Proposition 1

Let $x \in L$; denote by $R_x : y \mapsto x \wedge y$ the left multiplication by $x$ in the algebra $\Lambda(L)$. Then $d_u \circ R_x + R_x \circ d_u = u(x) \cdot 1_{\Lambda(L)} = u(x)_{\Lambda(L)}$.

Indeed $(d_u \circ R_x + R_x \circ d_u)(y) = d_u(x \wedge y) + x \wedge d_u(y)$; since $d_u$ is an antidifferentiation, $d_u(x \wedge y) + x \wedge d_u(y) = d_u(x) \wedge y = u(x) \cdot y$.