Given a compact homogeneous space $M = K/L$, consider its de Rham complex $(\Omega^\*,d)$. Will every cohomology class $[\omega] \in ker(d)/im(d)$ contain a representative $\nu$ which is invariant with respect to the left $K$-action on $\Omega^\*$?



Coinvariant representative of homogeneous space cohomology
Yes, assuming that $K$ is a connected compact Lie group. 


Indeed, fix $n$ such that $0\le n\le d={\rm dim}(M)$.
The group $K$ acts on the integral cohomology group $H^n(M,\Bbb Z)$ trivially, because $K$ is connected, while $H^n(M,\Bbb Z)$ is discrete. 
Therefore, $K$ acts trivially on the de Rham cohomology group


$$H^n\_{\rm dR}(M,\Bbb R)=H^n(M,\Bbb R)=H^n(M,\Bbb Z)\otimes\_{\Bbb Z} \Bbb R.$$


Consider the point $x=e\cdot L\in K/L=M$ with stabilizer $L$.
Then $L$ acts on the tangent space $T\_x(M)$. Since $L$ is compact, it preserves a positive definite quadratic form on $T\_x(M)$. It follows that $M$ admits a $K$-invariant Riemannian metric $g$. 
Therefore, $K$ acts on the vector space
${\mathcal H}^n(M,g)$ of harmonic differential $n$-forms on $M$ with respect to $g$. 


Since $M$ is compact, by a theorem of Hodge a cohomology class $\xi\in H^n\_{\rm dR}(M,\Bbb R)$ is represented by a *unique* harmonic form $\omega\in {\mathcal H}^n(M,g)$; see for instance the book by Jürgen Joost "Riemannian Geometry and Geometric Analysis", Theorem 2.2.1. 
Since $K$ acts trivially on $H^n\_{\rm dR}(M,\Bbb R)$, it acts trivially on ${\mathcal H}^n(M,g)$. Thus the harmonic differential form $\omega$ is a $K$-invariant representative of $\xi$.


