Suppose $M$ is a closed connected smooth manifold with fundamental group $\Gamma$, and $G$ is a simply-connected Lie group which acts smoothly on $M$. Then the Borel construction$$M//G = M \times_G EG$$ has fundamental group $\Gamma$ as well.

Can the 1-truncation (i.e., the map classifying the universal cover) $M//G \to B\Gamma$ be non-trivial in rational cohomology in degrees bigger than the dimension of $M$?

At first I thought perhaps one could produce a counterexample along the following lines:

Take $\Gamma < G$ a discrete group ($G$ some big simply-connected Lie group), and let $M = G/\Gamma$. Then $M//G = B\Gamma$, and the map in question is the identity.

But then I realized that the rational cohomological dimension of any such $\Gamma$ is at most $\text{dim}(G) - \text{dim}(K)$, where $K < G$ is a maximal compact subgroup, so this cannot possibly give a counterexample...

Let $M$ b a closed connected smooth manifold with fundamental group $\Gamma$. Suppose $G$ is a simply-connected Lie group that acts smoothly on $M$. Then the Borel construction  $$M//G = M \times_G EG$$ has fundamental group $\Gamma$ as well.

Can the 1-truncation (i.e., the map classifying the universal cover) $$M//G \to B\Gamma$$ be non-trivial in rational cohomology in degrees bigger than the dimension of $M$?

At first I thought perhaps one could produce a counterexample along the following lines:

Take $\Gamma < G$ a discrete group ($G$ some big simply-connected Lie group), and let $M = G/\Gamma$. Then $M//G = B\Gamma$, and the map in question is the identity.

But then I realized that the rational cohomological dimension of any such $\Gamma$ is at most $\text{dim}(G) - \text{dim}(K)$, where $K < G$ is a maximal compact subgroup, so this cannot possibly give a counterexample...