Let $S$ be a (small) symmetric monoidal category and $X$ a (small) category on which $S$ acts. $\pi\_0(S) = \pi\_0(BS)$ is naturally an abelian monoid, with $[A] + [B] := [A+B]$, where $[A]$ denotes the path component containing the 0-cell, i.e., object $A$.



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> $\pi\_0(S)$ acts on $H\_p(BX, \mathbb Z)$. How is this?
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Path components of a monoidal category acting on homology
The action of $S$ on $X$ induces an action of $BS$ on $BX$. One way to see it is to do it at the level of $p$ simplices and construct a map $B\_pS\times B\_pX\to B\_pX$ for all $p$. For any category $C$, the $p$ simplices $B\_pC$ are the functors $[p]\to C$. Since $S$ acts on $X$, we have a functor $S\times X\to X$. Mapping $[p]$ to this map, we construct a map: $B\_pS\times B\_pX\to B\_p X$ (note that $Fun([p],-)$ preserves product). It is formal to check that this map induce a map of simplicial sets $B\_\bullet(S)\times B\_\bullet(X)\to B\_\bullet(X)$. If you would rather work with topological space, you can use the fact that geometric realization is product preserving and you can turn this map into a map of topological spaces.


This action induces an action map $H\_0(BS)\otimes H\_p(BX)\to H\_p(BX)$ making $H\_p(BX)$ into an $H\_0(BS)$-module. But $H\_0(BS)=\mathbb{Z}[\pi\_0(BS)]$, therefore you have an action of the group algebra $\mathbb{Z}[\pi\_0(BS)]$ on $H\_p(BX)$ or equivalently an action of the group $\pi\_0(S)$ on $H\_p(BX)$.


