If $f, g : S^{n} \rightarrow S^{n}$ such that we have $f(x) \neq g(x)$ for every $x \in S^{n}$, then $f \sim a_{n} \circ g$.
(Where $a_{n} : S^{n} \rightarrow S^{n}$ is defined by $a_{n}(x)=-x$)
I tried to solve this by Brouwer degree concept. But there was no result. Any help would be great thanks.
Let $h=a_n\circ g$. Then $f(x)\ne -h(x)$ for all $x$. So in $\Bbb R^{n+1}$ the line segment joining $f(x)$ and $h(x)$ never passes through the origin. Then $\Phi_t(x)=(1-t)f(x)+th(x)$ is a homotopy from $f$ to $h$ within $\Bbb R^{n+1}-\{0\}$. We can scale that to a homotopy in $S^{n+1}$: $$\Psi_t(x)=\frac1{\|\Phi_t(x)\|}\Phi_t(x).$$