$f$ continuous over $S^{1}$ has a fixed point

Question

Let $f:S^{1}\rightarrow S^{1}$ be a continous map such that $f$ is homotopic to a constant function $c_{x_{0}}$. Then $f$ has a fixed point.

I already know how to prove this using fixed point theorem, but I would like to try another approach.

We can lift $f$ to a map $\tilde{f} :S^{1}\rightarrow\mathbb{R}$ by considering the classical covering map $$p:\mathbb{R}\rightarrow S^{1}.$$ The idea is that, since $p(\tilde{f}(x))=f(x)$, then $f(x)=x=e^{2\pi i\theta}$ if and only if $\tilde{f}(x)-\theta\in\mathbb{Z}$, but I don't see an easy way to prove this. Can someone help me?

Answer 1

Since each constant map $c : S^1 \to S^1$ trivially has a lift $\tilde c : S^1 \to \mathbb R$, the homotopy lifting theorem shows each $f$ which is homotopic to a constant map has a lift $\tilde f : S^1 \to \mathbb R$.

$\tilde f(S^1)$ is a compact connected subset of $\mathbb R$, thus a closed interval $J = [a,b]$. It may be degenerate ($a = b)$.

Consider the map $F : J \to J, F(t) = \tilde f(p(t))$. We have $$p(F(t)) = p(\tilde f(p(t)) = f(p(t)) . $$

$F$ has a fixed point $t_0 \in J$. For $a < b$ this is a consequence of the IVT. Let $z_0 = p(t_0)$. Then $$f(z_0) = f(p(t_0)) = p(F(t_0)) = p(t_0) = z_0 . $$

Answer 2

There is a different theorem along the lines of, if $f$ has no fixed points then $f$ has degree $1$.

For our task - if we want to explicitly use the covering:

Thinking of $f$ as a loop induced from some $\ell:I\to S^1$, we can lift to a path $\gamma:I\to\Bbb R$.

We know the integer $\gamma(1)-\gamma(0)$ completely classifies the homotopy type of this loop. Since we know the loop is trivial in $\pi_1$, we know $\gamma(1)=\gamma(0)$ is forced. Now you just need to show that any loop in $\Bbb R$ has a 'fixed' point (this is equivalent to the original problem because $f(e^{2\pi it})=e^{2\pi it}$ iff. $\ell(t)=e^{2\pi it}$ iff. $\gamma(t)-t\in\Bbb Z$). Note that $\gamma(0)$ need not be $0$. We can extend $\gamma$ to a $1$-periodic function $\Gamma:\Bbb R\to\Bbb R$ and now it really remains to show this has a fixed point in the true sense. This is straightforward with the intermediate value theorem.