# When is a local homeomorphism a covering map?

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 $\{ U_i \}$ such that $f(U_i)$ is open and $f |_{U_i}:U_i \to f(U_i)$ 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.

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For $y \in Y$, let $\{x_1, \dots, x_n\}= f^{-1}(y)$ (the $x_i$ all being different points). Choose pairwise disjoint neighborhoods $U_1, \dots, U_n$ of $x_1, \dots, x_n$, respectively (using the Hausdorff property).

By shrinking the $U_i$ further, we may assume that each one is mapped homeomorphically onto some neighborhood $V_i$ of $y$.

Now let $C = X \setminus (U_1 \cup \dots \cup U_n)$ and set $$V = (V_1 \cap \dots \cap V_n)\setminus f(C)$$

If I'm not mistaken this $V$ should be an evenly covered nbh of $y$.

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Seems right to me. f is a closed map (compactness+hausdoff), C is closed so f(C) is closed, so V is open, and $f^{-1}(y)\bigcap C = \emptyset$ so $y \in V$, and so it's indeed an open nbh of y. $f^{-1}(V) \subset U_1 \cup \ldots \cup U_n$ by definition, and so $f^{-1}(V) = (U_1 \cap f^{-1}(V)) \cup \ldots \cup (U_1 \cap f^{-1}(V))$ when $U_i \cap f^{-1}(V) \cong V$. –  Or Sharir Jun 17 '11 at 23:23
Right, looks like you finished the proof. :) –  Sam Jun 17 '11 at 23:31
@Sam why $f^{-1}(y)$ is finite? where is the compactness is nedded to argue that? –  Bunuelian Trick Feb 28 '13 at 6:05
@CityOfGod: Since $f$ is a local homeomorphism, the set $f^{-1}(y)$ consists of isolated points. Since $X$ is compact, this can only be the case, if $f^{-1}(y)$ is finite. –  Sam Feb 28 '13 at 6:13
thank you very much sam! –  Bunuelian Trick Feb 28 '13 at 6:37