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I'm studying Milnor's book "Dynamics in one Complex Variable", and he states this problem to the reader in the middle of the proof of Pick's Theorem:

If $S$ and $S'$ are two hyperbolic Riemann surfaces (that is, they are both universally covered by the unitary disk $\mathbb{D}$) and let $f: S \longrightarrow S'$ be a holomorphic function between them. Let $\phi_1: \mathbb{D} \longrightarrow S$ and $\phi_2 : \mathbb{D} \longrightarrow S'$ be their universal covering maps.

Making some choice of points, we can lift $f$ to a function $F: \mathbb{D} \longrightarrow \mathbb{D}$, such that the diagram below commutes:

$\require{AMScd}$ \begin{CD} \mathbb{D} @>{F}>> \mathbb{D}\\ @V{\phi_1}VV @VV{\phi_2}V\\ S @>{f}>> S' \end{CD}

The statement from Milnor's book is:

f is a covering map if and only if F is a conformal automorphism.

Supposing that f is a covering map, we can use the universal property of $\phi_2 $ to conclude that F must be a conformal automorphism.

The other side of this is bugging me. I've tried doing some arguments using that $\phi_1$ is a local homeomorphism or that $f$ is open (since it is holomorphic), but i couldn't get it right.

Hence, the question is how to prove this fact: If $F$ is a conformal automorphism then $f$ is a covering map.

Accepting any suggestions and insights to prove this.

Note: I don't know if this result is generalizable for greater dimensions or for the smooth case ($S$, $S'$ smooth manifolds and $f$, $F$ being differentiable). Some counterexamples in those directions would be nice too.

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I'll follow the proof as it is in Milnor's book. So we have Poincaré metric on the surfaces and the covering maps are Riemannian covering.

If $F$ is a conformal isomorphism, then $f$ is a local isometry.

Now, just use that a surjective local isometry with complete domain is always a Riemannian covering map.

Observation: This fact about Riemannian geometry used in the end can be found on Manfredo's book, for example, on the section about Hadamard theorem. The precise statement is the following.

Let $M$ be a complete Riemannian manifold and let $f:M \to N$ be a local diffeomorphism onto a Riemannian manifold $N$ which has the following property: For all $p\in M$ and all $v\in T_p M$, we have $|df_p(f)|\geq |v|$. Then $f$ is a covering map.

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The direction you are asking about (if $F$ is a conformal isomorphism, then $f$ is a covering map) follows from the definition of covering maps (without any use of the Poincaré metric).

You be have been missing the following fact.

Observation 1. A composition of covering maps is again a covering map.

This in turn follows directly from

Observation 2. $f\colon S\to S'$ is a covering map if and only if the following holds. For every simply-connected domain $D\subset S'$, and every connected component $U$ of $f^{-1}(S)$, the map $f\colon U\to D$ is a homeomorphism.

(The "if" holds by definition. The "only if" direction follows by noting that $f\colon U \to D$ is a covering map, and simply-connected domains have no nontrivial coverings.)

So now let $f$, $\phi_1$, $\phi_2$ and $F$ be as in your question. Let $D\subset S'$ be simply-connected, and let $U$ be a connected component of $f^{-1}(D)$.

If $\tilde{U}$ be a connected component of $\pi_1^{-1}(U)$, then $\tilde{U}$ isa connected component of $(f \circ \phi_1)^{-1}(D) = (\phi_2\circ F)^{-1}(D)$. Since $F$ and $\phi_2$ are covering maps, their composition is also, and
$$ f \circ \phi_1 = \phi_2 \circ F \colon \tilde{U} \to D $$ is a conformal isomorphism. Now $\phi_1 \colon \tilde{U}\to U$ is surjective (as it is a covering map.) It follows that $f\colon U\to D$ is a conformal isomorphism, as required.

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