First of all, I did some research on this question. I found the same question (but only for $\mathbb{R}^2$) here. I also found an article in Wikipedia about how to generalize spherical coordinates for n dimensions, which promptly leads to the answer.

My question is: Is there any other way of finding a homeomorphism that wouldn't rely on previous knowledge about this coordinate system?

  • $\begingroup$ I suppose you mean $S^n\times \mathbb{R}_{>0}$, right? Besides that I don't quite understand the question. Geometrically speaking this homeomorphism is natural, independently of the dimension $\endgroup$ – b00n heT Nov 10 '16 at 20:39
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    $\begingroup$ @b00nheT: $\Bbb R$ works just as well: after all, it’s homeomorphic to $\Bbb R_{>0}$. $\endgroup$ – Brian M. Scott Nov 10 '16 at 20:40
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    $\begingroup$ @b00nheT The Reals and the positive reals are homeomorphic, so it doesn't matter. $\endgroup$ – N. Owad Nov 10 '16 at 20:40
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    $\begingroup$ You are indeed right (Brian and Owad). But putting only positive reals makes the bijection more immediate, don't you think? $\endgroup$ – b00n heT Nov 10 '16 at 20:44
  • $\begingroup$ The question is about $S^n \times \mathbb{R}$, after all. Perhaps the solution of the problem will require a formula for a homeomorphism between $\mathbb{R}$ and $\mathbb{R}_{>0}$... $\endgroup$ – Lee Mosher Nov 10 '16 at 21:13

... And I myself have just found the answer I needed.

Let $x=(x_1,x_2,...,x_{n+1}) \in \mathbb{R}^{n+1}\backslash\{0\}$. Then, I could have simply made $f:\mathbb{R}^{n+1}\backslash\{0\}\rightarrow S^n\times\mathbb{R}$ as:


And it's inverse $f^{n-1}:S^n\times\mathbb{R}\rightarrow \mathbb{R}^{n+1}\backslash\{0\}$ as

$$f^{-1}(x_1,...,x_{n+1},z)=\left(e^z x_1,...,e^z x_{n+1}\right)$$


You have mistaken between R^(n+1)-{0} and R^(n+1), but this answer is very good. If you learn Algebraic of Topology, you will have some solutions to find a homeomorphism.

  • $\begingroup$ Thanks for the tip, I edited it. $\endgroup$ – AspiringMathematician Nov 16 '16 at 1:02

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