Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 115 / 2 / a / Solution

In local coordinates, forthe exterior derivative isApplying $d$ again gives symmetric second partial derivatives contracted with the antisymmetric wedge $d{x}^k\wedge d{x}^j$, hence $d^2=0$. Expanding coefficients and moving $d\alpha$ past a degree-$r$ form gives the graded Leibniz rule

For a function $f$ and smooth $F:X\to Y$, the chain rule givesEvery differential form is locally a sum of products $f_{0}d{f}_1\wedge \cdots \wedge d{f}_r$. Since pullback preserves products and wedge products, the function case and the graded Leibniz rule implyfor every form.