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For a discrete group $G$ I have the following two definitions, which I think are correct:

  • The nerve of $G$ is $NG$, a simplicial set whose $n$-simplices are $G^n$ ($G^0$ being the trivial group $\{1\}$) and whose face maps are $d_0(g_1,...,g_n)=(g_2,...,g_n)$, $d_n(g_1,...,g_n)=(g_1,...,g_{n-1})$, and $d_i(g_1,...,g_n)=(g_1,...,g_i g_{i+1},...,g_n$ for all $0<i<n$. The degeneracy maps are $s_0(g_1,...,g_n)=(1,g_1,...,g_n)$ and $s_i(g_1,...,g_n)=(g_1,...,g_i,1,g_{i+1},...,g_n)$ for $i>0$. The geometric realization of $NG$ is a space which may be viewed as a CW complex with $n$-cells that can be identified with the nondegenerate simplices of $NG$, that is, those that don't have $1$ as any of the $g_i$ in $(g_1,...,g_n)$. Generally this space is considered to be the (a?) classifying space for $G$.
  • The bar resolution of $G$ is the sequence $0\leftarrow \mathbb{Z}\leftarrow G^0 \leftarrow G^1 \leftarrow...$ with boundary maps $\partial_n(g_1,...,g_n)$$=g_1*(g_2,...,g_n)$$+\overset{n-1}{\underset{i=1}{\sum}}(-1)^i(g_1,...,g_i g_{i+1},...,g_n)$$+(-1)^n(g_1,...,g_{n-1})$.

I'm looking for a way to understand the cohomology groups of $G$, as defined using the bar resolution, in terms of the cellular cohomology groups of $BG$. It seems like there should be a very direct relationship between the face maps $d_i$ which can be thought of as determining the cellular structure of $BG$ and the bar resolution's boundary map $\partial_n$; in fact $\partial_n$ looks almost exactly like $\overset{n}{\underset{i=0}{\sum}}(-1)^i d_i(g_1,...,g_n)$, except that the first term is $g_1*(g_2,...,g_n)$ instead of just $(g_2,...,g_n)$. Essentially, my question is where this "extra multiplication by $g_1$" comes from. I also wonder if my definitions are missing something or my understanding of them has a gap. I'd like to understand this in a way that doesn't appeal to the "other" usual construction of $BG$ as a quotient of $EG$ if at all possible.

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    $\begingroup$ The free abelian group on the nerve is a simplicial abelian group; applying the Dold-Kan correspondence to it produces the bar resolution: ncatlab.org/nlab/show/Dold-Kan+correspondence $\endgroup$ Sep 27, 2020 at 21:29
  • $\begingroup$ I'm sure it's a matter of my own ignorance, and not a problem with your comment, but I honestly can't really grasp the relevance of your comment to my question. I generally have a lot of trouble understanding anything at all on nlab's pages. $\endgroup$
    – Xindaris
    Sep 28, 2020 at 15:26
  • $\begingroup$ Dold-Kan tells you how the face maps of the nerve produce the boundary maps of the bar resolution. Admittedly I haven’t worked this out in detail myself. $\endgroup$ Sep 28, 2020 at 16:12

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I believe I have settled this issue for myself, and the key ingredient is indeed the Dold-Kan correspondence. However, I find a synthesis of these two expositions to be much more comprehensible than anything nlab has to offer: (1), (2) They also both refer to the same book reference, namely this one, which is also helpful in understanding the result.

The basic idea is that any chain complex corresponds to a simplicial abelian group, and any simplicial abelian group $A$ has a corresponding chain complex $A_*$. The image of the degenerate simplices under this correspondence forms a subcomplex $DA_*$, and there is another subcomplex $NA_*$ which turns out to have the property that $NA_*\oplus DA_*\cong A_*$. In other words, $A_*/DA_*\cong NA_*$. The Dold-Kan correspondence theorem itself says that $A\to NA_*$ gives an equivalence of categories between chain complexes and simplicial abelian groups, which then gives a way to "normalize" certain chain complexes, including the chain complex defining the simplicial homology of a topological space.

In order to apply this to my situation, which involves cohomology rather than homology, it is necessary to "dualize" this result. The key insight, I think, is that actually any simplicial set $X$ can be "upgraded" into an abelian simplicial group $\mathbb Z (X)$, whose $n$-simplices $\mathbb Z(X)[n]$ are $\mathbb Z (X[n])$, which consists of formal sums $\sum_{x\in X}k_x x$ such that $k_x\in \mathbb Z$ and only finitely many of them are nonzero. Then any cochain complex $C^n(X,A)$ defined as $\lbrace f:X[n]\to A \rbrace$ can instead be thought of as $Hom_\mathbb{Z}(\mathbb Z(X)[n],A)$, and if $C_0^n(X,A)$ is the functions which are zero on any nondegenerate $n$-simplex, then $C^n(X,A)/C^0_n(X,A)\cong Hom_\mathbb{Z}(N\mathbb Z(X)[n],A)$ and the Dold-Kan correspondence gives a chain equivalence between $Hom_\mathbb{Z}(\mathbb Z(X)[n],A)$ and $ Hom_\mathbb{Z}(N\mathbb Z(X)[n],A)$, which makes the natural projection $C^n(X,A)\to C^n(X,A)/C^0_n(X,A)$ into a chain equivalence also.

After this, I found that it does appear necessary, or at least convenient, to go through $\mathcal E G$, the simplicial set with $\mathcal E G[n]=G^{n+1}$ and $d_i$, $s_i$ given by deletion and repetition of the $i$th element respectively, and whose geometric realization is $EG$. Even though this (or the bar resolution) may be viewed as a simplicial group, $G$ may not be abelian, so that isn't particularly helpful. It's better to think of it as a simplicial set and use a slightly modified version of the result above (essentially, the same formulation can be done for subcomplexes of $C^n(X,A)$ to "normalize" them). Then (in broad strokes) if $C^n(G,A)=\lbrace f:G^{n+1}\to A\text{ such that }gf=fg\text{ for all }g\in G \rbrace$, there is an injective map $C^n(G,A)/C^n_0(G,A)\to C^n_\text{cell}(EG,A)$, and an injective map $C^n_\text{cell}(BG,A)\to C^n_\text{cell}(EG,A)$; if $g$ acts trivially on $A$ then both of which have the same subcomplex as their image. The isomorphism follows from that, and if $G$ doesn't act trivially on $A$ then that first injective map can still be used to get an isomorphism to the cohomology with local coefficients instead.

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