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?
 A: 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
\begin{equation}
\bigtriangleup: M \rightarrow M\times M
\end{equation}
is the diagonal embedding, for singular cohomology classes $\alpha$ and $\beta$ they define the cup product as:
\begin{equation}
\alpha\cup \beta = \bigtriangleup^{*}(\alpha\otimes\beta)
\end{equation}
(I'm assuming this is the usual definition of cup product?). $\alpha\otimes \beta$ is defined via its action on singular homology classes as:
\begin{equation}
\alpha\otimes\beta (\sigma\times\tau) = \alpha(\sigma)\beta(\tau)
\end{equation}
($\sigma$ 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:
\begin{equation}
\int_{\sigma\times\tau}\pi_1^{*}\varphi\wedge\pi_2^{*}\psi = \int_{\sigma}\varphi\int_{\tau}\psi
\end{equation}
So $\pi_1^{*}\varphi\wedge\pi_2^{*}\psi$ is a form representing the cohomology class of $\alpha\otimes\beta$ Finally we have that
\begin{equation}
\bigtriangleup^{*}\pi_1^{*}\varphi\wedge\pi_2^{*}\psi = \varphi\wedge\psi
\end{equation}
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
