Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm new to Topology and need a little help in working out a general intuition of how to build proofs. I have a question that is asking me to show that a quotient map (from a topology onto it's quotient topology, defined by an equivalence relation) is a local homeomorphism.

I know the definition of a local homeomorphism and I realise that I need to somehow show that there exists a neighbourhood of each point in the topology that when the quotient map is restricted to it it becomes a local homeomorphism, but what would constitute a complete proof that that is the case?

Asre there any good examples or templates for questions like these that anyone can recommend?

I can post the question if needs be, but I don't want an answer, just an explanation of what needs to be done for the proof to be complete.

share|improve this question
1  
Is the problem to show this for a particular example? The quotient map is in general not open and therefore also not a local homeomorphism. –  Aleš Bizjak Sep 23 '12 at 17:45
    
There is, I didn't want to put it up as it is a homework problem and I want to solve it myself. It's a quotient map defined as the map from a manifold to the quotient topology on that manifold generated by a (free) group (with the discrete topology) action that acts continuously on the manifold. The action is also properly discontinuous. –  Fredo Baron Sep 23 '12 at 17:51

1 Answer 1

up vote 0 down vote accepted

The quotient map is certainly not in general a local homeomorphism. Consider the quotient $q:[0,1]\to \frac{[0,1]}{\sim} = \mathbb{S}^1$ given by identifying $\{0\}\sim\{1\}$. Then, for any sufficiently small $\epsilon$, $[0,\epsilon)$ is an open neighborhood of $0$, but $q[0,\epsilon)$ is not ever an open neighborhood of $q(0)$.

You alluded to a concrete question which defines a quotient of a manifold $M$ by the continuous action of a properly discontinuous, discrete group $G$. Here's a hint to see that the quotient map is a local homeomorphism. Let $M/G\ni [x] = Gx$ for some $x\in M$. A small open neighborhood of $[x]$ is exactly the image under $q$ of a union of open neighborhoods of points in the orbit of $x$, $\cup_{\gamma\in G}U_{\gamma x}$.

For $q$ to be a local homeomorphism, you need to find an open neighborhood $U$ about $[x]$ homeomorphic to each of the $U_{\gamma x}$. You should be able to do this using that the action is discrete and properly discontinuous.

share|improve this answer
    
I understand your first point, sorry for being overly general. The hint has also helped me on my way, so thanks for that. –  Fredo Baron Sep 23 '12 at 18:40
    
@FredoBaron No need to apologize! I'm glad the hint helped - you're welcome :) –  Neal Sep 23 '12 at 18:46

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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