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Let $B=\{x\in\mathbb R^n : ||x||<1\}$ the open unit ball with the subapce topology of $\mathbb R^n$. I want to show that $B^n\cong\mathbb R^n$ with the map $F(x)=\tan(\frac{\pi ||x||}{2})\frac{x}{||x||}$ for $x\not=0$ and $F(0)=0$

Well $\frac{x}{||x||}$ is more or less the direction of the vector $x$, the norm is a number out of $[0,1)$ and the scaled tangent is a Homeomorphism from $[0,1)$ to the positive real numbers.

I want to show that $F(x)$ is contionous, bijective and there is an inverse map.

Injectivity: Let $F(x)=F(y)\leftrightarrow ||F(x)||=||F(y)||\leftrightarrow \tan(\frac{\pi ||x||}{2})||\frac{x}{||x||}||=\tan(\frac{\pi ||y||}{2})||\frac{y}{||y||}||$

I know that the tangent is injective but I do not see how $x=y$ can be followed.

Surjectivity: Let $y\in\mathbb R^n$, now I need to show $\exists x\in B^n: F(x)=y$, i.e $\tan(\frac{\pi ||x||}{2})\frac{x}{||x||}=y$ I do not see how surjectivity can be followed form this.

Inverse map: I know that the tangent has an inverse map, namely the arctangent, but how can this be used here?

Continuity: Well I know that the tangent is only contin. on $\mathbb R\setminus\{(n+\frac{1}{2})\pi: n\in \mathbb Z\}$

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Your map works fine. For example for your beginning argument for injectivity, the two norms on the right of the tangents are each $1$ so after cancelling that and using injectivity of tangent you have $||x||=||y||$. Then since as you say the map preserves rays through the origin you have $x=y$. – coffeemath May 13 '13 at 20:22
babla da fatiye dichho toh! – Un Chien Andalou May 16 '13 at 13:47
up vote 4 down vote accepted


$F(x)=F(y)$, with $x\ne0$ and $y\ne0$ means

$$ \tan\biggl(\frac{\pi \|x\|}{2}\biggr)\frac{x}{\|x\|}= \tan\biggl(\frac{\pi \|y\|}{2}\biggr)\frac{y}{\|y\|} $$

So we can assume $x\ne0$ and $y\ne0$. Then, taking norms, $$ \tan\biggl(\frac{\pi \|x\|}{2}\biggr)=\tan\biggl(\frac{\pi \|y\|}{2}\biggr) $$ because $x/\|x\|$ has norm $1$ and the same for $y/\|y\|$; moreover $\tan t>0$ if $t>0$. By the injectivity of the tangent you get $$ \frac{\pi \|x\|}{2}=\frac{\pi \|y\|}{2} $$ so $\|x\|=\|y\|$ and, finally $x=y$.

It's clear that $F(x)=0$ if and only if $x=0$, so the proof is complete.


$0=F(0)$, so we only need to show that, for $z\ne0$, we can find $x$ with $F(x)=z$. We should have $$ z=\tan\biggl(\frac{\pi \|x\|}{2}\biggr)\frac{x}{\|x\|} $$ so $$ \|z\|=\tan\biggl(\frac{\pi \|x\|}{2}\biggr) $$ hence $$ \frac{\pi\|x\|}{2}=\arctan\|z\| $$ that is, $$ \|x\|=\frac{2\arctan\|z\|}{\pi}. $$ Thus the candidate is $$ x=\frac{2\arctan\|z\|}{\pi\|z\|}z $$ Verify it's the right one.


The continuity of $F$ is obvious in the points different from $0$, because it's obtained by continuous functions. Since your domain consists of vectors with norm $<1$, there's no problem with the tangent function, because you consider only arguments in $(0,\pi/2)$. The continuity of the inverse is obvious as well outside $0$

Are $F$ and its inverse continuous at $0$?

A well known fact is that $\lim_{x\to0}F(x)=0$ if and only if $\lim_{x\to0}\|F(x)\|=0$; now $$ \|F(x)\|=\biggl\|\tan\biggl(\frac{\pi\|x\|}{2}\biggr)\frac{x}{\|x\|}\biggr\| =\tan\biggl(\frac{\pi\|x\|}{2}\biggr) $$ that obviously satisfies the requested property. The same for the continuity at $0$ of $F^{-1}$.

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Thanks for you answer. I do not see why $x=\frac{2\arctan\|z\|}{\pi\|z\|}z$ should be unique (e.g the right one), isn't it possible to choose instead of $z/||z||$ something different, where the norm is equal to 1?. For continuity I set instead of $t$ $||x||$, but then $F(x)$ goes to $0$, independent if I come from left or right, is this ocrrect? – Babla May 13 '13 at 21:10
@Babla We're proving surjectivity, so finding one $x$ with $F(x)=z$ is sufficient. On the other hand, injectivity has already been proved, hasn't it? For continuity: that limit is $1$, so $\tan(\pi\|x\|/2)/\|x\|$ is bounded in a neighborhood of $0$. Hence... – egreg May 13 '13 at 21:12
Oh yes, you are right.Still, the continuity is not clear to me. – Babla May 13 '13 at 21:15
@Babla You can also use the fact that $\lim_{x\to0}F(x)=0$ if and only if $\lim_{x\to0}\|F(x)\|=0$. – egreg May 13 '13 at 21:18
Hmm, but $\lim_{x\rightarrow 0} ||F(x)||=0$ is clear because $\tan(x)=0$ for $x\rightarrow 0$, therefore $\lim_{x\rightarrow 0} F(x)=0$. Where do I need $\lim \frac{\tan t}{t}=1$ now? – Babla May 13 '13 at 21:22

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