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Let $M$ be a closed manifold and consider the constant sheaf $\mathbb{R}_M$. I have heard that the Yoneda product on $\operatorname{Ext}(\mathbb{R}_M, \mathbb{R}_M)= H^*(M; \mathbb{R})$ coincides with the usual cup product of cohomology. How does one prove this? More generally, I would like to show that the differential graded algebra $\operatorname{RHom}(\mathbb{R}_M, \mathbb{R}_M)$ coincides with the algebra of differential forms on $M$.

Is this easy to show? If not, what is a good reference? I guess that a natural thing to try is just to resolve $\mathbb{R}_M$ by de Rham cochains, but this didn't seem to work for me.

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    $\begingroup$ Any chain map between acyclic resolutions of sheaves gives rise to an element of an Ext group of the same degree (up to a minus sign depending on conventions), and the Yoneda product is given by composing these maps. Given a de Rham representative $\omega$, such a lift is given by multiplication with $\omega$, and the statement follows immediately. This should be essentially of the Eckmann-Hilton theorem applied to the Yoneda product and the one coming from the Hopf Algebra structure of $\Omega^*(M)$, which define commuting monoid structures on $\operatorname{Ext}(\Omega^*,\Omega^*)$. $\endgroup$
    – Bertram Arnold
    Commented Dec 24, 2020 at 0:45
  • $\begingroup$ @BertramArnold Thank you for this comment. Would you be willing to elaborate a bit more (possibly as an answer to the question), as I am a beginner in this. What do you mean by "a de Rham representative" of a map between acyclic resolutions? And what is "such a lift"? $\endgroup$
    – user155668
    Commented Dec 24, 2020 at 4:08

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