# Boundary Map of Bar Resolution vs. Face Map of the Nerve of a Group

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 $$0. The degeneracy maps are $$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.

• 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 Sep 27, 2020 at 21:29
• 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. Sep 28, 2020 at 15:26
• 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. Sep 28, 2020 at 16:12

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.