I know that if $f : X \rightarrow Y$ is a continuous bijection from a compact space $X$ to a Hausdorff space $Y$, then $f$ is an homeomorphism.
So I was thinking that if we relax the assumption $X$ compact to $X$ locally compact, it should be true as well. Using the above result, $f$ is an homeomorphism if we restrict it to a compact neighborhood of $X$. Since we can find a compact neighborhood around every point of $X$, $f$ should be a local homeomorphism. But a bijective local homeomorphism is a global homeomorphism, so that would be it.
Yet, if I'm not mistaken, the map $f : [0, 2\pi[ \rightarrow S^1, f(\theta) = e^{i\theta}$ is a counterexample. What is wrong with my reasoning ?
Best Answer
A simpler example is just the identity from a discrete $\mathbb{R}$ to the usual $\mathbb{R}$, both are locally compact metrisable, any map from a discrete space is continuous. The compact neighbourhoods in the first space are only the finite ones, so there we do have that there is a local homeomorphism (the finite sets are homeomorphic in both spaces), but they're not neighbourhoods in the image.
Likewise, for your example, the compact neighbourhoods $[0,r]$ of $0$ in $[0,2\pi)$ do have compact homeomorphic images in $S^1$ but these are not neighbourhoods of $f(0)$ in $S^1$ any more.
To go from something like a local homeomorphism to global one, we need a stronger condition. Something like: for every $x \in X$ and every compact neighbourhood $C$ of $x$ in $X$, we must have that $f[C]$ is a (necessarily homeomorphic) neighbourhood of $f(x)$ as well. And as we saw, this is by no means garantueed from just being continuous and bijective.