How to show that a map without fix point from annular region to annular region is homotopic to antipodal map

Question

$\Omega=\{x\in R^3: 1\le||x||\le2\}$

If $L:\Omega\rightarrow \Omega $ is continuous and without fix point , how to show $L$ is homotopic with antipodal map $x\rightarrow -x$?

Answer 1

The homology of $\Omega$ is $H_0(\Omega,\Bbb{Q})=H_2(\Omega,\Bbb{Q})=\Bbb{Q}$ and $H_i(\Omega,\Bbb{Q})=0$ if $i\ne 0,2$.

The Lefschetz formula tells you that a map f without fixpoints necessarily $0=Tr(H_0(f))+Tr(H_2(f)=H_0(f)+H_2(f)$. But $H_0(f)=H_0(id)=1$ for all $f$, hence in our case $H_2(f)=-1=H_2(-id)$. By the rational Hurewicz theorem and the naturality of the Hurewicz map this implies that $\pi_2(f)=\pi_2(-id)$, hence $f$ is homotopic to $-id$.