Is $\mathbb{S}^3 \backslash \mathbb{S}^1$ homotopic equivalent to $\mathbb{S}^1$? Is $\mathbb{S}^3 \backslash \mathbb{S}^1$ homotopic equivalent to $\mathbb{S}^1$?
I am reading this from some notes on algebraic topology.

*

*I am not even sure what does $\mathbb{S}^3 \backslash \mathbb{S}^1$ mean?


*what is $\mathbb{S}^1 \in \mathbb{R}^3$? Is it any one particular circle? Can anyone please define this as a set?


*How will $\mathbb{S}^2 \backslash \mathbb{S}^1$ look like? (homeomorphic to 2 open discs?)
 A: I assume that $S^1$ inside $S^3$ means some great circle of $S^3$, i.e. if
$$S^3=\big\{(x_0,x_1,x_2,x_3)\in\mathbb{R}^4\ \big|\ \sum x_i^2=1\big\}$$
then $S^1$ is for example $\{(x_0,x_1,x_2,x_3)\in S^3\ |\ x_2=x_3=0\}$ or any other combination of coordinates with two of them being $0$.
We most certainly cannot treat $S^1$ in $S^3$ as any homemorphic image, because the result then depends on the choice of such embedding.
If that's the case then we can apply the stereographic projection to $S^3\backslash S^1$ relative to some point in $S^1$, to obtain that this space is homeomorphic to $\mathbb{R}^3\backslash L$ with $L$ being a line, for example $L=\{(x,0,0)\ |\ x\in\mathbb{R}\}$.
And this space is homotopy equivalent to $S^1$. Indeed, we first construct a homotopy equivalence from $\mathbb{R}^3\backslash L$ to the tube $\mathbb{R}\times S^1$ via linear homotopy from the identity to $(x,y,z)\mapsto (x,c\cdot y, c\cdot z)$ map, where $c=1/\lVert (y,z)\rVert$. And then by contracting the $\mathbb{R}$ piece.
A: Another way to think about the case when $\mathbb{S}^1$ is embedded in $\mathbb{S}^3$ as an unknot (so for example as an equator of an equator) is the following. Treat $\mathbb{S}^3$ as formed by gluing two solid tori (see Complement of the Solid Torus in $S^3$ is Again a Solid Torus ). Then homotopically $\mathbb{S}^1$ is the same as the interior of one of that tori and hence complement has the same homotopy type as a solid torus. $\mathbb{S}^1$ is a deformation retract of solid torus, therefore in that case complement is homotopy equivalent to the circle.
As it was already noted, for different embeddings situation is far more complicated.
