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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 enter image description here

but I don't know my Idea is right or wrong?please help me with your knowledge,thank you very much.

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

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

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    $\begingroup$ 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. $\endgroup$
    – user98602
    Apr 1, 2014 at 16:22
  • $\begingroup$ @MikeMiller If I understood correctly, it also requires freeness on top of proper discontinuity (i.e. g.x = x => g is the neutral element) $\endgroup$
    – Evariste
    Feb 21, 2019 at 16:19
  • $\begingroup$ @Evariste Indeed. I guess me from 5 years ago assumed freeness was implicit from the question. $\endgroup$
    – user98602
    Feb 21, 2019 at 16:21

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