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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?

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Do we have to "go outside of $\mathbb R^3$" with the homeomorphism?

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.

I'm thinking yes, because they are both a quotient of the square.

Exactly. You just have to convince yourself that the identification is the same. You then have an explicit homeomorphism.

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    Another way of considering the issue of an isotopy is that an ambient homeo. of $\mathbb {R^3}$ taking one figure/space to the other would have to take boundary to boundary, but, if I get the drawing well, some boundary points on the left seem to be collapsed during the twist.2011-07-05