Rolfsen asked the question as to whether any knot is topologically isotopic to the unknot. Where a topological isotopy is a continuous path in $\operatorname{Emb}(S^1,\mathbb{R}^3)$.

From now on I will use isotopy to mean topological isotopy as stated above. Note this is not the same as smooth isotopy which is the standard equivalence relation used in knot theory.

For example the trefoil knot here is isotopic to the unknot. An animation of this is shown in Jim Belk's answer to Which two knots are isotopic but not ambient isotopic?.

It is currently known that every PL knot is isotopic to the unknot and in fact any wild knot that is locally flat at at least one point is also isotopic to the unknot (informally — by picking a tame neighbourhood and isotoping everything outside it to a point).

Let $K$ be a knot in $\mathbb{R}^3$ and consider locally flat neighbourhood $(U,K\cap U)\cong (\mathbb{R}^3,\mathbb{R})$ then $K$ pierces a disc at each point $x\in K\cap U$. Thus a contender for a counter example was conjectured to have to fail a disc at each point. The Bing sling is such an example. It is interesting that this condition is not sufficient. An example of a knot that fails to pierce a disk at each point that also bounds a disc (and so is isotopic to the unknot) was constructed by Gilliman in Sequentially 1-ULC tori. It is still open as to whether the Bing sling is isotopic to the unkot.

The Bing sling is the limit of a sequence of nested tori of which only the first is unkotted. My question (probably stupid) is why one can't define an isotopy of the Bing Sling and the unknot by contracting this torus to its centre. That is we pick a family of homeomorphisms of 3-space that fixes the boundary of the first unkotted torus but shrinks everything inside to its centre. The Bing sling lies inside this torus so we define the isotopy as the family of maps whose image is the composition of the standard embedding of the Bing sling into the first torus composed with the shrinking homeomorphism that fixes the boundary of the torus.

This defines a pseudo isotopy (a continuous map $F: S^1\times I\rightarrow \mathbb{R}^3$ where $F(-,t)$ is an embedding for $t\in [0,1)$ but only continuous for at $t=1$). I'm struggling to see what the problem to extending the isotopy is. The issue here seems like $F(-,1)$ will somehow fail to be injective but i am failing to describe it presicely. I would appreciate any help — I'm sure I must be missing something trivial.

  • $\begingroup$ Can you clarify Rolfsen’s question? It seems like there’s a missing hypothesis. Obviously there are knots that aren’t isotopic to the unknot. $\endgroup$
    – HJRW
    Dec 23, 2022 at 15:47
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    $\begingroup$ @HJRW Here topologically isotopic means that there is a path of topological embeddings from one knot to another. For tame knots, there is always a path to the unknot which is done by first isolating the ''interesting'' stuff into an $\epsilon$-ball, then limiting it to a point (in a way that preserves injectivity). What most people call knot theory is the study of smooth embeddings up to smooth isotopy which avoids this trickery. $\endgroup$ Dec 23, 2022 at 16:43
  • $\begingroup$ @ConnorMalin: Thanks for the reminder! $\endgroup$
    – HJRW
    Dec 23, 2022 at 16:51
  • $\begingroup$ In other words, the thing knot theory prevents you from doing is pulling on the ends of the string to make the knot smaller and smaller. The kind of things you do in real life when you tie a knot, you see? $\endgroup$
    – PseudoNeo
    Dec 23, 2022 at 17:40
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    $\begingroup$ I voted to close as the question is lacking details. Presumably the context is what might be called "weak" topological isotopy of embeddings, we are asking if the space of all topological embeddings Emb(S^1, R^3) with the compact-open topology is connected. The Alexander trick provides for this, when the knot is tame, so the issue is the non-tame knots. You should probably change the title, as well, as I imagine Rolfsen has asked more than one question in his career. $\endgroup$ Dec 23, 2022 at 18:26

2 Answers 2


First, I'd like to make some conventions. Knots $J$ and $K$ are non-ambiently isotopic if there is a level-preserving embedding $e:J \times [0,1]\to S^3 \times [0,1]$ such that $e(J \times \{0\}) = J \times \{0\}$ and $e(J \times \{1\}) = K \times \{1\}$. I think that's what OP means for topologically isotopy of embeddings. For historical reason, I prefer to call it non-ambient isotopy to distinguish from the one that knot theorists study the most.

A simple closed curve $J$ is said to pierce a disk $D$ if $J$ links $Bd D$ and $J \cap D$ is a single point.

Fact. Every knot that pierces a tame disk is non-ambiently isotopic to an unknot.

I stole the pictorial proof from Ric Ancel's preprint.


The Bing sling is a wild knot that pierces no disk and every point of it is wild. The following picture is quoted from P. 81 in Daverman-Venema's book, a similar one may find in Ancel's preprint as well. The Bing sling is the intersection of the nested tori. Although the Bing sling (denoted by) $\Sigma$ looks intimidating, it would be a fun exercise to show that the Bing sling (the innermost curve) is homeomorphic to the core $J$ of $T_1$. Or one may reembed the $J$ by infinitely many times so the limit is the Bing sling. enter image description here

A big open question is whether every knot is non-ambiently isotopic to an unknot. (I'm not sure if it was first due to Rolfsen). In addition, it is unknown if the Bing sling is non-ambiently isotopic to an unknot.

If I understand correctly, the approach described in OP about squeezing $T_1$ towards $\Sigma$ so there is a sequence of successive homeomorphisms sending the core $J$ of $T_1$ onto $\Sigma$ is carried out in Ancel's preprint. Actually, he proved something more general but unfortunately, a partial result to what OP hoped.

Theorem. Every knot is semi-isotopic to an unknot.

A knot $J$ is semi-isotopic to a knot $K$ if there is annulus $A$ is $S^3 \times [0,1]$ such that $\partial A = (J \times \{0\}) \cup (K \times \{1\})$ and there is a homeomorphism $e: S^1 \times [0,1) \to A - (K \times \{1\})$ such that $e(S^1 \times \{t\}) \subset S^3 \times \{t\}$ for every $t \in [0,1)$.
However, $e$ may not extend continuously to homeomorphism from $S^1 \times [0,1]$ onto $A$. I don't know how to push this idea further because knots like Bing sling are everywhere wild, one can't just contract the wild portion to a point. A brief sketch of Ancel's argument is the following: Every knot $J$ can be considered as a subset of the interior of an unknotted solid torus $T$. Let $K$ be the core of $T$. Then there is an obvious map $\pi: T \to K$ by squeezing $T$ towards the core. Since $\pi|J: J \to K$ is a homotopy equivalence, $\pi|J: J \to K$ is homotopic to an homeomorphism $\chi: J \to K$. Using the swirling map, one may build a mapping swirl $Swl(\pi|J)$ between $J$ and $K$. Roughly speaking, it's like a twisted mapping cylinder. It turns out that $Swl(\pi|J)$ is homeomorphic to a mapping cylinder $Cyl(\chi)$. Therefore, $Swl(\pi|J)$ minus the right end is homeomorphic to an annulus $A - (K \times \{1\})$, and every level of it is a homeomorphism. So things before $t =1$ can be lined up pretty nicely (in my view, it's like "combing" towards $K$), but it's unclear what will happen at the end $K$ in the mapping cylinder.

Remark. I think a philosophical point of view would help illustrate the difficulty. Note that in general $Homeo(X,Y)$ fails to be closed in $C(X,Y)$ -- the space of continuous function of $X$ to $Y$. An arbitrary Cauchy sequence of homeomorphisms in $Homeo(X,Y)$ may not converge to a homeomorphism, the injectivity is an issue. A commonly used approach to extend the homeomorphism to the limit is when constructing a sequence of homeomorphisms $h_k:X\to Y$ recursively, one need to impose control limiting the distance between $h_k$ and $h_{k+1}$ after $h_1,\dots,h_k$ are all specified. Because of the wildness of knots like the Bing sling, it's not easy to place such a desired control. Ancel's argument doesn't tell us how to find the next $h_{k+1}$.

  • $\begingroup$ Thanks for the answer, I was roughly aware of Ancels paper but not in detail. I can see loads of "intuitive" reasons as to why the end of the pseudo isotopy has some problems (breaking injectivity as you say is one of them). However I was wondering whether anyone had a precise argument. I was even considering the simpler question which could perhaps illuminate something here. Consider the double cover of the standard unknot which we will model by the centre of a solid torus. We can also view the double cover as living inside the interior of the solid torus. $\endgroup$
    – amd1234
    Dec 27, 2022 at 10:40
  • $\begingroup$ Repeat the process I describe above shrinking the solid torus onto its centre this defines a pseudo isotopy of the double cover of the circle onto the centre of the torus. Suppose we extend the pseudo isotopy naively just by taking the limit. Then the limit map is precisely projection from the double cover clearly breaking injectivity required for an isotopy. It feels like showing that there is no possible injective extention to this pseudo isotopy would be useful for the cas of the bing sling since something similar happens in my post (the naive limt map sends cantor sets to points) $\endgroup$
    – amd1234
    Dec 27, 2022 at 10:45

Okay so I think i may (hopefully) have an idea as to what is going wrong here. Let me attempt to give an outline:

First let us consider the example i wrote in the comments to Shijie Gu's answer.

Let $\pi:D\rightarrow S^1$ be the double cover of the circle by itself $D\cong S^1$. We can view $D\subset Int(T)\subset \mathbb{R^3}$ where $T$ is a solid torus with centre $C\cong S^1$.

Then we can define a pseudo isotopy (an isotopy but where one of the ends of the path traced through $Emb(S^1,\mathbb{R}^3)$ is not an embedding) as we have done in the question: by contracting the torus onto it's centre $C$. Suppose in fact this is extends to an isotopy $F_i:S^1\rightarrow \mathbb{R}^3$ for $i\in[0,1]$, with $F_0(S^1)=D$ and $F_1(S^1)=C$ (where the $F_i$ for $i\in[0,1)$ are given by the shrinking of the torus and $F_1\in Emb(S^1,\mathbb{R}^3)$ is the 'problematic' extension). Consider a line $l$ (or potentially path) intersecting $C$ in exactly one point but intersecting $F_i(S^1)$ in exactly two points for all $i\in[0,1)$


Consider $U=\mathbb{R}^3\setminus l$ and its preimage under the total isotopy $F:S^1\times I\rightarrow \mathbb{R}^3$. Where $F(x,i)=F_i(x)$ in the notation above. By construction $F^{-1}(U)$ is the subset of the cylinder $S^1\times I$ obtained by removing a closed path (homeomorphic to $[0,1]$ from one boundary ($S^1\times \{0\}$) to the other ($S^1\times \{1\}$) and a half open path (homeomorphic to $[0,1)$) from the boundary $S^1\times \{0\}$ with limit point (not attained) in $S^1\times \{1\}$. This is not open so $F_i$ cannot extend to an isotopy.

Now for the case of the Bing Sling the argument is very similar. Except when we pick a line $l$ it intersects the Bing sling in a Cantor set of points. However the argument is the same the preimage of $U=\mathbb{R}^3\setminus l$ under the total isotopy is a cylinder minus a Cantor set of disjoint half open arcs only one of which is an actual arc.

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    $\begingroup$ You might be on the right track, but I think the argument should take advantage of the fact that the Bing sling pierces no disk. $\endgroup$
    – Shijie Gu
    Dec 29, 2022 at 3:54
  • $\begingroup$ It may be a good idea to use this as you say (instead of using the complement of a line take the complement of a series of disks each do no pierce the knot until the end point of the isotopy) . However in the general case it seems like something else will be needed. Gillman's construction pierces no disc yet is isotopic to the unknot. $\endgroup$
    – amd1234
    Dec 29, 2022 at 8:00

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