The following problem is a culmination of a few questions I've asked the last two months, and it's still giving me some issues. I think I know the right way to solve it, but I'm having trouble with the details; my idea can be formulated in terms of selection theorems/selectors or fiber bundles.

Let $X \subset \mathbb{R}^n$ be any subspace, and let $I = [0,1]$. By a *proper isotopy* of $X$ I mean a continuous function $F: X \times I \rightarrow \mathbb{R}^n$ such that for each $t \in I$, $f_t := F|_{X \times \lbrace t \rbrace}$ is an embedding and $f_0 = \text{id}_X$. By an *ambient isotopy* I mean an isotopy on all of $\mathbb{R}^n$.

If $F$ is a proper isotopy of a tamely embedded copy $X$ of $\mathbb{S}^{n-1}$ in $\mathbb{R}^n$, does $F$ extend to an ambient isotopy?

As noted in the answer here, this will be true as long as $F$ can be extended in some neighborhood of $X$. Let $In(X)$ denote the inside, i.e. bounded complementary domain, of $X$. By extending in the sphere from both sides of $X$ to get a neighborhood extension, it's equivalent to:

If $F$ is a proper isotopy of $X$, does $F$ extend to $In(X)$?

I'm mostly interested in $\mathbb{R}^2$. There it's already known, but the proof is very difficult. There was a follow-up paper where they broadened the result with different techniques; between the preprint, the Annals paper and the follow-up they gave three different arguments, but all hinged on analytically controlling crosscuts using geometric function theory to explicitly construct $F$.

Using results from the Kirby-Edwards paper linked in the previous MO thread, and a 'canonical' Alexander-Pontryagin Duality Theorem in the plane, you can prove the isotopy extension theorem for compact, connected subsets of $\mathbb{R}^2$ in a different way from the case for the circle (it's still somewhat complicated, but much less-so).

What I'd like to do is get the case for the circle using some selection theorem, esp. the Michael Selection Theorem (or even better, a selection theorem whose proof is actually reasonable). To do this, let $D$ be a large, closed ball around the trace of $X$ under $F$. By the Annulus Theorem, each region between $\partial(D)$ and $f_t(X) := X_t$ is a closed annulus, call it $A_t$.

For any $A_t$ there are many ways to partition it into crosscuts, so that each crosscut has one endpoint on each boundary component. By a crosscut, I mean an embedded copy of $I$. Let $\mathcal{C}_t$ denote the collection of such partitions on $A_t$, and let $\mathcal{C} = \cup \mathcal{C}_t$. Then what we want is a continuous selection of cross-cut decompositions, one for each $A_t$. To be precise, we should probably consider a family of *parameterized* cross-cut decompositions, so that each has a time parameterization (that will be our way of getting around $0$-regular convergence issues).

Alternatively, we can directly look for a continuous selection of crosscuts of the disc as follows. By interpolating with an ambient isotopy of affine shifts in the plane, we can assume that $0 \in In(f_t(X))$ for all $t$. Then we would let $\mathcal{C}$ be the collection of parametrized crosscut decompositions whose elements go through $0$.

Long story short, the problem for the selection method is:

How do you topologize (a suitable subset of) $\mathcal{C}$ to apply the Michael Selection Theorem?

This is equivalently a fiber bundle problem on $\mathbb{A}^n \times I$ (or the closed disc $\times I$) in the following sense. We have two cylinders, one the usual smooth cylinder, and the other one just some Jordan mess, as the boundaries. Can we warp it into the standard, smooth representative in a way that's *slice*?

The problem for the fiber bundle method is:

If $\mathbb{A}^n \times I \subset \mathbb{R}^{n+1}$ is a bundle over $I$ whose slices $\mathbb{A}^n_t$ are contained in hyperplaces orthogonal to the $(n+1)$-axis, is there an isotopy to the standard smooth (thickened) cylinder that's slice? In other words, $f_s(\mathbb{A}^n_t) \subset \mathbb{A}^n_t$ for all $s$ and $t$.

Thanks, appreciate any help at all!