Klein bottle is a $K(G, 1)$, fundamental group is torsion-free?

Question

I have two questions.

1. How do I see that the Klein bottle is a $K(G, 1)$ with $G$ fundamental group?
2. Does it follow from this that the fundamental group of a Klein bottle is torsion-free?

Answer 1

The universal cover of a Klein bottle $K$ is $\mathbb{R}^2$, which has trivial higher homotopy groups.

There is a proposition (see for example Hatcher's book Proposition 4.1) which states that a covering space projection $p:(\tilde{X},\tilde{x})\to(X,x)$ induces isomorphisms $p_*:\pi_n(\tilde{X},\tilde{x})\to\pi_n(X,x)$ for $n\geq 2$. The proof of this consists only of elementary covering theory involving the lifting criterion. In any case, this means that the higher homotopy groups of the Klein bottle are trivial as well. Thus the only non-trivial homotopy group of $K$ is its fundamental group $G$, which makes it a $K(G,1)$.

Now there is another result which says that if a finite CW complex is a $K(G,1)$, then $G$ is necessarily torsion-free (see Hatcher's book Proposition 2.45). The proof is once again very straightforward using homology. Basically if $G$ has torsion, there will be a finite cyclic subgroup $\mathbb{Z}/m$. The cover corresponding to this subgroup is thus a $K(\mathbb{Z}/m,1)$ (by the proposition in the previous paragraph and the Galois correspondence for covering spaces). Since $K$ is a finite-dimensional CW complex, the same must be true of its covering space. In particular, the homology of $K(\mathbb{Z}/m,1)$ vanishes after a certain dimension. However, we know that the infinite dimensional lens space $S^\infty/(\mathbb{Z}/m)$ is a model of $K(\mathbb{Z}/m,1)$, and a calculation of the cellular homology of such spaces show that they have infinitely many non-trivial homology groups.