Given a bouquet of 2 circles $B_2$, is there a way to find a covering space of $B_2$ such that the deck transformation group of our covering space is isomorphic $F_2 / <R>$. i.e. the group $G = \langle x,y |R_1,R_2,.. \rangle$?
In general I want to find a covering space of $B_2$ whose deck transformation group is the alternating group $A_4$. I know that the fundamental group of $B_2$ is the free group on two generators and I know I can write $A_4$ as such $A_4 = \langle x,y | x^2 = y^3 = (xy)^3 = 1 \rangle$. But I'm not really sure how to go about finding a covering space that satisfies this. Could I perhaps relate this to the Cayley graph of $A_4$?
I've been searching everywhere to find some sort of connection between the deck transformations and the relations of the free group but I've come up empty. I've been using Hatcher's algebraic topology.
Is this fact true in general that given a group $F_n / <R>$ we can find a covering space of $B_n$ whose deck transformation group is $F_n / <R>$?
Yes, this is true in general. Suppose that $X$ is a wedge of $k$ circles so that its fundamental group $F_k$ is free of rank $k$. Orient each loop. These loops, $x_1, \dots, x_k$, define a basis. Suppose we are given a group $G$ (such as $A_4$) and a surjective homomorphism $f:F_k \to G$. This defines a generating set for $G$, namely $f(x_1), \dots, f(x_k)$. Let $C$ be the Cayley graph of $G$ with respect to this generating set. Thus, the vertices are $G$ and there is an oriented edge $(g,gs)$ for each $g \in G$ and generator $s$. The edge is labeled $s$. The group $G$ acts on $C$ on the left via $h.(g,gs) = (hg,hgs)$.
Prove this is a covering space action. This is essentially Exercise 17 in Hatcher. Proposition 1.40 relates the deck group to $G$: they are isomorphic
The picture for $A_4$ is pleasing. Try to visualize the deck maps defined by the generators in the presentation you give above.