I'm trying to calculate the fundamental group of two Möbius strips which have been identified along their boundary (which is a Klein bottle, I think). I've chosen an NDR pair $A,B$ where $A$ and $B$ are each of the original Möbius strips. The intersection $A \cap B$ is the boundary of the Klein bottle. I've heard that this boundary, and each of the Möbius strips, are contractible to $S^1$, but I'm having trouble visualizing this. Any help would be appreciated. Thanks
Why are the Möbius strip and the boundary of a Klein bottle homotopy equivalent to $S^1$
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algebraic-topology
klein-bottle
mobius-band
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10The Klein bottle does not have a boundary. – 2012-05-22
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1The Möbius strip deformation retracts to a circle drawn down the middle. Imagine pushing down uniformly toward the central circle all along the strip. – 2012-05-22
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0@JimConant Oops, sorry. I mean to ask why does $A \cap B$ deformation retract onto $S^1$? – 2012-05-22
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0Aha, I assume you mean, $A$, $B$ are slightly larger versions of the original Möbius strips. They intersect in something which is homeomorphic to a strip running along the boundary of a Möbius strip. This deformation-retracts to the boundary circle of the Möbius strip by pushing towards the boundary. – 2012-05-22
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0@JimConant Is there a way to visualise these on the unit square representations? – 2012-05-22