The stereographic projection $\sigma :\mathbb S^n(R) \setminus\{N\} \to \mathbb R^n$ is a conformal transformation .
To prove this theorem, we should show that for all $v\in T_q\mathbb R^n$ ($q\in \mathbb R^n $), we have $$(\sigma^{-1})^* \langle v,v\rangle=\langle T_q\sigma^{-1}(v),T_q\sigma^{-1}(v) \rangle=f\langle v,v\rangle_0,$$ with $\langle,\rangle$ the canonical riemannian metric in $\mathbb S_R^n$, $\langle,\rangle_0$ the Euclidean metric in $\mathbb R^n$ and $\sigma^{-1}$ is defined by $$\sigma^{-1}(u,v)=(\xi_1,\cdots\xi_n,\tau)=\left(\frac{2R^2u}{|u|^2+R^2},R\frac{|u|^2-R^2}{|u|^2+R^2}\right)$$
Let $\displaystyle v=\sum_{i=1}^n v_i \frac{\partial }{\partial u_i}$, so $$T_q\sigma^{-1}(v)=\sum_{i=1}^nv(\xi_i) \frac{\partial}{ \partial \xi_i}+v(\tau) \frac{\partial}{\partial \tau}$$ I don't know where this relation comes from !! I can prove the result directly by using the properties of pullback but my teacher use this one in the course.
Any help is highly appreciated !