# 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.

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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.

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It also works if $X$ is countably compact and $Y$ is $T_1$: Then the fiber $F_y=f^{-1}(y)$ is closed. If it were infinite, then it would have a limit point $p$ such that $f(p)=y$. Each neighborhood $U$ would contain $p$ and a point of $F_y\setminus\{p\}$, which would contradict the local injectivity. – Stefan Hamcke Aug 6 '13 at 17:00

Let me suggest a strategy. The problem is to show that every fiber $f^{-1}(y)$ is finite when $f$ is a local homeomorphism; if it's unclear what the role of this hypothesis is, try asking the same question without it: is it necessarily true that the fiber of any map $f$ is finite, when $X$ is compact? Obviously not: you could have, as a really silly example, the map $f \colon X \to \{\text{pt}\}$, whose fiber is all of $X$, which doesn't need to be finite. Though if such an $f$ were a local homeomorphism, then $X$ would be a discrete union of points and a discrete, compact space is finite. So the claim holds for maps from a compact space to a point.

It's crucial to recall the proof of the last claim there. If $D$ is a discrete compact space, then each point is, in itself, an open set (discreteness), and $D$ is the union of its points, so is a finite union of some of them (compactness). But since each singleton set contains only one point, this accounts for only finitely many points, so $D$ is finite.

So if we want to execute the same argument on a general map $f \colon X \to Y$, we need to cover the fiber $f^{-1}(y)$ by open sets each of which contains only one point of the fiber, and somehow argue by compactness that there is a finite subcover. Note that (without some kind of separation axiom) the fiber itself may not be compact, but this is not an impediment as one can always enlarge the covering by adding irrelevant open sets.

All the ingredients for this are in your outlined proof.

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I think you might be trying to prove too much here. As you say, there's a problem at the beginning. Even if $f$ is a covering map, it's not necessarily true that it evenly covers every neighborhood of $y_i$. So you would have to come up with a way to pick a small enough $U_i$. For what it's worth, I'm not sure how to do that, or even if it's possible.

There's a simpler approach, which is to forget about $y$ for a minute and apply the compactness of $X$ right away. If you can get a finite cover of $X$ on which $f$ is nice enough, it will make everything much easier...

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I think I might be confused about compactness then. A finite cover implies that there are finitely many sets that cover $X$, but how do I know that the set $f^{-1}(y)$ lies in is finite from that? I assume the local homeomorphism does something in that regard, but I'm not sure what. – Phdetermined Aug 6 '13 at 5:41
@user59699 I'm sorry, I'm being a little vague because I don't want to give the answer away. :) I'll try giving some stronger hints... You can use $f$ to get a certain covering of $X$ by open sets. Then you can use the compactness of $X$ to get a finite subcover. This finite subcover will lead you almost directly to the conclusion... – Chris Culter Aug 6 '13 at 5:45

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 ⊂ 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) ≡ 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.

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"They are a subset of X and by definition none of them are compact (being open)." There is a serious misconception here: open sets can certainly be compact. If the space is Hausdorff, then a compact subset must be closed, but of course a subset can be both open and closed. In general there are (nondiscrete) spaces in which every open subset is compact. Also, what is the "radius" in this context? Are you assuming the existence of a metric? – Pete L. Clark Aug 13 '13 at 3:32
I appreciate your bringing this to my attention. It looks as if my approach is promising where there is a metric, whether X is Hausdorff or not. The checked solution submitted later looks good. – Betty Mock Aug 14 '13 at 4:26