De Rham Cohomology of the open ball on $\mathbb{R}^n$

Question

I was recently in class and my professor suggested that one could compute the de Rham cohomology of the open unit ball in $\mathbb{R}^n$ in the following manner. Define the maps $f_{j-1}: \Omega^j(M) \to \Omega^{j-1}(M)$ for $j=1,\cdots , n$ with: $$ \sigma = x^{k_1} dx^{k_2} \wedge \cdots \wedge dx^{k_j} - x^{k_2} dx^{k_1}\wedge dx^{k_3} \wedge \cdots \wedge dx^{k_j} + \cdots $$ $$ \tau= g dx^{k_1} \wedge \cdots dx^{k_j}$$ Where we let: $$(f_{j-1} \tau) = \left(\int_0^1 s^{j-1} g(sx) dx \right) \sigma$$ And it can be show that these have the property that: $$ d \circ f_{j-1} + f_j \circ d = \mathrm{Id} $$ Where $d$ is the exterior derivative. Could anyone provide a reference or explain to me how to proceed in using this to compute the de Rham cohomology?

Answer 1

Suppose that $\omega$ is a closed form, so that $d\omega=0$. Since $d\circ f+f\circ d=\mathrm{Id}$, we have $d(f(\omega))+f(d(\omega))=\omega$. Now $d(\omega)=0$, so this in fact tells us that $\omega=d(f(\omega))$ is exact, that is, it is in the image of $d$.