Cohomology of a simplicial abelian group $X_bullet$, where $S_n$ acts on $X_n$

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.

MathOverflow Asked by Patrick Nicodemus on February 12, 2021

No. Let $$X$$ be the functor that takes $$[n]$$ to the group of maps $$[n]to mathbb Z$$. Then $$H_0X=0$$ while $$H_0X'congmathbb Z$$.

Correct answer by Tom Goodwillie on February 12, 2021

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