I've recently heard that the Möbius strip (without boundary) is a covering space of the Klein bottle. Honestly, I cannot see that. Any help to check this fact?
Möbius strip covering space of Klein bottle
algebraic-topologycovering-spacessurfaces
Related Solutions
Imagine the Möbius strip as the unit square, where $(0,x)$ is identified with $(1,1-x)$. The line $\{(x,\frac{1}{2}):x\in[0,1]\}$ is a circle as a subspace of the Möbius strip. Then the map $H:M\times[0,1]\to M$ given by $H((x,y),t)=(x,(1-t)y+\frac{t}{2})$ gives a strong deformation retract of $M$ to the circle.
Below is a picture of the Klein bottle cut into two Möbius strips (along the orange line), so you can see what's going on. To answer your last question: you can apply the above deformation retract to map one of those Möbius bands to the circle. In particular the orange line, which is your $A\cap B$, gets mapped to the circle by this deformation retract.
As I read in your comment to Miha Habič's answer, you already understood that the torus is covered universally by $\mathbb{R}^2$, so it remains to see how the Torus covers Klein's surface. Actually, you don't need four tori to cover Klein's surface. See the picture below.
Depending on the techniques available to you, there are several ways to proof that this indeed is a covering map and I think it's more appropriate to understand what happens, because this will yield a proof in any technique I guess. (Anyway, a straightforward computation using the indicated map between the quotient spaces of $[0,1]^2$ homeomorphic to the torus and Klein's surface definitely works.) Let me sketch a rather atypical argument. (It's the way I think of it.) Consider the so called orientation (double) cover of the (non-orientable) Klein's surface. To each manifold we can assign such an oriented double cover and it is roughly characterized (and defined, actually) by the property, that local sections correspond to local orientations. If the manifold is orientable, it will just be the disjoint union of two copies of the manifold because there is a way to glue these local orientations globally and there are exactly two of them. But in the non-orientable case, somewhere, the glueing isn't possible. Taking some non-contractible loop and trying to glue a local choice of orientation step by step will turn the orientation, so (in mind) we have to walk the loop twice to get the orientation we started with. Translating this to algebra, in our case, one may identify the index two subgroups of the fundamental group of Klein's surface.
This doesn't give the same picture as in the question, but a similar picture shows how the torus double covers itself and applying this double cover, you get the 4:1-cover of the torus over Klein's surface Olivier Bégassat suggests in the comment below Miha Habič's answer.
Best Answer
Think of the Klein bottle $K$ as $[0,1]\times[0,1]$ with its opposite sides identified, the top and bottom sides by $(x,0)\sim(x,1)$ and the left and right sides by $(0,x)\sim(1,1-x)$. Similarly, think of the open Möbius strip $M$ as $[0,1]\times\mathbb{R}$ where the left and right edges are identified by $(0,x)\sim(1,-x)$. There is then an easy covering map $M\to K$: just take the second coordinate mod $1$.