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

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

Najib Idrissi
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Farmer
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  • @MichaelHarrison I try to make homotopy $H(x,t)=\frac{tL(x)+(1-t)(-x)}{||tL(x)+(1-t)(-x)||}$, but when $f(x)=\lambda x $ at some $x$, it is not continuous. – Farmer Mar 04 '16 at 07:03
  • You are right. I was thinking of $\Omega = S^2$, in which case your answer works. – Michael Harrison Mar 04 '16 at 07:06
  • @MichaelHarrison But when it is annular region, I don't know how to deal it . – Farmer Mar 04 '16 at 07:09
  • What about $L(r,\theta)=(r,\theta+1)$, is this homotopicl with the antipodal map? Wouldn't that mean antipodal map is homotopic with the identity? – Marco Disce Mar 06 '16 at 11:53
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    I'm afraid some condition is missing. $L(x) = \dfrac{1}{\sqrt{\lVert x\rVert}}\cdot x$ is continuous without fixed point, but is homotopic to the identity (which has degree $1$) and hence not homotopic to the antipodal map (which has degree $-1$). – Daniel Fischer Mar 06 '16 at 11:59
  • @MarcoDisce Do you mean rotation? Rotation of $S^2$ has fix point . You can read this http://math.stackexchange.com/questions/1675676/fix-point-of-ls2-rightarrow-s2 – Farmer Mar 06 '16 at 12:04
  • @lanse2pty sorry I read "annular region" and considered the two dimensional case – Marco Disce Mar 06 '16 at 12:06
  • @DanielFischer Yes ,the $\Omega$ should be close domain. I have edit it .Thanks. – Farmer Mar 06 '16 at 12:09
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    Thinking about your space as $S^2\times I$, I think -- for the most part -- the usual proof of this fact for $S^2$ carries over to your case. The only hiccup is if there's a point whose spherical component is fixed: $L(x,t) = (x,s)$ with $t\neq s$. I'm inclined to believe this implies there's a fixed point (via Brouwer's theorem), but I don't have all the details. – Steve D Mar 06 '16 at 23:07
  • @SteveD I thought same with you, but I can't prove too. – Farmer Mar 07 '16 at 01:37

1 Answers1

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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 $-id$.

Daniel Fischer
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san
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