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This questions arises from working through the 2015 Heraeus lectures on gravity, specifically tutorial 2 exercise 2, so this is why I post it as a physics question although it is purely mathematical.

When we represent the Moebius strip as a rectangle, we draw opposite sense arrows on two opposite sides that describe how to glue the sides, i.e. Moebius strip as a rectangle. Now if I want to show it is a 2d topological manifold, I need to find a set of charts that cover it, so basically some open set(s) in the "Moebius rectangle" together with the corresponding homeomorphism(s) from each to $\mathbb{R}^2$. Here are my questions:

  1. Why no single chart can do this;
  2. why (and what) two charts would suffice;

I can see there is a problem with the twist on the top and bottom sides, like when I try to draw a ball there it gets ripped in two, but that doesn't help me answer the above.

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    The only reason is that it's part of a (mathematical) physics lecture series, that tackles the question of spacetime, and that I plan on asking other questions as work my way through it. I also saw other topologically minded spacetime questions on physics stack exchange, so I decided to target the physics stack. Also, the pointers you gave are very general, and my question is more specific but I admit that doesn't make it less mathematical. – 2017-01-17
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    Consider the usual physical method of tracing a pencil along a Möbius strip to demonstrate nonorientability. Now, imagine a normal vector whizzing along the strip... – 2017-01-18
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    Ok, so that would prove a single chart does not suffice, but can't you do that without orientability (was not introduced in the lecture at this point) ? Isn't there a way using open sets like when we show $S^1$ cannot be covered by a single chart because it violates the open set conservation ? – 2017-01-19

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The fact that two suffice is easy, just find two. (Look at a picture of it.)

To show that one chart does not suffice, note that if one chart could cover it, then it could be oriented by the orientation inherited from $\Bbb R^2$. However, it is known that it is not orientable.

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    So for the sufficient part, what mapping do we use there, identity? And how can we see that it solves any problem encountered by a single chart ? As for the single chart impossibility, how can you do that without orientability (was not introduced in the lecture at this point), isn't there a way using open sets like when we show $S^1$ cannot be covered by a single chart? – 2017-01-18