I've seen a couple papers (that I now can't find) that say that in his paper "On irreducible 3-manifolds which are sufficiently large" Waldhausen proved that the data $\pi\_1(\partial (S^3\setminus K)) \to \pi\_1(S^3\setminus K)$ is a complete knot invariant. However, the word "knot" doesn't appear in this paper (although the phrase "to avoid an orgy of notation" does :-). Is the claimed result a straightforward corollary of his main results? Or am I looking at the wrong paper?



Complete knot invariant?
As Ryan says, this follows from Waldhausen's paper, when appropriately interpreted. Sufficiently large 3-manifolds are usually called "Haken" in the literature, and as Ryan says, they are irreducible and contain an incompressible surface (which means that the surface is incompressible and boundary incompressible). An irreducible manifold with non-empty boundary and not a ball (ie no 2-sphere boundary components) is always sufficiently large, by a homology and surgery argument. By [Alexander's Lemma](http://en.wikipedia.org/wiki/Heegaard_splitting), knot complements are irreducible, and therefore sufficiently large (the [sphere theorem](http://en.wikipedia.org/wiki/Sphere_theorem_%283-manifolds%29) implies that they
are aspherical).


Waldhausen's theorem implies that if one has two sufficiently large 3-manifolds $M\_1, M\_2$ with connected boundary components, and an isomorphism $\pi\_1(M\_1) \to \pi\_1(M\_2)$ inducing an isomorphism $\pi\_1(\partial M\_1) \to \pi\_1(\partial M\_2)$, then $M\_1$ is homeomorphic to $M\_2$. This is proven by first showing that there is a homotopy equivalence $M\_1\simeq M\_2$ which restricts to a homotopy equivalence $\partial M\_1\simeq \partial M\_2$. Then Waldhausen shows that this relative homotopy equivalence is homotopic to a homeomorphism by induction on a hierarchy. The peripheral data is necessary if the manifold has essential annuli, for example the square and granny knots have homotopy equivalent complements. 


If $K\_1, K\_2\subset S^3$ are (tame) knots, and $M\_1=S^3-\mathcal{N}(K\_1), M\_2=S^3-\mathcal{N}(K\_2)$ are two knot complements, then Waldhausen's theorem applies. However, one must also cite the knot complement problem solved by [Gordon and Luecke](http://www.ams.org/mathscinet-getitem?mr=965210), in order to conclude that $K\_1$ and $K\_2$ are isotopic knots. Otherwise, one must also hypothesize that the isomorphism $\partial M\_1 \to \partial M\_2$ takes the meridian to the meridian (the longitudes are determined homologically). This extra data is necessary to solve the isotopy problem for knots in a general 3-manifold $M$, to guarantee that the homeomorphism $(M\_1,\partial M\_1)\to (M\_2,\partial M\_2)$ extends to a homeomorphism $(M,K\_1)\to (M,K\_2)$, since for example there are knots in lens spaces which have homeomorphic complements by a result of [Bleiler-Hodgson-Weeks](http://www.ams.org/mathscinet-getitem?mr=1734400). 


