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How to define an embedding $S^1\times B^2$ in $\mathbb R^3$ ?

We know that $\mathbb R^3$ can be represented as $\mathbb R \times \mathbb R^2$. There is a homemorphism between $B^2$ and $\mathbb R^2$. To embedd $S^1$ into $\mathbb R$ we can consider $f$, as inverse mapping to $g = (\cos2\pi t, \sin2\pi t)$, where $t \in [0, 1)$, which displays the point $(x, y) \in S^1$ to the point $t \in [0, 1)$. The mapping $g \times h$, where $h$ is a homeomorphism beetwen $B^2$ and $\mathbb R^2$, is a homeomorphism beetwen $S^1 \times B^2$ and $[0, 1) \times \mathbb R^2$, which is a subset of $\mathbb R^3$. Is my reasoning correct ? If not, then how to solve this problem correctly ?

kontsev_
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  • Define $B^2$? Is it the open disk ${(x,y) \in \Bbb R^2\mid x^2+y^2 < 1}$? – Henno Brandsma Nov 10 '21 at 17:53
  • @HennoBrandsma $B^2$ is closed disc of unit radius. As i understand i can consider the mapping $G(θ,u,v)=((1+u \cos v)\cos θ,(1+u \cos v) \sin θ,u\sin v)$, where $0≤u≤1$ and $−π≤v<π$ ? – kontsev_ Nov 11 '21 at 05:54

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There isn't any embedding of $\mathbb S^1$ into $\mathbb R$. Indeed, if $M$ is a compact $n$-dimensional manifold without boundary, there is no embedding of $M$ into $\mathbb R^n$ (see here).

Your mapping $f : [0,1) \to \mathbb S^1$, $f(t) = (\cos 2\pi t, \sin 2\pi t)$ is a continuous bijective map from $[0,1)$ to $\mathbb S^1$, but the inverse is not continuous. Indeed, $[0,1)$ is non-compact while $\mathbb S^1$ is compact, so there is no homeomorphism between these two spaces.

It is easier to solve this "geometrically". Think of any embedding $F= (F_1, F_2) : \mathbb B^2 \to \{ (x, z) : x >0\}$ (for example, $F(t, v) = (t+2, v)$. Then one can embed $\mathbb S^1 \times \mathbb B^2 \to \mathbb R^3$ by taking a revolution:

$$ G (\theta, u, v) = ( F_1 (u, v) \cos \theta, F_1(u, v)\sin \theta, F_2(u, v)).$$

Arctic Char
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    Words help: Solid torus. – Ted Shifrin Nov 10 '21 at 17:17
  • @Arctic Char can i also consider $F_1=1+u\cos v$ and $F_2=u\sin v$, where $0≤u≤1$ and $-\pi ≤ v<\pi$ – kontsev_ Nov 11 '21 at 04:52
  • What do you think? @kontsev_ – Arctic Char Nov 11 '21 at 09:09
  • if we consider such $F_1$ and $F_2$, then $G$ will be homeomorphism between $S^1 \times B^2$ and solid torus – kontsev_ Nov 11 '21 at 09:26
  • No it isnt. Indeed the G so defined is not even injective. The problem is when $u=1$ @kontsev_ – Arctic Char Nov 11 '21 at 09:30
  • @ArcticChar I consider $B^2$ as a closed disc of unit radius, what is the problem with $u =1$ ? – kontsev_ Nov 11 '21 at 10:49
  • @kontsev_ What is $G(1, -\pi, \theta)$ for your given $F$? – Arctic Char Nov 11 '21 at 14:16
  • @ArcticChar I see, my mistake, because, If the radius of the larger circle is 1, then the torus does not exist. In that case i can consider $F_1 = 2 + u\cos v$ and $F_2 = u\sin v$. If i'm not mistaken, $G$ will be homeomorphism between solid torus and $S^1_2 \times B^2$, where $S^1_2 = { (x,y): x^2 + y^2 = 4 }$, but $S^1_2 \times B^2$ is homeomorphic to $S^1 \times B^2$, since we can consider mapping $f \times id$, where $id: B^2 → B^2$ and $f: S^1_2 → S^1$, where $f(x, y) = (x/2, y/2)$ – kontsev_ Nov 11 '21 at 14:39