I have the following problem to solve.
Find a covering of $S^1\vee S^1$ corresponding to the commutant of $\pi_1(S^1\vee S^1)$.
Ok, we know (I do more or less) that the universal cover of this bouquet of circles is Cayley graph, which can be represented as generated by two elements $a,b$ (up,down) on a plane with every step of half of length of before one.
$\pi_1(S^1\vee S^1)=\mathbb{Z}\star\mathbb{Z}=<a,b>$
I think that the only elements that commute with $\mathbb{Z}\star\mathbb{Z}$ are loops of the form $aba^{-1}b^{-1}$, but that's the smallest one. We alse have loops of different shapes and length. I claim that these are elements of the commutator, but I have difficulties with formalizing it and finishing..
In the end I would divide $\pi_1$ of this group by the commutator and consider what kind of space could it be..
Thanks in advance
Let $F = \pi_1(S^1 \vee S^1)$ be the free group on two generators. If by commutant you mean the commutator subgroup $[F,F]$, then you're just looking for the abelianization $F/[F,F]$ of $F$, which is $\mathbb{Z} \times \mathbb{Z}$, the free abelian group on two generators. (One cheap way to see this is to note that the abelianization of $\pi_1$ is $H_1$.)
Consider the Cayley graph of $F$. By quotienting out the commutator subgroup, we are identifying the endpoint of the path $a^n b^m$ with the endpoint of the path $b^m a^n$. If I'm not mistaken, this turns out to give the space obtained by the "integral lattice grid" in $\mathbb{R}^2$. One way to see this is the answer is to note that this cover is regular and its Deck group is precisely $\mathbb{Z} \times \mathbb{Z}$. See exercise 1.3.18 in Hatcher's Algebraic topology.