# Stokes' Theorem from PMA Rudin. Confusing moment with simplex

I am reading the proof of Stokes' theorem from PMA Rudin but one moment seems to very weird.

Why Rudin considers the case when $\sigma=[0,\mathbf{e}_1,\dots, \mathbf{e}_k]$? After all $\sigma$ may take the form $[\mathbf{p}_0, \mathbf{p}_1,\dots, \mathbf{p}_k]$.

Can anyone explain this important moment?

I can't understand it by myself.

• He explains it in the following lines. With theorems 10.22 and 10.25, you can pull back the general case to that special case, so it suffices to prove it for the standard simplex. – Daniel Fischer May 7 '16 at 19:01
• @DanielFischer, Sorry but how theorems 10.22 and 10.25 can be used? In What sense? – ZFR May 7 '16 at 19:07
• Pull back via $T$. By 10.22, we have $(d\omega)_T = d(\omega_T)$. By 10.25, we have $\int_{\Phi} d\omega = \int_{\sigma} (d\omega)_T$ and $\int_{\partial \Phi} \omega = \int_{T(\partial \sigma)} \omega = \int_{\partial \sigma} \omega_T$. – Daniel Fischer May 7 '16 at 19:11
• I understood that moment. But I am interesting why he considers the case when simplex have form $\sigma=[0,\mathbf{e}_1,..,\mathbf{e}_k]$ but not some $\sigma=[\mathbf{p}_0,\mathbf{p}_1,..,\mathbf{p}_k]$. – ZFR May 7 '16 at 19:13
• The main question is: Why we can represent $\Phi$ as $T\circ \sigma$ where $\sigma=[0,e_1,\dots, e_k]$. That worries me :/ – ZFR May 7 '16 at 19:17