $Y$ is uncountable and indiscrete (only $\emptyset$ and $Y$ are open), so that $X$ has all "nice" separation axioms ($T_0$ up to and including $T_6$) while $Y$, its quotient image, has none of them. I gave a proof here.

We have a similar example with $X$ and the equivalence relation where the two only classes are $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$, and we map to a two point set $\{0,1\}$, (say $q(x) = 0$ for $x \in \mathbb{Q}$, and $1$ otherwise) which is also indiscrete (as for $\emptyset \neq O \subseteq \{0,1\}$ if the set $q^{-1}[O]$ is open, then being non-empty it contains rationals and irrationals, so that $\{0,1\} \subseteq q[q^{-1}[O]] = O$). Both maps above are open maps even.

If $X$ is $T_2$ (Hausdorff) and not regular, and the latter is witnessed by $x \notin A$, and $A$ closed. Then indentifying $A$ to a point, gives us a quotient map onto a space that is $T_1$ but not $T_2$. This shows that closed maps can also "kill" a separation axiom. A variation is given when $X$ is not normal but $T_3$, and we identify to witnessing closed sets to points, and we get a non-$T_2$ quotient again.

  • $\begingroup$ wow! Thank you so much! $\endgroup$ – MathCosmo Mar 31 '18 at 5:46

Wikipedia states:

In general, quotient spaces are ill-behaved with respect to separation axioms. The separation properties of $X$ need not be inherited by $X/\!\!\sim$, and $X/\!\!\sim$ may have separation properties not shared by $X$.

The Kolmogorov quotient gives a simple example of $X/\!\!\sim$ having separation properties potentially not shared by $X$. We start by defining an equivalence relation $\sim$ on $X$ wherein $x \sim y \iff$ $x$ and $y$ have the same open neighborhoods, i.e. whenever they are topologically indistinguishable. By passing to $X/\!\!\sim$, we arrive at a space where any two points are distinguishable. So when $X$ is not $T_0$, the Kolmogorov quotient on $X$ is $T_0$.

Of course, if $X$ already is $T_0$, then $X$ and $X/\!\!\sim$ are homeomorphic.

Though unsatisfying, note also that you can simply "glue together" all the points of any space into one. In this quotient, every separation property becomes vacuously true.

Just underneath the above quoted text, Wikipedia gives a hint$^\dagger$ for constructing a space $X$ and an equivalence relation $\sim$ such that we lose separation properties by passing to the quotient space:

$X/\!\!\sim$ is a $T_1$ space if and only if every equivalence class of $\sim$ is closed in $X$.

So simply construct a $T_1$ space together with a relation such that at least one of its equivalence classes is not closed in $X$. Because every equivalence relation on $X$ corresponds to a partition of $X$ and vice-versa, you can think of this as simply partitioning $X$ into subsets $\{P_\alpha\}$ where at least one of these subsets is not closed. We'll then lose $T_1$-ness by passing to $X/\!\!\sim$.

$^\dagger$Proof sketch of hint:

Let $f: X \rightarrow X/\!\!\sim$ be the quotient map. This result is a consequence of the following three facts, considered together:

  • $\begingroup$ Now I think that I'm a fool haha!. Thank you! $\endgroup$ – MathCosmo Mar 31 '18 at 5:18
  • 1
    $\begingroup$ Ha! Everything's trivial once you've seen it :) Glad I could help @PronayBiswas $\endgroup$ – Kaj Hansen Mar 31 '18 at 5:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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