a question in the proof of type $(p,q)$ torus knot and type $(q,p)$ torus knot are equivalent

Question

a proof says that if we remove a disk which doesn't touch the knot from the torus, deform it to get a surface $S$ which consists of two bands attached to each other, then we can turn each band inside out. So a type $(p,q)$ torus knot become a type $(q,p)$ torus knot. My question is, How do we know the deformation of $S$ can be extended to all of the ${R^3}$, isn't there topological obstruction?

Answer 1

There is no topological obstruction. The isotopy extension theorem gives you that if you have a smooth isotopy of embeddings, then there is a smooth ambient isotopy carrying the first embedding to the second.

We're not turning a torus itself inside out (there's definitely a topological obstruction to doing that inside of $\mathbb{R}^3$), but instead a punctured torus. Like this:

For an intuition for the isotopy extension theorem, imagine that there is some fluid filling all of $\mathbb{R}^3$ (like air), and imagine that we keep track of how this fluid is displaced during the movie. That's an ambient isotopy. Incidentally, this ambient isotopy is compactly supported in some region around the torus, the size of which is bounded I suppose by the speed of sound through the fluid.