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Let $f: M \to M$ define an automorphism on the smooth manifold M.

Given a differential form $\omega \in \Omega^k$ is it true that the de Rham cohomology class of $\omega$ and $f^*\omega$ are the same? That is, does $[\omega]=[f^*\omega]$.

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    $\begingroup$ Here's another type of example: Consider the antipodal map ($f(x)=-x$) on $S^n$ with $n$ even. $\endgroup$ – Ted Shifrin Mar 14 at 16:29
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No. One example: take the torus $X = \mathbb{R}^2/\mathbb{Z}^2$. The flip-flop on the factors interchanges the closed forms $dx$ and $dy$ which are linearly independent in $H^1(X)$.

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