Let $f: S^1 \to S^1$ be a continuous map, where $S^1$ is the 1-sphere. Prove that if there is an homotopy between $f$ and a constant map, then $f$ has a fixed point.
I don't know how to prove it. I try with the Brouwer fixed point theorem, but I think that it doesn't work in $S^1$
On
You are right that Brouwer fixed point theorem does not work as it is for $S^1$. You need a continuous map from a closed disk to itself to apply it. What you have for now is a continuous map $$ H : S^1 \times [0,1] \to S^1 $$ such that $H(-,0) = f$ and $H(-,1) = p$ (for $p$ a point of $S^1$).
Intuitively, you have a cylinder mapped into $S^1$, but all the points of the top of the cylinder are mapped to the same point. So you can squish the top of a cylinder and you end up with a map from a hollow cone (like a birthday hat), which is homeomorphic to the closed disk, to $S^1$ which you can think of as embedded in a disk. Can you now make that formal and finish the argument using Brouwer's theorem?
It's overkill, but this follows straight from the Lefschetz Fixed Point Theorem. In this case the action of $f$ on $H_1(S^1,\Bbb Z)$ takes a generator to zero, so $f_*$ has trace zero on $H_1$, and the alternating sum of traces equals $1$.