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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. – Dylan Moreland Sep 3 '11 at 3:14
@tomcuchta: Not really... the unit circle ($\partial D$) and the unit disk ($D$) are two different things. – leshik Sep 3 '11 at 3:16
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. – Jonathan Gleason Sep 3 '11 at 3:20
This is more a topological question than a complex-analysis one, but I suspect the problem was found in a complex-analysis book. – Arthur Sep 3 '11 at 3:32
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 $f=\mathrm{Id}$? – anon Sep 3 '11 at 3:40

Anyways, the way I solved it was through some drawings I can't show, but can hopefully explain.

Assume $f$ is not the identity. In that case there has to be at least one pair of points (A, A') being mapped to each other by $f$, reason being that $f$ is its own inverse.

Draw a simple, closed curve s through A and A', touching the edge at exactly two points (B and B'). Now, draw a simple curve t from A, out to the unit circle, ending at a point C on the edge, NOT intersecting s. Apply $f$, and see that the two must now intersect, hence $f$ can't be a bijection, contradicting the third assumption, that f has an inverse.

Now, I do not really like this proof, as I suspect the result applies to at least any finite-dimensional closed ball. Alas, I haven't been able to prove that just yet.

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Think you for your quick reply.I understand your idea of messing up a little bit the first curve so that in order to conserve the order of the images and to conserve the simplicity of the image of the first curb ,the second and the first curve has to intersect in more than 2 points. – user15502 Sep 3 '11 at 5:07

Here is a proof using algebraic topology.

Let $f$ be a continuous involution of the closed disc $D$ which acts as the identity on the boundary circle $S$. Suppose $f(x) = y \ne x$ for some $x$ in the interior of $D$. Then, since $f \circ f = \textrm{id}$, $f$ acts on the complement of $\{ x, y \}$. So let $X = D \setminus \{ x, y \}$. It is not hard to show that $X$ is homotopy-equivalent to the wedge union of two circles $S^1 \vee S^1$ by a deformation retract, so we see that the fundamental group of $X$ is the free group on two generators.

Now, consider the homomorphism $f_* : \pi_1(X) \to \pi_1(X)$ induced by $f$. By functoriality, $f_*$ is also an involution. Let $a$ be the homotopy class of a simple closed curve going counterclockwise around $x$, and let $b$ be the homotopy class of a simple closed curve going counterclockwise around $y$. It is clear that $a$ and $b$ generate $\pi_1(X)$, and it can be shown that $f_*(a) = b$ and $f_*(b) = a$. Let $c$ be the homotopy class of the curve $\gamma$ going counterclockwise on the boundary circle $S$. We must have $f_*(c) = c$, since $f \circ \gamma = \gamma$. We may assume without loss of generality that $c = a b$. But $f_*$ is a homomorphism, so this implies $a b = b a$, which contradicts our earlier assertion that $a$ and $b$ generate $\pi_1(X)$ freely.

We conclude that any continuous involution of $D$ fixing $S$ must in fact be the identity on all of $D$.

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why is $f_*(a) = b$ and $f_*(b) = a$? – Soarer Sep 3 '11 at 6:46
@Soarer: There is an open neighbourhood of $x$ consisting entirely of non-fixed-points of $f$. By contracting the loop into that neighbourhood we see that $f_*$ must take $a$ to $b$. – Zhen Lin Sep 3 '11 at 6:50
@Zhen: Why $ab=ba$ by homomorphism $f_*$? – user8186 Sep 3 '11 at 7:01
@group: If $c = a b$, and $f_*(c) = c$, $f_*(a) = b$ and $f_*(b) = a$... – Zhen Lin Sep 3 '11 at 7:02
@Zhen: Thanks! Nice Proof!! – user8186 Sep 3 '11 at 7:04

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