Lets's suppose we have a continuous map of the disk

$$f: \mathbb D \to \mathbb D,$$

where $\mathbb D := \{z \in \mathbb C\,:\, |z|\le 1 \}$.

We know by the Brouwer fixed point theorem that $f$ has a fixed point in $\mathbb D$. But lets's suppose further that $f$ preserves the unit circle $S^1 = \{z\,:\,|z| = 1\}$ (i.e. the boundary of $\mathbb D$):

$$f(S^1) = S^1$$

and that $f$ restricted to this circle is a degree $n \ge 2$ map. That is, it's action on the fundamental group,

$$f_\ast: \pi_1(S^1) \to \pi_1(S^1)$$

takes $[\gamma] \longmapsto [\gamma]^n\ $ for $[\gamma]\in \pi_1(S^1)$.


Is it true that $f$ has a fixed point in the interior of $\mathbb D$?

  • $\begingroup$ Well, we know $f$ has a fixed point somewhere in the disk, and since the degree of $f$ restricted to $S^1$ is 2 or higher, the fixed point can't be on the boundary circle... $\endgroup$
    – kamills
    Commented Dec 14, 2019 at 16:20
  • 1
    $\begingroup$ @kamilis are you quite sure? Consider $z \to z^n$, which fixes $1$ and ($n-2$) other points on the boundary (as well as $0$). $\endgroup$
    – Good Boy
    Commented Dec 14, 2019 at 18:31
  • $\begingroup$ Ah, you're right. Sorry for the mistake. Your counterexample almost finishes the question, though... if you post-compose with a map that shrinks the whole disk towards one of those fixed points (as in math.stackexchange.com/a/1396826/497007), I think it might be what you're looking for (degree $\geq 2$ on the boundary and no interior fixed point). $\endgroup$
    – kamills
    Commented Dec 14, 2019 at 18:40
  • $\begingroup$ You know, I'm not convinced it does the job: how would such a shrinking map preserve the boundary? $\endgroup$
    – Good Boy
    Commented Dec 14, 2019 at 19:12
  • $\begingroup$ I don't believe it would, if you're thinking about actually preserving the size of the disk (which is why I'm not sure about the answer to the question anymore). Certainly it would send the boundary circle to the boundary circle of the image, but the boundary circle of the image wouldn't consist of points all of length 1. $\endgroup$
    – kamills
    Commented Dec 14, 2019 at 19:18

2 Answers 2


This claim is simply false, here is an example. I will identify the circle $S^1$ with the unit circle in the complex plane. Then $D$ is the closed disk $\{|z|\le 1\}$.

First of all, for $n\ge 2$ consider the map $h: S^1\to S^1$ of degree $n$, $h(z)=z^n$.

I will define a map $f: D\to D$ by "coning off" from $1$: Consider the family of round circles $S_t$ (of radii $t\in [0,1]$) contained in $D$ and tangent to $S^1$ at $1$. For each $S_t$ let $g_t: S_t\to S^1$ denote the Euclidean dilation fixing $1$. Let $$ r(t)=t^2. $$ This map has no fixed points in the interval $(0,1)$.

For each $t\in [0,1]$ define the map $$ h_t= g_{r(t)}^{-1} \circ h \circ g_t: S_t\to S_{r(t)}. $$ Lastly, define $f: D\to D$ by combining the maps $h_t$: Every $z\in D$ belongs to a unique circle $S_t$. Then set $$ f(z)= h_t(z). $$ I will leave it to you to verify continuity to $f$. I claim that $f$ has no fixed points in the interior of $D$. Indeed, if $z\in S_t$ for $t\in (0,1)$ then $f(z)\in S_{r(t)}$. But $S_{r(t)}\cap S_t=\{1\}$ for $t\ne 1$. Hence, $z$ cannot be fixed by $f$ provided $|z|<1$.

Thus, all the fixed points of $f$ are on the boundary circle $S^1$. The restriction of $f$ to $S^1$ is the map $h$ which has degree $n$.


I believe I have an explicit counterexample which does the same thing: (a hyperbolic isometry squared)

$$f: z \longmapsto \left(\frac{3z+1}{3+z}\right)^2.$$

Maps the disk into the disk and fixes the boundary

Firstly, its a holomorphic map which preserves the circle and the disk, which can be seen by checking a few values:

$$(1,\ -1,\ i,\ 0) \longmapsto (1,\ 1,\ \frac7{25}-\frac{24}{25}i,\ \frac 19).$$

The first three mean it preserves the unit circle, the last means the interior of the disk is mapped to itself.

Has only one fixed point, which is on the boundary

$$f(z) - z = \frac{(1-z)^3}{(3+z)^2},$$

so the only fixed point of $f$ is $1 \in S^1$.

Restricts to a degree 2 map on the boundary

I'm not really sure how you check this in practical terms, besides a winding number computation:

$$\frac1{2\pi}\oint_{f \,\circ\, \gamma} \frac1z \mathrm dz = \int_0^{2\pi}\frac{f'(e^{2\pi i \theta})}{f(e^{2\pi i \theta})}e^{2\pi i \theta}\mathrm d \theta = \cdots =\int_0^{2\pi}\frac{8 }{3 \cos (2 \pi \theta )+5}\mathrm d \theta = 2$$

for $\gamma:[0,1]\to S^1$ is the "standard generator" $\gamma(t) = \exp (2\pi i t)$.

In other words, $f$ takes this curve of winding number 1 onto one of winding number 2; and is thus doubling.

  • $\begingroup$ Note that, by conjugation, any hyperbolic isometry has this property. It's also true for a parabolic isometry also: e.g. $$ z \longmapsto \left(\frac{1-(1-2 i) z}{(1+2 i)-z}\right)^2 $$ $\endgroup$
    – Good Boy
    Commented Dec 15, 2019 at 12:03
  • $\begingroup$ You can also check the degree by interpolating between the hyperbolic isometry and the identity map. $\endgroup$ Commented Dec 15, 2019 at 16:11

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