Pullback is a homomorphism of de rham cohomology algebras

Given: A smooth map $F:M\to N$ and closed homogeneous forms $\alpha ,\beta$ on $N$.

De rham cohomology is a contravariant functor: De Rham cohomology is contravariant: for smooth $F:M\to N$ and $G:N\to P$, $(G\circ F{)}^{\ast }=F^{\ast }\circ G^{\ast }$, and ${\mathrm{id}}_M^{\ast }={\mathrm{id}}_{H^k(M)}$.

De rham cohomology ring: The de Rham cohomology ring is $H_{\mathrm{d}\mathrm{R}}^{\bullet }(M)={⨁}_{k\in \mathbb{Z}}{H}_{\mathrm{d}\mathrm{R}}^k(M)$ with $[\alpha ][\beta ]=[\alpha \wedge \beta ]$ and unit $[1]$. By thm-wedge-product-descends-to-de-rham-cohomology it is a unital graded-commutative real algebra. On the empty manifold it is the zero algebra, with $1=0$; this convention allows the zero algebra among unital algebras.

Pullback of forms is smooth functorial and preserves wedges: For a smooth map $F:M\to N$, pullback sends smooth differential forms on $N$ to smooth differential forms on $M$, is functorial, and satisfies $F^{\ast }(\alpha \wedge \beta )=F^{\ast }\alpha \wedge F^{\ast }\beta \mathrm{.}$

Pullback induces a well defined map on de rham cohomology: For a smooth $F:M\to N$, the formula $F^{\ast }[\omega ]=[F^{\ast }\omega ]$ defines a linear map $H_{\mathrm{d}\mathrm{R}}^k(N)\to H_{\mathrm{d}\mathrm{R}}^k(M)$ for every integer $k$.