Hairy Ball Theorem and $SO(3)$ How to prove that $SO(3) \neq S^2 \times S^1$ using Hairy Ball Theorem? In other words, if assuming $SO(3) = S^2 \times S^1$, how to construct a non-vanishing vector field on $S^2$?
 A: I'm not exactly sure what you're looking for, but here's one approach.
Inside of the tangent bundle to $S^2$, $TS^2$, we can consider the vectors of unit length:  $T^1S^2 =\{(p,v)\in TS^2: |v| = 1\}$.  Then we have a map $\pi:T^1 S^2\rightarrow S^2$ given by $\pi(p,v) = p$, and this is a fiber bundle with fiber $S^1$.
Proposition  1: $T^1S^2\rightarrow S^2$ is bundle isomorphic to $\rho:SO(3)\rightarrow S^2$ with $\rho(A) = A_1$, the first column of $A$.
Proof:  Given $(p,v)\in T^1 S^2$, note that $p\in S^2$ so $|p| = |v| = 1$.  Further, $p\bot v$ since $v\in T_p S^2$.  Thus, using the cross product on $\mathbb{R}^3$, the triple $\{p,v,p\times v\}$ is an oriented orthonormal basis of $\mathbb{R}^3$.  So, we can create a map $\phi:T^1 S^2\rightarrow SO(3)$ by defining $\phi(p,v) = A$ where $A$ is the matrix whose columns are $A_1 = p, A_2 = v, A_3 = p\times v$.  A simple calculation verifies that $\det (A) = 1$, so this really does land in $SO(3)$.
It's also easy to write down the inverse - $\phi^{-1}(A) = (A_1, A_2)$.  
Finally, simply note that $\rho(\phi(p,v)) = p = \pi(p,v)$, so $\phi$ is a bundle map.$\square$
Note that under $\phi$, the projection map $T^1S^2\rightarrow S^2$ turns into $\rho:SO(3)\rightarrow S^2$ with $\rho(A) = A_1$.
Now, let $\psi:T^1 S^2\rightarrow S^1\times S^2$ be a diffeomorphism (or even just a homeomorphism).  Transporting the bundle structure on $SO(3)$ to $\psi$, we see that the map $\alpha:\rho \circ \psi^{-1}:S^1\times S^2\rightarrow S^2$ is a fiber bundle map with fiber $S^1$.  In particular, it is a fibration.
Proposition 2:  The bundle $\alpha:S^1\times S^2\rightarrow S^2$ has a section.  That is, there is a map $f:S^2\rightarrow S^1\times S^2$ for which $\alpha \circ f= Id_{S^2}$.
Proof:  The last portion of the long exact sequence in homotopy groups associated to the bundle $S^1\rightarrow S^1\times S^2\xrightarrow{\alpha} S^2$ looks like $$0=\pi_2(S^1)\rightarrow \pi_2(S^1\times S^2)\xrightarrow{\alpha_\ast} \pi_2(S^2)\rightarrow \pi_1(S^1)\rightarrow \pi_1(S^1\times S^2)\rightarrow \pi_1(S^2)= 0$$ which is the same as $$0\rightarrow \mathbb{Z}\xrightarrow{\alpha_\ast} \mathbb{Z}\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\rightarrow 0.$$  The last self map of $\mathbb{Z}$ must be surjective by exactness.  Since $\mathbb{Z}$ is Hopfian, this last self map is an isomorphism.  Now exactness shows that $\alpha_\ast$ is an isomorphism as well.
Because $\alpha_\ast$ is an isomorphism, it maps generators to generators.  So, let $g:S^2\rightarrow S^1\times S^2$ be a generator of $\pi_2(S^1\times S^2)$.  Then the composition $S^2\xrightarrow{g} S^1\times S^2\xrightarrow{\alpha} S^2$, perhaps after replacing  $\alpha$ with $a\circ \alpha$ with $a$ the antipodal map, has degree $1$, so is homotopic to the identity map.
Since $\alpha$ is a fibration, we may therefore apply the homotopy lifting property.  This yields a homotopy $g_t$ of $g$ for which $g_0 = g$ and $g_1$ has the property that $\alpha\circ g_1 = Id_{S^2}$.  Now simply take $f= g_1$.$\square$
Proposition 3:  Assuming $SO(3)$ is homeomorphic to $S^1\times S^2$, there is a non-vanishing vector field on $S^2$.
Proof:  Such a homeomorphism $\psi:SO(3)\rightarrow S^1\times S^2$ gives the map $\alpha = \rho \circ \psi^{-1}: S^1\times S^2\rightarrow S^2$ the structure of a fiber bundle over $S^2$ with fiber $S^1$.  Applying proposition $2$, there is a map $f:S^2\rightarrow S^1\times S^2$ with $\alpha \circ f = Id_S^2$.
Then the map $\psi^{-1}\circ f: S^2\rightarrow SO(3)$ has the property that $\rho\circ(\psi^{-1}\circ f) = (\rho \circ \psi^{-1})\circ f = \alpha \circ f = Id_{S^2}$.  That is, the bundle $SO(3)\rightarrow S^2$ has a section.  Since this bundle is isomorphic to the bundle $T^1 S^2\rightarrow S^2$ (Proposition 1), this bundle must have a section.  But a section is a unit length vector field on $S^2$, so is, in particular, a non-vanishing vector field.  $\square$
