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Question

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.

Answer 1

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.

Answer 2

Lemma 2.2 in the following paper contains the proof.

Synchronism of an incompressible non-free Seifert surface for a knot and an algebraically split closed incompressible surface in the knot complement