Locally homeomorphism on connected space

Question

I am wondering since locally constant maps $f$ on connected space are contant, what about locally homeomorphic maps on connected space?

I am trying to mimic the proof by Thomas Andrew here Locally Constant Functions on Connected Spaces are Constant showing locally constant + connected = constant, by replacing $\{x\ |\ f(x)=f(x_0)\}$ with the maximal neighbourhood of $x_0$ such that $f$ is homeomorphic on it. Is that the right way to do it? Or maybe they are just not homeomorphic.

Are you asking if a local homeomorphism on a connected space is always a homeomorphism? — Cheerful Parsnip, Feb 06 '22 at 17:51
Yes I am asking if a local hoemomorphism on a connected space is homeomorphic — Ja_1941, Feb 06 '22 at 17:52
There is a positive result along these lines, but you need to strengthen the assumption: Require the target to be simply connected and $f$ to be a covering map. You can find details in any algebraic topology textbook. — Moishe Kohan, Feb 06 '22 at 18:59

score 3 · Accepted Answer · answered Feb 06 '22 at 17:50

You cannot prove that every local homeomorphism with a connected domain is a homeomorphism, since that is not true. For instance, if $f\colon\Bbb R\longrightarrow S^1(=\{z\in\Bbb C\mid|z|=1\})$ is the map defined by $f(x)=e^{ix}$, then $f$ is a local homeomorphism, but it is not a homeomorphism.

score 2 · Answer 2 · answered Feb 06 '22 at 17:53

No it’s not true:

Take the parametrization of $\mathbb{S}^1$

$\gamma: \mathbb{R}\to \mathbb{S}^1$

sending $\alpha$ to $(cos(\alpha), sin(\alpha))$

In this case $\mathbb{R}$ is a connected space and $\gamma $ is a local homeomorphism (it’s a covering map and $\mathbb{R}$ is the universal cover of $\mathbb{S}^1$) but $\mathbb{R}$ is not homeomorphic to $\mathbb{S}^1$

Locally homeomorphism on connected space

2 Answers2