If a Klein bottle covers a manifold $M$, then $M$ is the Klein bottle

Question

I have to prove that if a Klein bottle covers a manifold $M$, then $M$ is the Klein bottle. Any suggestions? Thanks.

Answer 1

Suppose $K\to M$ is a covering, with $K$ a Klein bottle. Then $M$ is a compact surface. If the degree of the covering is $d$, we have that $\chi(K)=d\cdot\chi(M)$, with $\chi(M)$ and $\chi(K)$ the Euler characteristics of $M$ and $K$, respectively. Since $\chi(K)=0$ (and, of course, $d\neq0$) we have $\chi(M)=0$.

The classification of compact surfaces implies then that $M$ is either a torus or a Klein bottle.

But if $M$ were a torus, it would be orientable, and a standard property of covering spaces would then tell us that $K$ is itself orientable, which it isn't.

Therefore $M$ is a Klein bottle, as we wanted it to be.