# Does a knot and a tunnel exterior having free fundamental group imply it's an unknotting tunnel?

The title is just about it. Assume we have a nontrivial knot $$K$$ in $$S^3$$ and the exterior of $$K$$, $$E(K)$$, is $$S^3 \setminus N(K)$$. Here $$N(K)$$ is a regular neighborhood.

1. Let $$\tau$$ be a properly embedded arc in $$E(K)$$ and let $$M = E(K)\setminus N(\tau)$$. Now, if we know that $$\pi_1(M) = \langle x,y\vert \rangle$$, does this necessarily mean that $$\tau$$ is an unknotting tunnel and the tunnel number of $$K$$ is one, $$t(K)=1$$?
2. I would like to know this in a more general setting, but the simplest case is all I really need for now. Given $$T = \{\tau_1,\ldots,\tau_j\}$$ properly embedded disjoint arcs, and $$M = E(K)\setminus N(T)$$ with $$\pi_1(M)$$ free, does that mean that $$t(K) \leq j$$?

I am a little confused by the statements of Scharlemann and Thompson's Theorem 7.5 (first page of the pdf) about embedded graphs, but think that they are working in a more restrictive setting. But I could not find the answer to my question written down anywhere.

The answer is "yes". This is because the manifold $$M(K)$$ is (in both cases) a handlebody of the correct genus. To see this, you will need to apply the disk theorem several times. The end of the proof requires Alexander's theorem: the three-sphere is irreducible.