Tangent bundle of $S^2$ not diffeomorphic to $S^2\times \mathbb{R}^2$ I am trying to show that the tangent bundle of $S^2$ not diffeomorphic to $S^2\times \mathbb{R}^2$. This is from an exam, where there is a hint stating that this is more than showing that $TS^2$ is non-trivial.
I know how to show the hairy ball theorem, according to which $TS^n$ is non-trivial iff n is even.
I also know that a vector bundle $\pi:E\rightarrow M$ of rank $m$ on a smooth manifold $M$ is trivial (by definition) iff there exists a diffeomorphism $f:E\rightarrow M\times \mathbb{R}^m$ such that for every $p\in M$, $f$ induces a vector space isomorphism $f:\pi^{-1}(p)\rightarrow \{p\}\times \mathbb{R}^m$.
So I see that showing that $TS^2$ is non-trivial only guarantees that $TS^2$ is not diffeomorphic to $S^2\times \mathbb{R}^2$ via a diffeomorphism satisfying the property above, but it is not enough to conclude that there isn't any diffeomorphism.
How can I show this then?
 A: Here's a proof that relies on a theorem of Kervaire with a somewhat-involved proof (I think the particular case of $n=k=2$ is probably much easier, but I couldn't find such a proof to confirm my suspicions.) 

Theorem: The normal bundle to an $n$-sphere embedded inside $\Bbb R^{n+k}$ is trivial if $k> \frac{n+1}2$

This implies any 2-sphere embedded in $S^2 \times \Bbb R^2$ has a tubular neighborhood isomorphic (as fiber bundles) to $  S^2 \times \Bbb R^2$, since $S^2 \times \Bbb R^2$ embeds as an open manifold of $\Bbb R^4$. If $f:TS^2\rightarrow S^2 \times \Bbb R^2$ is a diffeomorphism, then the image of the zero section has a trivial tubular neighborhood  $N$ by the above disscussion, but then $f^{-1}(N)$ is a trivial tubular neighborhood of the zero section in $TS^2$.
References:


*

*Massey's  On the Normal Bundle of a Sphere Imbedded in Euclidean Space

*Kervaire's An Interpretation of G. Whitehead's Generalization of H. Hopf's Invariant
A: I think that if suppose that exist a diffeomorphism $F$ then there is an correspondence between sections in $S^{2}\times\mathbb{R}^{2}$ and vector fields on $TS^{2}$. Take the constant section $s(x)=(x,(1,0))$ and by diffeomorphism there is $X\epsilon\mathfrak{X}(S^{2})$ such that $dF(s)=X$, but this section vanishing in south pole under stereographic proyection, that means $dF$ isnt isomorphism
A: I think you can try this.
$S^2\times\mathbb{R}^2$ is trivial, while $TS^2$ is not trivial because only the spheres of odd dimension have non-trivial tangent bundle (Hopf's theorem). So they can't be diffeomorphic.
