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Let $D$ be the closed unit disk in $\mathbb{C}$ and let $f:D\to D$ be a function such that:

  • $f$ is equal to the identity function $\mathrm{Id}$ specifically on the unit circle ($\partial D$)
  • $f$ is continuous on $D$
  • $f\circ f=\mathrm{Id}$ on $D$

How do we show that $f=\mathrm{Id}$ on all of $D$?

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    Is $f$ holomorphic? Otherwise I don't understand the tag.2011-09-03
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    @tomcuchta: Not really... the unit circle ($\partial D$) and the unit disk ($D$) are two different things.2011-09-03
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    Well, $f$ is either holomorphic or bat-shit-crazy (as far as homeomorphisms go), because if $f$ is even $C^1$, then by Morera's Theorem, it is holomorphic. From there, an application of the Schwarz Lemma should do the trick.2011-09-03
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    This is more a topological question than a complex-analysis one, but I suspect the problem was found in a complex-analysis book.2011-09-03
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    @Arthur: I have converted your answer to a comment. Because you do not have 50 reputation points yet, [you can only comment on your own questions and answers](http://meta.stackexchange.com/questions/19756/how-do-comments-work/19757#19757). So, you didn't do anything wrong; the "add comment" button will only appear for you once you gain 50 points. By the way, here is an [explanation of reputation points](http://meta.stackexchange.com/questions/7237/how-does-reputation-work/7238#7238).2011-09-03
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    @Dylan: Schwarz's reflection principle is usually proved via Morera's theorem. But I can't tell what Jonathan had in mind specifically.2011-09-03
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    We'd have to have $f(0)=0$ for Schwarz to apply, and we can't assume holomorphicity right off the bat either. I agree with @Arthur that this looks more topological than complex-analytical. There should be a reason that $f\circ f=\mathrm{Id}$ is one of the assumptions; maybe we have to consider certain paths and their images and derive a contradiction unless $f=\mathrm{Id}$?2011-09-03

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