I wonder if someone can explicitly construct a proof for the following statement mentioned in Rudin chapter 10 differential form:
If $T_1, T_2$ are injective mappings of the n-simplex $Q^n$ into $\mathbb{R}^n$ of class $C''$ whose Jacobian is positive. Then if $T_1(Q^n) = T_2(Q^n)$, the following is true for any $(n-1)$ form $\omega$:
$\int_{\partial T_1}\omega = \int_{\partial T_2}\omega$
Is it necessary that the Jacobian is positive as compared to being nonzero? Otherwise, with Stokes' Theorem, i.e. $\int_{\partial T}\omega = \int_{T}d\omega$, I assume that this would imply that the integral of a form over a set in $\mathbb{R}^n$ is independent of the choice of parametrization, i.e. $\int_{T_1}\lambda = \int_{T_2}\lambda$ if $T_1, T_2$ have the same image of $Q^n$? (there could be problem if $\lambda$ is not exact, so we cannot directly get the result from statement above, any alternative ways to prove this?)
Also referring to Rudin 10.45 Green's theorem, if $\Omega$ is a closed set in $\mathbb{R}^n$, is there always a way to find an injective differentiable simplex $T$ such that $T(Q^k) = \Omega$?
Yes as the sign of the Jacobian determines whether the mappings are orientation preserving or reversing: for example, consider the unit interval $Q^{1} = [0,1]$, and set
$$T_{1} : [0,1] \rightarrow [0,1],\, t \mapsto t; \qquad T_{2} : [0,1] \rightarrow [0,1],\, t \mapsto 1-t,$$
(so $T_{1}$ is just the identity on $[0,1]$), so are injective mappings of $Q^{1} \subset \mathbb{R}^{1}$ to itself. The Jacobians are $d_{t}T_{1} = 1$, and $d_{t}T_{2} = -1$, so $T_{1}$ is orientation-preserving yet $T_{2}$ is orientation-reversing, and $\partial T_{1} = [1] - [0]$, and $\partial T_{2} = [0] - [1]$.