Q:

Difficulties on proving the continuity part of a homeomorphism

I am trying to prove that the open unitary disk $\mathbb{D}^n$ is homeomorphic to $\mathbb{R}^n$, so the way i am doing it is by showing that the function $$f(x)=\frac{1}{1-|x|}x$$ where $x$ is a vector in $\mathbb{D}^n$ and $|x|$ is its norm, is bijective, continuous and has inverse continuous.  To prove it's bijective it suffies to give its inverse function $f^{-1}(y)= \frac{1}{1+|y|}y$ but I'm kind of lost with the continuity part.  Using the definition i guess i should manipulate this expression: $$|\frac{1}{1-|x|}x - \frac{1}{1-|y|}y|< \epsilon$$ to get $$|x-y|< \delta$$ where $x$ and $y$ belong to $\mathbb{R}^n$ and $\delta$ issomething that depends on $\epsilon$ and $x$. I've tried some algebraic  manipulation but didn't get anything useful, perphaps there is some more direct way of doing it.  Any help would be much appreciated

A:

$f(x_1,...,x_n)=\dfrac {(x_1,...,x_n)}{1-\sqrt{(\sum_{i=1}^nx_i^2)}}=(\dfrac {x_1}{1-\sqrt{(\sum_{i=1}^nx_i^2)}},..., \dfrac {x_n}{1-\sqrt{(\sum_{i=1}^nx_i^2)}})=(g_1(x_1,...,x_n),...,g_n(x_1,...,x_n))$
If you know that $h(x_1,...x_n)=\sum_{i=1}^nx_i^2$ is continuous then $s(x_1,...,x_n)=\sqrt{(\sum_{i=1}^nx_i^2)}$ is also continuous as a composition of a square root function and $h$ and then $w(x_1,...,x_n)=1-\sqrt{(\sum_{i=1}^nx_i^2)}$ is continuous as a sum of constant $1$ and $-s$.
Each $g_k$ is continuous for $k=1,...,n$ as a quotient of continuous $x_k$ and continuous $w$.
So it follows that because each $g_k$ is continuous that then $f$ is continuous.