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Just to be sure because I can’t find the following on the nLab:
If A is a simplicial abelian group, that in addition is n-coskeletal for some n∈ℕ, then the normalized Moore complex is concentrated in degree ≤n, right?
That is because for an n-coskeletal simplicial set, we can say that any horn map ∂j:Am→∂jAm is bijective for all m≥(n+1). (Here ∂jAm means the j-horn set in dim m) Hence in particular the kernel of ∂m consist only of the identity in degree m and so the same holds for the kernels of the dj’s for all 0≤j≤(m−1). Consequently the normalization, i.e. the intersection ∩m−1j=0dmj is the trivial group in any dimension m≥(n+1) and that is what is called “concentrated in degree ≤n”.
Is that correct?
Mirco: your change to Moore complex does not make sense. You suggested an alternating last face, but the premise is that the group is a simplcial group and not a simplicial abelian group.
The process of truncations and its relationship with coskeleta was explored by Conduché in the simplcial group case, and probably in Illusie’s thesis for the abelian one. I think there is a discussion in the Menagerie (see the nLab entry on that). The answer to your question is, I think, yes.
There are different models of “the” Moore complex, all quasi-isomorphic but not all truncated, if one is.
What matters is not so much that the complex is truncated, but that its homology groups are concentrated in a given range. And this is the the case if the original simplicial group was n-truncated (aka n+1-coskeltal): by one of the statements of the Dold-Kan correspondence it in particular identifies the homotopy groups of a simplicial group with the homology groups of its corresponding chain complex.
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