Klein bottle as the union of two Mobius bands

**math8** · Dec 6th 2010, 03:26 PM

I am trying to calculate the homology groups of the Klein bottle. I want to use the Mayer-Vietoris sequence with the Klein bottle decomposed as the union of two Mobius bands (A and B which are homotopic equivalent to circles), now AUB is the Klein bottle, but I don't understand how according to

http://en.wikipedia.org/wiki/Mayer%E...e#Klein_bottle

AnB is also homotopic equivalent to a circle, I would think that the intersection is the disjoint union of two Mobius bands, so it is homotopic equivalent to the disjoint union of 2 circles, hence $H_n(A n B)$ should be $Z \oplus Z$ . But according to what's written in wikipedia, $H_n(A n B)$ is just $Z$ . Why?

Am I thinking this all wrong?

**Tinyboss** · Dec 6th 2010, 04:00 PM

On that Wikipedia page, the intersection of A and B is connected--look at the way the top and bottom edges are identified. When you go off the top of the left strip, you come back at the bottom of the right strip, and vice-versa.

**math8** · Dec 6th 2010, 04:16 PM

Duh, you're right, so AnB is also a Mobius band which is homotopic equivalent to a circle....

Thanks!

**Tinyboss** · Dec 6th 2010, 05:25 PM

Yup! One tiny detail: AnB has two twists in it, not one, so it's homeomorphic to a circle cross an interval, not a Mobius band.

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