Let $Q$ be a smooth manifold, the tautological 1-form is a 1- form on $T^{*}Q$, so a smooth map $M \rightarrow T^{*} Q$.
The definition is the following: A point $x \in T^{*}Q$ has projection $q= \pi(x) \in Q$ represents a linear 1-form \begin{align*} \alpha: T_qQ \rightarrow \mathbb{R} \end{align*} Now $\lambda_{can}$ is defined as \begin{align*} (\lambda_{can})_x(v) = \alpha(\pi_{*} v) \ \ \ \ \ \ (*) \end{align*} where $\pi : T^{*}Q \rightarrow Q $ is the bundle projection and $\pi_{*}: T(T^{*}Q) \rightarrow TQ$ its differential.
Then, in my lecture, it says
We choose local coordinates $(q_1,...,q_n) $ on open $U \subset Q$. Then locally on $Q$ any $1$-form can be written as \begin{align*} \beta= \sum_{j=1}^n p_j d q_j \text{ for some local functions } p_j \end{align*} The coordinates $(p_1,...,p_n)$ on the fibers of $T^{*} \alpha$ are called dual to the $q_j$. \ Together, the $q_j$ and $p_j$ give local coordinates on $T^{*}Q$. In such coordinates \begin{align*} \lambda_{can} = \sum_{j=1}^n p_j d q_j \in \Omega^{1}(\pi^{-1}(U)) \end{align*}
I do not really understand what is proven here. On $T^{*}Q$, we have some local coordinates $p_1,...,p_n,q_1,...,q_n$ and then it is clear that each one form is spanned by these coordinates, what is the claim here?
Thanks in advance for any help!