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Let a and b be the generators of $\pi_1$ $(S^1 \vee S^1)$ corresponding to the two $S^1$ summands. Draw a picture of the covering space of $S^1 \vee S^1$ corresponding to the normal subgroup generated by $a^2$, $b^2$, and $(ab)^4$, and prove that this covering space is indeed the correct one.

I know the standard way to do this is constructing an octagon, like so, but I was wondering if there are other ways or more interesting ways to do it.
enter image description here

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    $\begingroup$ The $\LaTeX$ command for the wedge sum is \vee: $S^1\vee S^1$ (probably so named because it looks like a "v"). Sadly, \wedge goes the wrong way, giving you the smash product: $S^1\wedge S^1$. $\endgroup$ – Arthur Mar 3 at 8:40
  • $\begingroup$ I didn't know that, thanks. $\endgroup$ – Issacg628496 Mar 3 at 18:37

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