2
$\begingroup$

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 Eilenberg-Maclane 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 $K(\pi_1(M),1)$?

Improve this question
$\endgroup$
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 Eilenberg--Mac Lane spaces. $\endgroup$
    – Mark Grant
    Commented Jun 2, 2016 at 8:29

1 Answer 1

Reset to default
8
$\begingroup$

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.

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .