I'm stuck with the following task, it's supposed to be easy to prove, but since I'm a physicist and don't have huge ambitions in topology, I dont know if I'm on the right track here, would appreciate any help. If someone could maybe comment my steps, suggest some improvements, or if I'm wrong, give a few pointers.
Let $p:E\rightarrow M$ be a $n$-dimensional real vector bundle over the manifold $M$. Then let $s_1,...s_n :M \rightarrow E$ be sections of the bundle $p$, such that the vectors $s_1(m),..., $ in $p^{-1}(m), \space m\in M$ are linearly independent, for each fibre $p^{-1}(m)$. Show that $p$ is isomorphic to $M\times\mathbb{R}^n $.
The above $n$ vectors are linearly independent and since $\dim(p^{-1}(m))=n$, they form a basis for $p^{-1}(m)$. So we choose $\{s_i(m)\}$ as a basis for $p^{-1}(m)$, additionally we choose $\{e_i\}$ as a basis for $\mathbb{R}^n$. In order to prove that $p$ is in fact isomorphic to $M\times\mathbb{R}^n$, we want to find a diffeomorphism:
$$\xi': E\rightarrow M\times \mathbb{R}^n,$$ $$\text{or alternatively, a isomorphism between } {p^{-1}(m)} \text{ and } {\mathbb{R}^n, \space\space \forall m}:$$ $$\xi=\xi'_{|p^{-1}(m)}:p^{-1}(m)\rightarrow\mathbb{R}^n.$$
We choose a linear map $\xi$ such that: $$\xi':(m,s_i(m))\rightarrow (m,e_i), \space E \rightarrow M\times\mathbb{R}^n $$ $$\text{or}$$
$$\xi=\xi'_{|p^{-1}(m)}:s_i(m)\rightarrow e_i,\space p^{-1}(m)\rightarrow\mathbb{R}^n. $$
Now we just need to show that $\xi$ is a linear bijection. It is linear by construction. Since it maps one basis to another it is trivialy to check that it's surjective and injective, therefore it is a linear bijection. Additionaly since the sections of a vector bundle are smooth functions, this hold for all $m \in M$ as required above. $p$ and $M \times \mathbb{R}^n$ are therefore isomorphic.