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Can any one give me any idea about how to solve this problem?
Suppose we have a simplicial complex G which is finite connected.
(1)The fundamental group of G is finite;
(2)The universal cover of G is compact.
If (1) is true, can we get (2)?
If (2) is ture, can we get (1)?

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up vote 1 down vote accepted

Consider a subgroup $H$ of $\pi_1(G)$. By classification of the covering spaces, the index of $H$ in $\pi_1(G)$ is the number of the sheets of the cover corresponding to $H$. Now, order of the group is obviously equal to the index of the trivial subgroup, and universal cover is the one that corresponds to the trivial subgroup, thus the order of $\pi_1(G)$ is equal to the number of the points in the fiber of the universal cover. Now, fibers of the covering maps are discrete. They are also closed, because they are preimages of the closed set (a point). Since the closed discrete subspace of a compact space is finite, the fibers are finite, so the fundamental group is also finite. This gives you $(2) \implies (1)$.

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I think this proof is correct, how about "(1) implies (2)"? Is it correct? – user63664 Mar 23 '13 at 14:43

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