Tautological 1-form: Lectures on Symplectic Geometry I am referring to page 10 of da Silva's Lectures on Symplectic Geometry. In particular, the exercise asks to show that $\alpha$ is tautological one-form if and only if for every one-form $\mu:X\to T*X$, we have $\mu^*\alpha=\mu$. Here $X$ is the base manifold, and $\alpha_p$ is a covector on the cotangent space of the cotangent bundle, i.e. $\alpha_p\in T_p^*(T^*X)$. I want to find a way to do this in coordinate-free and coordinate-dependent ways.
What I am struggling to see is the meaning of $\mu^*$. Usually, e.g. in case of vector fields, the tangent map $T_pF$ induced by a smooth map $F$ has then effect on vector field $v$ such that
\begin{equation}
T_pF(v)(f)=v(f\circ F)
\end{equation} 
which can be written as pull-back map
\begin{equation}
T_pF(v)=v\circ F^*
\end{equation}
In other words, $f\circ F=F^*\circ f$.
However in this case I cannot see what $\mu^*$ should be. I do know that one can see one-forms as just a map and hence we could imagine making a pull-back out of it. But it's a pullback of what to what?
For coordinate-dependent one, I am essentially stuck at the step which sources like Wikipedia does:
\begin{equation}
\beta ^{*}\theta =\beta ^{*}(\sum _{i}p_{i}\,dq^{i})=\sum _{i}\beta ^{*}p_{i}\,dq^{i}=\sum _{i}\beta _{i}\,dq^{i}=\beta .
\end{equation}
I do not see at all why $\beta^*p_i=\beta_i$. I suspect this is again a problem of identifying which one-form belongs to which space, but I am not clear about this. Would be great to see if this problem can be solved in both ways.
 A: For convenience we should also state the intrinsic definition of $\alpha$: We have the projection $\pi : T^*X \to X$. To define $\alpha$, one pick any $\xi \in T^*X$ and let $x = \pi(\xi)$. Consider the derivative of $\pi$, 
$$ (\pi_*)_\xi : T_\xi (T^*X) \to T_xX.$$
The one form $\alpha$ is defined by 
$$\alpha_\xi = \pi_\xi^*\xi \in T_\xi^* (T^*X).$$
Let $\mu$ be any one form on $X$ (Yes, we consider $\mu : X\to T^*X$ and $\mu^*$ is the pullback 
$$\mu^*_x : T^*_{\mu_x} (T^*X) \to T_x^*X.$$
induced by this map). To check that $\mu^*\alpha = \mu$, we check that for each $x\in X$. Let $\xi \in T_xX$ be arbitrary, then (write $\mu_x = \mu(x) \in T_x^*X$)
\begin{equation}
\begin{split}
 (\mu^* \alpha)_x (\xi) &= \alpha_{\mu_x} (\mu_* \xi) \\
&= \pi^*_{\mu_x}\mu_x(\mu_* \xi)\\
&= \mu_x (\pi_* \mu_* \xi) \\
&= \mu_x( (\pi\circ \mu)_*\xi)\\
&= \mu_x (\xi)
\end{split}
\end{equation} 
Thus we have $(\mu^* \alpha)_x = \mu_x$ for all $x$. 
If one uses the coordinate dependent definition, we have 
$$\alpha=\theta = \sum_{i=1}^n p_idq^i.$$
A one form is just a map 
$$ \beta : U \to U\times \mathbb R^n, (x^1, \cdots, x^n)\mapsto (x^1, \cdots, x^n, \beta_1 (x), \cdots \beta_n(x)).$$
Note that the coordinate on $U\times \mathbb R^n$ is denoted 
$$(q^1, \cdots, q^n, p_1, \cdots p_n).$$
Thus as pullbacks of functions, 
$$ \beta^*q^i = x^i, \ \ \ \beta^* p_i = \beta_i.$$
Then 
$$ \beta^*\theta = \sum_{i=1}^n \beta^* p_i d (\beta^*q^i) = \sum_{i=1}^n \beta_i(x) dx^i = \beta(x).$$
