# Problem 1.3.12 from Hatcher

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.

• 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: $S^1\wedge S^1$. – Arthur Mar 3 at 8:40
• I didn't know that, thanks. – Issacg628496 Mar 3 at 18:37