The question is this:
Consider the parabaloid $\{(x,y,z)|z=1-x^2-y^2\}$, let $A$ be the subset satisfying $z>0$.
Consider the plane $\pi$ given by $z=1$.
The functions $x$ and $y$ act as coordinates on A (not entirely sure what this means)
$x$ and $y$ act as coordinates on $\pi$
Define $F:A\rightarrow\pi$ by $F(x,y,z) = (\frac{x}{z},\frac{y}{z},1)$
In the coordinates $x,y$ on both A and $\pi$ compute the Jacobian for F, show that this determinant is non-zero everywhere.
Update
I am being confused by the choice on offer of coordinates to take! I can project an open ball in R^2 of radius 1 up to the parabaloid segment in question, then I take these into the plane given by F, then I project these onto R^2.
So I am mapping an open unit ball to the entire R^2 plane essentially. There are other ways to do this, and the question says "the" Jacobian
What have I tried
A few things, I think I'm getting muddled up on which coordinates are with respect to what.
A chart can be thought of as $(U,\phi)$ where $\phi:U\rightarrow\mathbb{R}^n$ and $\phi=(x^1(u),x^2(u),...,x^n(u))$ where $x^i:U\rightarrow\mathbb{R}$ by $x^i=r^i\composed\phi$
[After this what I say would be very woolly, I've backspaced it several times]
I want to map A to $\pi$ then $\pi$ to a more familiar $\mathbb{R}^2$ I think, but I have no confidence in saying this.
Using the information at hand I have come to:
$(\frac{1}{z}-\frac{2x^2}{z^2})(\frac{1}{z}-\frac{2y^2}{z^2})-\frac{4x^2y^2}{z^4}$ but I don't really know how to interpret this answer and I don't like how I got to it.
What I think I need to realise
I've seen that F maps a point on A to the entire plane (ones with small values with Z are tending off towards infinity), this is not a projection but I'm not sure where my coordinates are.
I'm quite muddled, I wont make this any longer by spewing my thoughts out!