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Given a real-analytic map $f : \mathbb{S}^1 \rightarrow \mathbb{S}^1$, where $\mathbb{S}^1 = \{z \in \mathbb{C} : |z| = 1\},$ does it admit a complex-analytic extension $\tilde{f} : U \rightarrow V$, where $U$ and $V$ are open subsets of $\mathbb{C}$ containing $\mathbb{S}^1$?

If so, how can you prove it? I'd appreciate a proof as elementary as possible (but complete).

Thanks.

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    Oh, I suppose finitely many disks isn't actually required.2011-11-26

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The map $\pi : \mathbb R \longrightarrow \mathbb S^1$, $\pi(x)=e^{2i\pi x}$ is the universal cover of the unit circle. Let's lift the map $f$ to $F : \mathbb R \longrightarrow \mathbb R$ which is now $1$-periodic. $F$ is also real-analytic. So for each $x_0 \in \mathbb R$ there exist $\varepsilon > 0$ such that $F(x) = \sum_{n \geq 0}a_n(x-x_0)^n$ (the sequence $(a_n)$ may depends on $x_0$). For simplicity, let's assume $x_0 = 0$.

$F(x) = \sum_{n \geq 0}a_nx^n$ on $(-\varepsilon,+\varepsilon)$. Let's extend $F$ to the disc centered at $0 = x_0$ of radius $\varepsilon$:

$\tilde{F}(z) = \sum_{n \geq 0}a_n z^n$ on $D(0,\varepsilon)$

It's obvious that the restriction of $\tilde{F}$ to the real line is $F$, and $\tilde{F}$ is holomorphic since it's analytic. We can go back to the circle.

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    That's true. I just wanted to emphasize the fact that this result is true for any non-empty open set of $\mathbb R$. In general it's easier to think about an open interval of $\mathbb R$.2012-08-26