Theorem 10.22 from PMA RUdin 
We know that $(dy_I)_T=dt_{i_1}\land \dots \land dt_{i_k}$ and using definition 10.18 we get $$d((dy_{I})_{T})=d1\land dt_{i_1}\land \dots \land dt_{i_k}=0$$ since $dc=0$ for any $c\in \mathbb{R}^1$. 
Are my reasonings true? Note that in my method i didn't use that $T\in C''$ and theorem 10.20.
I would be v ery grateful for any answer.
 A: Following my comment, observe that each $t_{i_j}$ is a function so that $dt_{i_1}\wedge\cdots \wedge dt_{i_k}$ is not of the form $\sum_J b_J(x)dx_J$.
Let's take an example to understand that point with $T:\mathbb R^2\rightarrow \mathbb R^3$ given by $$T(x)=(x_1,x_2)=(x_1x_2,x_1+x_2,x_1^3)$$
so that $$t_1(x)=x_1x_2, \ t_2(x)=x_1+x_2, \ t_3(x)=x_1^3.$$
Hence, $$\left\{\begin{array}{l}dt_1=x_2dx_1+x_1dx_2 \\ dt_2=dx_1+dx_2 \\ dt_3=3x_1^2dx_1\end{array}\right.$$
and $$dt_1\wedge dt_3 = \left(x_2dx_1+x_1dx_2\right)\wedge \left( 3x_1^2dx_1\right) = 3x_1^3dx_2\wedge dx_1$$ which is of the form $\sum_{J} b_J(x)dx_J$ so you can apply rule 10.18 : $$\begin{align*}
d(dt_1\wedge dt_3) & =d\left(3x_1^3dx_2\wedge dx_1\right)\\
& =6x_1^2dx_1\wedge dx_2\wedge dx_1\\
& =-6x_1^2dx_1\wedge dx_1\wedge dx_2\\
& = 0\end{align*}$$
since $dx_i\wedge dx_j=-dx_j\wedge dx_i$.
But, Rudin is actually saying that, using theorem 10.20 (a) and (b), one can directly notice that :
$$\begin{align*}
d(dt_1\wedge dt_3) & =d(dt_1)\wedge dt_3 - dt_1\wedge d(dt_3) \\
& = 0\wedge dt_3 - dt_1\wedge 0 \\ & =0\end{align*}$$
where the minus comes from the 1-form $dt_1$.
It works the same with any $T:\mathbb R^n\rightarrow \mathbb R^m$ and $dt_{_i1}\wedge\cdots\wedge dt_{i_k}$ instead of $dt_1\wedge dt_3$.
