# Show that $\{(x,y)\in\mathbb{R}^m\times \mathbb{R}^n ; |x|^2+|y|^2=1, y\neq 0\}$ is homeomorphic to $\mathbb{R}^m\times S^{n-1}$.

Show that $\{(x,y)\in\mathbb{R}^m\times \mathbb{R}^n ; |x|^2+|y|^2=1, y\neq 0\}$ is homeomorphic to $\mathbb{R}^m\times S^{n-1}$.

I tried to find an easy bijection (and then try to prove continuity and inverse's continuity) but I can't find this bijection.

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Suppose that $m = 1$ and $n=2$, hence you are in $\mathbb{R}^3$. You should see that the space $X \subseteq \mathbb{R}^3$ is the unitary sphere $S^2 \subseteq \mathbb{R}^3$ minus the two antipodal points $(-1,0,0)$ and $(1,0,0)$. It should be clear that $X$ is homeomorphic to the meridian $\mathbb{R}$ times the equator $S^2 \cap \{ x = 0 \} \simeq S^1$, i.e. $\mathbb{R} \times S^1$ as wanted. This is a sort of Mercator projection of the earth planet without the two poles onto an infinite cylinder which is tangent to the equator. Can you write down formulas in this particular case?
You mean sending $(x,y)$ to $\frac{1}{|y|}(x,y)\in\mathbb{R}^{m}\times \mathbb{S}^{n-1}$ right? In this case: Continuity and surjection are clear. Injection is not very easy but I think that I can prove it using $|x|^2+|y|^2=1$. But I can't see the form of inverse. Inverse's second componen it's obvious but inverse of first component I need to use that $|x|^2+|y|^2=1$ and is so algebraically difficult. – Gastón Burrull May 18 '12 at 23:20
Exactly! The inverse $\mathbb{R}^m \times S^{n-1} \to X$ is given by $(x,s) \mapsto (\vert x \vert^2 +1)^{-1/2} (x,s)$. You should be able to check that this is the inverse of $(x,y) \mapsto \vert y \vert^{-1} (x,y)$. – Andrea May 18 '12 at 23:42