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Let $F: \mathbb{S}^n \rightarrow \mathbb{S}^n$ be a continuous map such that $F(x)\ne -x$ for all $x\in \mathbb{S}^n$. Prove that $F$ is homotopic to the identity map $\mathbb{S}^n\rightarrow\mathbb{S}^n$.

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Here is the homotopy: $$ \Phi(t,x) = \frac{(1-t)F(x)+tx}{\lvert (1-t)F(x)+tx\rvert} $$ the condition $F(x)+x\neq 0$ guarantees that the denumerator is always positive.

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Geometrically, if $x,y$ are two distinct points on $S^n$ with $x\neq -y$ there is a unique "great circle" that contains $x,y$, and the homotopy follows that shortest path on the great circle from $F(x)$ to $x$. (When $F(x)=x$ the homotopy is constant on $x$.) –  Thomas Andrews Feb 20 '13 at 18:07

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