Theorem 10.22 Rudin

Question

Here's part of the proof of theorem 10.22 Theorem 10.22 from PMA RUdin

The rest part is

Assume now $\omega=fdy_{I}$. Then $\omega_{T}=f_{T}(x)(dy_{I})_{T}$

the preceding calculation leads to $$d(\omega_{T})=d(f_{T}) \land (dy_{I})_{T}=(df)_{T} \land (dy_{I})_{T}=((df) \land dy_{I})_{T}=(d\omega)_{T}$$

My question is that, Rudin said the first equality of the above equation follows from $(70)$, can someone help me explain why? $(70)$ said $d((dy_{I})_{T})=0$, how do we get the first equality?

Thanks in advance!

Answer 1

You know from a previous theorem (see your post Theorem 10.20 Rudin) that $$d(\omega_T)=d(f_T (dy_I)_T)=d(f_T)\wedge (dy_I)_T+f_T\wedge d((dy_I)_T),$$ and since $d((dy_I)_T)=0,$ we know that $d(\omega_T)=d(f_T)\wedge (dy_I)_T.$