Tautological 1-form, identifications in calculation of pullback In Lee's Intro to Smooth Manifolds, he introduces the (coordinate-free) definition of the tautological $1$-form by doing a pullback calculation. It appears that he is making an identification on the way, though, that I am having a hard time understanding. So I did the calculation from scratch, but I do not know how to relate it to what Lee did. 
For a manifold $M$, we have the following data:
$\pi:T^*M\to M:\ (q,\varphi)\mapsto q$
$d\pi_{(q,\varphi)}:T_{(q,\varphi)}(T^*M)\to T_qM;\ (d\pi_{(q,\varphi)}v)(f) =v(f\circ\pi_{(q,\varphi)})$ for smooth real-valued $f$.
$d\pi^*_{(q,\varphi)}:\ T^*_qM\to T^*_{(q,\varphi)}(T^*M);\ d\pi^*_{(q,\varphi)}(\omega_q)(v)=\omega(d\pi_{(q,\varphi)}(v)).$ 
So, in coordinates, if $\omega_q=\sum a_i(q)dx^i$, where it is understood that $dx^i$ acts on $T_qM$, then
$d\pi^*_{(q,\varphi)}(\omega_q)=\sum a_i\circ \pi(q,\phi) \ d\pi^*_{(q,\varphi)}(dx^i)=\sum a_i(q) \ d(x^i\circ \pi).$
In particular, if $\omega=dx^i$ then $d\pi^*_{(q,\varphi_)}(dx^i)=d(x^i\circ \pi)$, where it is understood that $d(x^i\circ \pi)$ acts on $T_{(q,\phi)}(T^*M).$
Lee has $d\pi^*(dx^i)=dx^i$, so I guess he is making some identification because since the domain and range of $d\pi^*$ are different, this cannot be literally true. I imagine this has to do with a "tautology" but I am not seeing it. 
Edit: To verify the correct identification, as prof Lee points out in the comments, and using the notation there, we calculate:
If $v=\alpha^i\frac{\partial}{\partial x^i}$ and $v'=\beta^i\frac{\partial}{\partial c^i}\in T_{(q,\varphi)}(T^*M),$ we have  $d\pi^*_{(q,\varphi_)}(dx^i)(v+v')=d(x^i\circ \pi)(v+v')=(v+v')(x^i\circ \pi)=\alpha^i$ and if $v=\alpha^i\frac{\partial}{\partial x^i}\in T_qM,$ then $dx^iv=\alpha^i$ as well so $d\pi^*_{(q,\varphi_)}(dx^i)$ acts in the same way on tangent vectors in $T_{(q,\varphi)}(T^*M)$ as $dx^i$ acts on vectors in $T_qM$. Hence, the identification.
 A: What I wrote in the comments seems to have clarified things for the OP. But I want to spell things out a little more carefully for anyone else who's interested.
Yes, there is an identification going on here -- I use the notation $x^i$ both for the given coordinates on $M$ and (some of the) corresponding natural coordinates on $T^*M$. 
In somewhat more detail, what's going on is this. For any choice of smooth local coordinates $(x^1,\dots,x^n)$ on an open subset $U\subseteq M$, you get natural coordinates $(x^1,\dots,x^n,\xi_1,\dots,\xi_n)$ on $\pi^{-1}(U)\subseteq T^*M$, defined as follows: any point in $\pi^{-1}(U)$ can be thought of as a pair $(q,\varphi)$, where $q\in U$ and $\varphi\in T^*_q M$. This covector $\varphi$ can be written in coordinates as $\varphi= \sum_i \xi_i dx^i|_q$, and we let $(x^1,\dots,x^n)$ be the coordinates of $q$ and take the $2n$ numbers $(x^1,\dots,x^n,\xi_1,\dots,\xi_n)$ to be the coordinates of $(q,\varphi)$.  As explained on page 277 of my smooth manifolds book, these provide smooth local coordinates on $T^*M$. Here is where the implicit identification occurs: when the notation $x^i$ is used in the context of $TM$, what it really means is $x^i\circ \pi$. 
Thus if $\tau$ denotes the tautological $1$-form on $T^*M$, a local coordinate expression for $\tau$ is going to be a linear combination of $dx^i$'s and $d\xi_i$'s, with coefficients that are functions of $(x^1,\dots,x^n,\xi_1,\dots,\xi_n)$. If you unwind the abstract definition of $\tau$ and express it in terms of these coordinates, what you get is $\tau = \sum_i \xi_idx^i$.
If you're being confused by the identifications, it might help to introduce a different notation for the coordinate functions on $T^*M$, such as $\tilde x^i = x^i\circ \pi$.
