Understanding how the pullback works in $f^*da_i = \sum_{k=1}^n\frac{\partial a_i}{\partial x_k}dx_k$ for a smooth $f:\mathbb{R}^n\to \mathbb{R}^n$ Let $f:\mathbb{R}^n\to \mathbb{R}^n$ be a smooth mapping and $da_i$ be a 1-form. In general, if $\nu$ is an $m$-form on $\mathbb{R}^n$, then the pullback of $\nu$ by $f$ is defined as $f^*\nu(z) \equiv \nu(f_*(z))$ where $z \in (\mathbb{R}^n)^m$ and $f^*z = (\partial f z_1,\dots, \partial f z_m)$ where $\partial f$ is the Jacobian of $f$.
My lecture notes say that $f^*da_i = \sum_{k=1}^n\frac{\partial a_i}{\partial x_k}dx_k$ for the 1-form $da_i$, but so far I have not been able to convince myself about this identity. Namely, if I am not mistaken, we use the partial derivatives $\frac{\partial}{\partial x_j}$ as a basis of the tangent space to expand the Jacobian as $(f^*da_i)(x_1,\dots,x_n) = da_i(\partial f(x_1,\dots,x_n)) = da_i\left(\sum_{j=1}^n\left(\sum_{k=1}^n\frac{\partial f_j}{\partial x_k}\cdot x_k\right)\frac{\partial }{\partial x_j}\right)$ where $\cdot$ between $\frac{\partial f_j}{\partial x_k}$ and $x_k$ is regular multiplication. Beyond this point I am quite lost how to proceed with the proof.
 A: Let $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$ where the coordinates on the first space are $\{x_k\}$ and the coordinates on the second space are $\{a_i\}$.
Since $\{dx_k\}$ form a basis for the space of $1$-forms on (the first) $\mathbb{R}^n$, you know that
$$
f^\ast da_i=\sum_{k=1}^n c_k dx_k
$$
for some coefficients functions $c_k$.  To find those coefficient functions, we need to apply $f^\ast da_i$ to $\frac{\partial}{\partial x_j}$ for various $x_j$'s.  In particular,
$$
(f^\ast da_i)\left(\frac{\partial}{\partial x_j}\right)=\left(\sum_{k=1}^n c_k dx_k\right)\left(\frac{\partial}{\partial x_j}\right)=c_j.
$$
Now, let's evaluate the right-hand-side.
$$
(f^\ast da_i)\left(\frac{\partial}{\partial x_j}\right)=da_i\left(f_\ast \left(\frac{\partial}{\partial x_j}\right)\right)
$$
We know that $f_\ast\left(\frac{\partial}{\partial x_j}\right)$ is a tangent vector in the tangent vector space spanned by $\left\{\frac{\partial}{\partial a_l}\right\}$.  Therefore,
$$
f_\ast\left(\frac{\partial}{\partial x_j}\right)=\sum_{l=0}^n b_l\frac{\partial}{\partial a_l}.
$$
To figure out the $b_l$'s, we need to apply this to a coordinate function $a_m$ since
$$
f_\ast\left(\frac{\partial}{\partial x_j}\right)(a_m)=\sum_{l=1}^n b_l\frac{\partial}{\partial a_l}(a_m)=b_m.
$$
Using the definition,
$$
f_\ast\left(\frac{\partial}{\partial x_j}\right)(a_m)=\frac{\partial}{\partial x_j}(a_m\circ f)=\frac{\partial}{\partial x_j}(f_m)=\frac{\partial f_m}{\partial x_j},
$$
where $f_m$ is the $m$-th coordinate function of $f$.
Starting to put this all together, we know that
$$
f_\ast\left(\frac{\partial}{\partial x_j}\right)=\sum_{l=1}^n\frac{\partial f_l}{\partial x_j}\frac{\partial}{\partial a_l}.
$$
Plugging this in, we have
$$
(f^\ast da_i)\left(\frac{\partial}{\partial x_j}\right)=da_i\left(\sum_{l=1}^n\frac{\partial f_l}{\partial x_j}\frac{\partial}{\partial a_l}\right)=\frac{\partial f_i}{\partial x_j}.
$$
Finally, putting it all together, we get that
$$
f^\ast da_i=\sum_{k=1}^n\frac{\partial f_i}{\partial x_k}dx_k.
$$
To me, the original formulation only makes sense if $f$ is a change of variables function.  In other words, $f(x_1,\dots,x_n)=(a_1,\dots,a_n)$.  In this case, $f_i$ would be $a_i(x_1,\dots,x_n)$.
