I am trying to find a homeomorphism between $\mathbb{R}^n\backslash\{0\}$ and $\mathbb{S}^{n-1}\times\mathbb{R}$. I read some related questions and am thinking of the map $f:\mathbb{R}^n\backslash\{0\}\rightarrow\mathbb{S}^{n-1}\times\mathbb{R}^+$ given by $$f(x)=(\dfrac{x}{||x||},||x||).$$ I understand that it is continuous but what is its inverse? I don't seem to see it... If I prove this is a homeomorphism, then since $\mathbb{R}^+\cong\mathbb{R}$ (by the exponential, right?) I get $$\mathbb{R}^+\cong\mathbb{R}\Rightarrow\mathbb{S}^{n-1}\times\mathbb{R}^+\cong\mathbb{S}^{n-1}\times\mathbb{R}$$ and by transitivity I get my result. Any help finding the inverse would be great!
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isn't $\mathbb{R^n}\cong\mathbb{S^n}$? – JMP Oct 16 '15 at 15:12
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2@JonMarkPerry No. Precisely one of those is compact. – Oct 16 '15 at 15:15
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i thought that was Poincare's conjecture - which has been proven – JMP Oct 16 '15 at 15:15
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3I'm sorry, but that has nothing to do with the Poincare conjecture. – Oct 16 '15 at 15:16
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okay, my mistake – JMP Oct 16 '15 at 15:17
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The inverse is of course $(s,r)\mapsto rs$. Note that $\frac x{\|x\|}\cdot \|x\|=x$.
Hagen von Eitzen
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