If an embedding of $S^1\times I$ has nice edges, are the edges ambiently isotopic? Let $f_t(x)$, $~t\in I$, $~x\in S^1$ be an embedding of $f:S^1\times I\to\Bbb R^3$. (Thus for example the images of $f_{t_0}$ and $f_{t_1}$ are disjoint if $t_0\ne t_1$.)
It's known that the images of $f_0$ and $f_1$ need not be ambiently isotopic. An easy construction of this is to "ribbon-ify" the Fox–Artin wild arc so that it's an embedded disc, and then delete a small disc in the interior so that it's an embedded cylinder. No knot containing the Fox–Artin wild arc, much less two of them, can be ambiently isotopic to the unknot.
But what if the functions $f_0$ and $f_1$ are smooth? (This doesn't mean that the function $f:S^1\times I\to\Bbb R^3$ is smooth at $t=0$ or $t=1$, because the $t$ coordinate can be very bad; only that they're smooth purely in the $x$-coordinate. It does mean, though, that the boundaries of the cylinder are tame knots.) In this case, must the images of $f_0$ and $f_1$ be ambiently isotopic?
So what we have is a cylinder with nice edges but a very crinkly, fractal-y, potentially Alexander horn-filled interior. A first attempt is to try to smooth $f$ somehow, but the worry is that this will have self-intersections that will be difficult to get rid of. A friend suggested that Dehn's lemma might be useful, but I'm not sure how to use it or else how to adapt its proof technique.
 A: Surprisingly, a much stronger result seems to hold:
Claim: Suppose there is a homotopy $f:S^1\times I\to\mathbb{R}^3$ such that $f_0$ and $f_1$ are smooth embeddings onto disjoint knots $K_0,K_1\subset \mathbb{R}^3$, respectively, and $f(S^1\times(0,1))$ is disjoint from $f(S^1\times\{0,1\})$. Then $K_0$ is isotopic to $K_1$.
(Or: "if two disjoint knots are freely homotopic in their complement, then they are isotopic.")  Actually, the knots cobound an embedded annulus! An equivalent conclusion is that $K_0\cup K_1$ is a 2-cabling of a knot.
All I'm doing in this answer is to explain how to eliminate the regularity condition from Moishe Kohan's answer.
Let $N\subset \mathbb{R}^3$ be an open tubular neighborhood of $K_0\cup K_1$, with $N_i$ the tubular neighborhood of $K_i$. Let $\Sigma_0,\Sigma_1\subset\partial(\mathbb{R}^3\setminus N)$ be the two boundary components, which are tori, with $\Sigma_i$ being the boundary of the closure of $N_i$.
A longitude of $K_i$ is an oriented simple closed curve in $\Sigma_i$ which is freely homotopic to $K_i$ in $N$ (respecting orientations).  I claim that there exist longitudes $\lambda_0\subset \Sigma_0$ and $\lambda_1\subset\Sigma_1$ that are freely homotopic to each other in $\mathbb{R}^3\setminus N$.  Here's one way to see it.  Consider the preimage of $N_0$ in $S^1\times I$. Since $S^1$ is compact, the tube lemma gives an open subset $S^1\times[0,\epsilon)$ of this preimage, so $f$ restricted to $S^1\times\{\epsilon/2\}$ is a loop in $N_0\setminus K_0$ that is freely homotopic to $K_0$ in $N_0$ and freely homotopic to $K_1$ in $\mathbb{R}^3\setminus K_0$ along $f$.  We can consider the other end, too, and so there is an interval $[a,b]\subset I$ such that the restriction of $f$ to $S^1\times[a,b]$ is a free homotopy through $\mathbb{R}^3\setminus (K_0\cup K_1)$ from a loop in $N_0\setminus K_0$ freely homotopic to $K_0$ in $N_0$ to a loop in $N_1\setminus K_1$ freely homotopic to $K_1$ in $N_1$.
Since $\overline{N_i}\setminus K_i$ deformation retracts onto $\Sigma_i$, we can get a free homotopy through $\mathbb{R}^3\setminus N$ between loops in $\Sigma_i$ that are freely homotopic to the respective $K_i$ in $\overline{N_i}$. Loops in $\Sigma_i$ that are freely homotopic to $K_i$ in $\overline{N_i}$ are themselves freely homotopic to longitudes, hence we may obtain a free homotopy from a longitude $\lambda_0$ to a longitude $\lambda_1$ through $\mathbb{R}^3\setminus N$ as claimed.
What we now have is that $M=\mathbb{R}^3\setminus N$ is a 3-manifold with a "singular annulus" $S^1\times I\to M$ from $\lambda_0$ to $\lambda_1$.  Moishe Kohan's JSJ argument should then apply from here.  (It seems like the annulus theorem could also be used, which is a wrapped-up gift of JSJ theory.)
A: With a bit of extra regularity, the answer is positive:

EDIT: Refer to Kyle Miller's answer for a way to get rid of this extra assumption.


Note also that, since the answers don't assume an embedding (only that the map is injective on the edges), they obtain a stronger result that is surprising even in the smooth case. -AW

Suppose that $F: S^1\times [0,1]\to S^3$ is a continuous map which is a smooth embedding near the boundary, i.e. there
exists $\epsilon> 0$ such that
$$
F|_{S^1\times ([0, \epsilon) \cup (1-\epsilon, 1])}$$
is an embedding in the sense of differential topology. Assume also that $F^{-1}(F(S^1\times \{0\}))= S^1\times \{0\}$ and
$F^{-1}(F(S^1\times \{1\}))= S^1\times \{1\}$. (This would be automatic if $F$ were 1-1.) Then the knots $K_0=F(S^1\times \{0\})$,
$K_1=F(S^1\times \{1\})$ are ambient-isotopic.
The proof is a bit long and requires some heavy 3d-topology machinery (JSJ theory). Namely, in view of the extra regularity assumption I made, there exist closed tubular neighborhoods $N_0, N_1$ of the knots $K_0, K_1$ in $S^3$ such that for $i=0, 1$,
$F^{-1}(N_i)$ is an annulus  in $S^1\times [0,1]$ where $F$ is still a smooth embedding. Now, remove from $S^3$ the interiors of $N_0, N_1$. The result is a compact 3-manifold with boundary $M$. The knots $K'_i=F(S^1\times [0,1])\cap \partial N_i$ are ambient-isotopic to the respective knots $K_i$. There are two cases to consider:

*

*$K_0'$ (hence, $K'_1$) is null-homotopic in $M$. Then, using Dehn Lemma, one proves that both knots $K_0, K_1$ are trivial, hence, ambient-isotopic.


*Thus, I will assume that both $K_i'$ are not null-homotopic. With a bit of thought, it follows that the manifold $M$ is irreducible.
Now, we have a compact irreducible 3-manifold $M$ and two freely homotopic simple loops $K_0', K_1'$ in $\partial M$. Then it follows from the JSJ (Jaco-Shalen-Johannson) theory that there exists a connected component $X_0$ of the characteristic submanifold $X\subset M$, such that $X_0$  contains both loops and a free homotopy between them (still in $X_0$). Here $X_0$ is an interval bundle over a compact connected surface $\Sigma$ (possibly with boundary), whose boundary consists of two parts, one of which, $\partial_0 X_0$, is contained in $\partial M$ and   contains $K_0'\cup K_1'$. In our case case one can prove that  $X_0=\Sigma \times [0,1]$ and, hence, $\partial X_0= \Sigma\times \{0, 1\}$. Since the inclusion $X_0\to M$ is $\pi_1$-injective, $\Sigma_0$ is an annulus, but we will not need this.
Lastly, using the free homotopy of $K_0', K_1'$ in $X_0= \Sigma \times [0,1]$, and $K_i'\subset \Sigma\times \{i\}, i=0, 1$, one constructs a smooth annulus $A'\subset X_0$ whose boundary is $K'_0\cup K'_1$. I leave this part to you as an exercise.  Combining this with the annuli in $N_i$ connecting $K_i, K'_i$, one obtains a piecewise-smooth annulus $A\subset S^3$ with the boundary equal to $K_0\cup K_1$. Lastly, you approximate this annulus by a smooth one (smoothing the corners at $K_0'\cup K_1'$).
