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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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    Ah, you're right, this is easier than I envisioned. So the pictures above are two non-equivalent embeddings of the same space. To see they are non-equivalent, one computes $\pi_1(\mathbb{R}^3-\text{boundary})$ (the boundary is a link of two circles) via the Wirtinger presentation. For the left complement, one gets $\langle a,b|\emptyset\rangle$ and for the right $\langle a,b|\text{some nontrivial relations}\rangle$, right?2011-07-05
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    I hate to repeat myself but: Exactly. My first idea was to use the linking number (http://en.wikipedia.org/wiki/Linking_number), but Wirtinger also does the trick. (You can also look at the class of each knot in the fundamental group of the complement of the other). I'm sure almost any link invariant works.2011-07-05
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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