Let $f : S^1 × [0,1] → S^1 × [0,1]$ denote a null homotopic map. Show that f has a fixed point. (hint: does f lift to a covering space?)
I know f may lift to $\mathbb{R}×[0,1]$ (Or not?), but how to proceed to prove it has a fixed point? Is this problem related to some homeomorphism to $B^2$(unit disk) so that we can use some fixed point theorems?
Thank you!
Best Answer
Solution 1: Observe that $S^1\times [0,1]$ is compact. So the lift of the null-homotopic map, lets call it $\bar{f}$, has its image in $[-n,n]\times [0,1]$ for some $n$. (Also I am assuming that the covering map $\mathbb R\to S^1$ is $x\mapsto e^{2\pi i x}$.) Thus we can think of $\bar{f}$ as a map from $[0,1]\times[0,1]\to [-n,n]\times [0,1]$. Also observe that $\bar{f}$ restrict onto $\{0\}\times [0,1]$ and $\{1\}\times [0,1]$ are identical. So infact you can extend $\bar{f}: [-n,n]\times [0,1]\to [-n,n]\times [0,1]$. And both of them are homeomorphic to $D^2$. So by Brower fixed point theorem, it has a fixed point. And now look at the image of this fixed point under the covering projection, and that will be a fixed point for $f$.
Solution 2: Use Lefschetz fixed point theorem (https://en.wikipedia.org/wiki/Lefschetz_fixed-point_theorem). Since $f$ is null homotopic, $f_*$ induces zero map on $H_1(S^1\times[0,1])$. Also since $S^1\times [0,1]$ is homotopic to $S^1$ thus $H_2=0$. And $f_*$ induces identity map on $H_0$. So $\Lambda_f=1\neq 0$. Thus have a fixed point.