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I'm trying to solve a question which asks me to construct a covering map from $\mathbb{R}^2$ to the Klein bottle K and use it to show $\pi_1(K)$ is isomorphic to the group whose elements are pairs of integers with the non-abelian group operation given by

$$(m,n) \space\star\space (x,y) = (m\space+\space(-1)^nx,\space n+y)$$

I've constructed the covering map and am pretty sure I've found $\pi_1(K)$ to be the group $< x,y \space|\space xyx^{-1}y >$, but I can't seem to spot how to construct an ismomorphism between this and the given group.

If anyone would be able to point out how I'd get such an isomorphism I'd really appreciate it.

vapof
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    In your $\pi_1(K)$ you have the relation $xyx^{-1}=y^{-1}$. In the other group you have $$(0,1)\star(1,0)\star(0,-1)=(-1,0).$$ This gives you the homomorphism. Showing that it is surjective is easy. Showing that it is injective needs a bit more work. – Jyrki Lahtonen Dec 24 '19 at 20:33

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There are exactly two different semidirect products $\Bbb Z\rtimes \Bbb Z$. They are given by a group homomorphism $\psi:\mathbb{Z}\rightarrow Aut({\mathbb{Z}}) \cong \mathbb{Z}/2\mathbb{Z}$. So we have $\mathbb{Z} \rtimes_{n \rightarrow id}\mathbb{Z}\cong\mathbb{Z}^2$ and $\mathbb{Z} \rtimes_{n \rightarrow (-1)^nId} \mathbb{Z}$. Since the fundamental group of the Klein bottle is a non-abelian group, which is a semidirect product of $\Bbb Z$ by $\Bbb Z$, it must be isomorphic to your group.

Reference: What are the semi-direct products of $\mathbb{Z}$ with itself? (Check my work please)

Dietrich Burde
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