Homeomorphism between open unit ball and $\mathbb R^n$ 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 $\mathbb R\setminus\{(n+\frac{1}{2})\pi: n\in \mathbb Z\}$
 A: Injectivity
$F(x)=F(y)$, 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.
Surjectivity
$0=F(0)$, 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.
Continuity
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 $F^{-1}$.
