Let $U, V \subset R^n$ be two open sets and $\phi : V \rightarrow U $ a diffeomorphism. For the outer derivative $d : \Omega^r(R^n) \rightarrow Ω^{r+1}(R^n)$ show that: $$d ◦ \phi^∗ = \phi^∗ ◦ d.$$
What I thought to do:
Consider the operator $d' := (\phi^{-1})^∗ ◦ d ◦ \phi^∗ : \Omega^r(U) \rightarrow \Omega^{r+1}(U)$ and verify that:
- d' is linear
- $d'(\omega ∧ \eta) = d'\omega ∧ \eta + (-1)^{deg\omega} \omega ∧ d'\eta$
- Propriety of cocycle: $d'^2= d' ∘ d' =0$
- For $f\in \Omega^0(M)= F(M):d'f$ ist equal to the differential of the function $f$
So after that we can say that $d'= d$ then is $d ◦ \phi^∗ = \phi^∗ ◦ d$ verified. But I'm not sure how to prove these points and if my reasoning can make sense...