is every path-connected covering of the Moebius strip a Galois cover? 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.
 A: 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.
