Tautological 1-form on the cotangent bundle is intrinsic using transformation properties I'm following this lecture in symplectic geometry and I'm trying to show the result stated at 31 minutes that the canonical 1-form on the cotangent bundle $M = T^*X$ is well defined regardless of which coordinates we choose, that is:
$$\alpha = \xi_j dx^j = \xi'_j dx'^j$$
From what I understand, we have each $M \ni p = (x,\xi)$ where $x \in X$ and $\xi \in T_x^*X$ so each "point" on $M$ is actually a tuple of a point in $X$ and a 1-form on $X$.
It starts to get confusing from here, but from what I understand $\alpha$ is a valid 1-form on $M$ because although $\xi_j$ are forms they are also coordinate maps on $M$ for a given chart, so even though it looks like we are multiplying forms nonsensically it's all well defined if we think of this construction on the manifold $M$, so we indeed have a differential form written as functions in front of the exterior derivative of some coordinate maps.
If we change coordinate chart from $(x^1, \ldots, x^n, \xi^1, \ldots, \xi^n)$ to $(x'^1, \ldots, x'^n, \xi'^1, \ldots, \xi'^n)$ then I know my forms will translate as $\displaystyle dx^j = \frac{\partial x^j}{\partial x'^i} dx'^i$ but I'm having trouble showing that the $\xi_j$ will transform as we want.
I was thinking that since each $\xi_j$ is a 1-form on $X$ I can write them as $\displaystyle \xi_j = a^j_i dx^i$ and then use the coordinate transformation rules to get $\displaystyle \xi_j = a^j_i \frac{\partial x^i}{\partial x'^k} dx'^k :=a'^j_kdx'^k$
This leads to $\displaystyle \alpha = \xi_j dx^j = a'^j_kdx'^k \frac{\partial x^j}{\partial x'^i} dx'^i$
Ideally I was hoping for the appearance of a term like $\displaystyle \frac{\partial x^j}{\partial x'^i}\frac{\partial x'^i}{\partial x^k} = \delta^j_k$ which would cancel out, but expanding out $a'^j_k$ gives me a second $\displaystyle \frac{\partial x^i}{\partial x'^k}$ which doesn't lead to anything.
Later I found this question which provides a solution that I don't quite understand, what is the meaning of $\xi_i(dx^j)$ of a form being evaluated on a form?
I'm looking for explanation of the answer in the linked post and what the error was in my attempted proof, it's all well if my approach doesn't go anywhere useful, but I am unsure why I get something that looks so wrong/ugly just by applying what I think are simple and correct rules.
 A: The $\xi_j$ are definitely not forms. They are functions defined on an open subset of $M = T^*X$ mapping into $\Bbb{R}$. Just to be clear, let me introduce the following notation. Let $\pi:T^*X \to X$ be the canonical projection (map each covector to its base point). Now, given a chart $(U, x)$ on the manifold $X$ (i.e $x:U \to x[U] \subset \Bbb{R}^n$ is the chart map, and we set $x^i := \text{pr}^i \circ x$), we obtain a chart for the cotangent bundle as follows: on $T^*U$, we get a coordinate chart $(x^1 \circ \pi, \dots, x^n \circ \pi, \xi_1, \dots, \xi_n)$, defined as follows: $\xi_i:T^*U \to \Bbb{R}$
\begin{align}
\xi_i(\lambda) := \lambda\left(\dfrac{\partial}{\partial x^i}\bigg|_{\pi(\lambda)} \right) \in \Bbb{R}
\end{align}
Now, observe what each object is.

*

*$x^i \circ \pi$ is a function $T^*U \to \Bbb{R}$ (people often abuse notation slightly and write simply $x^i$ when it really should be $x^i \circ \pi = \pi^*(x^i)$, where the RHS is the pull-back of a function).


*Next, $\lambda \in T^*U$ is a covector, which means $\lambda \in T_{\pi(\lambda)}^*X$ lies in this specific cotangent space.


*Next, $\frac{\partial}{\partial x^i}|_{\pi(\lambda)} \in T_{\pi(\lambda)}X$ is a tangent vector in this specific tangent space, so the evaluation of the covector on this tangent vector yields a number.


*Finally, $\xi_i$ is a function $T^*U \to \Bbb{R}$, so it makes sense to feed it a covector. Note that essentially what $\xi_i$ is doing is telling us what the $i^{th}$ component of $\lambda$ is with respect to the basis $\{dx^1|_{\pi(\lambda)}, \dots dx^n|_{\pi(\lambda)}\}$ of the cotangent space. In other words,
\begin{align}
\lambda &= \xi_i(\lambda) \cdot dx^i|_{\pi(\lambda)}
\end{align}
(this should be (hopefully) somewhat familiar from linear algebra).

Now, for the sake of precision, let me write the Tautological form as:
\begin{align}
\alpha := \xi_i \, d(x^i \circ \pi) = \xi_i \, d(\pi^*x^i) = \xi_i \, \pi^*(dx^i)
\end{align}
This currently is a form defined on $T^*U$ (because $\xi_i$ and $x^i \circ \pi$ are functions on $T^*U$, while $d(x^i \circ \pi)$ is a 1-form on $T^*U$, so their product is still a form on $T^*U$).
The objective is to show that this formula yields a globally well-defined $1$-form on the whole manifold $M=T^*X$. So, let's take another chart $(V,z)$ on the base manifold $X$, and then we "lift it" to a chart $(T^*V, z^1 \circ \pi, \dots, z^n \circ \pi, \zeta_1, \dots \zeta_n)$ (excuse me not using primes for the other coordinates, because I'll definitely make mistakes lol). To complete the proof, we really need to understand how the $\zeta_j$ are related to the $\xi_i$. This is simple: given any covector $\lambda \in (T^*U)\cap (T^*V) = T^*(U \cap V)$, we have by definition:
\begin{align}
\xi_i(\lambda) &:= \lambda\left(\dfrac{\partial}{\partial x^i}\bigg|_{\pi(\lambda)} \right) \\
&= \lambda\left(\dfrac{\partial z^j}{\partial x^i}\bigg|_{\pi(\lambda)}\cdot \dfrac{\partial}{\partial z^j}\bigg|_{\pi(\lambda)} \right) \\
&= \dfrac{\partial z^j}{\partial x^i}\bigg|_{\pi(\lambda)}\cdot\zeta_j(\lambda), \end{align}
where in the last line I used $\Bbb{R}$-linearity of the covector $\lambda$, along with the definition of $\zeta_j$. If we write this as an equality of functions on $T^*(U\cap V)$, we get
\begin{align}
\xi_i &= \zeta_j \cdot \left(\dfrac{\partial z^j}{\partial x^i} \circ \pi\right) = \zeta_j \cdot \pi^*\left(\dfrac{\partial z^j}{\partial x^i} \right)
\end{align}
Now, finally proving the well-definition is simple:
\begin{align}
\xi_i \cdot \pi^*(dx^i) &= \zeta_j \cdot \pi^*\left(\dfrac{\partial z^j}{\partial x^i} \right) \, \pi^*(dx^i) \\
&= \zeta_j \cdot \pi^*\left(\dfrac{\partial z^j}{\partial x^i}\, dx^i \right) \\
&= \zeta_j \cdot \pi^*(dz^j)
\end{align}

Remarks.
Typically, this final computation is presented with the following abuse of notation (usually for good reason, since with a bit of practice, it get very cumbersome to keep track of the $\pi$):
\begin{align}
\xi_i\, dx^i &= \zeta_j \cdot \dfrac{\partial z^j}{\partial x^i} \, dx^i = \zeta_j\, dz^j.
\end{align}
(so $x^i$ can mean either a coordinate function on the base manifold $X$ or its pullback to the bundle $T^*X$).
Also, you should take note that there is a completely chart-free definition of $\alpha$, which shouldn't be too hard to find (but also, try to construct it by yourself if you can).
