Let $\Sigma\_k$ be the $k$-th symmetric group and $B\Sigma\_k$ be its classifying space. How to prove:


for any $n\geq 1$ and the $n$-skeleton $sk\_n (B\Sigma\_k)$, there exists a finite dimensional $CW$-complex $K$ such that


(i). $sk\_n(B\Sigma\_k)\subseteq K\subseteq B\Sigma\_k$;


(ii). $H^\*(K;\mathbb{Q})$ is trivial?


Could I just let $K=B\Sigma\_k$?



rational cohomology of symmetric groups
Take the $n$-skeleton. It has trivial rational homology except possibly in degree $n$. Now add enough $n+1$-cells from the $n+1$-skeleton to kill this top homology. You won't have created any $n+1$-dimensional homology since the boundary operator $\partial\_{n+1}\colon C\_{n+1}\to C\_n$ is a rational isomorphism onto $\ker(\partial\_n)$. This construction works for any finite group, not just the symmetric group.


