9
$\begingroup$

I have this problem for Riemannian manifold, but think that it is just a topological problem. I know that this is probably a silly question, but it is since a while that I don't study general topology and algebraic topology..

Let $X$ be a compact topological space and assume that its universal covering space $\tilde{X}$ is also compact. How can I prove that the fundamental group of $X$ is finite?

Thanks!

$\endgroup$

2 Answers 2

12
$\begingroup$

Consider the covering map $\pi \colon \tilde{X} \rightarrow X$. Above any $p \in X$, the fiber $\pi^{-1}(p)$ is a discrete closed subset of a compact space $\tilde{X}$ and so must be finite. By the general theory of covering spaces, if we fix some $\tilde{p} \in \tilde{X}$ with $\pi(\tilde{p}) = p$ then we obtain a bijection between $\pi_1(X,p)$ and $\pi^{-1}(p)$ given by sending (the homotopy class of) a based loop $\gamma \colon [0,1] \rightarrow X$ at $p$ to $\tilde{\gamma}(1)$ where $\tilde{\gamma} \colon [0,1] \rightarrow X$ is the unique lift of $\gamma$ to $\tilde{X}$ satisfying $\tilde{\gamma}(0) = \tilde{p}$. Thus, $\pi_1(X,p)$ is finite.

$\endgroup$
5
$\begingroup$

The action of $\pi_1(X)$ on $\tilde X$ is proper and free. Suppose that $\pi_1(X)$ is infinite, consider $(f_n)$ pairwise distinct elements of $\pi_1(X)$, for $x\in \tilde X$ the sequence $f_n(x)$ has an accumulation point. Contradiction since the action of $\pi_1(X)$ on $\tilde X$ is proper.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .