Let $M$ be a finite $CW$complex. Let $\Sigma_k$ be the symmetric group acting on $k$letters. Suppose there is a free action of $\Sigma_k$ on $M$. Then we have a covering map $$ f:M\to M/\Sigma_k. $$ From $f$, we have an induced map on the fundamental groups $$ f_*:\pi_1(M)\to \pi_1(M/\Sigma_k). $$ From $f_*$, we have an induced map on the EilenbergMaclane spaces $$ F: K(\pi_1(M),1)\to K(\pi_1(M/\Sigma_k),1). $$ Question: Does there always exist an induced $\Sigma_k$action on $K(\pi_1(M),1)$ such that $F$ is a covering map obtained by modulo this $\Sigma_k$action on $K(\pi_1(M),1)$?

3$\begingroup$ This seems highly unlikely. Better to think of $f$ as the fibre inclusion in a fibration sequence $M\to M/\Sigma_k\to K(\Sigma_k,1)$ (where the last map classifies the covering). Then you see $F$ as the fibre inclusion in a corresponding fibration of EilenbergMac Lane spaces. $\endgroup$– Mark GrantJun 2, 2016 at 8:29
1 Answer
If $M$ is connected, then @MarkGrant's fibration sequence gives a long exact sequence on homotopy groups showing that $\pi_1(M/\Sigma_k)\to \Sigma_k$ is surjective. Now apply the $K(,1)$functor and obtain a map $K(\pi_1(M)/\Sigma_k,1)\to B\Sigma_k$. The corresponding $\Sigma_k$bundle over $K(\pi_1(M)/\Sigma_k,1)$ has as total space a $K(\pi_1(M),1)$, as can be seen again from the long exact sequence in homotopy.
So there exists a model of your map $F$ which is a covering map. You have defined $F$ only up to homotopy, so I think your question needs to be whether one can find a model of $F$ which is a covering map.