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Question: Let $f:S^2\to S^2$ be a continuous map that is homotopic to the identity. Does $f$ necessarily have a fixed point?

I thought about this question after learning the proof of Brouwer Fixed Point Theorem with de Rham cohomology. I hoped to be able to prove the affirmative with a similar proof, but all my attempts failed.

My intuition that it should be true started from two facts:

  1. If $f$ is a rotation of the sphere (i.e. $f\in SO(3)$) then $f$ has two fixed points.
  2. The Hairy Ball Theorem. Intuitively, if $f$ is homotopic to the identity we can imagine a flow from each $x\in S^2$ to its image $f(x)$. This flow must have a stationary point $x_0$, meaning that $f(x_0)=x_0$.
Najib Idrissi
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Patrick
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1 Answers1

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If a function $f:S^2\to S^2$ does not have a fixed point, then it is homotopic to the antipodal map. You can construct an explicit homotopy $$F(x,t)=\frac{(1-t)f(x)-tx}{\|(1-t)f(x)-tx\|}$$ The antipodal map of $S^2$ is not homotopic to the identity because it is a composition of three reflections.

Matt Samuel
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