Why is the stereographic projection bijective?

Question

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!

There are some pretty good pictures online. Think about it geometrically (e.g. in 2D): Stereographic projection takes a basepoint $e_{n+1}$. Then we take any point on the circle $p$, draw the line between $e_{n+1}$ and $p$ and the result of the stereographic projection is the intersection of this line with the $x$-axis. Maybe draw a picture yourself. You should be able to see that this is a bijective process. — Qi Zhu, May 29 '20 at 06:07
Related - inverse of a stereographic projection in the case of $\Bbb R^3$, that is $S^2$:. — Matcha Latte, Sep 25 '21 at 06:10

score 3 · Accepted Answer · answered May 29 '20 at 06:09

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.)

Thank you so much, I needed to see it analytically! – PCeltide May 29 '20 at 07:09 — PCeltide, May 29 '20 at 07:09

Why is the stereographic projection bijective?

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