baby rudin, chapter 10, (differential forms) theorem 10.27 I'm having difficulties with the reasoning in the proof of theorem 10.27 (regarding integration over oriented simplexes).
say $\sigma=[p_0,p_1,\dots,p_j,\dots,p_k],\bar{\sigma}=[p_j,p_1,\dots,p_0,\dots,p_k]$ ($\bar{\sigma}$ is obtained from $\sigma$ by interchanging $p_0,p_j$) and $\omega=f({\bf x}) dx_{i_1} \wedge dx_{i_2} \wedge \dots \wedge dx_{i_k}$ (the proof will then follow from linearity), I want to show that $\int_{\bar{\sigma}} \omega=-\int_{\sigma} \omega$.
I can see why the Jacobians $J_\bar{\sigma},J_\sigma$ have opposite signs (the column argument given in the text), which gives:
$\int_\bar{\sigma} \omega=\int_{Q^k}f(\bar \sigma( {\bf u})) J_\bar{\sigma} d {\bf u}=-\int_{Q^k} f(\bar{\sigma}({\bf u})) J_\sigma d {\bf u}$
I only miss the last part, where in the last expression $f(\bar \sigma({\bf u}))$ becomes $f(\sigma( {\bf u}))$.
if it matters, both Jacobians are constant, and can be moved outside the integrals.
Thanks in advance,
Michael
 A: For starters, we will solve the equation $\bar{\sigma}(u)=\sigma(v)$.
Say $u=u_1 e_1+\dots+u_k e_k$ and $v=v_1e_1+\dots+v_k e_k$, it follows that $\bar{\sigma}(u)=p_j+u_j(p_0-p_j)+\sum_{1 \le i \le k,i \neq j} u_i(p_i-p_j)$ and $\sigma(v)=p_0+\sum_{1 \le i \le k} v_i (p_i-p_0)$. Equating the two gives:
$(p_j-p_0)+u_j (p_0-p_j)+\sum_{1 \le i \le k,i \neq j} u_i(p_i-p_0+p_0-p_j)-\sum_{1 \le i \le k} v_i (p_i-p_0)=0$
$(1-u_j-\sum_{1 \le i \le k,i \neq j} u_i-v_j)(p_j-p_0)+\sum_{1 \le i \le k,i \neq j} (u_i-v_i)(p_i-p_0)=0$
A possible solution for $v$ is $v=\sum_{1 \le i \le k,i \neq j} u_i e_i+(1-\sum_{1 \le i \le k} u_i)e_j =:G(u)$. 
Effectively, we have shown that $\bar{\sigma}=\sigma \circ G$, for a $C^1$-primitive $G$ from the simplex to itself. Theorem 10.9 (change of variables) gives the result for functions with "well behaved" supports (notice that $|J_G|=|-1|=1$). The general case follows by approximating $f$ with such functions (or, equivalently using a stronger version of theorem 10.9).
I hope it's ok.
