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?


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


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, and the Poincaré conjecture. Hempel's book and Hatcher's notes on three-manifolds are standard references for the necessary background material.

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  • $\begingroup$ As I understand, by the Sphere Theorem if M is irreducible then $pi_2(M)=0$. But how does this also imply that $pi_2(M \setminus K) = 0$? $\endgroup$ – ali elgindi Dec 27 '14 at 17:56
  • $\begingroup$ Regarding typesetting: \pi gives $\pi$ when surrounded by dollar signs. $\endgroup$ – Sam Nead Dec 27 '14 at 18:01
  • $\begingroup$ That isn't making sense to me, maybe I don't understand your point. $K$ is cut out and $S$ bounds a ball in $M$, but the ball is now missing the knot. How is it still bounding a 3-ball? $\endgroup$ – ali elgindi Dec 27 '14 at 18:05
  • $\begingroup$ I understand now, thank you. I will work on the proof (which seems direct). Thanks again! $\endgroup$ – ali elgindi Dec 27 '14 at 18:24

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