To construct $\pi:TM \to M$ as vector bundle,
we need to find a local function $\Phi:\pi^{-1}(U)\to U\times \mathbb{R}^k$ as local trivilazation of $TM$ on $U$
We define this function $\Phi$ as follows on a smooth chart $(U,(x^i))$:
$$\Phi(v^i\frac{\partial }{\partial x^i}|_p) = (p,(v^1,…,v^n))$$
The question is how to prove it's linear on fibers.
I know by definition it's sufficient to check $\Phi(V_p +W_p) = \Phi(V_p)+ \Phi(W_p)$ where $V_p \in T_pM$ with expression $V_p = v^i\frac{\partial }{\partial x^i}|_p$
To do this
$$\Phi((v^i+w^i)\frac{\partial }{\partial x^i}|_p) =(p,(v^1+w^1,…,v^n+w^n)) \ne (2p,(v^1+w^1,…,v^n+w^n)) = \Phi(V_p)+ \Phi(W_p)$$
What's wrong with my proof, since no addition on $U\subset M$ so this is not well defined.I don't know how to show it's linear on fiber since the codomain of $\Phi$ is not a vector space,only part of it is vector space.
Best Answer
We don't have such a thing as $2p$ for $p\in M$.
The vector space structure of the trivial bundle $U\times\Bbb R^n$ is of course defined fibrewise: $$(p,v)+(p,w):=(p,v+w)$$