Let \(f, g : X \to S^2\) be continuous maps such that for all \(x\) in \(X\), \(f(x)\) is not antipodal to \(g(x)\). Show that \(f\) is homotopic to \(g\).
Solution
1. For \(x\in X\) and \(t\in I\), define\[ H(x,t)=\frac{(1-t)f(x)+t g(x)}{\|(1-t)f(x)+t g(x)\|}. \]
We regard \(S^2\) as the unit sphere in \(\mathbb R^3\).
2. The denominator never vanishes.
If \((1-t)f(x)+t g(x)=0\), then the two unit vectors \(f(x)\) and \(g(x)\) are collinear with opposite directions. Since their norms are equal, this forces \(t=1/2\) and \(g(x)=-f(x)\), contrary to the hypothesis.
3. Hence \(H\) is a continuous map \(X\times I\to S^2\) with\[ H(x,0)=f(x),\qquad H(x,1)=g(x). \]
Continuity follows from 2, and the endpoint formulas are immediate.
4. Therefore\[ \boxed{f\simeq g.} \]
The map \(H\) is the required homotopy.