Why $TM $ is trivial in this case?

Question

If I have a $X_1,...,X_n $ vectors fields and a basis of $T_pM $ for all $p \in M $, Why the tangent bundle $TM \cong M × \mathbb{R}^n $ ?

Answer 1

Choose a basis $v_i$ of $\mathbb{R}^n$. You can write down an explicit map $M\times\mathbb{R}^n\to TM$ by sending $(p, c_1v_1+\cdots + c_nv_n)\mapsto (c_1X_1 + c_2X_2 + \cdots + c_nX_n)_p$. The fact that the $X_i$ form a basis at each point implies that this is a diffeomorphism.

(You should convince yourself of this with a local coordinate computation.)