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Yes, if the Cantor set has measure zero. This is a consequence of the Measurable Riemann Mapping Theorem which guarantees that the map is quasiconformal, combined with the theorem that if a quasiconfo …

I will assume you mean that $U$ is a "deleted" neighborhood, i.e. that it does not contain $S$, since that is the more general assumption. Then there are two equivalence classes: one represented by th …

Here's an answer from a different point of view then Henri's. It may happen that $X'$ is disconnected but in that case I'll just argue one component at a time. The Riemann surface $X'$ is noncompact, …

If you are willing to quote the Schoenflies theorem and the classification of surfaces, a quick proof of this result is a standard exercise. First, using algebraic topology (as in the proof of the Jor …

This answer is an expansion of the answer of Yuri Bakhtin. Here is a kind of mime show. Silently write the formulas for $\cos(2x)$ and $\sin(2x)$ lined up on the board, something like this: $$\cos( …

If $G$ contains the linear group consisting of all functions of the form $h_m(x) = mx$ then obviously it satisfies your conditions. Conversely, by the definition of the derivative, (*) $lim_{m \to \ …

How about Goulden, I. P.; Nica, A. A direct bijection for the Harer-Zagier formula. J. Combin. Theory Ser. A 111 (2005), no. 2, 224–238. or one of the references therein?