I'm studying deRham cohomology using Madsen's From Calculus To Cohomology and the theorem 7.3 is the hairy ball theorem.
Given a tangent vector field $v:S^n\to\mathbb{R}^{n+1}$ such that $v(x)\neq 0$ forall $x\in S^n$, he extends it to a tangent vector field $w:\mathbb{R}^{n+1}-\{0\}\to\mathbb{R}^{n+1}$ by setting $w(x)=\frac{x}{||x||}$. Then he defines the homotopy $F:\mathbb{R}^{n+1}-\{0\}\times [0,1]\to \mathbb{R}^{n+1}-\{0\}$ by $$ F(x,t)=(\cos\pi t)x+(\sin\pi t)w(x) $$ between the identity on $\mathbb{R}^{n+1}-\{0\}$ and minus the identity. What I don't get here is why is $w(x)$ necessary, wouldn't any non-zero constant vector lead to the same kind of homotopy? More precisely, where in this homotopy is $w$ needed to be a tangent vector field?
In most proofs I've seen, the homotopy constructed is $F:S^n\times [0,1]\to S^n$, defined similarly as $ F(x,t)=(\cos\pi t)x+(\sin\pi t)v(x) $. In this latter case I think more assumptions should be made over $v$ in order to keep $F(x,t)$ in $S^n$, but I still don't see why being tangent is one of them.
Note: The fact that he extends the vector field to $\mathbb{R}^{n+1}-\{0\}$ is because at that moment he has only defined the cohomology over open euclidean sets. I'm not asking why is it necesary to extend $v$ to $w$, but why is $w$ the only valid vector for that homotopy.
A nonzero constant vector doesn't do the job. Otherwise, it could be possible that $F(x,t) = 0$, which is forbidden. More precisely, say you choose a constant $w$, then $F(-w,1/4) = 0$. So that settles that.
In fact, for all $x$, $w(x)$ cannot be a multiple of $x$. Otherwise, $t \mapsto F(x,t)$ will go through $0$ at some point by the intermediate value theorem. Therefore, if you restrict to $x \in S^{n-1}$, you can project $w(x)$ to the tangent space of $S^{n-1}$ at $x$, and the result will be nonzero because $w(x)$ is not colinear to $x$. So if you can find a continuous $w(x)$ that isn't colinear with $x$ for every $x$, then you have a never zero vector field on the sphere. And vice-versa (from a never zero vector field $v$ on $S^{n-1}$, let $w(x) = v(x / \|x\|)$ as Madsen does).
Strictly speaking, it's not true that $w$ is the "only valid vector field". You could have chosen to extend $v$ to $\mathbb{R}^n$ in another way, for example $w(x) = \|x\|^{12345} \cdot v(x / \|x\|)$. That works too. But no matter how you slice it, you must end up with a vector field $w$ such that $w(x)$ is never colinear with $x$. So why do something complicated when you can do something simple...