I'm thinking yes, because they are both a quotient of the square. But I can't figure out what the actual homeomorphism is. Do we have to "go outside of $\mathbb{R}^3$" with the homeomorphism?
ADDITION: is there an ambient isotopy between them?
2026-05-06 09:40:58.1778060458
Is 2x (360°) rotated (orientable) Möbius band homeomorphic to $S^1\times[0,1]$?
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Exactly. There is no homeomorphism of $\mathbb R^3$ sending one to the other (look at the boundary: in one case it is two parallel circles but in the other the circles are intertwined, even if that's not (yet) a formal proof), but as abstract topological space, they are homeomorphic.
Exactly. You just have to convince yourself that the identification is the same. You then have an explicit homeomorphism.