Homeomorphism between $\mathbb{R}^2$ and the open unit disc Given the function $f(x,y)=(\frac{x}{1+\sqrt{x^2+y^2}},\frac{y}{1+\sqrt{x^2+y^2}})$, I have to prove $f$ is a homemorphism between $\mathbb{R}^2$ and the open unit disc. Proving this function is conitunuous is trivial, proving it's a bijection is a bit harder (I wasn't able to find it's inverse). I did find the same function in a different notation - $f(z)=\frac{z}{1+\|z\|}$ where $z\in\mathbb{R}^2$, but since it's my first time dealing with two variable functions and homemorphisms, I can't see why $f(z)=f(x,y)$, other than $\|z\|=\|(x,y)\|$.
What is the best way to approach this with given notation ($f(x,y)$) - more generally how can I prove this function is a bijection and find its inverse.
(I was able to show that $lim_{(x,y)\rightarrow(\infty,\infty)}f(x,y)$ is (1,1) which I guess shows its a surjection but otherwise had no idea).
 A: It is easy to find the inverse in complex notation---
\begin{align}
&f(z) = \dfrac{z}{1+||z||}\\
\implies &||f(z)|| = \dfrac{||z||}{1+||z||}\\
\implies &||z|| = \dfrac{||f(z)||}{1-||f(z)||}\\
\implies &z = {f(z)}\times\left({1+\dfrac{||f(z)||}{1-||f(z)||}}\right)=\dfrac{f(z)}{1-||f(z)||}\\
\end{align}
So the inverse function is
\begin{equation}
f^{-1}(z) = \dfrac{z}{1-||z||}
\end{equation}
Now you can do whatever you wish with this. Note that proving this function  continuous is also trivial when $(x,y)$ takes values in open unit disk. Now just find the domain and range of the two functions and you're done.
A: We have $f(x,y)=(\frac{x}{1+\sqrt{x^2+y^2}},\frac{y}{1+\sqrt{x^2+y^2}})$.
It's easier to see in polar coordinates: $\vec x=(x,y)=(r\cos \theta,r\sin \theta);\ f(x,y)=(r'\sin \varphi,r'\cos \varphi)$. Then,
$r'(\cos \theta,\sin \theta)=\frac{r}{1+r}(\cos \varphi,\sin \varphi)$  so it is clear that $f$ maps $(x,y)$ into the disk along the radial line it makes with the origin. That is, $(x,y)\mapsto \frac{\|\vec x\|}{1+\|\vec x\|}(x,y).$ $f$ is clearly bijective.
It is also open: take a basic open set
$\vec x_0\in (x_0-\delta)\times ,x_0+\delta)\times (y_0-\eta,y_0+\eta):=U.$ If $\vec x\in U$ then, wlog $x_0,y_0\ge 0$, we have then
$\sqrt{(x_0-\delta)^2+(y_0-\eta)^2}< \|\vec x\|< \sqrt{(x_0+\delta)^2+(y_0+\eta)^2}$
and so
$\frac{\sqrt{(x_0+\delta)^2+(y_0+\eta)^2}}{1+\sqrt{(x_0+\delta)^2+(y_0+\eta)^2}}<\|f(x,y)\|<(x_0+\delta)^2+(y_0+\eta)^2.$
It follows that $f$ is a homeomorphism.
