I have been all over the internet but only got surprised in how no one mentions what the general equation(formula) is and how to derive it, for a map $\mathbb{R}^n \rightarrow S^n$ \ $P$where $S^n=\{(x_0,...x_n) \in \mathbb{R}^n+1| x_0^2+...+x_n^2=1\}$ and $P=(1,0...0)$.
So it is a stereographic projection but instead of the map "input;point on sphere" "output;Some point on$\mathbb{R}^n$" I am trying to find the opposite. So for some point $y \in Y=\{(0,y_1...,y_n) \in \mathbb{R}^{n+1}\}$, I want to find the point $x=(x_0,...,x_n) \in S^n$\ $P$ that $x,y,P=(1,0...,0)$ are all collinear.
I don't know how to derive this, I don't even know what the equation should look like. Can someone help me please?? Thank you....
It is known the stereographic projection $\pi : S^n\smallsetminus N\longrightarrow \Bbb R^n$ is given by mapping $$(x_1,\ldots,x_{n+1})\mapsto \frac{1}{1-x_{n+1}}(x_1,\ldots,x_n)$$
If you haven't proven this, consider the line $\lambda (N-x)+x$ that passes through a point $x$ in the sphere and the north pole $N=(0,\ldots,0,1)$. The equation $\lambda(x-N)+N=(y_1,\ldots,y_n,0)$ gives that $\lambda(x_{n+1}-1)=-1$ so that $\lambda =\dfrac{1}{1-x_{n+1}}$ and the rest is $\lambda x_i=y_i$ for $i=1,\ldots,n$, which is the above.
To find its inverse, set $y_i = \frac{x_i}{1-x_{n+1}}$. Note that $y_1^2+\cdots+y_n^2=\dfrac{ x_1^2+\cdots+x_n^2}{(1-x_{n+1})^2}=\dfrac{ 1-x_{n+1}^2}{(1-x_{n+1})^2}=\dfrac{1+x_{n+1}}{1-x_{n+1}}$ so that $$x_{n+1}=\frac{y_1^2+\cdots+y_n^2-1}{y_1^2+\cdots+y_n^2+1}$$
This gives the remaining equations for the $x_i$ since $y_i(1-x_{n+1})=x_i$.