$R^2-{0}$ is homeomorphic to $R×S^1$

Question

Show that $\mathbb{R}^2-\{(0,0)\}$ is homeomorphic to $\mathbb{R}×S^1.$

What I have done is the following: Define $\phi:\mathbb{R}^2-\{(0,0)\}$ by $\phi((x,y))=(\frac{(x,y)}{||(x,y)||},x+y)$ It's injective, surjective and continuous. But I have no idea what is the inverse.

What ? The map I have given can you give the inverse of this map? — Infinity, Oct 08 '23 at 13:34
I think you wanted to write $\phi(x,y) = (x+y,\frac{(x,y)}{||(x,y)||})$ instead of $\phi((x,y))=(\frac{(x,y)}{||(x,y)||},x+y)$ because you have written $\mathbb{R}\times S^{1}$ — user2582354, Oct 08 '23 at 13:35
Start by considering the map $\eta:(0,\infty)\times\mathbb{S}^1\to\mathbb{R}^2\setminus{(0,0)}$ given by $\eta(r,v)=rv$ — Lorago, Oct 08 '23 at 13:35
That's just the definition of injectivity, but how did you prove that this holds for your function? — Lorago, Oct 08 '23 at 14:09
I have understood my mistake now. I don't know how to prove this is injection! — Infinity, Oct 08 '23 at 14:17
It isn't injective. $\phi(1,-1)=\phi(2,-2)=((\theta,\theta),0)$, where $\theta=1/\sqrt{2}$. — , Oct 08 '23 at 14:24

score 3 · Answer 1 · edited Dec 12 '23 at 13:05

3

Define $$ f : \mathbb R^2 \setminus \{0\} \to \mathbb R \times S^1, f(p) = \left(\ln \lVert p \rVert,\frac{p}{\lVert p \rVert}\right) .$$ Its inverse is $$ g : \mathbb R \times S^1\mathbb \to \Bbb R^2 \setminus \{0\}, g(x,z) = e^xz .$$

edited Dec 12 '23 at 13:05

Anne Bauval

34,650

answered Oct 09 '23 at 15:09

Kritiker der Elche

4,417

(+1) Better answer imo than those in the duplicate, because not using (cautionless!) a parametrization of $S^1$ by $[0,2\pi)$. – Anne Bauval Dec 12 '23 at 13:04

$R^2-{0}$ is homeomorphic to $R×S^1$

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