# Manifold parallelizable equivalent condition

This is an exercise from Loring Tu's Introduction to Manifolds that I am stuck at. I know that a tangent bundle is trivial if it is isomorphic to the product bundle $$M \times \mathbb{R}^{n}$$. Here $$n$$ is the dimension of the (smooth of course) manifold $$M$$.

I think I have to construct a smooth frame from the the isomorphism and vice versa; but I cannot find a way to do so...Could anyone please help me?

On $$\mathbb{R}^n$$ there is the obvious basis $$e_1,e_2,\dots,e_n$$. Since $$TM\cong M\times\mathbb{R}^n$$ you have the $$p\mapsto e_i$$ for all $$p$$ defines a smooth section $$X_i$$ of $$TM$$. The $$X_1,\dots,X_n$$ defines a frame at every point.
Conversely, if you have global frame field $$X_1,\dots,X_n$$, then for $$v\in T_pM$$ we have $$v=v^iX_i(p)$$, and you send this $$v$$ to $$(p,(v^i))\in M\times\mathbb{R}^n$$. Check it works.
If there is a smooth frame $$X_1,\dots,X_n$$, then each tangent vector $$V$$ is uniquely written as $$V=v^1X_1+\dots+v^nX^n.$$
What can you say about the map $$(x,V)\mapsto(x,v^1,\dots,v^n)$$?