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We know that the fundamental group of $S^1\vee \Bbb RP^2$ is $\langle a,b\mid b^2\rangle$. I am attemping to find the covering spaces of $S^1\vee \Bbb RP^2$ corresponds to:

(a) the subgroup $\langle a\rangle$ of $\langle a,b\mid b^2\rangle$

(b) the subgroup $\langle b\rangle$

(c) the subgroup $\langle a, bab\rangle$

For (a) I think the double cover $S^1\vee S^1\vee S^2$ has the desired fundamental group. But it looks like a normal covering space and $\langle a\rangle $ is not a normal subgroup. So is it something wrong here?

For (b) I have found the covering space corresponds to the normal subgroup generated by $b$, it is a helix with an $\Bbb RP^2$ at each integer point. But no idea for the subgroup generated by $b$.

For (c) I have no idea.

May I please ask how can I find them? Thanks!

EDIT: Now I think for the covering space corresponds to $\langle b\rangle$, recall the universal cover of $S^1\vee \Bbb RP^2$, it is an infinite tree with an $S^2$ attaching on each vertices. I think the desired covering space is to link an $\Bbb RP^2$ with two branches linking to the tree which looks like the universal cover of $S^1\vee\Bbb RP^2$.

To construct a covering space corresponds to $\langle a\rangle$, first considering the universal cover of $S^1\vee\Bbb RP^2$, fix a "central" sphere $S^2$, it has four branches, keep two of them unchanged, and for the other two branches, remove them and attach an $S^1$ instead.

Is that correct? I think it is quite hard to discribe...

Thanks for patience of reading this. And thanks for any help.

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  • $\begingroup$ $\mathbb RP^1 \cong S^1$, so the universal cover should just be $F_2$ in your edit. $\endgroup$ Oct 20 '17 at 17:41
  • $\begingroup$ @AndresMejia Given the context, don't you think that was a typo? $\endgroup$ Oct 20 '17 at 19:35
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I like to work out covering spaces by enumerating cosets. Recall that the fiber above the base point for a covering space associated with a subgroup $H\subset \pi_1(X)$ is in one-to-one correspondence with the set of cosets $\pi_1(X)/H$. Let $X=S^1\vee\mathbb{R}P^2$.

(c) is the most concrete without pictures, so I'll start here. Notice $H=\langle a,bab\rangle$ is a normal subgroup since $bab=bab^{-1}$. In the quotient $\pi_1(X)/H$, $a$ is trivial and $b$ remains, so it is $\mathbb{Z}/2\mathbb{Z}$. This is saying that $a$ lifts to a loop but $b$ does not. This must correspond to the cover $S^1\vee S^2\vee S^1$ (unfurling the $\mathbb{R}P^2$).

For (b) what you say is correct. I drew a Scheier graph for the subgroup $\langle b\rangle$, and it shows that it is the universal cover with the $b$ direction folded back. So, an $\mathbb{R}P^2$ in the middle, with the universal covers in each $a$ and $a^{-1}$ direction.

For (a), it might be easier to describe it as an $S^1$ attached to the $b$ direction of the universal cover. I imagine it as the universal cover rolled up in the $a$,$a^{-1}$ directions.

Coset enumeration is the same thing as taking the universal cover and folding it up based on the subgroup, which can be useful for group theory.

Perhaps this drawing of the universal cover will help:

universal cover

This is obtained from the Cayley graph of the fundamental group generated by $a,b$ by replacing each $b$ edge with an $S^2$.

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