I'm reading some notes about topology and homeomorphisms and there is an example of a homeomorphism from the unit ball in $\mathbb R^n$ to $\mathbb R^n$. The map $x \mapsto {x \over 1 - |x|^2}$. I assume absolute value means Euclidean norm here. What I don't understand is the square in the definition. Wouldn't $x \mapsto {x \over 1 - |x|}$ also be an homeomorphism and a simpler example? Is there a reason why one would want to square the absolute value to get some sort of better homeomorphism?
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Apart from Thomas' argument, there is another advantage, from the computational point of view. Note that the Euclidean norm is usually defined through the Euclidean scalar product:
$$|x| = \sqrt{\langle x, x \rangle} = \sqrt{\sum_{k=1}^n x_k^2}.$$
Using $|x|^2$ instead of $|x|$ loses you square root, so $|x|^2$ is actually simpler than $|x|$.
Vedran Šego
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The only reason I can see to use $\dfrac{x}{1-|x|^2}$ is that it is differentiable, so that when you later discuss diffeomorphisms, you can refer back to this function.
BTW, a standard map is $x\to \dfrac{x}{\sqrt{1-|x|^2}}$.
This might seem more complicated, but there is actually a deep geometric meaning to this map.
Take the point $x=(x_1,x_2,\dots,x_n)$ in the open unit disk. Then $(x_1,\dots,x_n,\sqrt{1-|x|^2})$ is in $S^n$, and we can find the point on the line $x_{n+1}=1$ projected from the center of $S_n$. This gives $\dfrac{x}{\sqrt{1-|x|^2}}$.
Thomas Andrews
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Could I ask what you mean by $S^n$ and $S_n$? For people like me who're new to the subject. – Jack M Jun 17 '13 at 18:46
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$S^n$ is the $n$-dimensional sphere, the set of points $(x_1,\dots,x_{n+1})\in\mathbb R^{n+1}$ with norm $1$. So $S^1$ is the circle, $s^2$ is the normal sphere, etc. – Thomas Andrews Jun 17 '13 at 18:47