I'm trying to learn something about topology and category theory. Let us consider the category of compact Polish spaces. The category contains all quotients of all objects (wikipedia)

For an equivalence relation $\sim$ on $X\times X$, we denote with $X/\!\sim$ the quotient space and with $q:X\rightarrow X/\!\sim$ the quotient map.

I'm trying to convince myself that if $\sim$ is induced by a continuous function $f:X\rightarrow Y$ (i.e., $x\sim x^\prime$ iff $f(x)=f(x^{\prime}$), then $X/\!\sim$ is a retraction of $X$.

QUESTION 1 How do one formalize within category theory that a quotient is induced by a map in the category, and not by an arbitrary set-theoretic relation (which is looks as an external concept)?

QUESTION 2 As I said I'm trying to prove that $X/\!\sim$ is a retract of $X$. For this I should basically construct a continuous function (a "section") $s: X/\!\sim\rightarrow X$ such that $q\circ s=id$. The natural candidate is $s([x]_{\sim})=x$ for some chosen $x$ "representing" the equivalence class. Of course I can construct a set-function with this property. But how to prove that it is continuous?

QUESTION 3 Under what hypothesis a functor $F$ preserve quotients? Both in general and in the particular case of $F$ an endofunctor in compact Polish spaces.

An interesting functor is $\textbf{V}$, the Vietoris functor mapping $X$ to the space of its compact subsets which in Compact Polish spaces are precisely the closed sets with the hit-miss topology. This is interesting because the points in $\mathbb{V}(X)$ are closed sets, hence, $V(X)$ somehow topologies the (complement of the) topology on $X$ !

QUESTION 4 As a particular case of question 3 above, does $\mathbb{V}(X)$ preserve quotients?


  • 2
    $\begingroup$ It is certainly not true that every quotient is a retract. If you identify $0$ and $1$ in $[0,1]$ you get a circle, but this circle is not a retract of the interval. The equivalence relation is induced by the continuous function $f(x) = \exp(2\pi i x)$ $\endgroup$ – Martin Apr 14 '13 at 12:47
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    $\begingroup$ Question 1: You could say that it is an effective epimorphism, or even an extremal epimorphism. $\endgroup$ – Zhen Lin Apr 14 '13 at 12:51

Well, there are several ways to formulate quotient concept using category theory... But you can adopt one depending on your interest. But, first of all, I guess you have to understand the concept all by itself (in topology). Note that it is not true that all quotients are retractions... Let $ I = \left[ 0,1\right] $ be the closed interval and $ S^1 $ be the circumference. You have an obvious quotient map $ q: I\to S^1 $. Note that this is not a retraction!!

You can see that there is no section for $ q $, using the categorical fact that sections in the category of Top are point set injections, since Top is a concrete category. (using, here, a theorem such as weak Borsuk Ulam: there isn't any continuous injection $S^1\to\mathbb{R}$)

To answer question 3, you need to think in a way of answering question 1. One way of doing so is thinking a quotient as a particular pushout. So a functor preserves "quotient" when functors preserves this kind of pushout.

I haven't knew about this functor... But you can check if it is left adjoint or preserves some kind of pushouts.

To understand quotient in the categorical sense in a nontrivial way, you can do as following: If $q: X\to Y $ is a quotient map, let $U(X) $ be the space $X$ with the discrete topology. And let $A$ be the space of fibers of $q$ with discrete topology too, id est, $A = \left\{ q^{-1}(x) : x\in Y \right\} $ with discrete topology. Note that you have obvious morphisms $U(X)\to A$ and $U(X)\to X $. The pushout of these morphisms is what you want: the quotient (for us, here, it is $Y$).... It is, too, a nice way to see that the quotient is determined by the fibers and the total space (nothing more than this). But, as I said, there are many other ways to view this in categorical language: and, perhaps, this way doesn't satisfy what you want - the best way to see depends of what you want.

  • 1
    $\begingroup$ About the second question, take a look at coverings. If a covering is not a homeomorphism, it is a quotient map which is not a retraction. Why? Because retraction is preserved by functors - and, therefore, if a covering is a retraction, we would have a surjection between the fundamental groups. Since it is already a injection (because p is a covering), p induces a isomorphism of fundamental groups - and this implies that the covering is a homeomorphism. $\endgroup$ – Fernando Apr 15 '13 at 3:40
  • $\begingroup$ I forgot to sayt that coverings are quotient maps because they are (surjective) open maps. I proved, here, that if a covering is a retraction, then this covering is a homeomorphism. $\endgroup$ – Fernando Apr 16 '13 at 21:55

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