Help to prove bijection between subset of $S^2$ and $\mathbb{R}^2$ Well, I've been solving some exercises about bijections between subsets of $\mathbb{R}^n$ and $\mathbb{R}^m$ to get practice with this and I'm stuck with this one. Let $S^2 \subset \mathbb{R}^3$ be the unit sphere, and let $N = (0,0,1)$. I want to show that the map $f : S^2 \setminus \{N\} \to \mathbb{R}^2$ given by:
$$f(x,y,z)=\left(\frac{2x}{1-z},\frac{2y}{1-z}\right)$$
is a bijection. Well, I've tried going direct from the definition. First injectivity: I have to show that $f(x_1,y_1,z_1)=f(x_2,y_2,z_2)$ implies $x_1=x_2$, $y_1=y_2$ and $z_1=z_2$. The hypothesis gives
$$\begin{cases}\frac{x_1}{1-z_1}=\frac{x_2}{1-z_2}\\ \frac{y_1}{1-z_1}=\frac{y_2}{1-z_2}\end{cases}$$
From this, substituting $1-z_1$ on the second equation by $x_1(1-z_2)/x_2$ I've found that this implies that $x_1y_2=x_2y_1$, but I couldn't proceed any further from this.
To show that $f$ is surjective, I have to show that given $(a,b) \in \mathbb{R}^2$ there's some $(x,y,z)\in S^2\setminus \{N\}$ such that $(a,b)=f(x,y,z)$, in other words, such that:
I've tried proving surjectivity looking at the expression of $f$ and thinking "what $(x,y,z)$ I must put in there in terms of $(a,b)$ to get $f(x,y,z)=(a,b)$"? I've found then that $(a/2,b/2,0)$ does the job, but this point is not necessarily in $S^2\setminus\{N\}$. I should find $(x,y,z)$ such that $f(x,y,z)=(a,b)$ and $x^2+y^2+z^2=1$. But, just looking a long time to these two conditions, I find very difficult to take a guess. Is there a better way to procede?
Thanks very much in advance!
 A: Here's the injectivity argument:
You've shown that $\dfrac{x_1}{x_2} = \dfrac{y_1}{y_2} = \dfrac{1-z_1}{1-z_2}$. Let's call their common value $\lambda$, so that $$x_1 = \lambda x_2,$$ $$y_1 = \lambda y_2,$$ $$z_1 = \lambda z_2 - \lambda + 1.$$ Then $x_1^2 + y_1^2 + z_1^2 = 1$, of course, because it's on a sphere. But, plugging in the relationships above, we get $$\lambda^2 x_2^2 + \lambda^2 y_2^2 + (\lambda z_2 - \lambda + 1)^2 = 1,$$ which in turn gives $$\lambda^2 (x_2^2 + y_2^2 + z_2^2) - 2(\lambda-1)\lambda z_2 + (\lambda-1)^2 = 1,$$ and using the fact that $x_2^2 + y_2^2 + z_2^2 = 1$ (because it's on a sphere), this gives $$\lambda^2 - 2\lambda^2 z_2 + 2\lambda z_2 + \lambda^2 - 2\lambda + 1 = 1,$$ which finally simplifies to give $$2\lambda(\lambda - 1)(z_2 - 1) = 0.$$ Hence either $\lambda = 0$ (which gives $(x_1, y_1, z_1) = (0,0,1)$, which is explicitly disallowed), or $\lambda = 1$, or $z_1 = 1$ (hence $x_1 = y_1 = 0$, which is explicitly disallowed). So $\lambda = 1$.
Edit:
Another post that completed the surjectivity part was removed, so I'm adding that too:
If $f(x,y,z) = \left(\dfrac{2x}{1-z}, \dfrac{2y}{1-z}\right)$ is going to equal $(a,b)$, then $a/b = x/y$, so let's write $x = \mu a$, $y = \mu b$. But $\dfrac{2x}{1-z} = a = x/\mu$, so $2\mu = 1-z$. Substitute all of this into $x^2 + y^2 + z^2 = 1$ and solve for $\mu$ in terms of $a^2 + b^2$.
(Going back to Daniel's comment, notice that $f^{-1}$ only depended on $a^2 + b^2$, a sort of 'radius'-like term, and $a/b$, a sort of 'slope'-like term. This gives you a geometric picture of what stereographic projection does, but I'm sure Google can give you much better ones.)
