Show that $\pi \circ \phi=p_1$ (Vector bundles)? Corallary : Let $\pi:E\to M$ a vector bundle of rank $n$ and $\{s_1,\cdots,s_n\}$ $n$-sections such that : for all $x\in M, \{s_1(x),\cdots,s_n(x)\}$ linearly independent, so the vector bundle $\pi:E\to M$ is trivial.

Let $\phi: M\times \mathbb R^n \to E$ : $\phi(x,(y_1,\cdots,y_n))=\sum_{i=1}^n y_is_i(x)$.
I need to show that $\pi \circ \phi=p_1$ with $p_1 :M\times \mathbb R^n \to M$ the first proejction. I get stuck here! Any help please!
 A: Let $\pi\colon E\to M$ be a vector bundle and $s_1,...,s_n\colon E\to M$ be sections such that for each $x\in M$ the set $\big\{s_1(x),...,s_n(x)\big\}$ is a linearly independent subset of $\Bbb R$-vector space $\pi^{-1}(x)$. We need to show, $\pi$ is isomorphic to trivial bundle, i.e.,  there is a smooth map $\phi\colon M\times \Bbb R^n \to E$ with $\pi\circ \phi=\text{id}_M\circ p_1$ such that the restriction $\phi\big|p_1^{-1}(x)\to \pi^{-1}(x)$ is a $\Bbb R$-linear isomorphism for each $x\in  M$, where $p_1\colon M\times \Bbb R^n\to M$ is the projection vector bundle.
$\require{AMScd}$
\begin{CD}
 M\times \Bbb R^n @>\displaystyle\phi>> E\\
@Vp_1  V V @VV  \pi V\\
M @>>\displaystyle \text{id}_M> M\\
\end{CD}

$(1)$ Definition says that for a section $s\colon M\to E$ we have $\pi\circ s=\text{Id}_{M}$. In particular, $s(x)\in \pi^{-1}(x)$ for each $x\in M$.
$(2)$ Also, note that definition of vector bundle says that  (actually restriction of bundle chart on each fiber gives) there is a $\Bbb R$-linear isomorphism $f_x\colon\pi^{-1}(x)\to \{x\}\times \Bbb R^n\cong \Bbb R^n$, where $x\in M$. In particular, $\pi\circ f_x^{-1}(x,\mathbf c)=x$ for all $x\in M, \mathbf c\in \Bbb R^n$.

Write $f_x\big(s_i(x)\big)=(x,\mathbf c_i)$ for some $\mathbf c_i\in \Bbb R^n$ where $i=1,...,n$. Define $\phi\colon M\times \Bbb R^n\to E$ as $$\phi\big(x,(y_1,...,y_n)\big)=\displaystyle \sum_{i=1}^n y_i\cdot s_i(x)$$ where $x\in  M$ and $y_1,...,y_n\in \Bbb R$. It is easy to check that $\phi$ is a smooth map (as each $s_i$ is smooth) and the fiberwise restriction of $\phi$ is an linear isomorphism. So, we need only to check that $\pi\circ \phi=\text{id}_M\circ p_1=p_1$.  Now, $$\pi\circ \phi\big(x,(y_1,...,y_n)\big)=\pi\left(\sum_{i=1}^ny_i\cdot s_i(x)\right)=\pi\left(\sum_{i=1}^ny_i\cdot f_x^{-1}(x,\mathbf c_i)\right)$$$$=\pi\left( f_x^{-1}\left(x,\sum_{i=1}^ny_i\cdot\mathbf c_i\right)\right)\big[\text{as the inverse of the }\Bbb R\text{-linear map }f_x\text{ is also }\Bbb R\text{-linear}\big]$$$$=\pi\circ f_x^{-1}\left( x,\sum_{i=1}^ny_i\cdot\mathbf c_i\right)=x=p_1\big(x,(y_1,...,y_n)\big).$$
