# The universal property characterizing the quotient topology

1. Sep 5, 2008

### quasar987

I am trying to show that if X is a topological space, ~ an equivalence relation on X and q:X-->X/~ the quotient map (i.e. q(x)=[x]), then the quotient topology on X/~ (U in X/~ open iff q^{-1}(U) open in X) is characterized by the following universal property:

"If f:X-->Y is continuous and constant on each equivalence class of ~, then there exists a unique continuous map f':X/~-->Y such that f=f' o q"

That the quotient topology satisfies this is easy to show: just define f'([x]):=f(x). Then f' is continuous since for V open in Y, $q^{-1}(f'^{-1}(V)) = (f' \circ q)^{-1}(V)=f^{-1}(V)$, which is open in X.

But, for the other direction, we must show that if some topology T on X/~ satisfies the above universal property, then T is the quotient topology. I.e., we must show that $q^{-1}(U)$ is open in X <==> U is open in T.

I don't see how.

2. Sep 6, 2008

### jostpuur

If looked at this right, you get the ==> direction by choosing Y=X/~ with the quotient topology, and f=q. It follows that the identity map

$$\textrm{id}:(X/\sim,\; T)\to (X/\sim,\; T_{\textrm{quot}})$$

is continuous.

3. Sep 6, 2008

### jostpuur

I don't believe <== direction is true. For example, set the discrete topology on X/~. Then let Y be an arbitrary topological space, and f:X->Y some continuous function that is constant on each [x]. We can define f':X/~->Y with formula f'([x])=f(x), and now f' is continuous, and satisfies $f=f'\circ q$. But the discrete topology is not usually the quotient topology.

Or can you spot where there's a mistake, or did I understand something wrong about the original problem?

4. Sep 6, 2008

### quasar987

I believe you've found a counterexample. Thanks jostpuur!

I don't know why I was so sure the theorem was right. I apologize to everyone who tried to figure this out!

5. Sep 6, 2008

No problems

6. Sep 6, 2008

7. Sep 6, 2008

### quasar987

I took the wiki statement to mean

"If ~ is an equivalence relation on X, f:X-->Y is continuous and constant on each [x], q:X-->Z is a map, then there exists a unique continuous f':Z-->Y such that f=f' o q <==> Z is X/~ with the quotient topology and q is the quotient map."

But then if you set Z=X/~ (with some topology T) and q = the quotient map, then the statement becomes a characterization of the quotient topology:

"If ~ is an equivalence relation on X, f:X-->Y is continuous and constant on each [x], q:X-->X/~ is the quotient map, then there exists a unique continuous f':X/~-->Y such that f=f' o q <==> X/~ has the quotient topology."

No?!?

8. Sep 6, 2008

### jostpuur

This seems to be all about what the term "characterizing property" means. To me it seems that when the Wikipedia states

it is actually merely mentioning one property of the quotient space, which alone is not yet equivalent with the definition of the quotient space. I've never seen precise definitions of what this terminology is supposed to mean, but "characterizing property" does usually mean a property, which already alone is equivalent with some previous definition, right? If so, to me it seems that Wikipedia has mistake there then.

9. Sep 6, 2008

### jostpuur

Perhaps the Wikipedia's claim is assuming that the q is continuous (and surjective) already by definition? Then my counter example wouldn't work. I didn't yet check that that would make the claim correct, though.

Earlier, they say this:

So it could be, that we are never going to try to call a map a quotient map if it already isn't continuous (and surjective). The main focus would be on the condition that $f^{-1}(U)\subset X[/tex] being open implies [itex]U\subset Y$ being open?

IMO, when mathematics is well written, we shouldn't need this much interpreting. :tongue:

Last edited: Sep 6, 2008
10. Sep 6, 2008

### quasar987

Hey, that's true... If q is assumed to be a quotient map, then it implies that X/~ has the quotient topology induced by q and we can't assume that X/~ has the discrete topology.

But if q is assumed to be a quotient map, what we need to prove <== is that under the hypothesis that a continuous f':X/~-->Y such that f=f' o q exists, the topology on X/~ induced by q is actually the topology induced by $\pi:X\rightarrow X/$~, where $\pi(x)=[x]$.

Actually, the article says that the universal property characterizes both X/~ with the quotient topology and the quotient map $\pi$. So we would have to show the stronger condition that q is in fact $\pi$ !

But the fact alone that $f'\circ q = f'\circ \pi$ does not guarentee that does it? Damn it.

Last edited: Sep 6, 2008
11. Sep 6, 2008

### jostpuur

So you are interpreting the assumption like this:

q:X->X/~ is assumed to be a quotient map, but not necessarily the one given by q(x)=[x]?

That looks little far fetched to me. I was interpreting the assumption like this:

q:X->X/~ is assumed to be continuous and given by the formula q(x)=[x], but q is not necessarily assumed to be a quotient map.

I'm not sure if that's less far fetched, but that's what it came up with first, when I tried to figure out what Wikipedia is trying to say.

12. Sep 6, 2008

### jostpuur

This would be a kind of claim that I feel like being able to prove:

Let $X$ be a topological space, $\sim[/tex] be an equivalence relation in it, [itex]\mathcal{T}$ be some topology in $$X/\sim$$ such that the projection $\pi:X\to (X/\sim,\;\mathcal{T})$, given by the equivalence relation, is continuous. Then assume that for all topological spaces $Z$ and continuous mappings $f:X\to Z$ which are constant in each equivalence class, there exists a unique continuous mapping $f':(X/\sim,\;\mathcal{T})\to Z$ such that $f=f'\circ \pi$. Then $\mathcal{T}=\mathcal{T}_{\textrm{quot}}$, where $\mathcal{T}_{\textrm{quot}}$ is the quotient topology.

In post #2 I showed how to prove $\mathcal{T}_{\textrm{quot}}\subset\mathcal{T}$, and this part didn't need the assumption that $\pi:X\to (X/\sim,\;\mathcal{T})$ is continuous.

The claim $\mathcal{T}\subset\mathcal{T}_{\textrm{quot}}$ can be proven quite shortly when the continuity of $\pi:X\to (X/\sim,\;\mathcal{T})$ is assumed. Let $V\in\mathcal{T}$ be arbitrary. Now $\pi^{-1}(V)\subset X$ is open. By definition of the quotient topology, $V\in\mathcal{T}_{\textrm{quot}}$.

13. Sep 6, 2008

### jostpuur

When you put this together with the partial result you mentioned in the first post, we get this:

Let $X$ be a topological space, $\sim$ an equivalence relation, $\mathcal{T}$ be some topology in $X/\sim$ such that the projection $\pi:X\to (X/\sim,\;\mathcal{T})$, given by the equivalence relation, is continuous. Now $\mathcal{T}=\mathcal{T}_{\textrm{quot}}$ if and only if for all topological spaces $Z$ and continuous functions $f:X\to Z$ which are constant in each equivalence class, there exists a unique continuous mapping $f':(X/\sim,\;\mathcal{T})\to Z$ such that $f=f'\circ\pi$.

This looks like something that could be called a characterizing property.

14. Sep 6, 2008

### quasar987

The condition "such that pi is continuous" sounds funny but hey, your reasoning is good! And that is a caracterization by a universal property. Nice.

15. Sep 6, 2008

### jostpuur

Yeah, the claim would have more elegant visual appearance if you didn't mention it.

edit: I was trying to be humorous, but hmhmh..... actually the entire result would be a lot stronger and more impressive, if one didn't need to assume continuity of the projection initially. That is quite unpleasant assumption, that makes the result less usable.

Last edited: Sep 6, 2008
16. Sep 6, 2008

### Hurkyl

Staff Emeritus
The general notion of quotient object is as follows:

Let there be a map p:P-->Q, and let ~ be a binary relation on P. Then the following are equivalent:

1. Q is a quotient of P by ~, and p is the projection

2. For any map f:P --> R such that a~b implies f(a)=f(b), there is a unique map g:Q --> R such that f = g o p

In fact, the above is often taken as the definition of the idea of a quotient.

('map' means 'continuous function' in this particular context)

(Note that Q is a quotient of P by ~ iff Q is a quotient of P by the equivalence relation generated by ~. Also, the needed notion of 'binary relation' is subtle, but there are no relevant oddities in the case of topological spaces)

Does this make sense? Does it jive with how you were interpreting the problem at hand?

The other thing is that X/~ is supposed to denote a very specific topological space: it is the space whose underlying set of points you know, and whose topology is the quotient topology. If you were considering another topological space Z that satisfies |Z| = |X/~|, but it has a different topology, then it would be incorrect to use the notation X/~ for Z.

(However, it is eventually a convenient abuse of notation to let X/~ denote any quotient of X by ~, rather than this specific one)

17. Sep 6, 2008

### quasar987

Hi Hurkyl,

For b.==>a., jostpuur mentioned the counter-example where p is the projection map and Q is X/~ with the discrete topology. Then g:Q-->R defined by g([x]):=f(x) is continuous and satisfies f = g o p.

What are we missing?

18. Sep 6, 2008

### Hurkyl

Staff Emeritus
That p isn't a map of topological spaces.

Yes, the set |Q| is a quotient of the set |X|, but that isn't enough to be topologically interesting!

19. Sep 6, 2008

### quasar987

Ah, so we should add the requirement that p be continuous in your point b. In that case, this is exactly the characterization that justpuur proved.

20. Sep 6, 2008

### jostpuur

But the theorem that Hurkyl mentions is little bit stronger than the one I mentioned. I was thinking about starting with $\pi:\to (X/\sim,\;\mathcal{T})$, and focusing on the proof of $\mathcal{T}=\mathcal{T}_{\textrm{quot}}$, but Hurkyl is saying that we should start with $p:X\to Q$, and focus on proving $Q = X/\sim$ and $p=\pi$. I think the most difficult part of the proof, with topological stuff, was already gone through earlier in the thread. This should be a small addition now, and requires mainly checking injectivity and surjectivity related stuff.

Last edited: Sep 6, 2008