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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?

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

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