Do the results of (1/n)-surgery determine the link?... Knowing the result of knot surgery is often not enough to determine the knot.  Indeed, there are 3-manifolds admitting an infinite number of descriptions as surgery on a (1-component) knot in $S^3$.  However, if I know many surgeries, perhaps I can recover the knot?  Let me be specific:
Suppose I have a 2-component link $U_1 \cup U_2$ inside the 3-sphere $S^3$ which has linking number $0$ and such that each component $U_1$, $U_2$ is the unknot.
I'm interested in knowing how much surgery tells us in this situation.  If I do $1/n$-surgery on $U_2$ I get back $S^3$, but now $U_1$ sits inside $S^3$ as a knot $K(n)$.
Does the sequence $K(1), K(2), K(3), \ldots$ determine the original link $U_1 \cup U_2$ ?  Would it even be expected to?.
 A: Lackenby has a great result (MR1443548) which basically shows that the denominator of the surgery slope, if great enough in absolute value, determines the resulting 3-manifold and knot (subject to a few conditions on the manifold and knot).
A: If the orginal link $U_1 \cup U_2$ was hyperbolic, the answer is yes.  For large enough $n$,  $S^3 \setminus K(n)$ will also be hyperbolic, and will approach $S^3 \setminus (U_1 \cup U_2)$ in the Gromov-Hausdorff topology; so the sequence $K(n)$ determines the complement of $U_1 \cup U_2$.  The complement doesn't determine the link in general (unlike for knots), but we also have the marking of the component $U_1$ by its meridian, which I believe is enough.
The answer in general is also almost certainly yes, but I haven't thought through all the cases.  Note that this operation has a simple geometric description: arrange $U_1 \cup U_2$ so that $U_2$ sits as a flat unknot in a plane.  Then to get $K(n)$, remove $U_2$ and twist the bundle of strands that passed through $U_2$ by $n$ full twists.
(This is all much easier than Lackenby's result mentioned above.)
