I found what looks kind of like a proof that the pullback commutes with the exterior derivative, but it is so simple that I feel as if it is almost certainly wrong. Here is what I have:
Let $\omega = gdx_{1}\wedge\ldots\wedge dx_{k}$, so $d\omega = dg\wedge dx_{1}\wedge\ldots\wedge dx_{k}$ if $\omega$ is differentiable. Then we get that:
$$f^*(d\omega)= f^*(dg\wedge dx_{1}\wedge\ldots\wedge dx_{k}) $$
$$= d(g\circ f)\wedge df_{1}\wedge\ldots\wedge df_{k}$$
where $f_{i}$ is the $i$th component function of $f$. This last equality here seems a little sketchy to me, but if it does hold, we could then get the result very easily. I'm aware that $dg$ would behave somewhat differently to the $dx_{i}$, but it should still work out because we could simply pick a different basis for the exterior power (or should it)?
It seems like this follows immediately from the definition of a pullback, but I might be wrong here. Any help is appreciated.
Best Answer
They're using that $f^*$ distributes over wedge products and commutes with taking differentials of functions only. Namely, the steps being skipped are \begin{align}f^*({\rm d}g\wedge {\rm d}x_1\wedge\cdots\wedge {\rm d}x_k) &= f^*({\rm d}g)\wedge f^*({\rm d}x_1)\wedge\cdots\wedge f^*({\rm d}x_k) \\ &= {\rm d}(g\circ f)\wedge {\rm d}(x_1\circ f)\wedge\cdots\wedge {\rm d}(x_k\circ f) \\ &= {\rm d}(g\circ f)\wedge {\rm d}f_1\wedge\cdots\wedge {\rm d}f_k.\end{align}