Let $f: U \to \Bbb{R}^n$ be a differentiable function with differentiable inverse $f^{-1}: f(U) \to \Bbb{R}^n$. If every closed form over $U$ is exact, the same is true to $f(U)$.
I'm having trouble solving this problem. I know that if $\omega$ any $k$-form over $f(U)$, then we can find $\eta$ such that $f^{\ast}\omega = d\eta$, since $f^{\ast}$ maps $k$-forms over $f(U)$ into $k$-forms over $U$. Also, $$\omega = (f^{-1}\circ f)^{\ast}\omega = (f^{-1})^{\ast}f^{\ast}\omega = d(f^{-1})^{\ast}\eta = d\beta,$$ hence $\omega$ is exact. Is this correct?
There are lots of errors here. You must start with a closed form in order for it to have any hope of being exact. So start with a closed form $\omega$ on $f(U)$. Then $\phi = f^*\omega$ is closed, since $d\phi = df^*\omega = f^*(d\omega) = 0$. Therefore $\phi$ is exact and can be written as $\phi = d\psi$. We then claim that $$\omega = d((f^{-1})^*\psi),$$ as $$d((f^{-1})^*\psi) = (f^{-1})^*(d\psi) = (f^{-1})^*\phi =(f^{-1})^*(f^*\omega) = (f\circ f^{-1})^*\omega = \omega.$$