Show that the path integral of the pullback of a 1-form is equal to the integral of the 1-form over the transformation image of the curve This is from Edwards's Advanced Calculus of Several Variables, Exercise V2.15, page 321.  

Let $\omega$ be a continuous 1-form and $\gamma:[a,b]\to\mathbb{R}^{2}$
  be $\mathscr{C}^{1}.$ Show that
$$
\int_{\gamma}F^{\ast}\omega=\int_{F\circ\gamma}\omega.
$$

Based on discussions elsewhere in the text, I assume $F:\mathbb{R}_{uv}^{2}\to\mathbb{R}_{xy}^{2}$
is sufficiently well-behaved. So it is reasonable to write our 1-form
as
$$
\omega=\mathcal{P}\mathbf{d}x+\mathcal{Q}\mathbf{d}y,
$$
where $\mathcal{P},\mathcal{Q}:\mathbb{R}_{xy}^{2}\to\mathbb{R}.$
The pullback reparameterizes $\mathcal{P},\mathcal{Q}$ such that
$F^{\ast}\mathcal{P},F^{\ast}\mathcal{Q}:\mathbb{R}_{uv}^{2}\to\mathbb{R}.$
It also rewrites $\mathbf{d}x,\mathbf{d}y$ so that they take vectors
in $\mathbb{R}_{uv}^{2}$ as arguments. That is
$$
F^{\ast}\mathbf{d}x=\left(\frac{\partial\mathit{x}}{\partial u}\mathbf{d}u+\frac{\partial\mathit{x}}{\partial v}\mathbf{d}v\right),
$$
$$
F^{\ast}\mathbf{d}y=\left(\frac{\partial\mathit{y}}{\partial u}\mathbf{d}u+\frac{\partial\mathit{y}}{\partial v}\mathbf{d}v\right).
$$
So that the pullback of $\omega$ is
$$
F^{\ast}\omega=\mathcal{P}\circ F\left(\frac{\partial\mathit{x}}{\partial u}\mathbf{d}u+\frac{\partial\mathit{x}}{\partial v}\mathbf{d}v\right)+\mathcal{Q}\circ F\left(\frac{\partial\mathit{y}}{\partial u}\mathbf{d}u+\frac{\partial\mathit{y}}{\partial v}\mathbf{d}v\right)
$$
From this it is intuitively obvious that the assertion of the exercise
holds. That is, at every point along the path $\gamma$ the components of the 1-form $\omega$ are evaluated using the image $F\left(\gamma\left(t\right)\right)$, and $\omega$ operates on the image of the velocity vector, which is the velocity vector of the image curve. 
But I'm not sure how to state this in a rigorous way.
The definition Edwards gives for the path integral of a 1-form is
$$
\int_{\gamma}\omega=\int_{a}^{b}\omega_{\gamma\left(t\right)}\left(\gamma^{\prime}\left(t\right)\right)dt.
$$
The only thing that comes to mind is appeal to the Riemann sum definition
of the integral.
How might the assertion of the exercise be demonstrated rigorously?
 A: I think it is better to do this using the general definitions, of the line integral and pullback.  Going to coordinates clutters everything up. 
$\int_{\gamma}\omega:=\int_{a}^{b}\omega_{\gamma\left(t\right)}\left(\gamma^{\prime}\left(t\right)\right)dt=\int_{[a,b]}\gamma^*\omega\Rightarrow \int_{F\circ\gamma}\omega=\int_{[a,b]}(F\circ \gamma)^* \omega=\int_{[a,b]}\gamma^*F^* \omega=\int_{\gamma}F^*\omega.$
To do the calculation rigorously from scratch, (with the understanding that I do not know how Edwards introduces these ideas) note that 
$1).\ \gamma'(t_0)$ operates on smooth functions $f$ as follows: $\gamma'(t_0)f:=\left(\frac{df\circ \gamma(t)}{dt}\right )\big |_{t=t_0}$. So to distinguish the operator from the ordinary derivative, we write the latter with a dot above the functional symbol. So, $\overset{.}{\gamma}(t)$ means the ordinary derivative of $\gamma$ at $t$.
$2).\ $ The symbol $x$ is the $\textit{function}:(x,y)\mapsto x$. That is, the projection onto the first coordinate. Similarly for $y$. 
$3).\ $ We have $\gamma(t)=(\gamma^1(t),\gamma^2(t))=(x,y)$ and the differentials $dx$ and $dy$ operate on $\gamma'(t)$ as follows: $dx(\gamma'(t))=x\circ \gamma(t)=\gamma^1(t)$ and similarly for $dy$.
$4).\ $ the action of $F^*$ on $\omega$ will be clear in what follows. 
To start the calculation, we have $\omega_{\gamma(t)}=P(\gamma(t))dx+Q(\gamma(t))dy.$ Now
$F^*\omega_{\gamma(t)}=P(F\circ\gamma(t))d(x\circ F)+Q(F\circ\gamma(t))d(y\circ F)=$
$P(F\circ\gamma(t))dF^1+Q(F\circ\gamma(t))dF^2$  so
$F^*\omega_{\gamma(t)}(\gamma'(t))=P(F\circ\gamma(t))dF^1(\gamma'(t))+Q(F\circ\gamma(t))dF^2(\gamma'(t))=$
$P(F\circ\gamma(t))\overset{.}{(F\circ\gamma})^1(t)+Q(F\circ\gamma(t))\overset{.}{(F\circ\gamma)^2}(t))$ so

$\int_{\gamma}F^*\omega=\int_{a}^{b}F^*\omega_{\gamma\left(t\right)}\left(\gamma^{\prime}\left(t\right)\right)dt=\int_{a}^{b}P(F\circ\gamma(t))\overset{.}{(F\circ\gamma}^1)(t)+Q(F\circ\gamma(t))\overset{.}{(F\circ\gamma)^2}(t))dt$

On the other hand, 
$\omega_{F\circ\gamma(t)}(F\circ\gamma)'(t)=P(F\circ\gamma(t))dx(F\circ\gamma)'(t)+Q(F\circ\gamma(t))dy(F\circ\gamma)'(t)=$
$P(F\circ\gamma(t))(F\circ\gamma)'(t)x+Q(F\circ\gamma(t))(F\circ\gamma)'(t))y=$
$P(F\circ\gamma(t))(\overset{.}{F\circ\gamma)}^1(t)+Q(F\circ\gamma(t))(\overset{.}{F\circ\gamma)}^2(t))$
so 

$\int_{F\circ\gamma}\omega=\int_{a}^{b}\omega_{F\circ\gamma\left(t\right)}\left(F\circ\gamma)^{\prime}\left(t\right)\right)dt=\int_{a}^{b}(P(F\circ\gamma(t))(\overset{.}{F\circ\gamma)}^1(t)+Q(F\circ\gamma(t))(\overset{.}{F\circ\gamma)}^2(t)))dt.$

The items in blockquotes are equal, which finishes the proof.
