Continuous map $f:S^1 \to S^1$ with more than $k-1$ fixed points such that induced homomorphism $f_*$ is multiplication with $k$.

Question

For my Algebraic Topology class I have solved the following exercise:

Let $x_0 \in S^1$ and let $f: S^1 \to S^1$ be a continuous map with $f(x_0)=x_0$. Suppose moreover that $f_*:\Pi_1(S^1,x_0) \to \Pi_1(S^1,x_0):[g] \mapsto k[g]$ for some natural number $k>2$. (So $f_*$ is multiplication with $k$.) Show that there are certainly $k-2$ other fixed points for $f$, besides $x_0$.

Now I need to give an example of such a map $f$ with more than $k-1$ fixed points. I don't have to work it out in a strict mathematical way (a description in words is enough). Can someone help me to find such an example?

Answer 1

For the second part, note that it is easy to modify $f$ so that it is identity in a small neighborhood of $x_0$, thus giving it infinitely many fixed points.

For the first part: we may assume $x_0=1$ and thus there exists a continuous function $\tau: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(e^{it})=e^{i\tau(t)}$ and $\tau(0)=0$.

Since $f$ induces in the fundamental group the multiplication by $k$, $\tau(2\pi)=2k\pi$.

Now, define, for $1 \leq l < k-1$, $g_l(t)=-2l\pi+f(t)-t$. $g_l$ is continuous, negative at $0$, positive at $2\pi$, so has a root $0 < x_l < 2\pi$, and $e^{ix_l}$ is a fixed point of $f$. Besides, $l \longmapsto f(x_l)-x_l=2l\pi$ is injective, so the $0<x_l<2\pi$ are pairwise distinct, therefore so are the $e^{ix_l}$.