# Klein bottle contains Möbius band

I read the following: "The Klein bottle contains a copy of the Möbius band". I assume this means that there is a subspace of the Klein bottle that is homeomorphic to the Möbius band. How do we obtain the Möbius band from the fundamental polygon of the Klein bottle? • In fact every non-orientable two-dimensional manifold has a subspace that is homeomorphic to the Möbius strip. – Henning Makholm Nov 26 '14 at 17:05

## 2 Answers

Hint:

Here's a representation of a Mobius band. Hmmm, what does this suggest... • Can we somehow vertically cut out a strip from the Klein bottle? It'll look like the representation of the Mobius band – iwriteonbananas Nov 26 '14 at 17:04
• Since the red arrows in your diagram correspond to the blue arrows in the original diagram, rather than to the red arrows, I wonder why you made them red instead of blue. – MJD Nov 26 '14 at 17:08
• @iwriteonbananas, yes. If the original square is $[0,1] \times [0,1]$, the Klein bottle equivalence relation is $$(x,y) \sim (1-x,y)$$ So to get a Mobius strip, take for example $[1/3,2/3] \times [0,1]$ under that equivalence. – Simon S Nov 26 '14 at 17:17
• @SimonS Of course thank u – iwriteonbananas Nov 26 '14 at 17:21
• To be more careful, I should say that's one of the Kelin bottle equivalence relations, the one you need to make the Mobius strip. The other is $(0,y) \sim (1,y)$. – Simon S Nov 26 '14 at 17:26

If you don't join the red edges together you get what you want, so you can cut the Klein Bottle along the red edge (or what it becomes in the Klein Bottle when the two red edges are identified).

• Thank u. What do u mean by cut the Klein bottle along the red edge? – iwriteonbananas Nov 26 '14 at 17:09
• @iwriteonbananas You have the net for the Klein bottle. When you identify the two red sides, also draw a red line on the Klein Bottle where they join. If you cut along the line you get the net with the two blue edges identified (and not the red edges). And that is what you want. – Mark Bennet Nov 26 '14 at 17:37