# Proving a certain map on the closed unit disc must be the identity

Bounty expired. Will gladly re-create one if a satisfactory answer is posted in the future.

Prove: Let $f$ be a continuous function on the closed unit disc with two properties:

1. $f$ is the identity on the boundary, i.e., on the unit circle. That is, if $|z| = 1$, then $f(z) = z$.

2. $f^2$ is the identity, i.e., for all $z$ in the closed unit disc, we have $f(f(z)) = z$.

Then $f$ must be the identity function.

Motivation: I came across this question for the closed unit sphere here on MathOverflow. It seemed to me like considering the two-dimensional case might be a good place to start in trying to tackle the problem there. Ultimately, the MO question was resolved using some nontrivial results. I am wondering whether there is a proof of the question here, for the closed unit disc, which uses methods that don't extend beyond those of basic point-set topology or a first course in real analysis.

Remark: Please note that I am looking for a fundamentally different proof of the proposition above, that is, not a modification of either of the responses given at the MathOverflow link into more digestable language.

In response to a comment: If you have thought up a proof and are concerned the level of presentation is too high, then I hope you will still post it as an answer.

• Would you accept an answer that used the Jordan curve theorem? – Zach L. Dec 12 '13 at 22:36
• @ZachL. Quite possibly; do you have one? (See below the remark in my post.) – Benjamin Dickman Dec 16 '13 at 9:19

As pointed out in the comments on the MO question, this follows immediately from a 1931 theorem of M.H.A. Newman, which is Theorem 2 in this paper. Notice that Theorem 2 is basically a lemma (used to prove the main theorem 1) whose proof takes two pages, and uses nothing other than definition of manifold.

• It is quite clear to me that the answer given at MO extends to the case here. Once again, I am looking for a proof that could be given to students in a first point-set topology or real analysis class. Probably language like manifolds and simplicial (sub)divisions could be translated into something more understandable for these students, but what I am looking for here is a proof that, as it stands, is particularly easy to digest at the introductory (math major's) undergraduate level. I'd hoped that was clear from my motivation note and the education tag; if not, hopefully it is now. – Benjamin Dickman Dec 10 '13 at 1:17
• @BenjaminDickman Did you actually bother to read my answer or the linked to paper? If you had, you would see that he (M.H.A. Newman) does not use the word manifold, nor does he use simplicial subdivision, and the most sophisticated thing he uses is the least upper bound theorem (or whatever it is called). This is NOT, repeat NOT Andy Putman's argument which uses Smith theory. – Igor Rivin Dec 10 '13 at 1:21
• Dear Igor: Yes, and I think the Newman proof far easier to understand (for obviously reasons) than the original one supplied (using Smith theory). I am sure you can see that both of the example terms (manifold and simplicial division) occur in this paper; in the context of Theorem 2, see, for example, the first full paragraph on page eight. I am looking for a fundamentally different proof, and will edit my question appropriately. Sorry for any confusion. – Benjamin Dickman Dec 10 '13 at 1:31

I apologize for not making this a comment, but I lack the reputation. Emanuele Paolini's answer is incorrect. $(x_1, 0)$ need not be under the curve $\gamma$ (since it can "dip" below). Presumably one can choose the point more carefully so that the argument works, but it doesn't seem clear since $\gamma$ can oscillate wildly.

• Thanks for the comment-via-response. I have been trying to think about how to modify that answer to make it work; I'm still unsure. – Benjamin Dickman Dec 19 '13 at 1:33
• Uhm... you are right!... – Emanuele Paolini Dec 19 '13 at 11:02
• @EmanuelePaolini I am not sure your answer needs to be deleted; perhaps someone can see how to fix it! (Maybe just add a warning to the start, as another response did here.) iballa: Rather than letting the $50$ points go to waste, I think you should take the bounty for your careful reading (especially in light of your lack of reputation). – Benjamin Dickman Dec 19 '13 at 21:29
• Thank you very much! I promise to use this power to comment for good not evil :) – iballa Dec 20 '13 at 2:23

Consider the fact that $f(f(z))=z$. This shows that $f$ is its own inverse. Since $f$ is invertible, it is a bijection; a bijection from the compact space closed disk into the Hausdorff space of the closed disk, so that it is a homeomorphism (continuous bijection from compact to hausdorff is a homeomorphism). Since the closed unit disk is contractible, any two maps into it are homotopic, so $f$ is homotopic to the identity, so that , from the conditions on the boundary,$f(z)=z$ or $f(z)=-z$. From the conditions given, we must have $f(z)=z$

• Homotopic to identity is not the same as being the identity. – Igor Rivin Dec 10 '13 at 1:18
• No, but the condition $f(z)=z$ on the boundary forces the map to be $f(z)=z$, I believe. – User Some Number Dec 10 '13 at 1:19
• Why do you believe this? Take the map that takes each radius to a gently curved arc joining its two endpoints. – Igor Rivin Dec 10 '13 at 1:23
• The statement that f(z) = z or -z is simply not true. – Michael Menke Dec 10 '13 at 1:23
• Simply not true deserves a counterexample. – User Some Number Dec 10 '13 at 1:24

Warning: flawed argument below. (is it better to delete it? dunno how it works here)

As already pointed out by User Some Number in his answer, $f$ is clearly a self-homeomorphism of the unit disc $D$. By contradiction, assume $f$ is not the identity. Hence there are two distinct points $p$ and $q$ (necessarily in the interior of $D$) which get exchanged, i.e. $f(p)=q$ and $f(q)=p$ (this is because $f(f(z))=z$ for all $z\in D$). Consider the straight line segment $L$ in $D$ passing through $p$ and $q$ (cutting the disc into two parts), as well as its image $fL$ under $f$. They are both homeomorphic to the interval $[0,1]\subset\mathbb{R}$.

Now choose one of the endpoints of $L$ as your starting point (call it $x_0$), say the one which is closer to $p$ than to $q$. While moving along $L$ look at the corresponding image points on $fL$. Since $x_0$ is kept fixed by $f$, both $L$ and $fL$ start at $x_0$. After some time you meet $p\in L$, and the corresponding image on $fL$ must meet $q$ because $f(p)=q$. Later on you meet $q$, and the corresponding image on $fL$ must meet $p$ because $f(q)=p$. Now if you plot the map $f\colon L\to fL$ on a coordinate system you recognize that it can't be strictly increasing (or decreasing) because the graph must pass through the points $(x_0,x_0)$, $(p,q)$, $(q,p)$, $(x_1,x_1)$. In other words, it is not a homeomorphism (every homeomorphism between intervals either preserves or reverses the order).

I guess there is a more elegant (rigorous?) way to phrase this construction (maybe transporting the order of $[0,1]$ on $L$ and $fL$ via their respective homeomorphisms).

[Notice that this argument carries over to any dimension, i.e closed balls in $\mathbb{R}^n$]

• Given a homeomorphism $h_1 \colon L \to [0,1]$, a natural choice for the homeomorphism $h_2$ is $h_2 = h_1 \circ f$. Then you have $h_2 \circ f \circ h_1^{-1} = \operatorname{id}_{[0,1]}$, it does not exchange two points in the interior of $[0,1]$. – Daniel Fischer Dec 11 '13 at 12:43
• do you mean $h_2=h_1\circ f^{-1}$? you're right, this choice "untwist" the exchange. – johndoe Dec 11 '13 at 13:02
• what if I put an arbitrary order on $L$ compatible with its topology, do the same with $fL$ and consider $f\colon L\to fL$? – johndoe Dec 11 '13 at 13:04
• In re the first comment, we have $f = f^{-1}$ as a premise. In re the second, I don't understand what you're thinking of, so I can't comment on that, sorry. – Daniel Fischer Dec 11 '13 at 13:20
• static.inky.ws/image/4558/fL.png - how do you "plot the map on a coordinate system"? – aschepler Dec 11 '13 at 19:59

I am really not sure if this works but i thought I try. The picture in the end might help to understand my idea or even to correct it.

Let $f$ be as above i.e the identity on the boundary and a bijective. Assume for contradiction $f$ is not the identity. This means that the set $A:=\{x\in D^2:f(x)\neq x\}$ is not empty and is contained in the interior of $D^2$. Now consider some injective and straight $\phi_1:[0,1]\rightarrow D^2$ s.t $\phi(0)=x$ with $||x||=1$ i.e the "path" $\phi_1$ starts on the boundary and ends $\phi_1(1)=-x$ on the antipodal of $x$. Morever choose $\phi_1$ s.t $\phi_1([0,1])\cap A\neq\emptyset$. Just imagine a stright line going through the origin and hitting some element in A.

Consider now an arbitrary $x_0\in A$.Then $f(x_0)$ can't be an in $\phi_1([0,1])$ since otherwiese $(f\circ\phi_1)([0,1])$ would not be path connected any more as a consequence of continuity of $f$ and that $\phi(0)$ and $\phi(1)$ are fixed points of $f$. This yields that $f(x_0)$ lies somwhere in the interior of $D^2$ but not on the path $\phi_1$. Furthermore $(f\circ\phi_1):[0,1]\rightarrow D^2$ is a new path in $D^2$ which has the same start and end points but is different of $\phi$ at least for some small epsilon neighborhood of $x_0$. Finally consider $\phi_2:[0,1]\rightarrow D^2$ which is a straight line through $x_0$ and $f(x_0)$ starting and ending on the boundary and assume wlog that it hits $f(x_0)$ first. Then $(f\circ\phi_2)([0,1])$ contains $x_0$ since $\exists t \in [0,1]$ s.t $f(\phi(t))=f(f(x_0))=x_0$ but does not contain $f(x_0)$ anymore by the same argumentation as above. Moreover since it start on the boundary end ends on the antipodal point on the boundary, $(f\circ\phi_2)([0,1])$ cuts $(f\circ\phi_1)([0,1])$ in some additional point $z_0$. This implies that there exists some $y_1\in\phi_1([0,1])$ and some $y_2\in\phi_2([0,1])$ s.t $f(y_1)=f(y_2)=z_0$ for $y_1\neq y_2$ what contradicts that $f$ is bijective...