Proving that a solution is unique Let's say I have an $x\in \mathbb R^3$ with $x_1^2+x_2^2+x_3^2=1$. I want to proof that a function is well defined, which maps my $x$ to a point $$\mathcal E = \Big(\frac{-x_1}{x_3-1}, \frac{-x_2}{x_3-1},0 \Big)$$ where $x\neq (0,0,1)$
To show that I guess, I need to show that every $x$ in my domain gets mapped to exactly one point in the image. However, I am not sure how to formally show this for this function. Any help would be greatly appreciated!
Edit: To show surjectivity of this function $$f:\{x\in \mathbb R^3| x_1^2+x_2^2+x_3^2=1\}\setminus (0,0,1)\rightarrow E$$
with $E$ being the $2$d plane given by $x_3=0$, I calculated the inverse for arbitrary $\mathcal E \in E$, with $\mathcal E = (\mathcal E_1,\mathcal E_2,0)$. I get $$x_1 = \frac{2\mathcal E_1}{1+\mathcal E_1 ^2 +\mathcal E_2 ^2 }$$
$$x_2 = \frac{2\mathcal E_2}{1+\mathcal E_1 ^2 +\mathcal E_2 ^2 }$$
$$x_3 = 1-\frac{2}{1+\mathcal E_1 ^2 +\mathcal E_2 ^2 }$$
do these results suffice to show that $f$ is bijective? I found a well defined inverse function but I am not sure if that is enough?
 A: I don't see a lot of trouble here.
Let's define the set $$S = \{(x_1,x_2,x_3) \in \mathbb{R^3}|x_1^2 + x_2^2 + x_3^2 = 1\}$$
Then, define a function $$\phi: S-\{(0,0,1)\} \rightarrow E: (x_1,x_2,x_3) \mapsto (\frac{-x_1}{x_3-1},\frac{-x_2}{x_3 - 1},0)$$
Obviously, every element that gets mapped by $\phi$ gets mapped to a single element of $E$, so $\phi$ is a function.
Now, if you want to show that $\phi$ is surjective, you take an arbitrary $e \in E$, and show that there is an element in the domain that gets mapped on E.
So, let's take $e \in E$. Then, $e = (x,y,0)$. Since $$\phi(\frac{2x}{1+x^2 + y^2},\frac{2y}{1+x^2 + y^2}, 1 - \frac{2}{1 + x^2 + y^2}) = (x,y,0)$$ it follows that $\phi$ is surjective (so your reasoning is correct, well done!)
To show that $\phi$ is bijective, we also need to show that $\phi$ is injective though, which may be harder to prove since we have to solve a non linear system of equations for that.
Your idea to use the inverse function would work as well, but then you have to construct the function $g$ ($g(x_1,x_2,x_3) = ?$, use what you did to show surjectivity) and you have to show that $$f \circ g = 1_E$$ and  $$ g \circ f = 1_{S-\{(0,0,1)\}}$$ in order to conclude that $g = f^{-1}$
