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I am trying to prove that every tame knot is isotopic to the unknot. Here is my attempt. $\newcommand{\sone}{\mathbf{S}^1}$ $\newcommand{\rthree}{\mathbf{R}^3}$

isotopy from quotiented S1 to K

A (tame) knot is a locally flat embedding of $\sone$ in $\rthree$ and $\sone \subset \rthree$. Let $f \colon \sone \rightarrow K$ be an embedding of a knot $K$. Consider a locally flat neighbourhood $U_p \subset \rthree$ of a point $p \in K$, and a locally flat neighbourhood of $W_q \subset \rthree$ of a point $q \in \sone$. Now, $$f^{-1}|_{U_p \cap K} \colon U_p \cap K \rightarrow W_p \cap \sone$$ and $$f^{-1}|_{K \setminus (U_p \cap K)} \colon K \setminus (U_p \cap K) \rightarrow \sone \setminus (W_p \cap \sone)$$ are homeomorphisms.

So, $K \setminus (U_p \cap K)$ contains all the 'knotted' part and $f$ maps $\sone \setminus (W_p \cap \sone)$ to the 'knotted' part.

Let $\sim$ be the equivalence relation on $\sone$ defined by identifying all points of $\sone \setminus (W_p \cap \sone)$ to one single point. $\sone$ is isotopic to $\sone \setminus \!\sim$. Let $i \colon [0, 1] \times \sone \rightarrow \mathbf{R}^3$ be that isotopy, with $i_0(\sone) = \sone$ and $i_1(\sone) = \sone\setminus\!\sim$.

Now, is it correct that if one proves that $i \circ f^{-1}$ is an isotopy, then we are done?

I am not able to prove that $i \circ f^{-1}$ is an isotopy. Composition of two continuous and bijective functions is continuous and bijective. Thus, each $i_t \circ f^{-1}$ is continuous and bijective, for all $t \in [0,1]$.

By local flatness, I mean the following. A point $p \in K$ is said to be locally flat if there exists a neighbourhood $U_p \ni p$ such that the ordered pair $(U_p, U_p \cap K)$ is homeomorphic to $(B, d)$, where $B \subset \mathbf{R}^3$ is the unit ball around origin and $d \subset B$ is the set of points along a diameter of the ball. A knot $K$ is said to be locally flat if all points are locally flat.

By isotopy, I mean a homotopy $i$ such that all the individual $i_t$s are bijective. I distinguish between isotopy and ambient isotopy, although people usually mean ambient isotopy when they mean isotopy in a knot theory context. By ambient isotopy, I mean $a \colon [0, 1] \times \mathbf{R}^3 \rightarrow \mathbf{R}^3$. The domain in the ambient isotopy is not $\mathbf{S}^1$ but rather $\mathbf{R}^3$. My terminology follows Knots and Links by Cromwell.

$i \circ f^{-1}$ is defined as follows. $i \circ f^{-1} \colon [0, 1] \times K \rightarrow \rthree$, $(i \circ f^{-1}) (t, p) = i_t(f^{-1}(p))$, for all $t \in [0, 1]$ and $p \in K$.

Apoorv Potnis
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  • Why do you think this is true? – aschepler Dec 02 '22 at 18:16
  • @aschepler Because then all (tame) knots would be isotopic to $\mathbf{S}^1 \setminus! \sim$, which is isotopic to $\mathbf{S}^1$. Thus, all knots will be isotopic to $\mathbf{S}^1$, and isotopic to each other. – Apoorv Potnis Dec 02 '22 at 18:18
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    @MarianoSuárez-Álvarez By isotopy, I mean a homotopy $i$ such that all the individual $h_t$s are bijective. I distinguish between isotopy and ambient isotopy, although people usually mean ambient isotopy when they mean isotopy in a knot theory context. By ambient isotopy, I mean $a \colon [0, 1] \times \mathbf{R}^3 \rightarrow \mathbf{R}^3$. The domain in the ambient isotopy is not $\mathbf{S}^1$ but rather $\mathbf{R}^3$. My terminology follows Knots and Links by Cromwell. – Apoorv Potnis Dec 02 '22 at 18:36
  • "Because then all (tame) knots would be isotopic to $\mathbf{S}^1\setminus!\sim$, which is isotopic to $\mathbf{S}^1$. Thus, all knots will be isotopic to S1, and isotopic to each other." This what I want to prove. We can thus see that the notion of isotopy is not enough to distinguish knots and we need ambient isotopy. – Apoorv Potnis Dec 02 '22 at 18:38
  • @MarianoSuárez-Álvarez I have edited the question accordingly. – Apoorv Potnis Dec 02 '22 at 18:41
  • @MarianoSuárez-Álvarez Yes, I have seen it. I wanted a more rigorous proof. I have seen https://math.stackexchange.com/questions/215852/line-with-a-knot-in-mathbb-r3-isotopic-to-standard-embedding-mathbb-r-subs#comment9665032_215852 as well, which as I understand provides the explicit isotopy described in the animation. But I am struggling to prove that $F(x,t)$ is a homotopy, in the question I linked. So, I tried a different approach. – Apoorv Potnis Dec 02 '22 at 18:44
  • I now think that the whole question is nonsense. $i$ cannot be an isotopy as it is not injective. – Apoorv Potnis Dec 02 '22 at 22:39
  • You are correct in your last comment. But if you want to do so, what you could do is to edit your post to ask questions about details of the proof in https://math.stackexchange.com/questions/215852/line-with-a-knot-in-mathbb-r3-isotopic-to-standard-embedding-mathbb-r-subs#comment9665032_215852 – Lee Mosher Dec 08 '22 at 12:46
  • @LeeMosher Thank you for the clarification. I have asked a new question. Can you please have a look at https://math.stackexchange.com/questions/4594776/every-tame-knot-is-isotopic-to-the-unknot-bachelors-unknotting. – Apoorv Potnis Dec 09 '22 at 00:37

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