Diffeomorphism between $TM$ and $M\times R^n$ Let $(M,\mathcal{A})$ be a manifold with smooth structure $\mathcal{A}$. For any point $x\in M$, we define a tangent at x by the triplet $(c,x,h)$, where $c=(U,\phi)$ is a chart at $x$, $h\in R^n$ ($n$ is the dimension of the manifold). For two charts $c,c'$, define the following equivalence relation: $(c_1,x,h_1)\sim (c_2,x,h_2)$ if $D(\psi\circ\phi^{-1})(\phi(x))h_1=h_2$, where $c_1=(U,\phi), c_2=(V,\psi)$. Now, the tangent vectors at $x$ is defined by $T_xM=\{[c,x,h]: x\in M, h\in R^n\}$ and $[c,x,h]$ is the equivalence class. Finally, the tangent space is defined as $TM=\cup_{x\in M} T_xM$. My question is the following:
Can we show that $TM$ is diffeomorphic to $M\times R^n$ using this definition of tangent space? In particular, is it true when $M=\mathbb{S}^{n-1}$?
 A: In general this is not true. One can show that a vector bundle $(E, \pi,M)$(https://en.wikipedia.org/wiki/Vector_bundle) is trivial (i.e. $E \simeq M \times \mathbb{R}^k$) iff exists a global frame $\lbrace \sigma_j : M \rightarrow E \rbrace_{j=1,\ldots , k} $ (i.e. a family of sections of $\pi$) which never vanish on $M$ and form a pointwise base of $E_p\simeq \pi^{-1}(p)$ for all $p\in M$. The tangent bundle is an example of vector bundle. You can use this criterion for checking whether a vector bundle is parallelizable (that is $TM \simeq M \times \mathbb{R}^m$). 
As for the spheres it has been shown that only $S^1$, $S^3$ and $S^7$ are parallelizable. For $S^1 \hookrightarrow \mathbb{R}^2$ you just take the vector field $x_2 \partial_1- x_1 \partial_2$. For $S^3$ you note that it is a Lie group (quaternions with norm $1$), then Lie groups are always parallelizable with $TG\simeq G\times \mathfrak{g}$ with $\mathfrak{g} \simeq T_eG$ the Lie algebra. The fact that $S^7$ is parallelizable by a result of Milnor.
