Uniqueness of Preferred Framing of a Solid Torus in $S^3$

Question

One way to state my question tersely is: For a homeomorphism $f : S^1 \times \mathbb{D}^2 \rightarrow S^1 \times \mathbb{D}^2$, does $f|_{S^1 \times S^1}$ determine the isotopy class of $f$?

This is motivated by the question of why a "preferred framing" of a solid torus imbedded in $S^3$ is unique up to isotopy. 

In what follows I will try to give the background for this question and how it reduces to the first question. 

Let $V \subset S^3$ be a homeomorph of the solid torus ($S^1 \times \mathbb{D}^2$) continuously imbedded in the 3-sphere in such a way that the closure of $S^3 - V$ is a manifold (I'll call $X$)  with boundary equal to $\partial V$.  It can be seen from a Mayer-Vietoris sequence that there exists a simple closed curve $\lambda : S^1 \rightarrow \partial V$ which represents a generator of $H_1 (V)$ and is homologically trivial in $X$.  Call this a "preferred longitude for V". Moreover, for any other preferred longitude, $\lambda '$, we have $[\lambda]=  \pm [\lambda ']$ in $H_1 (\partial V)$. In addition there exists a simple closed curve $\mu : S^1 \rightarrow \partial V$ which is homologically trivial in V but not in $\partial V$.  Call $\mu$ a "meridian of V".  It is not hard to see that, like preferred longitudes, meridians fall into only two homology classes in $H_1 (\partial V)$, depending on orientation.

Now a choice of meridian and preferred longitude determines the isotopy class of homeomorphisms $f': S^1 \times S^1 \rightarrow \partial V$ which have [$S^1 \times 1] \mapsto [\lambda]$ and $[1 \times S^1] \mapsto [\mu]$ under $H_1(f')$ (here let's say $S^1$ has the counterclockwise orientation).  That all such $f'$ are isotopic is not obvious, but not very difficult to prove.

  Now because $1\times S^1$ is mapped to a meridian, any such $f'$ extends to a framing of $V$, i.e. a homeomorphism $f: S^1 \times \mathbb{D}^2 \rightarrow V$.  I would like to show that any two such $f$ are isotopic. I am allowed to assume that the two framings agree on the boundary because an isotopy of the boundary can be extended to an isotopy of the solid torus by means of a collar. This is why I believe an answer to the question at the top of the post would suffice.

Thank you for your time.

Answer 1

Consider two maps isotopic on the boundary, and quotient the meridian and longitude to a single point. This turns the solid torus into the ball. The question is then, if two maps from the ball to itself fixing a boundary point are isotopic on the boundary, are they isotopic on the ball? Since the ball is convex, I'm fairly certain this is true.

Answer 2

Elaboration of Brian's very clever solution:

As mentioned in the comments, we may assume one of the functions to be the identity and the other, $f$, to have $f|_{S^1 \times S^1} = id$. By means of a collar, we may then assume that $f$ is the identity in an open neighborhood of the $S^1 \times S^1$ (I believe this will resolve the issue I raised in my comment to Brian's answer). $f$ then descends to a homeomorphism, $f_1$ of the 3-ball which is the identity in a neighborhood of $S^2$. There exists an isotopy, $H_1$, from $f_1$ to the identity which is stable in an open neighborhood of $S^2$. Now we may define a function $$H: \{S^1 \times \mathbb{D}^2 \} \times I \rightarrow S^1 \times \mathbb{D}^2, \qquad H(p,t) = \left\{ \begin{array}{lr} H_1 (p,t) & : p \notin \ast \times S^1 \cup S^1 \times \ast \\ p & : p \in \ast \times S^1 \cup S^1 \times \ast \end{array} \right. $$

$H$ is clearly continuous whenever $p$ is not in the collapsed set. But it is also just the projection function in a neighborhood of $\{ \ast \times S^1 \cup S^1 \times \ast \} \times I$.