Let $f:\mathbb{CP}^1\setminus \{N\} \to \mathbb{C}$ be the stereographic projection of the complex projective line onto the complex plane. Similarly, let $U\subset\mathbb{C}$ be a closed rectangle. Consider a 1-form $\omega=g(z)dz$ defined strictly on $U,$ then how would the pullback $f^{*}\omega$ on a proper subset of $\mathbb{CP}^1$ (Namely the preimage of $U$) be computed explicitly?
Attempted Solution: We introduce spherical coordinates on $\Sigma$ (the Riemann sphere), i.e. the latitude $\phi,$ measured from the equator and ranging from $0$ to $\pi/2$ in the northern hemisphere and from $0$ to $-\pi/2$ in the southern hemisphere, and the longitude $\lambda,$ measured from the prime meridian (more exactly, from the point of intersection of the prime meridian with the positive $x$-axis) and ranging from $0$ to $\pi$ (including $\pi$) in the eastern hemisphere and from $0$ to $-\pi$ (excluding $-\pi$) in the western hemisphere. Under stereographic projection, we have $$argz=\lambda,|z|=tan\left(\frac{\pi}{4}+\frac{\phi}{2}\right).$$ Therefore, the point of the sphere with coordinates $\lambda$ and $\phi$ is the image of the complex number $$z(\lambda,\phi)=tan\left(\frac{\pi}{4}+\frac{\phi}{2}\right)(cos\lambda+isin\lambda),$$ i.e. the preimage of $z$ under $f:\mathbb{CP}^1\to \mathbb{C}.$ The 1-form $\omega$ can be written locally on $U$ as $\omega=g(z)dz.$ Consequently, the pullback on $\mathbb{CP}^1$ induced by $f$ is $$f^{*}\omega=g(f)\left(\frac{\partial f}{\partial \phi}d\phi+\frac{\partial f}{\partial\lambda}d\lambda\right)\equiv\frac{g(f)}{2}sec^2\left(\frac{\pi}{4}+\frac{\phi}{2}\right)(cos\lambda+isin\lambda)d\phi+ig(f)tan\left(\frac{\pi}{4}+\frac{\phi}{2}\right)(cos\lambda+isin\lambda)d\lambda.$$
Is this computation correct? Thanks in advance. I would also like to acknowledge Ted Shifrin for his help in reformulating this question.