Let $K$ and $J$ be disjoint knots in $S^3$ and let $D_1$ and $D_2$ be two immersed disks in $(S^3 - K) \times I$ with boundary $J$. Is it true that $D_1$ and $D_2$ are homotopic through a homotopy that fixes the boundary $J$?
I saw this mentioned in a paper where it was stated that it follows from the fact that complements of knots in $S^3$ have no $\pi_2$, but I do not see how to derive this conclusion. Thanks!
Gluing the two disks $D_1$ and $D_2$ together along their boundary defines a map $f:S^2\to (S^3-K)\times I$, which restricts to $D_1$ and $D_2$ on opposite hemispheres. Since $\pi_2((S^3-K)\times I)\cong \pi_2(S^3-K)$ is trivial, $f$ is nullhomotopic, so it extends to a map $g:D^3\to (S^3-K)\times I$. This map $g$ can be thought of as a homotopy from $D_1$ to $D_2$ rel the boundary, with the stages of the homotopy interpolating between the hemispheres of $S^2$ through the ball.