In the proof of the Hairy Ball theorem, we use the diffeomorphisms,
$F_t(x)=\cos(\pi t)x+\sin(\pi t)v(x)$
where $v(x)$ is the non-vanishing vector field, then use the fact that
$F_1^*=F_0^*$ as pullbacks on the cohomology to reach a contradiction.
However, I cannot see why this same proof would not work for $S^1$ for example>
Isn't $F_t$ a diffeomorphism (Actually, I think we just need it to be a smooth map) regardless of whether $n$ is even or odd, where $S^n$ is the $n$-sphere?
For example, considering $S^1$, and the usual non-vanishing tangent vector, we have
$F_t(\theta)=(\cos(\pi t+\theta),\sin(\pi t+\theta))$
and this looks like a diffeomorphism, basically, just rotating the point $(\cos(\theta),\sin(\theta))$ by the angle $\pi t$.
Where is the part where we use the evenness of the dimension?
The contradiction is reached in the following manner, without reference to the degree.
For any $\omega$, a $2m$-form on $S^{2m}$, we have that $F_0^*(\omega)=F_1^*(\omega)$ since $F:[0,1]\times S^{2m}\to S^{2m}$ is a smooth map, and thus, we would have that $[\omega]=-[\omega]$ since $F_0$ is the identity map and $F_1$ is the antipodal map, and since this holds for any $\omega$, we have that $H^{2m}(S^{2m})=0$. However, since $S^{2m}$ is a compact orientable manifold, then $H^{2m}(S^{2m})\neq 0$ (This is done by considering the non-vanishing volume form.)
I cannot see where the evenness is used here either.
You have $F_0=id_{\Bbb S^n}$ and $F_1=-id_{\Bbb S^n}$. Then $F_0$ has degree 1 and $F_1$ has degree $(-1)^{n+1}$, so $F_0^*\neq F_1^*$ if $n$ is even.
Edit: In the case where $n$ is odd, $F_1^*([\omega])\neq -[\omega]$ but you have $F_1^*([\omega])=[\omega]$ so you won't reach a contradiction.