Functor demonstrating universal property of quotient spaces The universal property of quotient spaces is, as quoted from wiki:

The quotient space $X/\sim$ together with the quotient map $q:X\to X/\sim$ is characterized by the following universal property: if $g:X\to Z$ is a continuous map such that $a\sim b$ implies $g(a)=g(b)$ for all $a,b\in X$ then there exists a unique continuous map $f:(X/\sim) \to Z$ such that $g=f\circ q$.

I wish to relate this to the formal definition of a universal property (e.g. the one on wiki), which means we have to find a suitable functor $F:\mathrm{Top} \to C$ to some category $C$.
I believe using the functor $F=\mathrm{Top}_\sim(X,-)$ (my notation) works, where
$$\mathrm{Top}_\sim(X,Z) = \{f\in\mathrm{Top}(X,Z)\mid f(a)=f(b) \textrm{ if } a\sim b\}$$
is the set of continuous maps $X\to Z$ constant on equivalence classes, and a morphism $f$ is sent to postcomposition, $f_*$. Then we can regard $q$ as a morphism $c_q:*\to \mathrm{Top}_\sim(X,X/\sim)$, where $*$ is the singleton, and $c_q$ means the function with image $q$. We get the universal property:

For any topological space $Z$ and morphism $c_g:*\to \mathrm{Top}_\sim(X,Z)$, there is a unique morphism (in $\mathrm{Top}$) $f:(X/\sim)\to Z$ such that $c_g=f_*c_q$.

This follows the definition of a universal property and is equivalent to quoted version in the top, so I think I'm happy with it. I suppose as well that the same sort of construction can be used for universal properties of other kinds of quotients. Questions:

*

*Do you agree that this works / did I make a mistake?

*Is this the natural way to see things, or did I miss something simpler? In particaular I feel iffy about being forced to use $*$ as my base object... Maybe there's actually a way to use $X$ directly?

 A: *

*I agree! The space $X/\sim$ is a representing object of your functor $\mathrm{Top}_\sim (X,{-})$. In other words you can exhibit an isomorphism
$$ \phi : y(X/\sim) \cong \mathrm{Top}_\sim (X,{-})$$
in the functor category $[\mathrm{Top},\mathrm{Set}]$, where y denotes the yoneda embedding.

*Now there are two correct ways to see $q$ (I'm not so sure about yours). The two correct ways are given by the yoneda lemma!
$$ \alpha : [\mathrm{Top},\mathrm{Set}](yZ,H) \cong HZ$$
(Take $Z = X/\sim$ and $H = \mathrm{Top}_\sim(X,{-}$)).
I like to call elements of both sets "yoneda gadgets".


*

*The first possiblity is to see q abstractly: it is your $\phi$ isomorphism.

*The second possiblity is to apply the isomorphism $\alpha$ from yoneda lemma to your abstract gadget $\phi$. This gives you... exactly the map $q\in Top_\sim(X,X/\sim)$. This is because $\alpha(\phi) = \phi_X(\mathrm{id}_X)$.

A: I think what you do works. In my opinion a universal property of an object is a description of the arrows which point out (or in) of that object. So what you really want to describe $X/R$ is a functor $F: \mathtt{Top} \to \mathtt{Set}$ and a natural isomorphism of functors $\mathtt{Top}(X/R,Y) \cong FY$. Since your space and your equivalence relation are fixed, it doesn't make much sense to mention $X$ in your functor $\mathtt{Top}_\sim(X,-)$. Your functor is not really a functor in $X$, is it? But else your description is correct. $X/R$ is a representing object of the functor $Y \mapsto FY = \{f: X \to Y\,|\, f \text{ respects the relation }R\,\} \subset \mathtt{Top}(X,Y)$. The quotient map is exactly the element of $F(X/R)$ that you get when you send the identity of $X/R$ along the natural isomorphism $\mathtt{Top}(X/R,X/R)\cong F(X/R)$. The data of a map $X \to X/R$ is exactly the same data as the data of a natural transformation $\mathtt{Top}(X/R,Y) \Rightarrow FY$ and the universal property of the quotient map translates into the statement that the natural transformation it induces is a natural isomorphism.
