Klein bottle as the union of two Mobius bands

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The discussion focuses on calculating the homology groups of the Klein bottle using the Mayer-Vietoris sequence, specifically by decomposing the Klein bottle into two Mobius bands, A and B. The user, Mat, questions the homotopy equivalence of the intersection A ∩ B, which he believes should correspond to the disjoint union of two circles, leading to the conclusion that H_n(A ∩ B) should equal Z ⊕ Z. The confusion arises from the properties of the Mobius bands and their intersection, which requires a deeper understanding of their topological characteristics.

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  • Understanding of homology groups in algebraic topology
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  • Knowledge of Mobius bands and their properties
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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–Vietoris_sequence#Klein_bottle

AnB is also homotopic equivalent to a circle, I would think that since the intersection is the disjoint union of two Mobius bands, it is h.e. to the disjoin union of 2 circles, hence H_n(AnB) should be Z \oplus Z .

Am I thinking this all wrong?
 
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It's been a long time since I looked at this stuff but I think that it has something to do with wrapping around twice.

Mat
 

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