I'm looking for a reference for a proof of the following fact:

Let $G$ be a compact, connected Lie group acting on a smooth manifold $M$. Then inclusion of the differential forms invariant under the action of $G$ give an isomorphism with the de Rham cohomology group.

I could not find this in Lee or Spivak, which are my go-to references.

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    $\begingroup$ Invariant cohomology is actually discussed in Volume 5 of Spivak, among other places. $\endgroup$ – Ted Shifrin Dec 15 '13 at 23:07
  • $\begingroup$ Ah, thank you! (I only looked in Volume 1) $\endgroup$ – user908123 Dec 15 '13 at 23:19
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    $\begingroup$ Basic idea, not sure where it is written down: Forms are cohomologous to its shifts under the group action. Averaging all of these shifts gives an invariant form cohomologous to the original. $\endgroup$ – Steven Gubkin Dec 15 '13 at 23:34
  • $\begingroup$ To be precise, in Spivak vol. 5, Look at chapter 13, section 16 pg 308,309 $\endgroup$ – Warlock of Firetop Mountain Dec 20 '18 at 15:18

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