Group cohomology topologically with simplicial sets I have a question about the usual formula for the differential in the usual projective resolution of $\mathbb{Z}$ as a $G$-module for a finite group $G$ 
Recall that for a $G$-module $A$, $C^i(G,A)$ is defined as the abelian group of functions $G^i\to A$ and the differential is $d(f)(g_0,...,g_i)=$ $$g_0\cdot f(g_1,...,g_i) + \displaystyle\sum_{j=1}^i(-1)^jf(g_0,...,g_{j-1}g_j, ..., g_i) + (-1)^{i+1}f(g_0,...,g_{i-1})$$
I was trying to find an interpretation for this formula, and for this I found out that there was a nice topological point of view on group cohomology. Pick a $K(G,1)$ space $X$, and let $p:\tilde{X}\to X$ be its universal covering space. 
Then $\pi_1(X)=G$ acts on $\tilde{X}$ and thus on the singular complex $C_*(\tilde{X})$, making $C_*(\tilde{X})\to \mathbb{Z}$ a projective (actually free since the action on $C_*(\tilde{X})$ comes from an action on the simplices and the action of $G$ on $\tilde{X}$ is free) resolution of $\mathbb{Z}$ as a trivial $G$-module ( the complex $C_*(\tilde{X})\to \mathbb{Z}$ is exact because $\tilde{X}$ is simply connected and has zero higher homotopy groups (because so does $X$), thus is contractible, because it's a CW-complex)
Now we also know that $|BG|$ is a $K(G,1)$ space, where $BG$ is a the nerve of $G$ (seen as a category), and we already know its universal cover, it's $|EG|$  (I don't know if it's standard notation, so let me make it clear) where $(EG)_n = G^{n+1}$ and $d_i(g_0,...,g_n) = (g_0,...,\widehat{g_i},...,g_n)$ and $s_i(g_0,...,g_n) = (g_0,...,g_i,g_i,...,g_n)$; which gives another explicit cochain complex whose cohomology is $H^*(G,\mathbb{Z})$. I had already seen this complex and now this gives me a topological interpretation for it.
But my trouble is with the first one I introduced. Indeed the formula for $d$ seems like a twisted version of $\displaystyle\sum_{j=0}^i(-1)^j d_j$ where $d_j$ is the boundary of $BG$; twisted by the action of $G$ but only on the first summand. I'd like to understand this connection with BG (and not $|BG|$, that I understand, I think) more precisely than "it looks a bit similar":

How do we get this formula for $d$ from the nerve $BG$ ? What is the topological interpretation of this formula ? 

 A: Consider the simplical set $EG$. It has a free $G$-action and is contractible, so that $C_*(\mathbb Z[EG])$ is a complex of free $\mathbb Z[G]$-modules with a quasi-isomorphism $C_*(\mathbb Z[EG]) \to \mathbb Z[0]$.
$EG$ in degree $n$ is $G^{n+1}$ so this chain complex is $\mathbb Z[G^{n+1}]$ in degree $n$, and the differential is $\sum_i (-1)^i d_i$.
($d_i$ is the one I mentioned in the question, and the $G$-action is coordinate-wise by translation)
The point is that the writing $\mathbb Z[G^{n+1}]$ doesn't make the free-ness apparent, and one way to do so will introduce the $G$-action on $d_0$.
Namely, if $X$ is an arbitrary $G$-set, then $G\times X^{triv}\cong G\times X$ via $(g,x)\mapsto (g,gx)$ (this is obviously a bijection, and it is a simple computation to check that it is $G$-equivariant). In particular, $G^{n+1}\cong G\times (G^n)^{triv}$ via $\varphi: (g,x)\mapsto (g, g^{-1}x)$. One issue with this is that it doesn't respect the simplicial structure, but of course, it shouldn't, and can't ! $EG$ is not $G\times -$ as a $G$-simplicial set. A bigger issue is that the formula for $d_0$ we get is not that good if we do this. But luckily, in this case, we can trivialize the $G^n$ part in several steps.
Indeed, $G^{n+1} = G^{n-1}\times G\times G\cong G^{n-1}\times G\times G^{triv}$, and then you can again remove one $G$ by doing $G\times G\times G^{triv}\cong G\times (G\times G^{triv})^{triv}$, and so on and so forth. If you work this out, you'll see that the isomorphism is given, on $G^{n+1}$, by $G\times (G^n)^{triv}\cong G^{n+1}, (g_0,...,g_n)\mapsto (g_0,g_0g_1,g_0g_1g_2,...., g_0...g_n)$ (you can check a posteriori that this is a $G$-map, and a bijection). Call this $\varphi^{-1}$.
If we compute what $d_i$ looks like with this presentation, we see that for $0\leq i< n$, $\varphi \circ d_i \circ \varphi^{-1} (g_0,...,g_n) =\varphi\circ d_i (g_0, g_0g_1,...,g_0...g_n) = \varphi (g_0,g_0g_1, ...,\widehat{g_0...g_i} ,..., g_0...g_n) = (g_0,...,g_ig_{i+1},...,g_n)$
For $i=n$, $\varphi \circ d_n \circ \varphi^{-1} (g_0,...,g_n)= \varphi\circ d_n(g_0,g_0g_1,...,g_0...g_n) = \varphi(g_0,g_0g_1,...,g_0...g_{n-1}) = (g_0,...,g_{n-1})$.
The important point is to observe that for $i=0$, this does not send $\{e\}\times G^n$ to $\{e\}\times G^{n-1}$
So if we write the $n$-term of our chain complex as $\mathbb Z[G]\otimes_\mathbb Z\mathbb Z[(G^n)^{triv}]$ to witness that it is indeed a free $G$-abelian group, and compute the differential on $\hom_G(\mathbb Z[G]\otimes_\mathbb Z[G^n],A)$, we see that if we view elements of this as set maps $G^n\to A$, then for all differentials except for $d_0$, we have nothing to do, but precomposition by $d_0$ makes us go out of $G^n$ so there is something to say.
Precomposition by $d_0$ looks like this : if $f: G^n\to A$, then we can view $f$ as a $G$-map $\tilde f: G\times G^n\to A$ by sending $(g_0,...,g_n)$ to $g_0\cdot f(g_1,..,g_n)$.
So now, the map $G^{n+1}\to A$ associated to $\tilde f\circ d_0$ is defined by $\tilde f\circ d_0 (e,...,g_{n+1}) $, and this is $\tilde f(g_1,..., g_{n+1}) = g_1\tilde f(e,g_2,...,g_{n+1}) =g_1 f(g_2,...,g_{n+1})$.
So this gives us the desired $d_0$, where the other $d_i$'s preserve $\{e\}\times G^n$, so there is nothing to say.
