Finitely generated group acting cocompactly on a Manifold of bounded geometry

Question

Want a justification of the following fact.

If G is a finitely generated group, then there exist a Riemannian manifold $M$ such that the action of $G$ is cocompact isometric properly discontinuous.

To give such an example the author justify that if $G$ is finitely generated say by $k$ generators then there exist a epimorphism from $\phi:\pi_1(F_k)\rightarrow G$,where $F_k$ is genus $g$ riemann surface.Then G acts isometrically and cocompactly on the covering space M of $F_k$ so that $\pi_1 (M) = ker(\phi)$

The line in bold I did not understood properly.

Answer 1

To expand on my comment, here are three theorems of covering space theory which you can look up. Learn that theory and you'll have your answer.

Let $X$ be any path connected, locally path connected, semilocally simply connected space. Let $p \in X$ be a base point.

1. For every covering map $f : M \to X$ and base point $q \in M$ such that $f(q)=p$, the induced map on fundamental groups $f_* : \pi_1(M,q) \to \pi_1(X,p)$ is injective.
2. For every subgroup $H < \pi_1(X,p)$ there exists a covering map $f : M \to X$ and base point $q \in M$ such that $M$ is path connected, $f(q)=p$ and $H = \text{image}(f_*)$.
3. Furthermore, in 2, if $H$ is a normal subgroup then the deck transformation group of the covering map $f:M \to X$ is isomorphic to the quotient group $\pi_1(X,p)/H$.

In your question, take $X=F_k$ and $H = \text{kernel}(\phi)$, take $M$ to be the covering space in item 2. By 3, $G = \pi_1(X)/H$ acts on $M$ as the deck transformation group.