Let $\mathbb{W^2} = \{x \in \mathbb{R^3} \mid q(x,x) = -1, x_3 > 0\}$ where $q(x,y) = x_1y_1 + x_2y_2 -x_3y_3$ and $\mathbb{B^2} = \{x \in \mathbb{R^3} \mid x_3 = 0, x_1^2 + x_2^2 < 1\}$.
The metric on $\mathbb{W^2}$ is $g_x(v,w) = q(v,w) $ whereas the metric on $\mathbb{B^2}$ is $g_x(v,w) = \dfrac{4}{(1-\| v\|^2)^2}\langle v,w\rangle_{Euclidean}$
Let $f: \mathbb{W^2} \rightarrow \mathbb{B^2}$ given by requiring that for each $p \in \mathbb{W^2}$ the points $f(p)\in \mathbb{B^2}$ and $p$ are collinear with the point $(0,0,−1)$ (i.e. $f$ is a projection from this point to the plane $\{z= 0\}$).
I have shown that $f(X,Y,Z) = (\dfrac{X}{Z+1}, \dfrac{Y}{Z+1})$.
I have that for a specific chart that: $Df(p)(\dfrac{\partial}{\partial x}) = \dfrac{\partial}{\partial y}$, where $\dfrac{\partial}{\partial x_i} \in T_p\mathbb{W^2}, $$\dfrac{\partial}{\partial y_i} \in T_{f(p)}\mathbb{B^2}$.
Now I want to show that $\left<\dfrac{\partial}{\partial x_i}, \dfrac{\partial}{\partial x_j}\right>_p = \left<\dfrac{\partial}{\partial y_i}, \dfrac{\partial}{\partial y_j}\right>_{f(p)} $.
However, I'm having some difficulty:
$Df(p) = \begin{pmatrix} \dfrac{1}{Z+1} & 0 & \dfrac{-X}{(Z+1)^2} \\ 0 & \dfrac{1}{Z+1}& \dfrac{-Y}{(Z+1)^2} \end{pmatrix}$
so, $\left<\dfrac{\partial}{\partial y_i}, \dfrac{\partial}{\partial y_j}\right>_{f(p)} = \dfrac{4}{(1 - (\| f(p)\|)^2)^2} \left<\dfrac{\partial}{\partial y_i}, \dfrac{\partial}{\partial y_j}\right>_{Euclidean} $.
I then have $\| f(p)\|^2 = \dfrac{X^2 + Y^2}{(Z+1)^2} = \dfrac{Z^2-1}{(Z+1)^2} = \dfrac{Z-1}{Z+1} $.
And so: $\dfrac{4}{(1 - (\| f(p)\|)^2)^2} = (Z+1)^2 $
$\Rightarrow \dfrac{4}{(1 - (\| f(p)\|)^2)^2} \left<\dfrac{\partial}{\partial y_i}, \dfrac{\partial}{\partial y_j}\right>_{Euclidean} \\= (Z+1)^2 \left<Df(p)\dfrac{\partial}{\partial x_i}, Df(p)\dfrac{\partial}{\partial x_j}\right>_{Euclidean}$.
However, this doesn't seem to be giving me the $\left<\dfrac{\partial}{\partial x_i}, \dfrac{\partial}{\partial x_j}\right>_p$.
I would massively appreciate a hint, or a correction if I've made some huge error. As for the assumed results, I'm fairly certain they're correct.