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Suppose a sequence of analytic maps $f_n: \mathbb{D} \to \mathbb{D}$ from the unit disk to itself, each of which is a topological covering map to its image, converges locally uniformly to an analytic map $f$. Under what conditions will the limit $f$ also be a topological covering map to its image? (I am also interested in other possibilites for domain and range of the $f_n$.)

Using Rouche's theorem, one can show that the local uniform limit of injective analytic functions is again injective, provided that the limit is not constant. One thing I would like to know is if a similar result holds if we replace "injective" with "covering map".

This sort of question comes up if you try to prove the uniformization theorem in the special case of a plane domain $U\subset \mathbb{C}$.

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Does $\mathbb{D}$ stand for the unit disk? If so, then an analytic map $\mathbb{C}\to\mathbb{D}$ is bounded and hence constant. – algori Feb 20 '12 at 0:39
Yeah, sorry, $\mathbb{D}$ is the unit disk, and both domain and range are supposed to be $\mathbb{D}$. – Ben Dozier Feb 20 '12 at 0:50

There are essentially two cases when the limit is not a covering map: (a) the limit is constant, (b) the images $S_n:=f_n(D)$ are annuli which collapse to a circle. There could be a more elementary explanation, but the one below uses Fuchsian groups and a bit of hyperbolic geometry.

Sketch of the proof. Each surface $S_n$ can be identified with the quotient $D/G_n$, where $G_n$ is a Fuchsian group (a discrete torsion-free isometry group of the unit disk $D$, which we will regard as the hyperbolic plane). Thus, I will think of $S_n$ as complete hyperbolic surfaces. Now, the sequence of discrete groups $G_n$ will subconverge in "Chabauty topology" (or "geometrically" in Thurston's terminology) to a closed subgroup $G$ of $PSL(2,R)$, the isometry group of $D$. Most the the time the group $G$ will be discrete and torsion-free. The one exception is when the groups $G_n$ converging to $G$ are elementary, i.e., cyclic, then the limit could be 1-dimensional (this is where you discover the annular example). Suppose now that the group $G$ is discrete and consider the quotient surface $D/G=S$. Then (after you choose base-points) the surfaces $S_n$ converge to $S$ in "Gromov-Hausdorff topology." (You may have to read something like "Lectures on Hyperbolic Geometry" by Benedetti and Petronio if the terminology is unfamiliar.) This means that large metric balls in $S$ (in the hyperbolic metric of $S$) are nearly conformal to corresponding metric balls in $S_n$ (i.e., quasiconformal with small quasiconformality constants). Thus, your limiting holomorphic map $f: D\to D$ (which I assume to be nonconstant) would factor as a composition of $p: D\to S$ (the universal covering map) and a holomorphic map $F: S\to D$, which I claim is 1-1. If the map $F$ were the limit of a sequence of conformal maps $S\to D$, you would be done (say, by Rauche's theorem), as you observed in your question. Instead it is a limit of almost conformal quasiconformal maps above ($S\to S_n$) defined on larger and larger compact subsets of $S$. Now, you have to check that the same conformal arguments that you know would go through in the quasiconformal situation as well (since they are nothing but degree arguments and work for local homeomorpisms as well), which concludes the proof.

Edit: Here is a better and more general argument which works in (smooth) topological setting:

Theorem. Let $M, N$ be smooth manifolds of the same dimension, so that $M$ is connected and $N$ is compact. Let $f_i: M\to N_i\subset N$ be a sequence of (locally diffeomorphic) covering maps, which converges on compacts in $C^1$-topology to a local diffeomorphism $f_0: M\to N$, whose image is a domain $N_0\subset N$. Then $f_0: M\to N_0$ is also a covering map.

Proof. Let $g$ be a Riemannian metric on $N$. Define function $\rho_i$ on $N_i$ to be the (minimal) distance function from $x\in N_i$ to the frontier of $N_i$ in $N$, $i\ge 0$. Then the sequence of functions $\rho_i$ is semicontinuous in the sense that for every $x\in N_0$, $\lim inf \rho_i(x) \ge \rho_0(x)$. Now define (Lipschitz) continuous Riemannian metrics $g_i=g/\rho_i$ on $N_i$. These are my replacements of Poincare metrics. used in the original argument One verifies easily that every $g_i$ is complete (here and below completeness is in Cauchy sense since exponential map need not be defined even locally).

It is a standard fact of Riemannian geometry that a local diffeomorphism of smooth connected manifolds is a covering map iff the pull-back of some (every) complete Riemannian metric on the target is complete as a Riemannian metric on the domain. It is easy to check that the proof goes through in the case of continuous Riemannian metrics.

Now, the continuous Riemannian metric $\tilde{g}_i=f_i^*(g_i)$ is complete for every $i>0$. On the other hand, for every $i$, $$ f_i^*(g_i)\le f_0^*(g_0). $$ Thus, completeness of the metrics $\tilde{g}_i$ for $i>0$ implies completeness of $\tilde{g}_0$. Hence, $f_0: M\to N_0$ is a covering map. Qed.

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Thanks for the answer and the reference! I think I understand the general idea, but the topologies on groups and surfaces that you speak of are unfamiliar to me. – Ben Dozier Feb 25 '12 at 18:59
Misha can you please tell some reference(s) regarding your second argument (general one). – dushya Aug 5 '12 at 6:59
@dushya: The only reference you need is about characterization of coverings in terms of completeness: I read it in Kobayashi-Nomizu (1st volume, I think). It is typically used in the proof of Cartan-Hadamard theorem, so you should be able to find it in any book where this theorem is proven. – Misha Aug 5 '12 at 7:16
Thanks a lot :) – dushya Aug 5 '12 at 8:26

If $f_n:U\to\mathbb{C}$ are holomorphic and $f=\lim_{n\to\infty} f_n$ uniformly on each compact subset of $U$, then $f'$ has no zeros, provided none of the $f'_n$'s have any and $f$ is not constant. Indeed, then $f'_n\to f'$ uniformly on all compact subsets of $U$ and if $f'(z)=0$, then we can apply Rouch\'e's theorem to $f',f'-f'_n$ and a disk $D\subset U$ such that $z\in D$, $f'$ has no zeros on $\partial D$ and $sup_{\partial D}|f'-f_n'|<inf_{\partial D}|f'|$ to conclude that $f'_n$ has a zero in the interior of $D$ as well.

On the other hand, it may happen that $f$ is constant: take $U$ to be the unit disk and $f_n(z)=z/n$.

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I think I understand what you're saying, but I don't quite see the relevance of the first point. – Ben Dozier Feb 20 '12 at 18:09
Ben -- in my answer I assumed that by a "covering" you meant a map with a nowhere vanishing differential. Did you in fact mean an honest (topological) covering? If so, apologies. I'll then edit my answer to include this case as well. – algori Feb 20 '12 at 22:16
@algori: I mean a topological covering map. I'm not familiar with the other definition, but if it's somehow related, perhaps I should look into it. – Ben Dozier Feb 20 '12 at 23:43

Any covering from a disc to a disc is a homeomorphism. (Because the disc is simply connected). Thus every analytic covering of a disc to itself is fractional linear. The limit is either fractional-linear or a constant.

On my opinion, this topics should be closed; it does not belong to this list.

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@Alexandre: The maps are only assumed to be covering maps to their images, not to the entire disk. The solution is not very hard but is not completely trivial either, see my answer. – Misha Aug 4 '12 at 15:47
In such case why to use the technical term "covering", (which has a precise meaning), rather than to say that they are injective. The limit of injective holomorphic map is a constant or an injective holomorhic map. What else are you asking? – Alexandre Eremenko Aug 4 '12 at 17:28
@Alexandre: Alex, I am personally not asking for anything here. However, the question that Ben asked is natural: Is it true that the limit of a sequence of covering maps is a covering map (under certain conditions)? Why is he interested in this particular question (and why did he consider only maps from disk to disk which is not very natural) I do not know, you should ask him and not me. If you read his question you notice that he knows that limit of injective maps is injective. In any case, as I said, the proof that is not hard, but is not immediate either, so the question does belong to MO. – Misha Aug 5 '12 at 4:08

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