Why is a triangle with edges identified according to $aab$ a Mobius strip?

Question

Suppose we've got a fundamental polygon with edge-word $aab$. It's not too hard to use classification of compact surfaces to show that this is a Mobius strip (e.g. $\mathbb{R}P^2 - D^2$, the projective plane minus a disc). Simply note that its Euler characteristic is $0$, it's non-orientable because of the $aa$ pair, and its got one boundary component.

But, I'm failing to come up with a simple cutting and pasting argument that actually visually displays the homeomorphism. Since the normal form polygon of a projective plane with one hole should be $aabcb^{-1}$, there should be a way to bring it to this form, but I'm not seeing one.

Answer 1

We start by introducing a third edge-pair, $c$:

Now, take the left triangle, and mirror it along the vertical axis. At the same time, turn the right triangle $180^\circ$ upside-down. Then glue the $a$'s together. The final shape might look more like a Mobius strip you're familiar with:

Note that the two $b$'s aren't supposed to be glued together. They're just supposed to meet up end-to-end.

One can also do this the other way: If we take a mobius strip and cut it straight across (i.e. along $c$), we only get a rectangle with a twist. If, however, we cut once around the strip, ending the cut in the same place we started it, we get your $aab$ triangle. You can see this from the diagonal line that $a$ has become in the last drawing: going once around the strip, from one point on the edge to the same point.