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Is $\mathbb{S}^3 \backslash \mathbb{S}^1$ homotopic equivalent to $\mathbb{S}^1$?

I am reading this from some notes on algebraic topology.

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

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

  3. How will $\mathbb{S}^2 \backslash \mathbb{S}^1$ look like? (homeomorphic to 2 open discs?)

MUH
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  • As to 1 and 2 I suppose they mean some fixed circle inside $\Bbb S^3$, like all points of the form $(x,y,0,0)$ in the 3-sphere. 3 would be the 2-sphere minus the equator which indeed is homeomorphic to the disjoint union of two open disks (each of which is homeomorphic to $\Bbb R^2$ again). – Henno Brandsma Feb 15 '21 at 12:43
  • Indeed, one cannot pick just any homeomorphic image of $\Bbb S^1$ in the larger space - you want to avoid knots. Picking the (apparently not so) obvious circle is intended (while on the other hand many other choices would work as well). The standard model of $\Bbb S^n$ has an "equator" that is a $\Bbb S^{n-1}$. So here we want the equator of the equator ... – Hagen von Eitzen Feb 15 '21 at 12:46
  • What about the homotopy equivalence of the two spaces? – MUH Feb 15 '21 at 12:59

2 Answers2

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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.

freakish
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  • I am trying to visualize the spaces but I am having a hard time.
    1. Could you give any reference for "stereographic projection to $S^3\backslash S^1$ relative to some point in $S^1$, to obtain that this space is homeomorphic to $R^3∖L$ with $L$ being a line"?
    2. Could you translate the object $S^3, S^1,R^3, L $ to a lower dimension analogue?
     – MUH Feb 15 '21 at 14:50
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    @MUH you won't visualize $S^3$ because it doesn't fit into $\mathbb{R}^3$. And we (human beings) are not used to anything higher dimensional. But $S^3$ without a point is an ordinary $\mathbb{R}^3$ via the stereographic projection. The choice of the point doesn't matter, but it has to be done. So $S^3\backslash X$ is homeomorphic to $\mathbb{R}^3\backslash (s(X)\backslash{p})$, where $s$ is the projection. We loose one point. And the projection maps circles passing through $p$ to lines. – freakish Feb 15 '21 at 15:01
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    As for the lower dimensional intuition: consider the standard sphere $S^2$. Then we consider the zero dimensional sphere $S^0$ which is two points. Then $S^2\backslash S^0$ is homeomorphic via the projection to $\mathbb{R}^2\backslash{p}$, we loose one point, which can then be deformed to $\mathbb{R}\times S^1$ (well, in the zero dimension these are homeomorphic, but in higher dimensions these are only homotopy equivalent) and finally to $S^1$. I'm pretty sure that the same method works for $S^n\backslash S^{n-2}$ for any $n\geq 2$. – freakish Feb 15 '21 at 15:06
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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.

Shingle
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