Let $X$ be the unit sphere in $\mathbb R^3$, with the poles at $(0, 0, 1)$ and $(0, 0, -1)$ removed.
Let $H$ be the surface in $\mathbb R^3$ defined by the equation $x^2+y^2-z^2=1$.
We define a function $F: X \to H$ as follows. For each $p \in X$, there exists a unique line $l$ through $p$ that intersects the $z$-axis at right angles. This line $l$ intersects $H$ at two points. We define $f(p)$ to be the point of intersection between $l$ and $H$ such that $p$ lies between the $z$-axis and $f(p)$.
I have tried to express that mathematically but I haven't been able to. I want to conclude the solution which is $$F(x,y,z)=\left(x\sqrt\frac{1+z^2}{1-z^2},y\sqrt\frac{1+z^2}{1-z^2},z\right).$$ Any hint? Thanks.
Here is a diagram of the construction. The space is three-dimensional, but I can only draw a two-dimensional cross-section. You should visualise $X$ and $H$ as the "surfaces of revolution" that you get by rotating everything I've drawn about the $z$-axis.
Now let's tackle the problem. Suppose that $p = (x_0, y_0, z_0)$ is some point on $X$. Then the line $l$ is given by $$ l = \{ (0, 0, z_0) + t(x_0, y_0, 0) : t \in \mathbb R \} = \{ (tx_0, ty_0, z_0) : t \in \mathbb R \}.$$ To spell this out, $(0, 0, z_0)$ is the point where the line $l$ intersects the $z$-axis. Meanwhile, $(x_0, y_0, 0)$ is a direction vector along the line $l$; in fact, $(x_0, y_0, 0)$ is the displacement vector from $(0, 0, z_0)$ to $p$.
The points where $l$ and $H$ intersect are points of the form $(tx_0, ty_0, z_0)$, where $t$ satisfies the condition $$ (tx_0)^2 + (ty_0)^2 - z_0^2 = 1.$$ This condition on $t$ is equivalent to $$ t= \pm \sqrt{\frac{1 + z_0^2}{x_0^2 + y_0^2}}.$$ Since $(x_0, y_0, z_0)$ lies on the unit sphere, we have $x_0^2 + y_0^2 + z_0^2 = 1$, so our condition on $t$ is equivalent to $$ t = \pm \sqrt{\frac{1 + z_0^2}{1 - z_0^2}}.$$
Thus $l$ intersects $H$ at the two points $$ \left(x_0\sqrt{\frac{1 + z_0^2}{1 - z_0^2}}, y_0\sqrt{\frac{1 + z_0^2}{1 - z_0^2}}, z_0 \right), \qquad \left(-x_0\sqrt{\frac{1 + z_0^2}{1 - z_0^2}}, -y_0\sqrt{\frac{1 + z_0^2}{1 - z_0^2}}, z_0 \right).$$
The point $\left(x_0\sqrt{\frac{1 + z_0^2}{1 - z_0^2}}, y_0\sqrt{\frac{1 + z_0^2}{1 - z_0^2}}, z_0 \right)$ lies on the same side of the $z$-axis as $p$, whereas the point $\left(-x_0\sqrt{\frac{1 + z_0^2}{1 - z_0^2}}, -y_0\sqrt{\frac{1 + z_0^2}{1 - z_0^2}}, z_0 \right)$ lies on the opposite side of the $z$-axis from $p$. So $$ F(p) = \left(x_0\sqrt{\frac{1 + z_0^2}{1 - z_0^2}}, y_0\sqrt{\frac{1 + z_0^2}{1 - z_0^2}}, z_0 \right).$$