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Let $X$ be a compact and path connected space and let $p:\tilde X\rightarrow X$ be its universal cover. I can show that if $\tilde X$ is compact then $\pi_{1}(X)$ is finite:

$\pi_{1}(X, x_{0})$ acts on $p^{-1}(x_{0})$ via $[\gamma].y=\tilde\gamma(1)$ where $\tilde\gamma$ is the unique lifting strating at y. Another observation is that Stab(y)={$[\gamma]$$\in\pi_{1}(X, x_{0})$ : $\tilde\gamma$ is a closed loop}=$p_{*}(\pi_{1}(\tilde X, \tilde x_{0}))$ where $p_{*}$ is the induced homomorphism of $p$. Since this action is transitive, there is a bijection between $p^{-1}(x_{0})$ and $\pi_{1}(X, x_{0})$/$p_{*}(\pi_{1}(\tilde X, \tilde x_{0}))$. In our case $\tilde X$ is simply connected. So $p^{-1}(x_{0})$ and $\pi_{1}(X, x_{0})$ has the same cardinality. But $p^{-1}(x_{0})$ is closed and discrete subspace of compact space $\tilde X$. so it is finite, as desired.

But I couldn't show the converse. I am new in algebraic topology and I follow Hatcher's book chapter 1. Can any one give a hint or answer? Thanks!

Ergin Süer
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    Well, the fibres of the universal cover are going to be finite, so the projection will be open. So you can pushforward open sets and do things to them on the compact base... – Zhen Lin May 29 '13 at 16:19
  • Sorry, why should the fibres of the universal cover be finite? – Ergin Süer May 29 '13 at 16:33
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    Because they're in one-to-one correspondence with $\pi_1(X)$, as you yourself said above. :) But I don't understand @ZhenLin's comment. Covering maps, being local homeomorphisms, are always open maps. – Ted Shifrin May 29 '13 at 16:35
  • Sorry, that was a thinko. I confused it with the fact that local homeomorphisms (over a connected base) with finite fibres are covering maps. – Zhen Lin May 29 '13 at 16:40
  • @Zhen Lin still I didn't understand your last statement in your first comment. Can you explain it more please? – Ergin Süer May 29 '13 at 16:58
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    @ZhenLin, I believe your last remark is also flawed. This is one of my favorite trap questions in topology. You need more than finite fibers — e.g., compact domain. – Ted Shifrin May 29 '13 at 17:02
  • True. Local homeomorphism + finite fibres + surjective + connected base. – Zhen Lin May 29 '13 at 18:06

2 Answers2

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I find the following a useful way of thinking about it.

We may as well prove the statement that a covering map with finite fibers is a proper map (this means that the pre-image of any compact set is compact). (This argument is nice because it easily generalizes to prove the statement that a locally trivial fiber bundle with compact fibers is proper).

First note that when the covering map is just the projection $U \times F \rightarrow U$ then this map is proper because the preimage of a compact set $K \subset U$ is just $K \times F$ which is compact (being the product of compact spaces). In general, suppose that $\widetilde{X} \rightarrow X$ is a cover with finite fiber, $F$, then we may take a cover $\{U_\alpha\}$ of $X$ so that the restriction of the map to the cover just looks like $U_\alpha \times F \rightarrow U_\alpha$, and we've already seen that this is a proper map.

It is an easy exercise to see that properness (as I've defined it above) is local on the target, i.e. that the map $\widetilde{X} \rightarrow X$ is proper if and only if there is a cover of $X$ where all the restrictions are proper. (Indeed, any compact subset of $X$ is covered by finitely many $U_\alpha$, and so the preimage of $X$ is the union of the preimages of those specific $X \cap U_\alpha$. A finite union of compact sets is compact, QED.)

It follows that $\widetilde{X} \rightarrow X$ is a proper map. But now, if $X$ is compact, then the preimage of $X$, which is $\widetilde{X}$, must be compact, which was to be shown.

Dylan Wilson
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Let $\{ \tilde{U}_i : i \in I \}$ be an open cover of $\tilde{X}$. For each point $\tilde{x}$ in the fibre $p^{-1} \{ x \}$, choose an open neighbourhood $\tilde{U}_{\tilde{x}}$ of $\tilde{x}$. Since $p$ is a local homeomorphism, we may choose an open neighbourhood $\tilde{V}_{\tilde{x}} \subseteq \tilde{U}_{\tilde{x}}$ of each $\tilde{x}$ that $p$ maps homeomorphically onto an open neighbourhood $V_{\tilde{x}}$ of $x$, and then $V_x = \bigcap_{\tilde{x} \in p^{-1} \{ x \}} V_{\tilde{x}}$ is an open neighbourhood of $x$ such that $p^{-1} V_x$ is a disjoint union of homeomorphic copies of $V_x$, each contained in some $\tilde{U}_i$.

Observe that $p$ is surjective, so $\{ V_x : x \in X \}$ is an open cover of $X$, so has a finite subcover. Consider $\{ p^{-1} V_x : x \in X \}$. This is an open cover of $\tilde{X}$, but not necessarily subordinate to $\{ \tilde{U}_i \}$. Nonetheless, by construction, there is a finite refinement of $\{ p^{-1} V_x : x \in X \}$ that is subordinate to $\{ \tilde{U}_i \}$, and this implies the existence of the desired finite subcover. I leave the details to you.

Zhen Lin
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