# a special type of 2 component link complement

It is well know that a 2 component link complement may doesn't detect the link type. My question is whether the following type of 2 component links detect their link types?

Such a link is composed of a knot with its meridian circle. Here its meridian circle means a meridian of the boundary of a solid torus neighborhood of the knot. Thank you.

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The technique of a proof goes as follows. Since now "the complement of" means "the complement of an open tubolar neighborhood of". Let $K$ be a knot and $m$ an encircling meridian. We want to prove that $K\cup m$ is the only 2-component link having its complement. Gordon-Luecke says that the complement of $K$ is a compact manifold $M$ with one torus boundary, which has only one slope $s$ whose Dehn filling gives $S^3$.
The complement of $K\cup m$ is homeomorphic to the union $N=M\cup(P\times S^1)$ of $M$ and one copy of $P\times S^1$ where $P$ is the pair-of-pants. The fiber of $P\times S^1$ is glued to $M$ along the slope $s$. As usual with link complements, the problem of finding links in $S^3$ whose complement is $N$ translates into studying the Dehn fillings of $N$ giving $S^3$. By Gordon-Luecke, the only way to get $S^3$ is that $P\times S^1$ transforms into a solid torus with meridian $s$. One studies and easily classifies the fillings of $P\times S^1$ giving a solid torus with fiber-parallel meridian, and see that they all give rise to the same link.
Yes, exactly. If you use meridian-longitude coordinates for the two boundary tori of $P\times S^1$, the Dehn fillings that give a solid torus with fiber-parallel meridian are: $(1,q),(0,1)$, and $(0,1),(1,q')$. So you have two families of infinitely many fillings each. But the varying of $q$ or $q'$ actually does not affect the link. –  Bruno Martelli Dec 2 '11 at 18:35