A proof of the fact that $\mathbb{S}^2$ is biholomorphic to $\mathbb{CP}^1$ uses the concept of divisors on Riemann surfaces. The proof is the following.
Take $D:= \{p\}$, $\mathcal{M}$ the space of meromorphic functions, $Z(f)$ the zeros of $f$ and $P(f)$ the poles of $f$.
Then $\mathcal{O}_D:= \{f \in \mathcal{M} \, | \, \text{Div}(f) +D \ge 0\}= \{f \in \mathcal{M} \, | \, Z(f)-P(f) +p \ge 0\}$ then $f \in \mathcal{O}_D $ can have a unique pole in $p$. A theorem states that if $f: \mathbb{S}^2 \rightarrow \mathbb{CP}^1$ admits only a pole in $p$ then it is a biholomorphism, then $ \mathbb{S}^2\simeq \mathbb{CP}^1$.
I don't understand why the choice of a particular divisor make the proof globally valid. Every $D$ I choose I am able to say that there exists a meromorphic function on $M$ with a certain number of poles and zeros.
Am I right?