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The function is as follows: $$F:D^2(0,1) \rightarrow \Bbb R^2$$ $$ F(x)=\begin{cases}\tan\left(\frac{\pi||x||}{2}\right)\frac{x}{||x||}&, x\neq0 \\ 0 & ,x=0 \end{cases}$$

I am trying to show this is function is 1:1 and onto on the open disk of radius 1 and centered at 0 in 2 dimensions to a disk of infinite radius. I have tried to represent the magnitude using coordinates but I don't think I am approaching the problem correctly. Any advice would be appreciated

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    I hope the edit is correct. Feel free to roll back if not – 2017-02-16

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Informally speaking, the key is that this function acts radially: it only scales each vector $x$ by a numerical factor along the same line. So as long as it's one-to-one and onto on each line thru the origin, by mapping the "$[0,1]$" segment of that ray emanating from the origin onto the entire ray, the same is true for the map as a whole. (These lines have the origin in common, but it's dealt with separately in the definition of this function anyway.)

More rigorously, we could switch to polar coordinates: represent each $x\in\mathbb{R}^2$ as $x=(r\cos\theta,r\sin\theta)$ with $0\le r$ and $0\le\theta<2\pi$ to avoid ambiguity. The domain of this function is then $D^2(0,1)=\{x=(r\cos\theta,r\sin\theta) \mid 0\le r<1,0\le\theta<2\pi\}$. And the function simplifies into

$$F(x)=F[r,\theta]=\begin{cases}\tan\left(\frac{\pi r}{2}\right)\cdot(\cos\theta,\sin\theta)&, r\neq0 \\ 0 & ,r\end{cases}$$

or even purely in polar coordinates (using brackets for polar coordinates to avoid confusion with cartesian coordinates)

$$F(x)=F[r,\theta]=\begin{cases}\left[\tan\left(\frac{\pi r}{2}\right),\theta\right]&, r\neq0 \\ 0 & ,r\end{cases}$$

Now we can see that points with different polar angles $\theta$ will have different images under this function. And for a fixed $\theta$ and assuming $r\neq0$, the function $r\mapsto\tan\left(\frac{\pi r}{2}\right)$ maps $(0,1)$ onto $(0,+\infty)$ in a one-to-one fashion.

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    Hm interesting I didn't think to use polar coordinates. This simplifies it a lot. It still is not obvious to me how to get an inverse formula though. – 2017-02-16
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    Using $\arctan$. – 2017-02-16
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    is it simply [2arctan(y)/pi,theta] – 2017-02-16
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    It should be (when y doesn't equal 0). I verified it by plugging each one into the other and checking it was the identity. Thanks for the help and insight. – 2017-02-16
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    @numtheory1555242: Yes, that's right (the inverse, I mean). And then it can also be expressed in terms of $y$ similarly to how the original function was given: $F^{-1}(y)=\left(\frac{2\arctan(\|y\|)}{\pi}\right)\frac{y}{\|y\|}$ for $y\neq0$. You're very welcome! – 2017-02-17