Show $f$ has a fixed point if $f\simeq c$

Question

I have the following problem:

Show that if $f:S^1\to S^1$ is a continuous map, and $f$ is homotopic to a constant, then $\exists p\in S^1 : f(p)=p$.

My approach is to show that if for all $p, \ $ $f(p)\neq p$, then $f$ is homotopic to $\mathrm {id}_{S^1}$. To prove this I thought of using the parametrization of the segment $pf(p)$, and projecting outwards to $S^1$. However this creates problems if the points are antipodes. I suppose then, that this assumption is too strong, and I should either prove that $f$ is homotopic to $z^n$ (by contradiction) for non zero $n$, or bring in some theorem such as Borsak-Ulam, which applies here because the map is non-surjective, and can be thought of as a map into $\mathbb R$, but I can't think of how this would help. I would appreciate some help.

Answer 1

An alternative argument would be to show that if $f$ has no fixed point, then $f$ is homotopic to the antipodal map $h(x)=-x$.

Answer 2

There is a proposition that you will want to use here which is the following:

Proposition. A map $g\colon S^1\to X$ is null-homotopic if and only if there exists a map $\tilde{g}\colon D^2\to X$ such that the restriction satisfies $\tilde{g}|_{S^1}=g$. That is, $g$ can be extended to a map on the disk.

Now, if $f\colon S^1\to S^1$ is null-homotopic, let us suppose that it does not have a fixed point. Let $i\colon S^1\to D^2$ be the inclusion of the circle into the disk $D^2$. Let $\tilde{f}\colon D^2\to S^1$ be the extension of $f$ that must exist by the above proposition. Clearly if $f$ has no fixed points, then $$i\circ \tilde{f}\colon D^2\to S^1 \to D^2$$ also has no fixed points but this contradicts Brouwer's fixed-point theorem.