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Let $G$ be the group of homeomorphisms of $\mathbb{R^{2}}$ generated by the two elements: $g:(x,y)\rightarrow (x+1,y),h:(x,y)\rightarrow(-x,y+1)$.

Then it is clear that this group is isomorphic to $G=\langle g,h | h^{-1}ghg\rangle$.

This group acts on the group $\mathbb{R}^{2}$ and therefore the projection $\phi:\mathbb{R^{2}}\rightarrow \mathbb{R^{2}/}G$ is a covering space for the Klein bottle. where $\mathbb{R^{2}/}G=\lbrace G*(x,y)|(x,y)\in \mathbb{R^{2}}\rbrace$.

My question is there any way to calculate the fundamental group of the Klein bottle using the previous covering space.

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The concept you're probably looking for is the deck transformation group of a covering space. In this case, your group $G$ happens to be an invariant composite of the covering $\phi$, that is, for all $\rho\in G$, the composition $\phi\circ\rho=\phi$. It follows that, as $\mathbb{R}^2/G$ is the full orbit space, it must be that $G$ is the full deck transformation group for the covering $\phi$.

Now, $\mathbb{R}^2$ is a simply connected topological space (in fact it's contractible) and so it is not just a cover but a universal cover of the Klein bottle. Universal covers have the nice property of always being regular coverings (the induced action of the deck transformation group on the fiber of the cover is both free and transitive) and so the deck transformation group $G$ is isomorphic to the fundamental group $\pi_1(\mathbb{R}^2/G)$ of the Klein bottle.


The justification of the last claim is easiest to describe using the theory of principle $G$-bundles although this might be a bit out of the scope of your current course so feel free to ignore this if you get lost (I believe Hatcher gives a very good elementary explanation of the isomorphism without reference to bundles, so feel free to check his textbook). For completeness though, we'll show that $G\cong\pi_1(\mathbb{R}^2/G)$. Let $F$ be the fiber of the map $\phi$. The principle $G$-bundle $\phi$ induces a short exact sequence in homotopy $$\pi_1(\mathbb{R}^2)\stackrel{\phi_*}{\longrightarrow}\pi_1(\mathbb{R}^2/G)\longrightarrow\pi_0(F)\longrightarrow\pi_0(\mathbb{R}^2).$$ Now $\pi_1(\mathbb{R}^2)$ is trivial, and $\mathbb{R}^2$ is path connected so $\pi_0(\mathbb{R}^2)$ is trivial. We know that $\pi_0(F)$ is actually a group and is isomorphic to $G$ because $\phi$ is a principle $G$-bundle. It follows that we have the short exact sequence $$0\stackrel{\phi_*}{\longrightarrow}\pi_1(\mathbb{R}^2/G)\longrightarrow G\longrightarrow 0$$ and so we have an induced isomorphism $\pi_1(\mathbb{R}^2/G)\stackrel{\cong}{\longrightarrow} G$.

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  • $\begingroup$ this is very nice, thanks a lot. $\endgroup$ – kiranovalobas Feb 28 '13 at 15:09

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