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Apr 15, 2016 at 23:32 comment added Chris Gerig I am very confused on how doubling works here, because 1) we change the complements, and 2) I don't see what stops the knots from moving through the double to become isotopic. Am I correct in the setup that we first take $X:=S^3-\text{nbhd}(J)$ where $J$ is one component of the Whitehead link $J\cup K$ and then consider $K$ (along with some homotopic knot $K'$) inside $M=D(X)$? If so, then granted $J\cup K$ is not isotopic to $J\cup K′$ in $S^3$, I don't see how to use this to prove that $K$ is not isotopic to $K′$ in $M$. That is, why $K\cong K'$ in $M\;\Rightarrow\;K\cong K'$ in $X$?
Apr 8, 2016 at 4:12 comment added Carl I believe that doing a Dehn filling of slope 0, at the very least, does not work. You can slide the knot across the filled meridian disk to get the same effect as a twist would have, and therefore you have an isotopy. I am not sure how doubling works in this case since you need to end up with a closed manifold and it is not clear to me that the complements will still be homeomorphic.
Apr 8, 2016 at 2:29 comment added Ryan Budney I think you can make these examples work as well -- just embed these 3-manifolds in a closed 3-manifold in a way that doesn't allow the knots to become isotopic. The Whitehead example works fine if you double the manifold (the complement of one component of the link). I suspect the Borromean rings one will work similarly. Alternatively do a Dehn filling of slope 0 on the manifolds.
Apr 7, 2016 at 22:21 comment added Carl Thanks for the observation. The Whitehead link (among others) does seem to work for the case where $M$ has boundary. However, I'm mainly interested in when $M$ is closed, preventing that kind of twisting operation.
Apr 7, 2016 at 21:40 history answered Ryan Budney CC BY-SA 3.0