I know this statement is true and I can see the reasoning behind it by the Classification Theorem, but I am still having trouble seeing why it holds. I know the form of the Klein bottle using a square and defining an equivalence relation which I suppose is the starting point. Then I can turn the square into a homeomorphic hexagon and I am stuck. Can you help me?
2026-03-25 19:02:49.1774465369
Klein bottle being homeomorphic to the surface with 2 crosscaps
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Since I now know the answer, might as well post it.
We know that $xy^{-1}xy$ and $xxyy$ give the same surface up to homeomorphism by cutting along the diagonal of the square and hence the claim is proven.