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 $[\gamma]\in \pi_1(S^1)$.
Question:
Is it true that $f$ has a fixed point in the interior of $\mathbb D$?
Best Answer
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 $n$.