Thinking about the fact that $f:\left(\frac{-\pi}{2},\frac{\pi}{2}\right) \rightarrow \mathbb{R}$ where $f(x)=\tan(x)$ for all x in $\operatorname{dom}(f)$ is a homeomorphism between $\left(\frac{-\pi}{2},\frac{\pi}{2} \right)$ and $\mathbb{R}$ gives me a bijective function $j:B \left(O,\dfrac{\pi}{2} \right) \rightarrow \mathbb{R}^{n} $ provided that $O$ is the origin of $\mathbb{R}^{n}$ and $j(x)=\dfrac{\tan(|x|)}{|x|}x$ for x $\in B \left(O,\dfrac{pi}{2} \right)$ if it is not the origin, $j(0)=0$.
But I have question about its continuity. Since I wanted to prove continuity of this function with the definition of continuous function, I wanted to use triangular inequality. But it fails to prove that j is homeomorphism between two metric spaces with usual metric in $\mathbb{R}^{n}$. Is it really continuous function? If it is true, how can I prove continuity of this function and the inverse function $j^{-1}:\mathbb{R}^{n} \rightarrow \left( 0,\dfrac{\pi}{2} \right)$ ?
A simpler example of a homeomorphism from $\mathbb R^{n}$ to $B(0,\pi/2)$ is $f(x)=(\pi/2 ) \frac x {1+|x|}$. Its inverse is $g(x)=\frac x {\pi /2 -|x|}$.
Your function is clearly continuous at $x \neq 0$. For continuity at the origin use the fact that $\tan |x| \to 0$ and $|\frac x {|x|}|=1$. [You have to define $j(0)$ to be $0$]. The inverse function is $j^{-1}(x)=\frac {\arctan |x|} {|x|} x$ which is also continuous by a similar argument.