(June ’05) Show, using covering spaces, that the fundamental group of the Klein bottle is not abelian.
Solution
Use the standard universal covering \(\mathbb R^2\to K\) of the Klein bottle. Its deck group is generated by \[ T(x,y)=(x+1,y), \qquad S(x,y)=(-x,y+1). \] These transformations generate the usual Klein-bottle identifications, and \(\mathbb R^2\) is simply connected, so the deck group is naturally isomorphic to \(\pi_1(K)\).
A direct computation gives \[ STS^{-1}=T^{-1}. \] Indeed \(S^{-1}(x,y)=(-x,y-1)\), hence \[ STS^{-1}(x,y) =S(-x+1,y-1) =(x-1,y)=T^{-1}(x,y). \] Since \(T\ne T^{-1}\), the transformations \(S\) and \(T\) do not commute. Therefore the deck group, and hence \(\pi_1(K)\), is nonabelian.