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We know that if we want to construct a space with a given fundamental group $G$ ,we can use cells and attaching maps, or fundamental domains and attaching maps, as in : How to determine space with a given fundamental group.

But there is a different way: if $M$ is, I think a manifold, and $G$ acts proper-discontinuously on $M$, then the quotient $ M/G$ has $G$ as a fundamental group. EDIT: As Seirios points out, $M$ may have to be simply-connected.

Now, can we always do this for any group $G$, i.e., can we always construct a space with fundamental group $G$ by defining a proper discontinuous action $M/G$? Can we always find $M$ so that the action is of the necessary type? Thanks.

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    $\begingroup$ I don't know about non-finitely-presented $G$; I suspect you must make some requirements on the size of $G$. Every finitely presented group is the fundamental group of a closed 4-manifold $M$ (and every closed manifold has finitely presented fundamental group); so take the universal cover of $M$ and consider the action of $\pi_1(M)$ on $\widetilde{M}$ by deck transformations. $\endgroup$ – user98602 Dec 5 '14 at 17:29
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    $\begingroup$ Given a countable atlas with convex charts, and a countable dense set $S \subset M$ every path is homotopic to chain of finitely many paths that are each a line segment in some chart between points of $S$. This is a countable set. So that cardinality restriction is there. $\endgroup$ – JHance Dec 5 '14 at 18:14
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    $\begingroup$ @MikeMiller: Therefore, any subgroup of $\pi_1(M)$, not necessarily finitely-presented, will work. In that way, we deduce from Higman's embedding theorem that any recursively presented group can be represented as the fundamental group of a (possibly open) manifold of dimension at least four. $\endgroup$ – Seirios Dec 6 '14 at 7:31
  • $\begingroup$ In your second paragraph, $M$ must be simply connected. $\mathbb{Z}$ acts proper-discontinuously on $S^1 \times \mathbb{R}$, for example, but the quotient has fundamental group $\mathbb{Z} \times \mathbb{Z}$. $\endgroup$ – MartianInvader Dec 8 '14 at 23:01
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You don't need any hypotheses on $M$. For any topological group $G$, we can construct a weakly contractible space $EG$ on which $G$ acts freely. The resulting quotient $BG$ is called the classifying space of $G$, and when $G$ is discrete it is an Eilenberg-MacLane space $K(G, 1)$, and in particular has fundamental group $G$. Both $EG$ and $BG$ can be constructed using the bar construction.

If you want $M$ to be a reasonable manifold then there are some size conditions on $G$: in particular, if you want $M/G$ to be a closed manifold then $G$ must be finitely presented. With this caveat, as Mike Miller says in the comments, it's known that any finitely presented group occurs as the fundamental group of some closed $4$-manifold, and you can take $M$ to be its universal cover.

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  • $\begingroup$ Do you know an easy example of a topological space $X$ such that $\pi_1(X)=S_3$? $\endgroup$ – rafa Jan 11 '17 at 17:24

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