Edit to add: The reason I believe there should be a more direct and intuitively satisfying proof is that a pullback is conceptually fairly simple. For example: A mapping of joystick coordinates to screen display coordinates is a kind of pullback.
That informs my intuition that there is some perspective from which this proof becomes simple. To me, my proof seems like swatting a fly with a sledgehammer.
This question regards Advanced Calculus of Several Variables, by C.H. Edwards, Jr.
I've used my own notation because I have trouble following the notation used by Edwards. It may not be satisfying to the mathematical purist, but I believe it is effective enough to communicate my meaning. My questions pertain to Lemma V.2.2, and in particular part (c). First I must give enough context to make it meaningful, and to explain my notation.
The first question: have I got this right? Edit: Barring typos, I believe my derivation is correct. So the question remains:
The second question: is there a shorter, more direct means of proving the assertion that $\overset{*}{\left[\mathbf{d\omega}\right]}=\mathbf{d}\overset{*}{\mathbf{\omega}}.$
Here's the background:
Edit to add:
Square brackets indicate either parameter lists, or matrices, depending on context. For example: $$\mathfrak{\mathfrak{x}\left[\mathfrak{u}\right]}=\begin{bmatrix}x\left[\mathfrak{u}\right]\\ y\left[\mathfrak{u}\right] \end{bmatrix}\in\mathcal{X}\text{ and }\mathfrak{u}=\begin{bmatrix}\mu\\ u \end{bmatrix}\in\mathcal{U},$$ Means a vector valued "function" of a vector argument equals a two component, real number column vector ($2\times{1}$ matrix) each component of which is a single valued function of the vector $\mathfrak{u}$, etc.
The expression $\overset{*}{\left[\mathbf{d\omega}\right]}$ means that $*$ applies to the entire sub-expression between the square brackets.
The "back-arrows" indicate objects for which I have no formal name, but seemed interesting enough to call attention to. They might be called the $\mathcal{U}$-space components of $\overset{*}{\mathbf{\omega}}.$
$$\overset{*}{\mathbf{\omega}}=\overleftarrow{P}\mathbf{d\mu}+\overleftarrow{Q}\mathbf{d\nu}.$$
Introduce the $\mathscr{C}^{1}$ mapping $\mathfrak{x}:\mathbb{R}_{\mathcal{U}}^{2}\to\mathbb{R}_{\mathcal{X}}^{2}$, where $\mathfrak{\mathfrak{x}\left[\mathfrak{u}\right]}=\begin{bmatrix}x\left[\mathfrak{u}\right]\\ y\left[\mathfrak{u}\right] \end{bmatrix}\in\mathcal{X}$ and $\mathfrak{u}=\begin{bmatrix}\mu\\ u \end{bmatrix}\in\mathcal{U}$.
The pullback $\overset{*}{\phi}:\mathcal{U}\to\mathbb{R}$ of the 0-form $\phi:\mathcal{X}\to\mathbb{R}$ is given by $$ \overset{*}{\phi}\equiv\phi\circ\mathfrak{x}. $$ That is $\overset{*}{\phi}\left[\mathfrak{u}\right]=\phi\left[x\left[\mu,\nu\right],\left[\mu,\nu\right]\right].$
The pullback $\overset{*}{\omega}$ of the 1-form $\mathbf{\omega}_{\mathfrak{x}}=P\left[\mathfrak{x}\right]\mathbf{dx}+Q\left[\mathfrak{x}\right]\mathbf{dy}$ is given by $$ \overset{*}{\mathbf{\omega}}\equiv\overset{*}{P}\overset{*}{\mathbf{dx}}+\overset{*}{Q}\overset{*}{\mathbf{dy}}. $$
The pullbacks of $\mathbf{dx}$ and $\mathbf{dy}$ are given by $$ \overset{*}{\mathbf{dx}}\equiv\frac{\partial x}{\partial\mu}\mathbf{d\mu}+\frac{\partial x}{\partial\nu}\mathbf{d\nu}, $$
$$ \overset{*}{\mathbf{dy}}\equiv\frac{\partial y}{\partial\mu}\mathbf{d\mu}+\frac{\partial y}{\partial\nu}\mathbf{d\nu}. $$
So, written out in detail $$ \overset{*}{\mathbf{\omega}}_{\mathfrak{u}}=P\left[\mathfrak{x}\left[\mathfrak{u}\right]\right]\left(\frac{\partial x}{\partial\mu}\mathbf{d\mu}+\frac{\partial x}{\partial\nu}\mathbf{d\nu}\right)_{\mathfrak{u}}+Q\left[\mathfrak{x}\left[\mathfrak{u}\right]\right]\left(\frac{\partial y}{\partial\mu}\mathbf{d\mu}+\frac{\partial y}{\partial\nu}\mathbf{d\nu}\right)_{\mathfrak{u}}. $$
Or
$$ \overset{*}{\mathbf{\omega}}=\left(\overset{*}{P}\frac{\partial x}{\partial\mu}+\overset{*}{Q}\frac{\partial y}{\partial\mu}\right)\mathbf{d\mu}+\left(\overset{*}{P}\frac{\partial x}{\partial\nu}+\overset{*}{Q}\frac{\partial y}{\partial\nu}\right)\mathbf{d\nu}. $$
$$ \overset{*}{\mathbf{\omega}}=\overleftarrow{P}\mathbf{d\mu}+\overleftarrow{Q}\mathbf{d\nu}. $$
The products $\overset{*}{\mathbf{dx}}\overset{*}{\mathbf{dx}},\overset{*}{\mathbf{dx}}\overset{*}{\mathbf{dy}},$ etc., are given by $$ \overset{*}{\mathbf{dx}}\overset{*}{\mathbf{dx}}\equiv\overset{*}{\mathbf{dy}}\overset{*}{\mathbf{dy}}\equiv0, $$
$$ \overset{*}{\mathbf{dx}}\overset{*}{\mathbf{dy}}\equiv-\overset{*}{\mathbf{dy}}\overset{*}{\mathbf{dx}}, $$
$$ \overset{*}{\mathbf{dx}}\overset{*}{\mathbf{dy}}=\left(\frac{\partial x}{\partial\mu}\mathbf{d\mu}+\frac{\partial x}{\partial\nu}\mathbf{d\nu}\right)\left(\frac{\partial y}{\partial\mu}\mathbf{d\mu}+\frac{\partial y}{\partial\nu}\mathbf{d\nu}\right) $$
$$ =\left(\frac{\partial x}{\partial\mu}\frac{\partial y}{\partial\nu}-\frac{\partial x}{\partial\nu}\frac{\partial y}{\partial\mu}\right)\mathbf{d\mu}\mathbf{d\nu} $$
$$ =\left|\left|\frac{d\mathfrak{x}}{d\mathfrak{u}}\right|\right|\mathbf{d\mu}\mathbf{d\nu}. $$
The pullback $\overset{*}{\alpha}$ of the 2-form $\mathbf{\alpha}_{\mathfrak{x}}=a\left[\mathfrak{x}\right]\mathbf{dx}\mathbf{dy}$ is given by $$ \overset{*}{\mathbf{\alpha}}\equiv\overset{*}{a}\left|\left|\frac{d\mathfrak{x}}{d\mathfrak{u}}\right|\right|\mathbf{d\mu}\mathbf{d\nu}. $$
Where $\left|\left|\frac{d\mathfrak{x}}{d\mathfrak{u}}\right|\right|$ is the absolute value of the determinant of the transformation's differential. So
$$ \overset{*}{\mathbf{\alpha}}_{\mathfrak{u}}=a\left[\mathfrak{x}\left[\mathfrak{u}\right]\right]\left(\overset{*}{\mathbf{dx}}\overset{*}{\mathbf{dy}}\right)_{\mathfrak{u}}. $$
The differential $\mathbf{d\omega}$ of the 1-form $\mathbf{\omega}=P\mathbf{dx}+Q\mathbf{dy},$ is given by $$ \mathbf{d\omega}\equiv\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\mathbf{dx}\mathbf{dy}. $$
The objective:
With the assumption that the mapping $\mathfrak{x}$ is orientation-preserving, Edwards leaves the proof that $\overset{*}{\left[\mathbf{d\omega}\right]}=\mathbf{d}\overset{*}{\mathbf{\omega}}$ to his reader. So here is what I came up with:
First pullback the 2-form $\mathbf{d\omega}\equiv\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\mathbf{dx}\mathbf{dy}$.
$$ \overset{*}{\left[\mathbf{d\omega}\right]}_{\mathfrak{u}}=\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\left[\mathfrak{x}\left[\mathfrak{u}\right]\right]\left|\left|\frac{d\mathfrak{x}}{d\mathfrak{u}}\right|\right|\mathbf{d\mu}\mathbf{d\nu}. $$
Using the chain rule for partial differentiation gives
$$ \overset{*}{\left[\mathbf{d\omega}\right]}=\left(\left(\frac{\partial\overset{*}{Q}}{\partial\mu}\frac{\partial\mu}{\partial x}+\frac{\partial\overset{*}{Q}}{\partial\nu}\frac{\partial\nu}{\partial x}\right)-\left(\frac{\partial\overset{*}{P}}{\partial\mu}\frac{\partial\mu}{\partial y}+\frac{\partial\overset{*}{P}}{\partial\nu}\frac{\partial\nu}{\partial y}\right)\right)\left|\left|\frac{d\mathfrak{x}}{d\mathfrak{u}}\right|\right|\mathbf{d\mu}\mathbf{d\nu}. $$
The inverse of a matrix is given by the transpose of the matrix of cofactors devided by the determinant of the original matrix.
$$ \begin{bmatrix}\frac{\partial x}{\partial\mu} & \frac{\partial x}{\partial\nu}\\ \frac{\partial y}{\partial\mu} & \frac{\partial y}{\partial\nu} \end{bmatrix}^{-1}=\frac{1}{\left|d\mathfrak{x}/d\mathfrak{u}\right|}\begin{bmatrix}\frac{\partial y}{\partial\nu} & -\frac{\partial x}{\partial\nu}\\ -\frac{\partial y}{\partial\mu} & \frac{\partial x}{\partial\mu} \end{bmatrix}. $$
It also happenes to be the case that $\frac{d}{d\mathfrak{x}}\left[\mathfrak{x}\left[\mathfrak{u}\left[\mathfrak{x}\right]\right]\right]=\left\{ \frac{\partial x^{i}}{\partial u^{b}}\frac{\partial u^{b}}{\partial x^{j}}\right\} =\left\{ \delta_{j}^{i}\right\} .$ So
$$ \begin{bmatrix}\frac{\partial x}{\partial\mu} & \frac{\partial x}{\partial\nu}\\ \frac{\partial y}{\partial\mu} & \frac{\partial y}{\partial\nu} \end{bmatrix}\begin{bmatrix}\frac{\partial\mu}{\partial x} & \frac{\partial\mu}{\partial y}\\ \frac{\partial\nu}{\partial x} & \frac{\partial\nu}{\partial y} \end{bmatrix}=\left\{ \delta_{j}^{i}\right\} . $$
Since the inverse is unique, it follows that $$ \begin{bmatrix}\frac{\partial y/\partial\nu}{\left|d\mathfrak{x}/d\mathfrak{u}\right|} & -\frac{\partial x/\partial\nu}{\left|d\mathfrak{x}/d\mathfrak{u}\right|}\\ -\frac{\partial y/\partial\mu}{\left|d\mathfrak{x}/d\mathfrak{u}\right|} & \frac{\partial x/\partial\mu}{\left|d\mathfrak{x}/d\mathfrak{u}\right|} \end{bmatrix}=\begin{bmatrix}\frac{\partial\mu}{\partial x} & \frac{\partial\mu}{\partial y}\\ \frac{\partial\nu}{\partial x} & \frac{\partial\nu}{\partial y} \end{bmatrix}. $$
Because the mapping $\mathfrak{x}$ is orientation-preserving, the determinant of its derivative is positive.
$$ \overset{*}{\left[\mathbf{d\omega}\right]}=\left(\left(\frac{\partial\overset{*}{Q}}{\partial\mu}\frac{\partial y/\partial\nu}{\left|d\mathfrak{x}/d\mathfrak{u}\right|}-\frac{\partial\overset{*}{Q}}{\partial\nu}\frac{\partial y/\partial\mu}{\left|d\mathfrak{x}/d\mathfrak{u}\right|}\right)-\left(-\frac{\partial\overset{*}{P}}{\partial\mu}\frac{\partial x/\partial\nu}{\left|d\mathfrak{x}/d\mathfrak{u}\right|}+\frac{\partial\overset{*}{P}}{\partial\nu}\frac{\partial x/\partial\mu}{\left|d\mathfrak{x}/d\mathfrak{u}\right|}\right)\right)\left|\left|\frac{d\mathfrak{x}}{d\mathfrak{u}}\right|\right|\mathbf{d\mu}\mathbf{d\nu}. $$
$$ =\left(\left(\frac{\partial\overset{*}{Q}}{\partial\mu}\frac{\partial y}{\partial\nu}-\frac{\partial\overset{*}{Q}}{\partial\nu}\frac{\partial y}{\partial\mu}\right)+\left(\frac{\partial\overset{*}{P}}{\partial\mu}\frac{\partial x}{\partial\nu}-\frac{\partial\overset{*}{P}}{\partial\nu}\frac{\partial x}{\partial\mu}\right)\right)\mathbf{d\mu}\mathbf{d\nu}. $$
Now write out $\mathbf{d}\overset{*}{\mathbf{\omega}}$
$$ \mathbf{d}\overset{*}{\mathbf{\omega}}=\left(\frac{\partial\overleftarrow{Q}}{\partial\mu}-\frac{\partial\overleftarrow{P}}{\partial\nu}\right)\mathbf{d\mu}\mathbf{d\nu} $$
$$ =\left(\frac{\partial}{\partial\mu}\left[\overset{*}{P}\frac{\partial x}{\partial\nu}+\overset{*}{Q}\frac{\partial y}{\partial\nu}\right]-\frac{\partial}{\partial\nu}\left[\overset{*}{P}\frac{\partial x}{\partial\mu}+\overset{*}{Q}\frac{\partial y}{\partial\mu}\right]\right)\mathbf{d\mu}\mathbf{d\nu} $$
$$ =\left(\left(\frac{\partial\overset{*}{P}}{\partial\mu}\frac{\partial x}{\partial\nu}+\overset{*}{P}\frac{\partial^{2}x}{\partial\mu\partial\nu}+\frac{\partial\overset{*}{Q}}{\partial\mu}\frac{\partial y}{\partial\nu}+\overset{*}{Q}\frac{\partial^{2}y}{\partial\mu\partial\nu}\right)-\left(\frac{\partial\overset{*}{P}}{\partial\nu}\frac{\partial x}{\partial\mu}+\overset{*}{P}\frac{\partial^{2}x}{\partial\nu\partial\mu}+\frac{\partial\overset{*}{Q}}{\partial\nu}\frac{\partial y}{\partial\mu}+\overset{*}{Q}\frac{\partial^{2}y}{\partial\nu\partial\mu}\right)\right)\mathbf{d\mu}\mathbf{d\nu} $$
$$ =\left(\left(\frac{\partial\overset{*}{P}}{\partial\mu}\frac{\partial x}{\partial\nu}+\frac{\partial\overset{*}{Q}}{\partial\mu}\frac{\partial y}{\partial\nu}\right)-\left(\frac{\partial\overset{*}{P}}{\partial\nu}\frac{\partial x}{\partial\mu}+\frac{\partial\overset{*}{Q}}{\partial\nu}\frac{\partial y}{\partial\mu}\right)\right)\mathbf{d\mu}\mathbf{d\nu} $$
$$ =\left(\left(\frac{\partial\overset{*}{Q}}{\partial\mu}\frac{\partial y}{\partial\nu}-\frac{\partial\overset{*}{Q}}{\partial\nu}\frac{\partial y}{\partial\mu}\right)+\left(\frac{\partial\overset{*}{P}}{\partial\mu}\frac{\partial x}{\partial\nu}-\frac{\partial\overset{*}{P}}{\partial\nu}\frac{\partial x}{\partial\mu}\right)\right)\mathbf{d\mu}\mathbf{d\nu} $$
$$ =\overset{*}{\left[\mathbf{d\omega}\right]} $$
as advertised.