Wedge product descends to de rham cohomology

Given: Closed forms $\alpha ,\beta$ of degrees $p,q\ge 0$.

De rham cohomology: The real de Rham cohomology is $H_{\mathrm{d}\mathrm{R}}^k(M)=Z^k(M)\mathrm{/}{B}^k(M)$, with $Z^k,B^k$ as in def-closed-and-exact-differential-forms. This is def-cohomology-object-of-a-cochain-complex in real vector spaces. Only a closed form $\omega$ represents a class $[\omega ]$. For closed forms $\omega ,\omega \mathrm{\prime }$, equality $[\omega ]=[\omega \mathrm{\prime }]$ means precisely $\omega \mathrm{\prime }-\omega =d\eta$ for some $(k-1)$-form $\eta$. Addition and real scalar multiplication are induced by those of forms. All groups on the empty manifold are zero.

Wedge with a closed form preserves exactness classes: If $\alpha \in {\mathrm{\Omega }}^p(M)$ and $\beta \in {\mathrm{\Omega }}^q(M)$ are closed, then $d\eta \wedge \beta =d(\eta \wedge \beta )$ for $\eta \in {\mathrm{\Omega }}^{p-1}(M)$ and $\alpha \wedge d\theta =(-1{)}^{p}d(\alpha \wedge \theta )$ for $\theta \in {\mathrm{\Omega }}^{q-1}(M)$.

Differential forms form a graded commutative algebra: The graded vector space ${\mathrm{\Omega }}^{\ast }(M):={⨁}_{k\ge 0}{\mathrm{\Omega }}^k(M)$ with the wedge product is an associative graded-commutative algebra.

The exterior derivative is a graded derivation: Let $M$ be a smooth manifold. The exterior derivative is an $\mathbb{R}$-linear map $d:{\mathrm{\Omega }}^{\ast }(M)\to {\mathrm{\Omega }}^{\ast }(M)$ of degree one. For homogeneous smooth forms $\alpha \in {\mathrm{\Omega }}^p(M)$ and $\beta \in {\mathrm{\Omega }}^q(M)$, $d(\alpha \wedge \beta )=d\alpha \wedge \beta +(-1{)}^{\mathrm{deg}\alpha }\alpha \wedge d\beta \mathrm{.}$