Suppose $f:X\to Y$ is a bijective continuous map between compact spaces $X$ and $Y$. $f$ is also a local homeomorphism, which (here) means that for every $x\in X$, there exists an open set $U$ containing $x$, such that the restriction $f\upharpoonright U:U\to f(U)$ is a homeomorphism.
My question is: can we prove that $f:X\to Y$ is a homeomorphism, or does there exist a counterexample?
Notes:
- "Local homeomorphism" here has a slightly different meaning, as it doesn't require $f(U)$ to be open as well;
- The condition "compact" cannot be omitted. Otherwise, consider the map $f:[0,2\pi)\to S^1$, $f(x)=e^{ix}$;
- It is appreciated as well if you can prove the proposition with some stronger conditions(e.g., Hausdoff, second counterable, etc.).
Best Answer
Let $X = \{0,1\}$ with the discrete topology and $Y = \{0,1\}$ with the trivial topology. Then $f = id : X \to Y$ satisfies your assumptions, but is no homeomorphism.