# Covering space calculation of figure eight.

I'm trying to do this calculation in Hatcher.

So for the (1). I imagine cutting the loop at $a$ on the left call this Y and cutting the loop $a$ on the right call this Z.

This will give you two subsets that you can apply Van Kampen to. The intersection would be homotopy to the loop generated by $b^2$.

As if you do the intersection of both then you are just essentially cutting out the loop two sides loops leaving you with the middle.

So tracing the possibility I think the answer is this

$\pi_1(Y \cup Z)=\mathbb{Z}\langle b^2,bab,ba^{-1}b,b^{-1}a^{-1}b^{-1},\ldots \rangle \ast \mathbb{Z}\langle b^2,a,a^{-1},b^{-2}\rangle /\mathbb{Z}\langle b^2\rangle.$

See I don't see how he gets it like that picture with only three generators. Surely, for example that you should have $bab$ as one of the generators on it i.e. go through the b, then go through a, then go back through b. Why isn't this allowed?

Sorry, but I'm confused on how he actually does this calculations. I know how to show for example $\pi_1(S^{1}\vee S^{1}) \cong \mathbb{Z} \ast \mathbb{Z}$ so I'm not a complete noob. But, this just confuses me.

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$bab = (bab^{-1})(b^2)$, so that's already accounted for. And note that $bab^{-1} = (bab)(b^2)^{-1}$. –  Dylan Moreland May 5 '12 at 20:29
@Dyland I can see that. Where does $a^{-1}$ go as I'm not factoring that out. Wonder where that disappears. I suppose you are assuming that the if you have a word a you can get the inverse of the word so then it's really is a group. So I'm assuming if you have a generator $a$ you get $a^{-1}$ in the language, but Hatcher isn't including it as it's redundant. –  simplicity May 5 '12 at 20:33