Let **FinCar** denote the category whose objects are the finite cardinal numbers $[n]=\{0,\dots, n\}$ and whose morphisms are all functions between them, and let $X$ be a a contravariant functor from **FinCar** into **Ab**, the category of Abelian groups. The morphisms of **FinCar** are generated by the co-face and co-degeneracy maps of the subcategory **FinOrd** of finite ordinals and monotonic maps, together with the symmetric group $S\_{n+1}$ which is a subset of $Hom([n],[n])$ for each $n$. Therefore, $X$ can be regarded as a simplicial Abelian group together with an action of $S\_{n+1}$ on $X\_n$ for each $n$.


I define the group $X'\_n$ to be the subgroup of $X\_n$ which is invariant/fixed under the action of $S\_{n+1}$. I notice that each face map $d\_i$ carries $X\_n'$ into $X\_{n-1}'$; so I can regard $X'$ as a semi-simplicial Abelian group; I don't think $X'$ is a simplicial Abelian group as I can't figure out what the degeneracies would be.


In any case by taking alternating sums of face maps both $X,X'$ become chain complexes. Do they have the same homology? Is the inclusion map a quasi-isomorphism?


I have some basic competency with spectral sequences if this helps.



Cohomology of a simplicial abelian group $X_\bullet$, where $S_n$ acts on $X_n$
No. Let $X$ be the functor that takes $[n]$ to the group of maps $[n]\to \mathbb Z$. Then $H\_0X=0$ while $H\_0X'\cong\mathbb Z$.


