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I would like to see why the following two statements in Kirby's list of problem are equivalent:

Statement 1:

If $K_1$ and $K_2$ are knots in a closed oriented 3-manifold $M$ whose complements are homeomorphic via orientation-preserving homeomorphism then there exists an orientation-preserving homeomorphism of $M$ taking $K_1$ to $K_2$.

Statement 2:

Let $M$ be an oriented 3-manifold with torus boundary. If $M(r_1)\cong M(r_2)$ for inequivalent slopes, then the homeomorphism is orientation-reversing.

(Here $M(r_1)$, $M(r_2)$ are the result of the Dehn filling of $M$ with slope $r_1$, $r_2$.)

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I guess, if one is doing a surgery of slope $r$ on a manifold with torus boundary, then we can consider the core of the surgery solid torus as a knot in the resulting manifold. This would do the equivalence. –  christian Nov 11 '12 at 19:32

1 Answer 1

You are right about this equivalence. It is also worth noting why the hypotheses are interesting. These question have now become the cosmetic surgery conjecture. As stated in Ni-Wu (http://arxiv.org/pdf/1009.4720.pdf), the conjecture (Conj 1.1) is as follows:

Suppose K is a knot in a closed oriented three-manifold Y such that Y − K is irreducible and not homeomorphic to the solid torus, then for all pairs of slopes $r_1$,$r_2$. If $Y(r_1)\cong Y(r_2)$. The homeomorphism is orientation reversing.

First, there are plenty of knot complements $Y\cong S^3-K$ that are amphichiral which implies $Y(r_1) \cong Y(r_2)$ for all but 2 slopes (if $\partial Y$ is framed in the standard way these slopes should be 1/0 and 0/1). The figure 8 knot complement is a good example of this.

Finally, there are plenty of knot complements in $Y\cong S^1\times D^2$ that admit two $S^1\times D^2$ fillings and one that even admits 3 fillings. These are called the Berge-Gabai knots.

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