Proving that the pullback map commutes with the exterior derivative
You're missing only one item. Why is $d(\phi^*(dx^{\mu_{1}}\wedge\cdots\wedge dx^{\mu_{r}}))=0$?
Proving that the pullback map commutes with the exterior derivative
You're missing only one item. Why is $d(\phi^*(dx^{\mu_{1}}\wedge\cdots\wedge dx^{\mu_{r}}))=0$?