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.


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.


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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