Let \(f,g \rightarrow S^n\) be such that \(\forall x\in X, f(x) \neq -g(x)\). Show that \(f \simeq g\).
Solution
1. Define\[ H:X\times I\longrightarrow S^n, \qquad H(x,t)=\frac{(1-t)f(x)+t g(x)}{\|(1-t)f(x)+t g(x)\|}. \]
We regard \(S^n\) as the unit sphere in \(\mathbb R^{n+1}\). The only issue is whether the denominator can vanish.
2. The denominator in 1 is nonzero for every \((x,t)\).
If \[ (1-t)f(x)+t g(x)=0, \] then, since both \(f(x)\) and \(g(x)\) have norm \(1\), taking norms gives \(1-t=t\), hence \(t=1/2\), and then \(g(x)=-f(x)\). This is excluded by hypothesis.
3. Thus \(H\) is a continuous homotopy from \(f\) to \(g\).
By 2 the normalization map is defined continuously everywhere. At the endpoints, \[ H(x,0)=f(x),\qquad H(x,1)=g(x). \] Therefore \[ \boxed{f\simeq g}. \]