Let $R_2$ the rose with $2$ petals, that is the wedge of $S^1$ with itself. We know its fundamental group is the free group with two elements, $\pi_1(R_2)=F_2=\langle a,b\rangle$. Now given some subgroup $H=\langle a^3,ba,aba^2,a^2b^2a \rangle$. I want to find the covering space $X_H$ of $X$ associated to $H$ by the Galois Correspondence of subgroups and covering spaces. I know the universal covering space $\widetilde{X}$ of the rose of $n$-petals is the Cayley Graph of the free group on $n$ elements. Moreover, I know that to get $X_H$ we quotient the $\widetilde{X}$ by $[\gamma]~[\gamma']$ whenever $\gamma(1)=\gamma'(1)$ and $[\gamma\gamma'^{-1}]\in H$, since we may see $\widetilde{X}$ as the space of paths to $R_2$. Still, I was not able to use this fact to obtain a concrete graph $X_H$. I wonder if there's a general procedure to obtain our space $X_H$ given a graph $H$, or, at least, in this specific case.
Given a subgroup of a free group, find the associated covering space.
algebraic-topologycovering-spacesfree-groupsgroup-theory
Related Solutions
Subdivide the universal covering space $\mathbb R^2$ in the usual manner as squares, with vertical lines $x=m$ and horizontal lines $y=n$ for integers $m,n \in \mathbb Z$.
To visualize the desired covering space, draw two vectors based at the origin: $v = \langle 3,0 \rangle$ corresponding to $a^3$; and $w = \langle 2,1 \rangle$ corresponding to $a^2 b$. Let $P$ be the parallelogram determined by the vectors $v,w$, which form two sides of $P$, the other two sides then being determined. Now glue opposite sides of $P$ to form a quotient space $S$. And as usual, gluing opposite sides of the unit square $Q = [0,1] \times [0,1]$ gives the covering space $T$.
You can then use the pattern of intersections of the parallelograph $P$ with unit squares $[m,m+1] \times [n,n+1]$ to define the desired covering map $Q \mapsto T$.
It is theoretically more straightforward to view this construction using orbit spaces of deck transformations. If I have $(a,b) \in \mathbb R^2$ let me use $\tau_{(a,b)}$ to represent the translation $\tau_{a,b}(x,y) (x+a,y+b)$. Thus we can think of $T$ as the quotient of $\mathbb R^2$ by the action of the deck group $\langle \tau_{(1,0)},\tau_{(0,1)} \rangle$ (with fundamental domain $[0,1] \times [0,1]$), so $$T = \mathbb R^2 / \langle \tau_{(1,0)},\tau_{(0,1)} \rangle $$ and we can think of $S$ as the quotient of $\mathbb R^2$ by the action of the deck group $\langle \tau_{(3,0)}, \tau_{(2,1)} \rangle$ (with fundamental domain $P$), so $$S = \mathbb R^2 / \langle \tau_{(3,0)}, \tau_{(2,1)} \rangle $$ This way, the desired quotient map $S \mapsto T$ can be precisely defined as the map $$S = \mathbb R^2 / \langle \tau_{(3,0)}, \tau_{(2,1)} \rangle \mapsto = \mathbb R^2 / \langle \tau_{(1,0)},\tau_{(0,1)} \rangle = T $$ that is induced by the identity map $\mathbb R^2 \mapsto \mathbb R^2$.
I am afraid that this answer might not be what you are looking for, because I am thinking algebraically rather than topologically.
In your example, the subgroup is the commutator subgroup $[F,F]$. The vertices of the Schreier graph can be labelled by a transversal to the subgroup, and the obvious choice is $\{a^ib^j: i,j \in {\mathbb Z}\}$. You have edges labelled $a$ and $b$ from $a^ib^j$ to $a^{i+1}b^j$ and $a^ib^{j+1}$, respectively.
The associated free subgroup generators (often called the Schreier generators) are $\{ a^ib^jab^{-j}a^{-i-1} : i,j \in {\mathbb Z}, j \ne 0 \}$.
In a general problem of this type, the vertices of the Schreier graph correspond to the elements of the quotient group $F/\langle S^F \rangle$ of $F$, which is the group defined by the presentation $\langle a,b \mid S \rangle$.
Best Answer
There is a general procedure, which you can work out on a blackboard. It's a fair amount of work, but fun! ... if you like that kind of thing... which I do!
EDITED: I've rewritten my answer to simplify the procedure.
Step 1: Using the first word $a^3 = a a a$, draw a loop with base point $p$ subdivided into three oriented edges labelled $a$. Let $G_1$ denote this labelled graph.
Step 2: Enfold the second word $ba$ into $G_1$. To do this, first draw a second loop based at $p$ subdivided into two oriented edges labelled $ba$. But now, at $p$, there are two oriented edges labelled $a$ that terminate at $p$: one in $G_1$ and the other in the new loop. Fold these two edges together. The result is a new oriented graph $G_2$, containing $G_1$. One of the original points of $G_2$ will now be labelled $q$, and in $G_2$ we have: an edge from $p$ to $q$ subdivided into two oriented edges labelled $a$, $a$; and an oriented edge from $q$ to $p$ labelled $a$; and an oriented edge from $p$ to $q$ labelled $b$.
Step 3: Enfold the third word $aba^2$ into $G_2$. To do this, first draw a new loop based at $p$ and subdivided into four oriented edges labelled $a$, $b$, $a$, $a$. At $p$, there is an oriented edge in $G_2$ pointing away from $p$ labelled $a$, and an oriented edge in the new loop pointing away from $p$ labelled $a$; fold those edges together. Also at $p$, there is an oriented path in $G$ pointing towards $p$ labelled $aa$, and an oriented path in the new loop pointing towards $p$ labelled $aa$; fold those paths together. The result is a new labelled graph $G_3$, containing $G_2$. The last unnamed vertex of $G_2$ now gets a name, call it $r$, and at $r$ there is an oriented loop attached to $r$ at both ends labelled $b$.
Step 4 (and final step): Enfold the fourth word $a^2 b^2 a$ into $G_3$. To do this, draw a new loop based at $p$ subdivided into five oriented edges labelled $a$, $a$, $b$, $b$, $a$. In both $G_3$ and the new loop there is an oriented path starting from $p$ and labelled $a$ $a$; fold those two paths together. Also, in both $G_3$ and the new loop there is an oriented path ending at $p$ and labelled $b$ $a$; fold these two paths together. The effect is to create a new labelled graph $G_4$, which contains $G_3$, and is obtained form $G_3$ by attaching an oriented edge from $q$ to $p$ labelled $b$.
You can now check that the label map $G_4 \to X$ is an (irregular) covering map of degree $3$, and that the induced fundamental group monomorphism has image equal to $H$. So $G_4$ is the desired $X_H$.
And, by the way, as a bonus you get a proof that $H$ is a non-normal subgroup of index $3$ in $\langle a,b \rangle$.
EPILOGUE: We got lucky on this problem, in that $G_4$ was indeed a covering space of $X$. When we carry this process out on a general (finitely generated) subgroup, the resulting graph $G$ will have a locally injective map $G \mapsto X$, but it may fail to be a local homeomorphism and hence will not be a covering map. As it turns out, though, it will always be the "core" of a covering map: by attaching infinite trees to the vertices in an appropriate manner we will always get a covering map. This situation happens when the given subgroup has infinite index in $\langle a,b \rangle$.