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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?

and

  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?

wormram
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    No because there is no injection $\Bbb Z/2\to\Bbb Z$ – pancini Jun 03 '21 at 20:57
  • Are you talking about (1)? Why does it follow from that? – wormram Jun 03 '21 at 21:30
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    @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 $\mathbb{Z}/2$. – wormram Jun 03 '21 at 22:51
  • I meant $\Bbb Z^2$ (and any homomorphism must pass through the abelianization) – pancini Jun 03 '21 at 23:00

2 Answers2

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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 ^_^

Chris Grossack
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For the second question, here's a slick way to show that $\mathbb{Z}/3$ doesn't inject into $\pi_1(K)$ which doesn't require you to know anything about semidirect products. (This proof was suggested to me by Kate Ponto.)

There is a two-sheeted cover of the Klein bottle by the torus, which gives rise to a short exact sequence

$$ \mathbb{Z} \times \mathbb{Z} \to \pi_1(K) \to \mathbb{Z}/2 \,. $$

Now if we have a homomorphism $\mathbb{Z}/3 \to \pi_1(K)$, it factors through $\mathbb{Z} \times \mathbb{Z}$ since

$$ \mathbb{Z}/3 \to \pi_1(K) \to \mathbb{Z}/2 $$

is zero. But $\mathbb{Z}/3$ only maps trivially to $\mathbb{Z} \times \mathbb{Z}$, so the map $\mathbb{Z}/3 \to \pi_1(K)$ is trivial.

Justin Barhite
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