Given a hyperbolic plane $\mathcal{H}^2=\{(t,x,y)\in\mathbb{R}^3|t^2=x^2+y^2+1, t>0\}$, I had to construct a function $f$ that maps as follows: For a point $P$ in $\mathcal{H}^2$, draw a line $l$ from this point to $(-1, 0, 0)$. The function $f$ maps every point $P$ to the intersection of $l$ with the $xy-$plane. I figured that the $l$ was given by $$ l: \begin{pmatrix} -1 \\ 0 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} t+1 \\ x \\ y \end{pmatrix}.$$
Then $f$ will be defined as \begin{align*} f:\;\mathcal{H}^2&\to\mathcal{D}=\{(x,y)\in\mathbb{R}^2|x^2 + y^2 < 1\} \\ (t,x,y) &\mapsto \big(\frac{1}{t+1}x, \frac{1}{t+1}y\Big). \end{align*}
Assuming this function does as I require I tried to proof bijectivity by proving injectivity and surjectivity, but failed on both fronts.
For injectivity I tried: Injective when $f(t_1, x_1, y_1)=f(t_2, x_2, y_2)\implies(t_1,x_1,y_1)=(t_2,x_2,y_2)$. So we want to show: $\big(\frac{1}{t_1+1}x_1, \frac{1}{t_1+1}y_1\Big)=\big(\frac{1}{t_2+1}x_2, \frac{1}{t_2+1}y_2\Big)\implies (t_1,x_1,y_1)=(t_2,x_2,y_2)$. From here I don't know how to continue because $x$ is a based on both $x_1$ and $y_1$ (based on $x_1$ because it's based on $t_1$). I tried filling in $t_1$ and $t_2$ as terms of $x_1,y_1$ and $x_2,y_2$ respectively, but this yielded me equations that I don't know how to work with: $$ x_1(t_2 + 1)=x_2(t_1+1)\\ y_1(t_2 + 1)=y_2(t_1+1). $$
For surjectivity I tried: Let $(x_1,y_1)\in\mathcal{D}$, then we want to show that there is a $q\in\mathcal{H}^2$ such that $f(q)=(x_1,y_1)$. I tried finding such a $q$, but didn't know how to incorporate the restrictions on $t$.
EDIT: I then thought that for all $(x_1,y_1)$, we could have $f(t, x_1(t+1), y_1(t+1))=(x_1,y_1)$. Which is true, but how do I confirm that $(t, x_1(t+1), y_1(t+1))$ is in $\mathcal{H}^2$.
Some things I noted:
Because for every element $(t,x,y)\in\mathbb{H}^2$ we have that $t^2=x^2+y^2+1$, we have that $t=\pm\sqrt{x^2+y^2+1}$, but because we also require $t>0$, we have $t=\sqrt{x^2+y^2+1}$, so $t$ is a function of $x$ and $y$.