I had an exercise that goes like this that i was able to solve but was wondering if it was true for more Riemann surfaces , hopefully something not isomorphic to $\mathbb{C}_{\infty}$.
Let $f$ and $g$ be two meromorphic functions on $\mathbb{C}_{\infty}$ such that their set of poles and zeroes are disjoint , then $\prod_{p\in \mathbb{C}_{\infty}}f(p)^{ord_p(g)}= \pm \prod_{p \in \mathbb{C}_{\infty}} g(p)^{ord_P(f)}$.
I have come up with a way to do this, that uses the fact that if $f$ is a meromorphic function on the Riemann sphere then if $\{\lambda_i\}_{i=1}^{n}$ denotes its set of poles and zeroes and $e_i =ord_{\lambda_i}(f)$, then $f=a\prod_{i=1}^{n}(z-\lambda_i)^{e_i}$, for some constant $a$ and then we just do the substituitions in each side and get the result we want. My quesiton is if there is an alternative way to do this, something a little more abstract, if there are other Riemann surfaces where this result is true and why ?
I have tried to see if this was even true for $\mathbb{P}^1$ because they are isomorphic but i still havent been able to do it, in fact i think its easy to construct a counterexample, wich is odd because the surfaces are isomorphic, does anyone know if this is true or not and give some insight on why ?Thanks in advance.