Let $ X $ be compact and let $f: X \rightarrow Y$ be a local homeomorphism. Show that for any point $y \in Y$, $f^{-1}(y)$ is a finite set. PROBLEM: Let $ X $ be compact and let $f: X \rightarrow Y$ be a local homeomorphism. Show that for any point $y \in Y$, $f^{-1}(y)$ is a finite set.

I was trying out this problem from Massey, but got stuck. 

MY ATTEMPT: My thought was to pick an open neighborhood $U_i$ of arbitrary but fixed $y_i \in Y$, then show that $U_i$ is evenly covered by $f$, so $f^{-1}(U_i)$ is a disjoint union of open sets in $X$,

$f^{-1}(U_i) = \underset{\alpha \in X} \amalg V_{{y_i}_{\alpha}} $ such that for each $\alpha$, $f|_{V_{{y_i}_{\alpha}}}: V_{{y_i}_{\alpha}} \rightarrow U$ is a surjective homeomorphism. This also implies $f$ is a covering map.

Then $X = \underset{\alpha \in X}\amalg V_{{y_1}_{\alpha}} \cup \underset{\alpha \in X}\amalg V_{{y_2}_{\alpha}} ...$, which is an open covering. Since $X$ is compact, that means there is a finite subcover. Specifically, this means only finitely many $f^{−1}(U_i) = \underset{\alpha \in X} \amalg V_{{y_i}_{\alpha}}$ cover $f^{-1}(y_i)$ for each $y_i$.

Does this outline seem correct? I have a problem at the beginning, not knowing how to use the local homeomorphism to show that $f$ evenly covers $U_i$, and I'm not sure about if my conclusion appropriately proves the statement.
Without going into evenly covered maps etc., it might be simpler: Suppose for some $y$, $F_y := f^{-1}[\{y\}]$ is infinite. As $X$ is compact (in fact we need only countably compact) there is a limit point $p$ of $F_y$. But if $U$ is a neighbourhood of $p$ such that $f \restriction U: U \to f[U]$ is a homeomorphism, then $U$ contains at least 2 (in fact infinitely many; here I'm assuming $X$ is $T_1$ as well) points of $F_y$, which makes $f$ not even injective on $U$, as both map to $y$ under $f$.
So in fact, for a locally injective map $f$ on a countably compact $T_1$ space, all fibres are finite.
So if some $F_y$ were infinite, by compactness it would have an $\omega$-accumulation point $x \in X$. If $O$ were the witnessing neighbourhood of $x$ wrt local homeomorphism, a contradiction to injectivity ensues; no sep. axioms needed.
I don't have a full answer for you but here are some thoughts.  If f-1(Ui) is a disjoint union of open sets in X, that does not mean that they are a cover for X (and we don't care).  They are a subset of X and by definition none of them are compact (being open).  So I think you should close them, forming a compact set V &sub; X (V is compact because X is).  Now you can conclude as you say that a finite set of the f-1Ui) cover V.

Since you obviously can form all the f-1(Ui) you want, the problem comes down to whether you can make them disjoint.  For this I think you need to pay attention both to the fact that f is a local homeomorphism, and  that it is (therefore) continuous.  The former rules out cases such as X = [0,1] on the real line and f(x) &equiv; 1.  The latter allows you to choose 
each Ui so that yi is unique.

You need to do this so that f-1(yi) is unique in f-1(Ui), which I think it will be because f is 1-1; otherwise having a finite cover doesn't necessarily imply a finite number of f-1(yi).

I still don't quite see how to get from there to disjoint, but I think that plus the compactness of X should be the right approach. 

I'll keep thinking about it.  If you solve it, please post the solution so I can see it.



Further thought: First just to be sure we are clear, I am using the set {yi} to be all the y's which have the same value.  Following up on the idea of there being a Ui for each yi which contains only one yi, it must be that there is a there is a constant a > 0 such that the "radius" of each Ui can be a while still preserving just one yi for each Ui.   Otherwise there is a convergent sequence of yi which goes to say yc.  I believe that f-1 would not be continuous at yc, so that f would not be a homeomorphism. (I haven't quite shown this, but maybe you can do that).

Having gotten that far, maybe the disjointness of f-1(Ui) will follow.