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Let $K$ be a knot. For an embedding of $K$ into $\mathbb{R}^3$, let $N_1$ be the tubular (open) neighborhood of $K$ in $\mathbb{R}^3$. Similarly, one can embed $K$ in $S^3$ with corresponding tubular neighborhood $N_2$.

Is it true that the spaces $\mathbb{R}^3\setminus N_1$ and $S^3\setminus N_2$ are homotopy equivalent?

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  • $\begingroup$ Usually by a knot one means a tame embedding of the circle to the sphere (up to isotopy). Then it makes no sense to ask about an embedding of a knot. As it stands, your question is very unclear. Did you mean that $K$ is the circle? In this case, your question has negative answer. $\endgroup$ – Moishe Kohan Feb 7 '18 at 11:41
  • $\begingroup$ Yes, I meant the $K$ to be the circle. $\endgroup$ – piotrmizerka Feb 7 '18 at 12:21
  • $\begingroup$ Ok, then the answer to your question is that these spaces are never homotopy equivalent since one is always aspherical and the other is never aspherical. $\endgroup$ – Moishe Kohan Feb 7 '18 at 12:22
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No, clearly not. Take the unknot for example. Then $S^3 \setminus N_2$ is homotopy equivalent to a circle $S^1$, whereas $\mathbb{R}^3 \setminus N_1$ is homotopy equivalent to a wedge sum of a circle with a sphere $S^1 \vee S^2$.

In fact $\mathbb{R}^3 \setminus N_1$ is equal to $S^3 \setminus N_2$ with a point removed. Removing a point from a compact 3-manifold with boundary always changes its homotopy type ($M \setminus \{\mathrm{pt}\} \simeq M \vee S^2$).

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  • $\begingroup$ When I say with boundary I really mean with nonempty boundary. $\endgroup$ – Najib Idrissi Feb 7 '18 at 13:09

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