Let $K$ be a knot. Embed $K$ in $S^3$ with (open) tubular neighborhood $N$. I want to show that $\pi_2(S^3\setminus N)=0$ using the Papa Sphere Theorem.
I think that it may be true that there is no embedding of $S^2$ in $S^3\setminus N$ with nontrivial homotopy type and then, by the Sphere Theorem, one could conlude that $S^3\setminus N$ has trivial second homotopy group. I don't see however why should it be true. Maybe my reasoning is not correct...
I would be thankful for any advice.
Preliminary fact: $S^3\setminus N$ is irreducible, meaning every embedded $S^2$ bounds a ball. Let $\Sigma\subset S^3\setminus N$ be an embedded $S^2$. Now think of $S^3\setminus N$ as a submanifold of $S^3$. By what Hatcher calls Alexander's theorem (or what others call the 3D Schönflies theorem), $\Sigma$ bounds two embedded balls $B_1,B_2\subset S^3$ with $S^3=\operatorname{int}B_1\cup\Sigma\cup\operatorname{int}B_2$ a disjoint union. Since $N$ does not intersect $\Sigma$, by connectivity $N$ is contained in either $B_1$ or $B_2$; assume $N\subset B_2$. Then, $\Sigma$ bounds $B_1\subset S^3\setminus N$.
If $\pi_2(S^3\setminus N)$ were nontrivial, the sphere theorem in the orientable case implies there is an embedded $S^2$, but $S^3\setminus N$ is irreducible hence the $S^2$ bounds a ball, and that ball can be used to produce a nullhomotopy.