Let $S^n \subset \mathbb{R}^{n+1}$ be a unit sphere. Let $a=(a_1,\ldots,a_{n+1}) \in S^n$ and $v_1=(v_{1,1},\ldots,v_{1,n+1}),\ldots,v_n=(v_{n,1},\ldots,v_{n,n+1}) \in T_aS^n$. Define $\omega(a) (v_1,\ldots,v_n) = \operatorname{det} \begin{pmatrix} a_1 & \ldots & a_{n+1}\\ v_{1,1} & \ldots & v_{1,n+1}\\ \vdots & \ddots & \vdots\\ v_{n,1} & \ldots & v_{n,n+1} \end{pmatrix}$. Let $f\colon S^n \smallsetminus \{(0,\ldots,0,1)\}\to \mathbb{R}^n$ be a stereographic projection to the hyperplane given by the equation $x_{n+1}=0$. Compute the $n$-form $(f^{-1})^*\omega$ on $\mathbb{R}^n$.
I have found explicit formulas for direct and inverse stereographic projection: $$f\colon S^n\smallsetminus \{(0,\ldots,0,1)\}\to \mathbb{R}^n$$ $$f(a_1,\ldots,a_{n+1}) = (\frac{a_1}{1-a_{n+1}},\ldots,\frac{a_n}{1-a_{n+1}},0)$$ $$f^{-1}\colon \mathbb{R}^n\to S^n\smallsetminus \{(0,\ldots,0,1)\}$$ $$f^{-1}(x_1,\ldots,x_n) = (\frac{2x_1}{1+\sum\limits_{i=1}^n x_i^2},\ldots,\frac{2x_n}{1+\sum\limits_{i=1}^n x_i^2},\frac{\sum\limits_{i=1}^n x_i^2 - 1}{1+\sum\limits_{i=1}^n x_i^2})$$ The pullback of differential form in coordinates is given by $(f^{-1})^* \omega = (f^{-1})^* \omega (f(a)) (d_af(v_1),\ldots,df_a(v_n))$, since $f(v_i) = \frac{v_i}{1 - v_{n+1}}$ and $df(v_i) = \frac{1}{1 - v_{n+1}}dv_i + \frac{v_i}{(1 - v_{n+1})^2}dv_{n+1}$ the explicit formula for pullback is $$(f^{-1})^* \omega = \operatorname{det} \begin{pmatrix} \frac{a_1}{1 - a_{i+1}} & \ldots & \frac{a_n}{1 - a_{n+1}}\\ \frac{1}{1-v_{1,n+1}}+\frac{v_{1,1}}{(1-v_{1,n+1})^2} & \ldots & \frac{1}{1-v_{1,n+1}}+\frac{v_{1,n}}{(1-v_{1,n+1})^2}\\ \vdots & \ddots & \vdots\\ \frac{1}{1-v_{n,n+1}}+\frac{v_{n,1}}{(1-v_{n,n+1})^2} & \ldots & \frac{1}{1-v_{n,n+1}}+\frac{v_{n,n}}{(1-v_{n,n+1})^2} \end{pmatrix}$$ Could please anyone verify my calculations? Am I right or are there any mistakes?