Möbius strip covering space how can we describe the universal covering space of the Möbius strip?
the Möbius strip is a square $[0,1]\times [0,1]$ with identifications $(0,y)\sim (1,1-y)$.
So my guess is that the universal covering is an infinite strip $\Bbb R\times [0,1]$. How can we describe the map?
I am confused by the twist in the Möbius strip.
 A: The Möbius band is the quotient of $R\times [0,1]$ by the map defined by $f(x,y)= (x+1,1-y)$ which generates a freely and proper action on $R\times [0,1]$ this implies that the universal cover of the Möbius band is $R\times [0,1]$
A: Let me add the following variation to the argument provided above. 
If you consider $\theta_0:\mathbb{R}^2 \to \mathbb{R}^2$ by $\theta_0(x,y) = (x+1/2,-y)$then, the group $G = \{\theta_0^n:n \in \mathbb{Z}\}$ acts properly and discontinuously over $\mathbb{R}^2$, a fundamental domain is $[-1/2,1/2] \times [-1/2,1/2]$ and you can see that the identification performed over this fundamental domain correspond precisely to the identifications performed on a square to obtain the Moebius strip. 
Since the action is properly discontinuous, $\pi: \mathbb{R}^2 \to \mathbb{R}^2/G$ will be a covering map. Since, $\mathbb{R}^2$ is simply connected it is the universal cover of $\mathbb{R}^2/G$. Now by the properties of fundamental domains, what remains is to use the that every identification space is isomorphic to a quotient space so that you can prove that $\mathbb{R}^2/G \cong M$ where $M$ is the Moebius strip. 
Please feel free to complete the full explicitation of this identification exercise. 
