Fundamental group of a compact space with compact universal covering space 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!

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.
