What does $\Bbb S^{n-1}\times \mathbb{R}$ stand for? > Let $n\geqslant 1,f: \Bbb R^n - \\{ 0 \\} \to S^{n - 1} \times \Bbb R,x\to(\frac{x}{||x||},\ln(||x||))$ is a homeomorphism which the inverse is $f^{-1}:S^{n-1}\times\mathbb{R}\to\mathbb{R}^n-\\{0\\},(y,t)\to e^{t}y$. I was reading this example but I cannot understand what $S^{n-1}\times\mathbb{R}$ means. I suspect that $S^{n-1}$ is the counter dominion of the function. y **Questions:** **1)** Is $S^{n-1}$ the counter-dominion of f? **2)** Why is the counter dominion $S^{n-1}$? Where does the $n-1$ comes from? Thanks in advance!

Answer $\Bbb S^{n-1}$ is the unit sphere in $\Bbb R^n$ ie $$\Bbb S^{n-1} =\\{\xi\in\Bbb R^n: \|\xi\|=1\\}$$ $n-1$ represent the DIMENSION OF $\Bbb S^{n-1}$ as a MANIFOLD precisely, $$\dim \Bbb S^{n-1}= n-1, ~~~~~ \dim \Bbb R^{n} =n$$

For instance, in dimension 2, i.e in $\Bbb R^{2}$ the unit circle is defined $$\Bbb S^{1} =\\{\xi\in\Bbb R^2: \|\xi\|=1\\} \equiv \\{e^{i\theta}: \theta\in[0,2\pi)\\}$$ is of dimension one. $\dim\Bbb S^1 =1$ roughly speaking you see $\Bbb S^1$ as the real line $\Bbb R$.