If $f(x) \neq g(x)$ for every $x ∈ S ^n$, then $g$ is homotopic to $a ◦ f$

Question

Is this generalization that any map $f: S^n → S^n$ with no fixed points is homotopic to the antipodal map true?

Let $f , g : S ^n → S^n$. Show that if $f(x) \neq g(x)$ for every $x ∈ S ^n$, then $g$ is homotopic to $a ◦ f$ where $a$ is the antipodal map.

My attempt: Looking at the proof of the original theorem I think we must consider that if $f(x) /neq g(x)$ then the line joining $a(f(x))$ and $g(x)$ can be projected from the origin on to $S^n$

Sharing your thoughts on the problem will better enable other users to answer your question appropriately. What is the context, what do you know, what have you tried already? — Ben Sheller, Oct 20 '15 at 21:22

score 3 · Answer 1 · answered Oct 20 '15 at 21:26

Yes, and a similar proof works. Define a homotopy $$\frac{tg(x)+(1-t)\alpha f(x)}{||tg(x)+(1-t)\alpha f(x)||}.$$ Using the fact that $||g(x)||=||f(x)||=1$, we see that the denominator can only be $0$ if $t=1/2$, which would force $$g(x)+\alpha f(x)=0.$$ But $\alpha(u)=-u$, so $g(x)=f(x)$, contradicting the hypothesis.

If $f(x) \neq g(x)$ for every $x ∈ S ^n$, then $g$ is homotopic to $a ◦ f$

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