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Let $D$ be the open unit disc in $\mathbb{C}$ and let $D^* = D\setminus\{0\}$. Now, Picard's Great Theorem implies the following fact.

If $f\colon D^\ast \to \mathbb{C}$ is a holomorphic function with finite fibres, then $f$ does not have an essential singularity at $0$.

Here is the simple argument (using Picard's Great Theorem). The fibre $f^{-1}(0)$ is finite and so is the fibre $f^{-1}(1)$. Thus, upon replacing $D$ by a disc of radius $r$ for some suitable $0<r<1$ and $D^*$ accordingly, we may assume that $f$ takes values in $\mathbb{C}\setminus \{0,1\}$. Then Picard's Great Theorem says that $f$ has no essential singularity. QED

Is there a proof of the above fact that does not use Picard's Great Theorem? I'm hoping to find an argument which uses simpler results (e.g., Casorati-Weierstrass, existence of logarithm after shrinking $D^*$, etc.)

Harry
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    This is about covering projections, I think. Simple case: $D^=D\setminus U,$ where $U$ is discrete non-empty and $\frac{df}{dz}|D^ \neq 0$ and $|f^{-1}(z)|\le 1.$ Then $f$ is biholomorphic onto open subset of $\mathbb{C}$ and the result follows from Casorati-Weierstrass (image cannot be dense).

    General case: assume $\frac{df}{dz}|D^* \neq 0,$ find evenly covered $V\subseteq D\setminus U$ and also use Casorati-Weierstrass.

     – dsh Mar 04 '24 at 06:28
  • I see. If $f$ has finite fibres, then also $f' = df/dz$ has finite fibres. Thus, $f'$ has only finitely many zeroes. Outside of these, $f$ is a local biholomorphism. Thus, after shrinking $D^*$ (just throwing out the zeroes of $f'$), the holomorphic function $f$ is just a local biholomorphism onto its image. I think that this should imply that it does not have an essential singularity at $0$, but I don't see yet how. – Harry Mar 05 '24 at 15:21
  • If $V\subseteq \mathbb{C}$ is evenly and finitely covered by map $f:D\setminus U\to \mathbb{C}.$ Shrinking $V$ if necessary (still evenly and finitely covered), one finds $W \subseteq D\setminus U,$ punctured neighborhood of 0, such that $W\cap f^{-1}(V)=\emptyset.$ It means that $f|W$ does not have dense image (does not contain open $V$). Derivative of $f$ is zero on discrete set, not sure if this set is finite. – dsh Mar 06 '24 at 07:04

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