Consider a closed $3$-manifold $M$ and a knot $K$ in $M$.


Is it necessarily true that $\pi\_2 (M \setminus K) = 0$?


If not, are there any conditions on $M$ and/or $K$ to ensure the above 2nd homotopy group of the knot complement is trivial?


Thanks!


(Note: This is, of course, true when $M$ is simply connected --> $S^3$)



Knots in 3-manifolds
EDIT - I've rewritten my previous answer in an attempt to remove everything except the answer to your question. All submanifolds are assumed to be smooth. 



> 
> Suppose that $M$ is a closed, connected, oriented, irreducible three-manifold (and $M$ is not the three-sphere). Suppose that $K$ is a knot in $M$. Then $\pi\_2(M - K)$ is non-trivial if and only if $K$ is contained in an embedded three-ball $B^3 \subset M$. 
> 
> 
> 


The proof is an exercise using Alexander's theorem (every embedded two-sphere in $S^3$ bounds balls on both sides), the [sphere theorem](http://en.wikipedia.org/wiki/Sphere_theorem_%283-manifolds%29), and the [Poincaré conjecture](http://en.wikipedia.org/wiki/Poincar%C3%A9_conjecture). Hempel's book and Hatcher's notes on three-manifolds are standard references for the necessary background material. 


