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I don't have a precise answer, but the genus of $S$ has to grow like $n!$. To see this, note that when $n$ is large enough, $S$ cannot be the sphere or torus. So $S$ admits hyperbolic metrics. By Ni …

There is a well-developed theory in the punctured case, and a reasonably well-developed theory in the “flaring” case. See here.

Here is a more "topological" counterexample. Let $C = \mathbb{C}^\times$ be the punctured plane. Let $D = \{ z \in \mathbb{C}^\times \mid |z| < 1 \}$ be the punctured disk. Define $\iota \colon D \ …

Fred Gardiner's book Teichmüller theory and quadratic differentials is a good reference. He (a) deals with the punctured case (called finite analytical type: see the first page of Chapter 2), (b) cov …

I think that there are counterexamples, but I am not an expert.... Here is my attempt. Let $S$ be the unit square in the complex plane. We obtain $R$ by gluing opposite sides by translation. So $R …

For question one: one possible qualitative approach is to consider the "thick-thin" decomposition of the given metric on $S$. For question two: the hyperbolic metric on a surface and the "moduli of a …

No, certainly not. Biholomorphic maps are very rare and very rigid. Generically, given Riemann surfaces $X$ and $Y$, there will be no biholomorphic maps between them. Yes. You will want to read u …

I have been told (rather forcefully, regarding this question) that phrases like "original source" or "first example" or "earliest proof" are wrong-headed. The person who scolded me explained (more or …

Here is Moishe Kohan’s comment turned into an answer. Suppose that $S$ is a closed, connected, oriented surface of genus at least two. Let $\rho \colon \mathrm{Homeo}^+(S) \to \mathrm{MCG}^+(S)$ be …

The answer is "yes" if $S$ is the Riemann sphere. This is because a map $f$ of degree $d$ from the sphere to itself is homotopic to $z \mapsto z^d$. The answer is "basically no" if $S$ has genus two …

Below is a cartoon showing one "solution" for $S_{0, 4}$. I'd encourage you to think about $S_{0, 6}$ next, then deal with $S_{1,4}$, then deal with $S_{2, 6}$, and then ponder the general case. Not …

I interpret your question as follows: Suppose that a Riemann surface $X$ is given as a flat surface (say, by gluing together euclidean polygons by local isometries, satisfying a few nice conditions). …

Edit: as Moishe points out, my answer (just below) is for a different, and easier, question. I will leave the answer up, as it does feel “related”. The answer is no. I find it conceptually easier to …

The answer is "no". For consider the case where $L_1 = L_2$ is a single simple closed geodesic. Now you've added the hypothesis that $L_1$ and $L_2$ have no common leaf, the answer becomes "yes". H …

Unless I misunderstand your question, you can take $r = \infty$. This is because we can choose representatives of marked conformal classes to be "Riemann surface structures" on $S_g$. Doing this, th …