Find the rank and the free generators Consider the homomorphism$ \ $ $f:\ F\{x,y\} \to  <x,y|x^2, y^3, xyx^{-1}=y^{-1}>$, find the free generators of $kerf$.
I know that we should first consider the wedge sum of circles whose fundamental group is $F\{x,y\}$, then consider the covering space of the wedge sum of the circles which corresponds to the subgroup $kerf$. But how should I find the corresponding covering space, is there a general algorithm for this sort of problem?
Edit: I tried the following approach.
Step 1: Draw the Caley graph of the cosets 

Step 2: finding the free generators according to the caley graph.
If I start at Hx, then the free generators should be $y^3,x^2,yx^2y^{-1},yxy^2x^{-1},y,yxyxy^{-1},xyx^{-1}y^{-1}$
Am I right?
 A: In general, let $S$ be a set, and $X_S$ a wedge sum of circles labeled with $S$ (which we see as a graph). Then for any group $G$ generated by $S$ (as a monoid), you have a covering map $\Gamma_G\to X_S$ where $\Gamma_G$ is the Cayley graph of $G$ (of course $\Gamma_G$ depends on $S$) given by sending all vertices of $\Gamma_G$ to the only vertex of $X_S$, and all edges labeled by $s\in S$ to the only edge of $X_S$ labeled with $s$. Then the group of this cover is $G$.
Then in your case, $S=\{x,y\}$, and you have two groups generated by $S$ : your $G$, and the free group $F_S$. Clearly the Cayley graph $Y_S$ of $F_S$ is the universal cover of $X_S$, since it's a tree. So you have a tower of coverings $Y_S\to \Gamma_G\to X_S$.
Since the group of $Y_S\to X_S$ is $F_S$ and the group of $\Gamma_G\to X_S$ is $G$, the group of $Y_S\to \Gamma_S$ is $\ker(f)$, so $\ker(f) = \pi_1(\Gamma_G)$. 
Now you have to find generators of $\pi_1(\Gamma_G)$. As explained for instance in the Lyndon-Schupp, this is done by choosing a maximal subtree in $\Gamma_G$ and contracting it to a point : the edges that remain are your generators. Looking at the labels along each one should give you explicit elements of $F_S$ that generate $\ker(f)$.
