Given two k-surfaces $\sigma:[-a, a]^k\rightarrow\mathbb{R}^n$ and $\sigma':[-b, b]^k\rightarrow\mathbb{R}^n$, if $\sigma([-a, a]^k) =\sigma'([-b, b]^k)$ and $\exists\, g\, :\sigma'=\sigma\circ g$, for all k-forms $\omega$ it is: $$\int_{\sigma} \omega=\int_{\sigma'} \omega$$ So, the proof states that $$\int_{\sigma} \omega=\int_{[-b, b]^k} \omega(\partial_{z_1}(\sigma\circ\omega),\dots,\partial_{z_k}(\sigma\circ\omega))dz_1\dots dz_k$$ And it is perfect, now we can, given $k$ numbers $\{i_1,\dots,i_k\}$, define $k$ functions $$\bar{\sigma} (x) =(\bar{\sigma}_1(x),\dots,\bar{\sigma}_k(x)):=(\sigma_{i_1}(x),\dots,\sigma_{i_k}(x))$$ and the same for $\bar{\sigma}'$. Now it is: $$dx_{i_1}\wedge\dots\wedge dx_{i_k} (\partial_{z_1}(\sigma\circ\omega),\dots,\partial_{z_k}(\sigma\circ\omega))=det(D(\bar{\sigma} \circ g)) =det((D\bar{\sigma}) \circ gDg) $$
I don't understand why is necessary to define $\bar{\sigma}$ and what does it means the definition. And, in consequence, I can't understand the two equality above.
Edit:
g is an invertible function $g:[-b, b]^k \rightarrow [-a, a] ^k$ and $detDG>0$ as said in a comment.
I think that the subscripts $i_j$ indicate somehow a "position", but I'm not sure what is the meaning of this in this case.
Thanks and sorry for bad english