# Holomorphic maps of the Riemann sphere

I'm trying to understand a proof that the holomorphic maps of the Riemann sphere are the rational functions from this book:

All rational maps are analytic from the extended plane to itself. For the converse, suppose $$f(z) \in \mathbb{C}$$ for all $$z \in \mathbb{C}_\infty$$. Then $$f$$ is entire and bounded and thus constant. We can therefore assume that $$f(z_0) = \infty$$ for some $$z_0 \in \mathbb{C}$$ (consider $$f(1/z)$$ if necessary). By continuity of $$f$$ the point $$z_0$$ cannot be an essential singularity of $$f$$. In other words, $$z_0$$ is a removable singularity or a pole. By the uniqueness theorem, the poles cannot accumulate in $$\mathbb{C}_\infty$$. Since the latter is compact, there can thus only be finitely many poles. Hence, after subtracting the principal part of the Laurent series of $$f$$ around each pole in $$\mathbb{C}$$ from $$f$$, we obtain an entire function which grows at most like a polynomial. By Liouville's theorem, such a function must be a polynomial and we are done.

I have a few questions regarding this:

1. Using the continuity argument can we also conclude that $$z_0$$ must be a pole?
2. Why does the uniqueness theorem imply the poles cannot accumulate in $$\mathbb{C}_\infty$$?
3. Perhaps slightly less relevant to this proof, does $$f : \mathbb{C}_\infty \to \mathbb{C}_\infty$$ holomorphic imply $$f$$ is meromorphic? I'm using Conway's definition that $$f$$ is meromorphic iff it is analytic except for poles.

2. If they did accumulate, then their accumulation point would be an accumulation point of points on the domain where $$f$$ is $$\infty$$. The function which is $$\infty$$ everywhere is a holomorphic function with such an accumulation point, and since it is unique, $$f$$ must be this same constant function. This would make $$z_0$$ neither a pole, nor removable, nor essential, by the way. But it would make $$f$$ rational, depending on how you define rational. But I admit that this could have been communicated better by the author.
3. No, meromorphic functions make a distinction between $$\infty$$ and other points, while holomorphic ones do not. The function which is a constant $$\infty$$ is not meromorphic because the singularities are not isolated, for instance, but it is holomorphic as a function $$\mathbb C^\infty\to\mathbb C^\infty$$.