I was recently introduced to the de Rham cohomology and while we were given a basic overview, there were no hints to a bigger context. As I found the little insight we got extremely interesting, I'm trying to delve deeper (recommendations for literature will be highly appreciated).
My question relates to the cross product which we defined as follows:
Let $M,N$ be smooth manifolds. With $\pi_M$ and $\pi_N$ denoting the natural projections for the product space $M \times N$ we can define a well-defined product by:
\begin{align*}
\times : H^*(M) \times H^*(N) &\rightarrow H^*(M \times N) \\
(\omega, \eta) &\mapsto [\omega] \times [\eta] := [\pi^*_M \omega \wedge \pi_N^* \eta ]
\end{align*}
Further we defined the wedge product via
\begin{align*}
\wedge : H^*(M) \times H^*(M) &\rightarrow H^*(M) \\
(\omega, \sigma) &\mapsto [\omega] \wedge [\sigma] = [\omega \wedge \sigma]
\end{align*}
Now given the standard diagonal embedding of $M$ via \begin{align*} \Delta: M &\rightarrow M \times M \\ p &\mapsto (p,p) \end{align*} I want to prove that for $\omega, \eta \in H^*(M)$ $$ \omega \wedge \eta = \Delta^* (\omega \times \eta) $$ holds true. Those are my thoughts so far:
\begin{align*} (\omega \times \eta) &= (\pi^*_M \omega \wedge \pi^*_M\eta) \\ \Rightarrow \Delta^*(\omega \times \eta) &= \Delta^*(\pi^*_M \omega \wedge \pi^*_M\eta) = \Delta^*(\pi^*_M (\omega \wedge \eta)) \end{align*} Now while I see that $$ \Delta^*\pi^*_M: \Omega^*(M) \rightarrow \Omega^*(M \times M) \rightarrow \Omega^*(M) $$ My question is, if the above is correct and if so, why $$ \Delta^*\pi^*_M = id_{\Omega^*(M)}$$ holds true. Maybe I'm just missing something obvious here ...
$\Delta$ is a section of $\pi_M$, ie., $\pi_M \circ \Delta = \text{id}_M$. Use contravariance of pullbacks to conclude that $\Delta^*\pi_M^* = \text{id}_{\Omega^*(M)}$.