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Suppose we have a cover of $X = S^1 \vee S^1$ by the set $\widetilde{X} = \{(x,y) \in \mathbb{R}^2 \; \vert \; x \in \mathbb{Z} \text{ or } y \in \mathbb{Z} \}$, the integer grid in the plane. The covering map $p$ wraps vertical lines around the left circle, and horizontal lines around the right circle, say.

We have $\pi_1(X) = \langle a, b \rangle$ and $\pi_1(\widetilde{X})$ is the free group on countably many generators.

Say $\pi_1(\widetilde{X}) = \langle \gamma_{m,n} \rangle$ where $\gamma_{m,n}$ is the path that goes $n$ units up, $m$ units right, traverses the square at lower left corner $(m,n)$ and returns.

Then we have $$ p_*(\gamma_{m,n}) = a^n b^m (aba^{-1} b^{-1}) b^{-m} a^{-n} $$

I want to show that $p_* \pi_1(X)$ is the commutator subgroup of $\pi(X)$, but I cannot see how I can show that given the presentation by generators of the form written above. Certainly $p_*$ sends every generator to an element of the commutator subgroup, but I have not been able to see that its image actually is the whole commutator subgroup.

(If I can get it to be the commutator subgroup, then I can obtain that the automorphism group of the covering space is $\mathbb{Z} \times \mathbb{Z}$, which makes sense because this would correspond to integer shifts of the grid -- this is why my intuition suggests that I should be able to get $p_* \pi_1(X)$ to be $\langle a b a^{-1} b^{-1} \rangle$).

Any hints would be greatly appreciated.

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You have a group homomorphism $p_*\colon \pi_1(\tilde X)\to [\pi_1(X),\pi_1(X)]$. Notice that the commutator subgroup on the right is generated by elements $[a^n, b^m]$.

In order to show that the map is surjective, it suffices to see that each generator is hit. But $[a^n,b^m]$ is hit by the path that goes $n$ units up, $m$ units right and then runs back along the opposite edges to make a rectangle.

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  • $\begingroup$ Thanks, this was enough for me to flesh out into a proof. I appreciate it! $\endgroup$ – cemulate Nov 16 '16 at 0:50

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