If the complement of a knot $K$ fibers over the circle is $K$ necessarily fibered?

Question

Let $K \subseteq S^3$ be a knot in the $3$-sphere and assume there exists a smooth map $p \colon S^3\setminus K \to S^1$ which is a fiber bundle.

For every point $\require{enclose} \enclose{horizontalstrike}{t \in S^1}$ the fiber $\enclose{horizontalstrike}{p^{-1}(t)}$ together with $\enclose{horizontalstrike}{K}$ forms a compact 2-dimensional submanifold $\enclose{horizontalstrike}{F_t}$ of $\enclose{horizontalstrike}{S^3}$ such that $\enclose{horizontalstrike}{\partial F_t = K}$, though $\enclose{horizontalstrike}{F_t}$ may not be orientable/connected(?).

Question: Does there exists a smooth fiber bundle $q \colon S^3\setminus K \to S^1$ such that for every $t \in S^1$ the fiber $p^{-1}(t)$ together with $K$ forms an orientable connected compact 2-dimensional submanifold $F_t$ of $S^3$ with $\partial F_t = K$?

Edit: What I get from the comments:

• For every $t \in S^1$ the fiber $p^{-1}(\{t\}) = F$ is orientable as locally we have that $(0,1) \times F$ is diffeomorphic to an open subset $U$ of $S^3\setminus K$. Since $S^3\setminus K$ is orientable, so is $U$ and then also $F$.

• We can construct a smooth map $q \colon S^3 \setminus K \to S^1$ such that the fiber is connected: Assume $F$ has $n$ components. From the exact sequence $$ 0 \to \pi_1(F) \to \pi_1(S^3 \setminus K) \to \pi_1(S^1) = \mathbb Z \to \pi_0(F) = \mathbb Z_n \to 0 $$ we obtain a lift: $$ \require{AMScd} \begin{CD} S^3\setminus K @>\exists q>> S^1 \\ @V=VV @VVx \mapsto x^nV \\ S^3 \setminus K @>p>> S^1 \end{CD} $$ and $q$ is a smooth fiber bundle with connected fiber.

• It remains to show that $\partial (F \cup K) = K$.

Answer 1

Suppose that $K$ is a smooth knot in the three-sphere $S^3$. We set $X = S^3 - K$. Suppose that $p \colon X \to S^1$ is a smooth fiber bundle. Let $F_z = p^{-1}(z)$ be the fiber lying over $z$. Exercise: the inclusion of $F_z$ into $X$ is $\pi_1$-injective.

We now take $U$ to be a small, closed tubular neighborhood of $K$. Note that $U - K$ is homeomorphic to a torus cross a ray. Let $T = \partial U$. We perturb $p$ to make the intersections of the surfaces $F_z$ and $T$ "as transverse as possible". Let $\mathcal{F}$ be the resulting foliation of $T$ given by the sets $T \cap F_z$. This foliation is possibly singular, but it has only finitely many singularities. Exercise: If $\mathcal{F}$ has center singularities, then we can alter $p \colon X \to S^1$ by a diffeotopy to remove them. (The solution to the exercise uses the fact that the $F_z$ are $\pi_1$-injective.)

Since the Euler characteristic of $T$ is zero, once the foliation $\mathcal{F}$ has no centers, it has no singularities at all. We now "foliate" $U$ by annuli $A_z$ where $\partial A_z = K \cup (T \cap F_z)$ and where $A_z \cap A_w = K$ iff $z \neq w$.

The surfaces $(F_z - U) \cup A_z$, after smoothing, have the desired properties.