# Method for showing a quotient map is locally homeomorphic

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.

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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

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.

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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