# Meromorphic Functions and Holomorphic Functions on the Riemann sphere

I have been reading Rick Miranda's Book on Algebraic Curves and Riemann Surfaces, and he states that if we have a meromorphic function $$f$$ on a Riemann surface $$X$$ we can construct a holomorphic function $$F$$ to the Riemann sphere, intuitively this seems alright we just send the poles to $$\infty$$, but i cant seem to prove this in a very rigorous way,and they way he states it seems like he implies that is trivial, so any tips or help is appreciated. Thanks in advance.

Let $$F$$ be a meormorphic function on $$X$$. The there exists discrete set $$P$$ of poles such that we have an actual function $$F\colon X\setminus P \rightarrow \mathbb{C}$$. Now take $$Z$$ the discrete set of zeros of $$F$$. We also have a function $$\dfrac{1}{F}\colon X\setminus Z \rightarrow \mathbb{C}$$. Note that $$(X\setminus Z) \cup (X\setminus P) = X$$.
Recall that the Riemann Sphere is constructed by glueing two copies of $$\mathbb{C}$$ with coordinates $$z$$ and $$w$$, respectively, by making the identification $$z= \dfrac{1}{w}$$. Hence we can glue these two maps we have defined in an obvious manner.
Another way to realize this map is by considering the projective line $$\mathbb{P}^1$$. Around each point of $$X$$ we have a representation $$F=\dfrac{f}{g}$$ where $$f$$ and $$g$$ are holomorphic. Then we set a map $$X\ni x \mapsto (f(x):g(x)) \in \mathbb{P}^1$$. To prove that it is well defined we only need to show that it does not depend on the representation and it is indeed the case. If $$F = \dfrac{f_1}{g_1}$$ is other representation then $$f_1g = f_1g_1$$ which ensures that $$(f:g) = (f_1:g_1)$$.