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Let $p : E → M$ be a covering of $M$ the Moebius Sttrip such that $E$ is path connected.

Is this a Galois covering?

My intuition is there must be some non locally path connected coverings that are not Galois. I know little about coverings of Moebius band. I know that its universal covering is $\Bbb R\times [0,1]$.

Thanks for your help.

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    $\begingroup$ Since $M$ deformation retracts onto $S^1$ and a space's theory of covering spaces is determined by its fundamental group, I think you can equivalently ask the same for $S^1$ to simplify things. $\endgroup$ – William Feb 22 at 18:23
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    $\begingroup$ Locally, a covering will look like the Moebius strip, hence it can't be nonlocally path connected. Now by William's comment, the fundamental group is abelian, so path connected coverings are all Galois $\endgroup$ – Max Feb 22 at 18:53
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Hint: The inclusion $S^1 \to M$ as the zero section induces a bijection

$$ Cov(M)/\cong \to Cov(S^1)/\cong $$

where $Cov(X)/\cong$ is the set of isomorphism classes of covering spaces. Moreover a covering space of $M$ is connected (respectively regular) iff its restriction to $S^1$ is connected (resp. regular).


Solution to problem over $S^1$:

The connected covering spaces of $S^1$ are isomorphic to quotients of $\mathbb{R}$ by subgroups of $\mathbb{Z}$. The $n$-fold cover $\mathbb{R}/n\mathbb{Z} \to S^1$ has automorphism group $\mathbb{Z}/n\mathbb{Z}$ which acts transitively in each fibre.

I think you can run a similar argument on the universal cover $\mathbb{R}\times [0,1]$ of $M$. The covering is the quotient map of the relation $$(x, t) \sim (x + n, f^{(n)}(t))\text{ for }n\in\mathbb{Z}$$ where $f$ is the "flip" homeomorphism of the interval. The deck transformations are again $\mathbb{Z}$, which acts by $n\cdot(x, t) = (x + n, f^{(n)}(t))$.


Update: as per Max's comment to the original question, this also follows from the general theory: if $X$'s fundamental group is abelian then all its connected covers are regular.

The idea is that all the connected covers are quotients of the universal cover $p\colon\tilde{X}\to X$ by subgroups of $G=\pi_1(X)$. If $H$ is a subgroup of $G$ then the fibres of $\tilde{X}/H \to X$ will have an action of $G$ that looks like its action on the quotient set $G/H$; if moreover $H$ is normal then this action descends to the group $G/H$ and the fibres are "$G/H$-torsors" (i.e. have a free, transitive action of $G/H$) and in fact $G/H$ is isomorphic to the group of deck transformations, hence the covering is regular. Therefore if all subgroups are normal then all connected covers will be regular.

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  • $\begingroup$ Thank you @William. I am trying to understand why connected covers are quotients of the universal cover by subgroups of $G=π_1(X)$. Is it because of the bijection between coverings isomorphism classes and $π_1(X)$ subgroups conjugacy classes? the theory behind is not clear in my head. $\endgroup$ – PerelMan Feb 23 at 0:12
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    $\begingroup$ Yes it is because of this bijection. Given a subgroup $H$ we get a connected covering $\tilde{X}/H$, and if we had chosen a conjugate subgroup $H'$ the resulting covering spaces would be isomorphic. Conversely, given a connected covering space $p\colon P\to X$ it is isomorphic to the quotient $\tilde{X}/p_*(\pi_1 P)$. For a reference on the relation between normality of $H$ and regularity of $\tilde{X}/H$, check out Proposition 1.39 in the Fundamental Group chapter of Hatcher (page 71). $\endgroup$ – William Feb 23 at 2:40
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    $\begingroup$ Actually both of the sections "Classification of Covering Spaces" and "Deck Transformations and Group Actions" in Hatcher (starting page 63) are very relevant to this discussion. $\endgroup$ – William Feb 23 at 3:17

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