Let $M$ be a compact manifold and $\varphi:M\rightarrow M$ a diffeomorphism. The invariant differential forms
$$
\Omega^{k}\_{inv}(M)=\{\alpha\in\Omega^{k}(M):\varphi^{\*}\alpha=\alpha\}
$$
form a subcomplex $(\Omega\_{inv}(M),d)$ of the de Rham complex. Denote by $H\_{inv}(M)$ its cohomology.


**Question:** Is there an example where the map $H^{1}\_{inv}(M)\rightarrow H^{1}(M)$ is not injective?


It seems like the map is injective when forms can be averaged with respect to $\varphi$ in some way. For instance if $\varphi$ has finite order, or more generally if $\varphi$ is an isometry for some Riemannian metric $g$, in which case one can average over the closure $\overline{\{\varphi^{n}\}}$ in the compact Lie group $Isom(M,g)$. But I think there should be an example where the map is not injective.



Cohomology of invariant differential forms
There is no such thing: any $\phi$-invariant exact
1-form is a differential of $\phi$-invariant function.


Indeed, let $\alpha$ be an exact $\phi$-invariant
form, $\alpha=df$, where $f$ is not $\phi$-invariant.
Then $d(\phi^\* f -f)=0$, hence
$\phi^\* f = f + C$, where $C$ is a constant.
This gives $\sup f = \sup \phi^\* f = \sup f + C$, which is
a contradiction.


For higher cohomology I think there are some examples.


