# Can embedding non-equivalent knots in manifold produce equivalent knots?

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 $$M$$?

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.)