I'm brushing up on covering spaces for an upcoming exam, and I came across the following problems related to the Klein bottle:

  1. Can there exist a cover of the torus by the Klein bottle?


  1. Let $X$ be a connected and locally path connected space with $\pi_1(X)=\mathbb{Z}/3\mathbb{Z}$. Show the Klein bottle cannot be covered by $X$.

I'm mainly looking for hints here, not necessarily full solutions. For (1), should I be looking at the Galois correspondence for covers of the torus? And for (2), do I also want to look at the Galois correspondence, or would it be easier to show that $\mathbb{Z}/3\mathbb{Z}$ does not embed into the fundamental group of the Klein bottle $\langle a,b\mid abab^{-1}\rangle$?

Am I on the right track for these, or should I be doing something else?

  • 2
    $\begingroup$ No because there is no injection $\Bbb Z/2\to\Bbb Z$ $\endgroup$
    – pancini
    Jun 3, 2021 at 20:57
  • $\begingroup$ Are you talking about (1)? Why does it follow from that? $\endgroup$ Jun 3, 2021 at 21:30
  • 1
    $\begingroup$ @AndresMejia I know that, but I don't see what that has to do with $\mathbb{Z}/2$ and $\mathbb{Z}$. The fundamental group of the torus is not $\mathbb{Z}$, and the fundamental group of the Klein bottle is not $\mathbb{Z}/2$. $\endgroup$ Jun 3, 2021 at 22:51
  • $\begingroup$ I meant $\Bbb Z^2$ (and any homomorphism must pass through the abelianization) $\endgroup$
    – pancini
    Jun 3, 2021 at 23:00

1 Answer 1


Since you ask for hints:

The easiest way (in my mind) to see the answer to $1$ is "no" is that covering spaces of an orientable space are orientable. If you like, you can consider fundamental groups instead. If $K$ covered $T$, then by the galois correspondence we would need to have $\pi_1 K$ as a subgroup of $\pi_1 T$. But every map from $\pi_1 K = \langle a, b \mid abab^{-1} \rangle$ to $\pi_1 T$ sends $a \mapsto 0$. Do you see why?

For your second question, I think your idea with embeddings of fundamental groups is a good one. It's "well known" that $\pi_1 K$ is torsion free. In particular, there are no elements of order $3$. Can you show this? It's a bit tricky, so I'll leave a hint under the fold:

Can you show $\pi_1 K = \mathbb{Z} \ltimes \mathbb{Z}$? Is it clear that this is torsion free? It might be helpful to rewrite $\pi_1 K = \langle a, b \mid bab^{-1} = a^{-1} \rangle$.

I hope this helps ^_^


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