Why must a biholomorphic automorphism of $\mathbb{C}$ have a pole at infinity? In the proof of the first proposition in this document, why must a biholomorphic function $f \colon \mathbb{C} \to \mathbb{C}$ have a pole at infinity? I can't retrace the topological argument given.
 A: I'm not entirely sure what is meant by the first paragraph (it is a bit vague in my opinion), but its purpose is to indicate that $f$ has either a pole or an essential singularity at $\infty$.
I have addressed the question below without reference to the linked document in an effort to be more precise.

There are only two other possibilities:

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*$f$ is bounded in a neighbourhood of $\infty$.
Recall that a neighbourhood of $\infty$ (inside $\mathbb{C}$) is a open subset of $\mathbb{C}$ containing $\{z \in \mathbb{C} \mid |z| > R\}$ for some $R \geq 0$. So if $f$ is bounded in a neighbourhood of $\infty$, there is some $R \geq 0$ and $L \geq 0$ such that $|f(z)| \leq M$ for all $|z| > R$. As $f$ is continuous and $\{z \in \mathbb{C} \mid |z| \leq R\}$ is compact, there is $K \geq 0$ such that $|f(z)| \leq K$ for all $|z| \leq R$. But then $|f(z)| \leq M := \max\{K, L\}$ for all $z \in \mathbb{C}$; i.e. $f$ is bounded. By Liouville's Theorem, $f$ is constant which contradicts the fact that $f$ is injective.


*$f$ has an essential singularity at $\infty$.
By Picard's Great Theorem, $f$ takes on all but possibly one value infinitely many times in a neighbourhood of $\infty$ which contradicts the injectivity of $f$.
Therefore, $f$ must have a pole at $\infty$.
