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

1 AnswersNo. 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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