Showing $f: S^1 \rightarrow S^1$ has certainly $k-2$ other fixed points, besides $x_0$

Question

I have a similar question to what was asked already here

But I do not really understand the answer there.

The problem is: Let $x_0 \in S^1$ and let $f: S^1 \rightarrow S^1$ be a continuous map with $f(x_0) = x_0$. Suppose moreover that the induced map $f_{*} : \pi_1 (S^1, x_0) \rightarrow \pi_1 (S^1, x_0): [g] \mapsto k [g]$ for some $k > 2$.

(i) Show that there are certainly $k-2$ other fixed points for $f$ besides $x_0$. (hint: consider $f$ as being a map $f^{'}: I \rightarrow S^1$ with $f^{'} (0) = f^{'} (1) = x_0$ and study the lifts of $f^{'}$ to the universal covering space $\mathbb{R}$.)

(ii) Give an example of such an $f$ with precisely $k-1$ fixed points (of which $x_0$ is one).

I do not understand the hint really. How can we consider $f$ as the map $f^{'}$? And why would we do this?

I know that the fundamental group of the circle is $\mathbb{Z}$, and that every covering of $S^1$ is regular. If $p: \mathbb{R} \rightarrow S^1$ is the standard covering map, then the covering transformations are the homeomorphisms $\mathbb{R} \rightarrow \mathbb{R}: x \mapsto x + n$, for $n$ an integer. But I'm not sure how this will help me.

An elaborate answer is appreciated.

Answer 1

Let $p:t\mapsto e^{2\pi it}$ be the usual covering map of $S^1$. Since the induced map $f_*$ takes $[g]$ to $k[g]$, it is not hard to show that the degree of the map

$$g:I\to S^1: t\mapsto e^{2\pi it}\mapsto f(e^{2\pi it})\ \text{is}\ k,$$ which means that the lift of $g,$ namely, $F:I\to \mathbb R$ satisfies

$$F(1)-F(0)=k\ \text{and of course,}\ e^{2\pi i F(t)}=f(e^{2\pi i t}).$$

Suppose that there is an $s\in I$ such that $F(s)-s=j$ for some integer $j$. Then,

$$f(e^{2\pi i s})=e^{2\pi i F(s)}=e^{2\pi i (j+s)}=e^{2\pi i s}\ \text{so}\ e^{2\pi i s}\ \text{is a fixed point of}\ f.$$

Now as $s$ goes from $0$ to $1$, $F(s)-s$ maps onto an interval containing the $k-2$ integers between the first integer greater than $F(0)$ and the greatest integer less than $F(1)$. It follows that $f$ has at least $k-2$ fixed points.

Answer 2

The general idea here is that loops with a fixed class in the fundamental group are uniquely associated to paths between a fixed pair of lifts in the universal cover. For this answer, let's have $S^1$ be the unit circle in the complex plane and the universal cover $p:\mathbb R\rightarrow S^1$ be $p(t)=e^{2\pi i t}$ and $I=[0,1]$ so that we can talk in terms of concrete examples.

Suppose we wanted to think about the map $f(z)=z^2$ from $S^1$ to $S^1$, for instance. This, when thought of as a loop based at $1$, defines a curve which winds twice around the origin. It is, generally, reasonable to try to study a lift of this loop, but this is not possible because there is no map $\tilde f$ from $S^1$ to $\mathbb R$ so that $p\circ \tilde f = f$ - because, if we tried to lift $f$, we might start at $f(1)=0$ and then note that $f(e^{2\pi i t})$ has to be $2t$ plus some integer - hence $f(e^{2\pi i t})$ must equal $2t$ if we base $f(1)=0$. This obviously is not well-defined because $f(1)$ and $f(e^{2\pi i})$ are given as different values. The failure of $f$ to lift is precisely because $f$ is not contractible as a loop.

Thus, instead, we define $f':I\rightarrow S^1$ by $f'(t)=f(p(t))$. This changes our loop into a path - which we know that we can always lift. In particular, we can find that $$\tilde f'(t)=2t$$ is a lift of $f'$ - so we have solved the issue. You can imagine that $f'$ is obtained by taking the domain $S^1$ of $f$ and "cutting" it at $1$, then unrolling the circle into a line segment and that this is done to avoid the issue of loops not lifting into the cover.

Once we've done that, however, we will have a much easier time; in particular, if we forget about the example $f$ we've been using and just assume it has the desired property, we can, by knowing about the fundamental group of the circle, see that $f$ has to be homotopic (assuming, without loss of generality, that $x_0=1$) to $f(z)=z^k$. Then, $f'$ must be homotopic to $kt$ relative the endpoints of the path. In particular, this means that if we lift $f'$ to a path $\tilde f':I\rightarrow\mathbb R$ with $\tilde f'(0)=0$, then $\tilde f'(1)=k$.

To finish, we just note that $f$ has a fixed point (other than $1$) exactly when $f(t)=t$ which lifts to when $\tilde f'(t) = t + m$ for some $m\in\mathbb Z$. However, the intermediate value theorem implies that such at $t$ exists for each $m\in \{1,2,\ldots,k-2\}$.