if X and Y are Hausdorff spaces, $f:X \to Y$ is a local homeomorphism, X is compact, and Y is connected, is f a covering map?
It seems to be, and I almost have a proof, but I'm stuck at the very end of it:
I've already proved that f is surjective (using the connectedness), and that for each $y \in Y$, $f^{-1}(y)$ is finite. Because X is compact, there exists a finite open cover of X by $ \lbrace U_i \rbrace $ such that $f(U_i)$ is open and $ \left. f \right|_{U_i}:U_i \to f(U_i) $ is a homeomorphism. For each $y \in Y$, we choose the subset $ \lbrace U_{i_j} \rbrace $ such that $y \in U_{i_j}$, and then define $V = \bigcap_{j=1}^k f(U_{i_j})$, and $U'_j = U_{i_j} \bigcap f^{-1}(V)$.
... and this is were I got stuck. I really want to write that $f^{-1}(V) = \bigcup_{j=1}^k U'_j$ (more or less proving it's a covering map), but I can't justify that, and I actually think that it's not true. I think I might need an extra step, and to take an even smaller neighborhood of y, in order to make sure that extra sets from $ \lbrace U_i \rbrace $ didn't sneak into $f^{-1}(V)$.
Any help would be greatly appreciated as I've already spent several hours working on this problem.