Given 2-tangles $T\_1,T\_2\subset B^3$ with their endpoints at some fixed points NW, NE, SW, SE of $\partial B^3$ we can glue them along $\partial B^3$ to obtain a link $L=T\_1\cup T\_2\subset S^3.$


**Q:** Does that link $L$ together with $T\_1$ determine the homeomorphism class of $(B^3,T\_2)$?


Remarks


1. Note that $L$ and $T\_1$ do not determine the isotopy class of $T\_2$ (within the space of tangles with endpoints at NW, NE, SW, SE, fixed on $\partial B^3$).
2. More generally, any 3-manifolds $M\_1,M\_2$ glued along their boundaries or partial boundaries, $S\hookrightarrow M\_1, S\hookrightarrow M\_2$, form a 3-manifold $N=M\_1\cup\_S M\_2$. However $N$ and $M\_1$ do not determine the homeomorphism class of $M\_2$ in general.



Glueing two 2-tangles
No, $L$ and $T\_1$ together do not determine $T\_2$. Take a non-trivial knot $K$ and suppose that $L$ is $K\#K$, and $T\_1$ is the trivial tangle. Then $T\_2$ can certainly be:


* a 2-tangle with one boundary-parallel component, and one with a boundary-parallel component connected-summed with $L$;
* a 2-tangle with two boundary-parallel components, each connected-sum with a copy of $K$.


I think you can do similar examples when $L$ and $T\_2$ are prime, for instance by cutting a 2-bridge knot in the usual way or in a more "silly" way where all the knotting is in one of the two tangles.


