Let $X$ be the Riemann sphere with local coordinates $z$ in one chart and $w=\frac{1}{z}$ in the other. Let $\omega$ be a meromorphic 1-form on $X$. Show if $\omega = f(z)$d$z$ in the coordinate $z$, then $f$ must be a rational function of $z$.
So I am super new to this stuff and pretty stuck, does this translate to , there are two charts on $X$ call them $\phi$ and $\psi$ where $\phi(z):= z$, $\psi(z) := \frac{1}{z}$.
Do I use a transformation to send this form $\omega$ to some other form $\omega'$? I know that there's a transformation $T$ mapping from domain of $\psi$ to domain of $\phi$ such that $g(w) = f(T(w))T'(w)$. Am I going anywhere or am I really lost? Any hints would be appreciated.
Do I take the two charts and take the composition $\phi \circ \psi^{-1}$ or? Also can I assume their domains overlap or?
Does meromorphic 1-form mean that in any chart $\omega=f_u du$ with $f_u$ meromorphic? If so then $\omega=f(z)dz$ where $f$ must be meromorphic on the Riemann sphere minus $\infty$ and $\omega=f(1/w)d(1/w)=-f(1/w)w^{-2}dw$ where $f(1/w)w^{-2}$ must be meromorphic at $0$ so $f$ is meromorphic on the whole Riemann sphere, which implies it is a rational function.