# Induced de Rham map is a ring map

The de Rham Theorem states that for a smooth manifold $M$ the cochain map $R: \Omega^*(M) \to C^*(M;\mathbb{R})$ from differential forms to singular real cochains defined by $R(\omega)(\sigma)= \int_{\sigma}\omega$ induces an vector space isomorphism in cohomology. How can one prove that the induced map $R^*: H^*_{dR}(M) \rightarrow H^*(M;\mathbb{R})$ takes the wedge product to the cup product?

• How do you integrate the pullback of a differential form along an arbitrary continuous map $\sigma \rightarrow M$? – Aaron Mazel-Gee Mar 12 '13 at 5:31

I don't know much about singular cohomology, but on the top of page 60 of Griffiths and Harris 'Principle of Algebraic Geometry' they discuss exactly this. If $$\bigtriangleup: M \rightarrow M\times M$$ is the diagonal embedding, for singular cohomology classes $\alpha$ and $\beta$ they define the cup product as: $$\alpha\cup \beta = \bigtriangleup^{*}(\alpha\otimes\beta)$$ (I'm assuming this is the usual definition of cup product?). $\alpha\otimes \beta$ is defined via its action on singular homology classes as: $$\alpha\otimes\beta (\sigma\times\tau) = \alpha(\sigma)\beta(\tau)$$ ($\sigma$ and $\tau$ are cycles on $M$) Now if $\varphi$ and $\psi$ are differential forms representing $\alpha$ and $\beta$ respectively, and $\pi_1$ and $\pi_2$ are projections onto the first and second copy of $M$ in $M\times M$ respectively, observe that: $$\int_{\sigma\times\tau}\pi_1^{*}\varphi\wedge\pi_2^{*}\psi = \int_{\sigma}\varphi\int_{\tau}\psi$$ So $\pi_1^{*}\varphi\wedge\pi_2^{*}\psi$ is a form representing the cohomology class of $\alpha\otimes\beta$ Finally we have that $$\bigtriangleup^{*}\pi_1^{*}\varphi\wedge\pi_2^{*}\psi = \varphi\wedge\psi$$ and since the de Rham isomorphism is functorial (discussed at the top of page 45 in Griffiths and Harris) we know that if $\pi_1^{*}\varphi\wedge\pi_2^{*}\psi$ represents $\alpha\otimes\beta$, $\bigtriangleup^{*}(\pi_1^{*}\varphi\wedge\pi_2^{*}\psi)$ will represent $\bigtriangleup^{*}(\alpha\otimes\beta)$. Hope this helps