In Baby Rudin's theorem 10.22, the proof requires that for $$d(\omega_T)=(d\omega)_T$$ we need $\omega \in \mathcal C'$ and a transformation $T\in \mathcal C"$ where the transform of the k-form $\omega=\sum_I \mathbf{b}(\mathbf{x})dy_I$ ($dy_I$ a basic k-form) is denoted and defined as $$\omega_T=\sum_I \mathbf{b}(T(\mathbf{x}))dt_I$$ and the $t$'s are the components of the transformation. My question is, why is this so?
The proof he gives relies on proving that the result is true for a 0-form $f$ (i.e. $d(f_T)=d(f_T)$). Then he uses a previously proved result that says that if the k-form $g\in \mathcal C"$ then $d^2g=0$ to show that if $dy_I=dy_{i_1}\wedge\cdots \wedge dy_{i_k}$ then $(dy_I)_T=dt_{i_1}\wedge\cdots \wedge dt_{i_k}$ and therefore that $d((dy_I)_T)=0$ since $T\in \mathcal C"$.
Here I do not see the need to invoke the previous result, is it not sufficient to use the definition of the derivative of a k-form to show that $$d((dy_I)_T)=(d1)\wedge dt_{i_1}\wedge\cdots \wedge dt_{i_k}=0$$?
And why would that previous result be applicable here since $T$ is not a $k$-form?
Rudin then goes on to say that if $\omega =fdy_I$ then $\omega_T=f_T(dy_I)_T$ and that to obtain $d(\omega_T)=d(f_T)\wedge (dy_I)_T$ we need the "wedge product rule" derived before. But why does this not follow directly from the definition of the derivative of differential forms? There is no wedge product in $\omega_T=f_T(dy_I)_T$ so that theorem should also not be applicable. What am I missing?
Change of variables with differential forms
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