My question relates to the problem discussed here: https://math.stackexchange.com/a/3429032/342834 I will propose a different approach based on the proposition that a pull-back is really nothing more than a composition of mappings.
I admit that I find pull-backs confusing. They appear to be a complicated way of doing something I have done for years without ever uttering the term pull-back. But, I really want to understand the literature where they appear. And learning a new persepctive is probably good.
Background
I begin with some background in order to communicate what pull-backs are to me.
Assume everything is $\mathscr{C}^{\text{enough}}.$ This is my understanding of how pull-backs work for one-forms.
In this lecture, https://www.youtube.com/watch?v=q8s14TJVTEk Dr. Shifrin presents a discussion of the proposition that the pull-back of the differential of a function is the differential of the pull-back. This is my poetically licensed interpretation. Let there be
$$\begin{aligned} \vec{y}&:\mathbb{R}^{n}\to\mathbb{R}^{m}\text{; }\mathbf{x}=\left\{ \mathrm{x}^{\iota}\right\} \in\mathbb{R}^{n};\\ f&:\mathbb{R}^{m}\to\mathbb{R}\text{ ; }\mathbf{y}=\left\{ \mathrm{y}^{i}\right\} \in\mathbb{R}^{m}\text{ ; and }\\ f\circ\vec{y}&:\mathbb{R}^{n}\to\mathbb{R}. \end{aligned}$$
Using "the grunge found in physics and engineering," I write, with Einstein summation
$$ df=\frac{\partial f}{\partial\mathrm{y}^{i}}dy^{i}=\frac{\partial f}{\partial\mathrm{y}^{i}}\frac{\partial y^{i}}{\partial\mathrm{x}^{\iota}}dx^{\iota}=\frac{df}{d\mathbf{y}}\frac{d\vec{y}}{d\mathbf{x}}d\vec{x}, $$
and wonder why anybody talks about pull-backs.
Note: IIRC, C. H. Edwards intends a determinant when writing $\frac{df}{d\mathbf{y}}.$ Following Wolfgang Pauli, I intend the obvious matrix. Roman script $\mathrm{x}^{\iota}$ and bold-faced $\mathbf{x}$ indicate parameter names, or free elements of their respective sets. Italic $x^{\iota},$ and over-arrow $\vec{x}$ indicate dependent values (functions and mappings).
Show the differential of the pull-back is the pull-back of the differential
Now, for a (single valued) function, the pull-back really is the composition of the function $f$ with the transformation $\vec{y}.$ So
$$ \vec{y}^{*}f=f\circ\vec{y}. $$
The differential of the pullback is
$$ d\left(\vec{y}^{*}f\left(\mathbf{x}\right)\right)=d\left(f\circ\vec{y}\left(\mathbf{x}\right)\right)=\frac{\partial f}{\partial\mathrm{y}^{i}}\left(\vec{y}\left(\mathbf{x}\right)\right)\frac{\partial y^{i}}{\partial\mathrm{x}^{\iota}}\left(\mathbf{x}\right)dx^{\iota}. $$
And the pullback of the differential is (with a bit of abbreviation)
$$ \vec{y}^{*}df=\vec{y}^{*}\left(\frac{\partial f}{\partial\mathrm{y}^{i}}dy^{i}\right)=\left(\frac{\partial f}{\partial\mathrm{y}^{i}}\circ\vec{y}\right)\vec{y}^{*}dy^{i}=\frac{\partial f}{\partial\mathrm{y}^{i}}\left(\vec{y}\right)\frac{\partial y^{i}}{\partial\mathrm{x}^{\iota}}dx^{\iota}. $$
If any of the above reveals a misunderstanding, please correct me.
The Integral of a one-form
Given the curve $\vec{\gamma}:\left[t_{0},t_{L}\right]\to\mathbb{R}^{n}$ and the differential one-form field (in my grungy notation, where $\backepsilon$ means such that)
$$ \tilde{\omega}=\omega_{\iota}dx^{\iota}\backepsilon\omega_{\iota}:\vec{\gamma}\to\mathbb{R}, $$
the integral along $\vec{\gamma}$ is defined in equation V1-14 of Edwards's Advance Calculus of Several Variables as
$$ \int_{\vec{\gamma}}\tilde{\omega}=\int_{t_{o}}^{t_{L}}\tilde{\omega}_{\vec{\gamma}\left(t\right)}\left(\vec{\gamma}^{\prime}\left(t\right)\right)d\mathit{t}. $$
Again using my Kurt Cobain notation
$$ \int_{\vec{\gamma}}\tilde{\omega}=\int_{t_{o}}^{t_{L}}\omega_{\iota}dx^{\iota}\left(\vec{\gamma}^{\prime}\right)d\mathit{t}=\int_{t_{o}}^{t_{L}}\omega_{\iota}\gamma^{\iota\prime}d\mathit{t}. $$
The exercise where my question arose
Here's my restatement of the problem linked above. Given the continuous one-form $\tilde{\omega}$ and the $\mathscr{C}^{1}$ mappings
$$ \vec{\mathit{r}}:\mathbb{R}_{\mathfrak{u}}^{2}\leftrightarrow\mathbb{R}_{\mathfrak{r}}^{2}\text{ and }\vec{\gamma}:\left[t_{0},t_{L}\right]\to\mathbb{R}_{\mathfrak{u}}^{2}, $$
show that
$$ \int_{\vec{\gamma}}\vec{\mathit{r}}^{*}\tilde{\omega}=\int_{\vec{\mathit{r}}\circ\vec{\gamma}}\tilde{\omega}. $$
Treating the pull-back of a one-form as a composition of mappings
Since $\vec{\mathit{r}}\circ\vec{\gamma}=\vec{\rho}:\left[t_{0},t_{L}\right]\to\mathbb{R}_{\mathfrak{r}}^{2}$ is $\mathscr{C}^{1}$ it is a curve in $\mathbb{R}_{\mathfrak{r}}^{2}$ parameterized by $t\in\left[t_{o},t_{L}\right].$ Now, this is the essence of my question: It seems correct to equate the $\vec{\mathit{r}}$-pull-back of $\tilde{\omega}$ with the composition of mappings as
$$ \vec{\mathit{r}}^{*}\tilde{\omega}=\tilde{\omega}\circ\vec{\mathit{r}}=\left(\omega_{i}d\mathit{r}^{i}\right)\circ\vec{\mathit{r}}=\omega_{i}\circ\vec{\mathit{r}}\frac{\partial\mathit{r}^{i}}{\partial u^{j}}d\mathit{u}^{j}. $$
So I write
$$ \int_{\vec{\gamma}}\vec{\mathit{r}}^{*}\tilde{\omega}=\int_{\vec{\gamma}}\tilde{\omega}\circ\vec{\mathit{r}}. $$
I'm thinking of the one-form $\tilde{\omega}$ as a covariant tensor field whose components are point-functions of position in $\mathbb{R}_{\mathfrak{r}}^{2}$. The pull-back is converting the one-form to a point-function of position in $\mathbb{R}_{\mathfrak{u}}^{2}.$ So not only do we need to transform the components, we also need to transform the basis which consists of the coordinate differentials $d\mathit{r}^{i}.$
At the same time, I am thinking of $\tilde{\omega}$ and its transformed version $\tilde{\omega}\circ\vec{\mathit{r}}$ as a geometric object which I shouldn't have to "look inside of" in order to integrate it. That is, if we accept the expression of the pull-back as a composition, then all that matters is what it means to integrate $f\circ\vec{\mathit{r}}$ along $\vec{\gamma}.$ Where $f$ is anything integrable over the relevant domain.
Evidently have
$$ \int_{\vec{\gamma}}\tilde{\omega}\circ\vec{\mathit{r}}=\int_{\vec{\rho}}\tilde{\omega}=\int_{\vec{\mathit{r}}\circ\vec{\gamma}}\tilde{\omega}. $$
Which by the above handwaving, reduces to the problem of proving
$$ \int_{\vec{\gamma}}f\circ\vec{\mathit{r}}=\int_{\vec{\mathit{r}}\circ\vec{\gamma}}f, $$
where $f$ is anything integrable over the curve $\vec{\mathit{r}}\circ\vec{\gamma}.$
My question:
Is $\vec{\mathit{r}}^{*}\tilde{\omega}=\tilde{\omega}\circ\vec{\mathit{r}},$ which reduces the problem to proving $\int_{\vec{\gamma}}f\circ\vec{\mathit{r}}=\int_{\vec{\mathit{r}}\circ\vec{\gamma}}f,$ where $f$ is integrable?
Proof that looks inside of $\tilde{\omega}$
The following is what I believe to be a valid proof of the assertion made in the exercise. But does not use the abstraction suggested above. $$\begin{aligned} \vec{\rho}^{\prime}&=\frac{\partial\vec{\mathit{r}}}{\partial\mathrm{u}^{j}}\gamma^{j\prime}=\left\{ \rho^{i\prime}\right\} =\left\{ \frac{\partial\mathit{r}^{i}}{\partial\mathrm{u}^{j}}\gamma^{j\prime}\right\}\\ \omega_{i}\circ\vec{\mathit{r}}\circ\vec{\gamma}&=\omega_{i}\circ\vec{\rho}=\mathit{w}_{i}:\left[t_{o},t_{L}\right]\to\mathbb{R}\\ \mathit{w}_{i}\left(t\right)&=\omega_{i}\circ\vec{\rho}\left(t\right)=\omega_{i}\circ\vec{\mathit{r}}\circ\vec{\gamma}\left(t\right)\\ \int_{\vec{\gamma}}\vec{\mathit{r}}^{*}\tilde{\omega}&=\int_{\vec{\gamma}}\tilde{\omega}\circ\vec{\mathit{r}}\\ &=\int_{t_{o}}^{t_{L}}\mathit{w}_{i}\frac{\partial\mathit{r}^{i}}{\partial u^{j}}d\mathit{u}^{j}\left(\vec{\gamma}^{\prime}\right)\\ &=\int_{t_{o}}^{t_{L}}\mathit{w}_{i}\frac{\partial\mathit{r}^{i}}{\partial u^{j}}\gamma^{j\prime}\\ &=\int_{t_{o}}^{t_{L}}\mathit{w}_{i}\rho^{i\prime}\\ &=\int_{t_{o}}^{t_{L}}\tilde{\omega}_{\vec{\rho}}\left(\vec{\rho}^{\prime}\right)\\ &=\int_{\vec{\rho}}\tilde{\omega}\\ &=\int_{\vec{\mathit{r}}\circ\vec{\gamma}}\tilde{\omega}. \end{aligned}$$
From Hermann Weyl's Space-Time-Matter published in 1922:
Various attempts have been made to set up a standard terminology in this branch of mathematics involving only the vectors themselves and not their components, analogous to that of vectors in vector analysis. This is highly expedient in the latter, but very cumbersome for the much more complicated framework of the tensor calculus. In trying to avoid continual reference to the components we are obliged to adopt an endless profusion of names and symbols in addition to an intricate set of rules for carrying out calculations, so that the balance of advantage is considerably on the negative side. An emphatic protest must be entered against these orgies of formalism which are threatening the peace of even the technical scientist.