fibered knots in $ S^3$

Question

Given a fibered knot $k$ in $S^3$, we have the decomposition of $S^3$ as union of $M$ and $S^1\times D^2 $, where M is a fiber bundle over $S^1$, with fiber $F$ such that its boundary is the knot $k$. We also have decomposition of $S^3$ as two copies of $S^1\times D^2$. Does this mean that M is actually diffeomorphic to $S^1\times D^2$?

Answer 1

No. A theorem of Gordon and Luecke states that knots are determined by the homeomorphism type of their complement. Therefore, $M$ is only equal to $S^1\times D^2$ for the trivial knot.

Answer 2

Consider the knot complement $M$. Then a homeomorphism $M \cong S^1 \times D^2$ induces an isomorphism on fundamentalgroups, hence $\pi_1(M) \cong \mathbb{Z}$. But this contradicts to the early pages of every knot theory book, where it is stated that only the unknot has this trivial group.