Why is the stereographic projection bijective? I know this might be a very basic question, but I am just not able to wrap my head around it.
Why is the map 
$$ S:\mathbb{S}^n-\{e_{n+1}\}\rightarrow \mathbb{R}^n \quad \textrm{such that } \bar{x}\mapsto (\frac{x_1}{1-x_{n+1}},...,\frac{x_n}{1-x_{n+1}})$$
a bijective function? I tried to understand it geometrically as well, but I just could not see how it was formulated in that manner. 
Here $\bar{x}=(x_1,...,x_n)$ and $e_{n+1}=(0,...0,1)$.
Any help will be much appreciated!

It's certainly not obvious (except, like many things in mathematics, in hindsight). One can show directly that this map is both injective and surjective. It's easier perhaps to simply write down the inverse function $S^{-1}\colon \Bbb R^n \to \Bbb S^n - \{e_{n+1}\}$:
$$
S^{-1}\big( (y_1,\dots,y_n) \big) = \frac1{1+y_1^2+\cdots+y_n^2}\big( 2y_1,\dots,2y_n,y_1^2+\cdots+y_n^2-1 \big).
$$
(Of course one has to verify that the domain/codomain of this function are correct and that both $S\circ S^{-1}$ and $S^{-1}\circ S$ are the identity maps.)
