Differential forms theorem reference request

Question

Let $f: A\subset\mathbb{R}^n\rightarrow A$ be smooth ($A$ not necessarily open) and homotopic to the identity map ${\rm id}_A$. If $s^k$ is a singular $k$-chain with image set $A$ and such that $\partial s^k =0$, then for a smooth $k$-form $\omega^k$ \begin{eqnarray} \int_{s^k}\omega^k = \int_{f\circ s^k}\omega^k \end{eqnarray}

Answer 1

This is false as stated. Let $A = \{(x,y,z): x^2+y^2=1, z=0\}$, and let $f\colon \Bbb R^3\to\Bbb R^3$ be given by $f(x,y,z) = (x,y,z+3)$. If $\omega = xz\,dy$, then $\int_A \omega = 0$ and $\int_{f(A)} \omega = 3\pi$.

Upshot: You need $\omega$ to be a closed form (i.e., $d\omega=0$).

EDIT: My apologies. I originally read the problem too quickly. Because $s^k$ is a singular $k$-cycle ($\partial s^k = 0$) with image $A$, and because $f\colon A\to A$ is homotopic to the identity, we can apply Stokes's Theorem. Let $F$ be the homotopy and consider the singular $k+1$-chain $S(x,t) = F(s(x),t)$.

Here's how we want the proof to go: By Stokes's Theorem, $$\pm\left(\int_s f^*\omega - \int_s \omega\right) = \int_{\partial [0,1]^{k+1}} F^*\omega = \int_{[0,1]^{k+1}} d(F^*\omega) = \int_{[0,1]^{k+1}} F^*(d\omega) = \int_S d\omega = 0.$$

However, since you are not given that $d\omega = 0$, you have to argue that since $A$ is the image of $s^k$, $d(\omega|A) = 0$, as there cannot be $k+1$ linearly independent vectors in the image of the derivative of $s^k$ on which to evaluate this $(k+1)$-form. This is a bit non-standard, and I'm at this stage not sure what you know and what you don't.