A cellular wild arc I am currently trying to explicitly show that the Fox-Artin arc (example 1.2 in the paper https://maths.dur.ac.uk/users/mark.a.powell/fox-artin.pdf) is cellular.
This should definitely be true (in two ways) since firstly it is locally polyhedral apart from one point and secondly it is a compact subset of $S^3$ whose complement is an open 3-cell (however none of these arguments help with the actual construction of decreasing balls).
The following archived post https://wildandnoncompactknots.wordpress.com/tag/wild-arcs/ claims that we can take a ball around the wild point and then a tubular neighbourhood of the rest of the arc lying outside the point. This results in a ball with some handles. The author then claims these handles can be filled in. This is true however it seems that when filling a handle naively another handle is created.
In fact since the arc outside the initial ball around the wild point passes under the handle I do not see a way of filling it in without intersecting the arc and thus creating another handle.
Fox and Artin claim that example 1.1 is not cellular, but it seems the same words from the archived post would show that it is indeed cellular. It seems crucial that the other end point of the arc is not wild however I am not really sure how to use this in my construction.
Another reason why the existence of such a construction seems strange is that the penetration index of the curve is 3, so any open neighbourhood homeomorphic to a 3-cell will have boundary sphere intersecting the arc in 3 components. Thus will create at least one handle and the rest of the arc will pass under that handle at least once (or at least I cannot find an example where it does not - one can find a homeomorphism of $S^3$ to itself that takes the arc to itself and makes the arc no longer pass under the handle however I do not see how this helps since to prove cellularity we need to intersect balls that are already embedded). So since the arc passes under the handle, filling the handle in will create another one.
Am I misunderstanding the something or is there a clever construction here?
 A: Actually I think I have a construction if anyone is interested I will leave it here as I have not found a detailed explanation anywhere so far.
Denote the wild arc by X and take concentric balls V_n around the wild end point, such that $(V_n\setminus V_{n-1})\cap X=s_n\cup r_n\cup h_n$ looks like the three arcs $K_0, K_-, K_+ $(respectively) in the cylinder used to construct the arc (as they are referred to in Fox and Artin’s original paper).
Then $(S^3\setminus V_n)\cap X=h_n \cup a_n$ where $h_n$ is a handle and $a_n$ is a tame arc. Let $H_n$ and $A_n$ be tubular neighbourhoods of $h_n$ and $a_n$ respectively with radius say $\frac{1}{4}^n$ and let $X_n = H_n\cup A_n\cup V_n$. We can further suppose that we have drawn the arc in such a way that $H_n$ may be filled using a cube $C_n=[0,1/2^{n-1}]^3$(this filling will intersect A_n right at this moment so we will not fill the handle just yet).
Next, there is a family of isotopies $f_n^t$ taking $A_n$ to a straight line segment and restricting to the identity on $H_n\cup V_n$ (this is precisely the unknotting operation you describe). Then we may fill the the handle $f_n^1(H_n)=H_n$ using $C_n$ without intersecting $f_n^1(A_n)$. The result will clearly be a ball $B_n=C_n\cup f_n^1(X_n)$.
Now we pick another family of isotopies $g_n^t$ that return $f_n^1(A_n)$ to its original position $A_n$. The cube $C_n$ gets deformed as this happens (perhaps becoming thinner so it can fit inside a tubular neighbourhood of radius $\frac{1}{4}^{n-1}$) . Let $S_n$,$R_n$ be tubular neighbourhoods of $s_n$ and $r_n$ with radius $\frac{1}{4}^{n-1}$
We require the following:

*

*$g_n(C_n)\subset V_{n-1}\setminus V_n $ for $t\in [0,1/2^n]$

*$g_n(C_n)\subset H_{n-1}$ for $t\in [1/2^n,1/2^{n-1}]$

*$g_n(C_n)\subset R_n $ for $t\in [1/2^{n-1},1/2^{n-2}]$

*$g_n(C_n)\subset A_{n-1}$ for $t\in [1/2^{n-2},1]$
Notice that we can simultaneously fulfil the same conditions for $g_n^t(A_n)$ since informally $g_n^t(C_n)$ acts like a hat that surrounds $g_n^t(A_n)$ without intersecting it. In fact for $A_n$ we can require condition 2 to be changed so that $g_n(A_n)\subset S_n$.
Also note that in order for condition 3 to be fulfilled $C_n$ will be deformed in a way that it forms a hat over itself.
Thus the balls we want are: $E_n=g_n^1(B_n)$. As n tends to infinity the intersection of $X_n$ tends to the arc $X$ but $E_n\subset X_{n-1}$ by construction of isotopy so the intersection of $E_n$ is indeed $X$ as desired.
I hope I have not made any mistakes with indices. There is a need to be careful about the number of times the tame arc wraps around a handle.
This construction generalises but one still needs to argue carefully as the arc may wrap around distinct handles multiple times in different ways.
Visualisation of the second step
