TITLE: Hairy Ball Theorem: homotopy from the identity to the antipodal map
QUESTION [5 upvotes]: I'm having trouble to prove the following statement:

If the $n$-sphere $S^n\subset\mathbb{R}^{n+1}$ admits a (continuous) nonvanishing tangent vector field, then $n$ is odd.

The idea of the proof is pretty clear on my mind: we use the fact that the antipodal map has degree $(-1)^{n+1}$ and that the Brouwer degree is invariant under homotopies.
We have a (continuous) nonvanishing tangent vector field $v\colon S^n\to\mathbb{R}^{n+1}$ that can be supposed unitary (otherwise, just consider $u=v/\|v\|$).
Paraphrasing Hirsch: "a homotopy of $S^n$ from the identity map to the antipodal map is obtained moving each $p\in S^n$ to $-p$ along the great semicircle in the direction $v(p)$".
My question is: is there an explicit formula for this homotopy?

REPLY [6 votes]: The application $H\colon S^n\times[0,1]\to S^n$ given by
$$H(x,t) = \cos(\pi t)x + \sin(\pi t) v(x)$$
is what I was looking for.