# what is the covering space of figure eight which is corresponding to commutator subgroup.

Let $F$ be the free group on two generators and let $F^{'}$ be its commutator subgroup. Find a set of free generators for $F^{'}$ by considering the covering space of the graph $S^{1} \vee S^{1}$ corresponding to $F^{'}$ .

I know that this covering space should be free because any subgroup of free groups is also free.because of commutators I have a feeling that this covering space most be

Edit:there is something that make me a little confused,for $S^{1} \vee S^{1}$ the universal covering space $\widetilde{X}$ is

and if we take $\frac{\widetilde{X}}{F^{'}}$, this is abelian covering space of $F$,is it true that $\pi_{1}(\frac{\widetilde{X}}{F^{'}})=F^{'}$?

• It's a theorem that if you have a group $G$ acting properly discontinuously on a simply connected space $Y$, then $\pi_1(Y/G)=G$. See for instance Hatcher, Algebraic Topology, proposition 1.40.
– user98602
Apr 1, 2014 at 16:22
• @MikeMiller If I understood correctly, it also requires freeness on top of proper discontinuity (i.e. g.x = x => g is the neutral element) Feb 21, 2019 at 16:19
• @Evariste Indeed. I guess me from 5 years ago assumed freeness was implicit from the question.
– user98602
Feb 21, 2019 at 16:21