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We have a non-trivial knot $K$. Can we embed it in some 3-manifold $M$, so that $K$ becomes a trivial knot? Can embeddings even change the type of a knot anyhow?

The answer to this post points that, well, if we take the unknot, and we send it to a solid torus in such a way that it circles around the torus hole, then it is, of course, not equivalent to the unknot in the solid torus.

But this is not a very interesting example, so let's make our question formal, in such a way that we will avoid that:

  1. We have two knots $K_1, K_2 \subset \mathbb{R}^3$, a 3-manifold $M$, and continuous $h: \mathbb{R}^3 \to M$ which is homeomorphic onto its image. If $h(K_1)$ and $h(K_2)$ are equivalent knots in $M$, does it mean that $K_1$ and $K_2$ are equivalent?

(Of course, we can take two orientations of the trefoil and send it to $\mathbb{RP}^3$ – but are there any less trivial examples?)

We can add two questions to that:

  1. The same as 1., but with $K_2$ being the unknot (this is obviously the special case, but the interesting one).
  2. Generalization: now we have two 3-manifolds $N$ and $M$, knots $K_1, K_2 \subset N$ with a common point $*$, and $h: N \to M$ which is a homeomorphism onto its image. Are $K_1$ and $K_2$ equivalent in $N$, if and only if they induce the same element of $\pi_1(N)$ and $h(K_1), h(K_2)$ are equivalent in $M$?
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The answer to 2 is fairly straightforward: if $K\subset \mathbb{R}^3$ is embedded in $M$ and is isotopic to the unknot in $M$ then it is isotopic to the unknot in $\mathbb{R}^3$. A way to see this is that there are two important submanifolds at play. The first is a ball $B\subset M$ whose interior contains $K$ (which comes from embedding such a ball from $\mathbb{R}^3$) and the second is a disk $D\subset M$ whose boundary is $K$ (which proves $K$ is an unknot). We may assume that $D$ is transverse to $\partial B$ by the usual density arguments, and so $D\cap \partial B$ is a collection of simple closed curves in the sphere $\partial B$. Then one does the usual sort of innermost disk argument: we assume $D$ is chosen such that $D\cap \partial B$ has the least number of curves and, for sake of contradiction, that it has at least one curve, then we take a disk in $\partial B$ whose boundary is one of these curves and whose interior is disjoint with $D$, and with it we compress $D$. By throwing away the $S^2$ component after compression, then we've reduced the number of intersection curves by at least one. Hence we may assume $D\cap \partial B$ is empty, and therefore that there is such a $D$ that lies in the interior of $B$. That is, $K$ is unknotted in $\mathbb{R}^3$.

The answer to 1 is "no" when $M$ is not orientable, since it's possible that $K_1$ and $K_2$ are non-isotopic mirror images, so then by dragging $K_1$ around $M$ one would be able to see that $K_1$ is isotopic to $K_2$.

If $M$ is compact and orientable, then the answer is (perhaps surprisingly) "yes," but I'm not sure if there's a quite so elementary proof. Let $K$, $B$, and $M$ be like before, and then consider the manifold $M-\nu K$, which is the knot exterior (a compact manifold). The boundary of $B$ is a decomposition sphere that gives $M-\nu K$ as a connect sum of $M$ and $S^3-\nu K$, which is prime. By uniqueness of prime decompositions, every prime decomposition of $M-\nu K$ will have an $S^3-\nu K$ part with the remaining summands connect summing to $M$. Now suppose that $K_1$ and $K_2$ are knots in $\mathbb{R}^3$ that have been embedded in $M$ and are isotopic. By the above, we can conclude that $S^3-\nu K_1$ and $S^3-\nu K_2$ are orientation-preserving homeomorphic. By the Gordon-Luecke theorem, $K_1$ and $K_2$ are isotopic in $S^3$. Hence they are isotopic in $\mathbb{R}^3$.

The answer to 3 is "no." Consider $K\subset D^2\times S^1$ being a component of the Whitehead link in the exterior of the other. It is not unknotted, but if you embed $D^2\times S^1$ in $S^3$ then it is. Both $K$ and the unknot induce the same element of $\pi_1(D^2\times S^1)$, the identity. (Normally I would have suggested asking your third question as a new question, since you're not supposed to ask more than one question per question here, but this was quick enough to answer.)

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  • $\begingroup$ Excellent answer! $\endgroup$ Jun 20, 2022 at 13:43

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