Show that if \(f: S^n\to S^n\) has no fixed points \(\iff \deg f = (-1)^{n+1}\) and \(f\) is homotopic to the antipodal map.
If \(f(x)\neq x\), the line segment \(L(-x, f(x))\) does not contain \(0\).
If \(f(x) \neq -x\), the line segment \(L(x, f(x))\) does not contain \(0\).
Solution
The straight line homotopy \(H(t, x) = (1-t)f(x) + t(-x)\) is a homotopy between \(f\) and the antipodal map in \(\RR^{n+1}\). Use that \(H(t, x) = 0 \iff t=1/2 \iff f(x) = x\), so \(H(t, \wait)\) is always a line from \(x\) to \(f(x)\) not passing through \(0\), so \(H(t, \wait) \neq 0\). So this descends to a homotopy of \(\RR^{n+1}\smz \homotopic S^n\), or explicitly one can project \[ H: I\cross S^n &\to S^n \\ (t, x) &\mapsto {H(t, x) \over \norm{H(t, x)}} .\]
Less explicitly: if \(f(x) \neq x\) for all \(x\in S^n\), there is a unique geodesic through these two points, so let each point flow along its corresponding geodesic.