5 hopefully remove all typos

The "trick" you are referring to, of replacing $G$-linear morphisms $f:\mathbb ZG^{i+1}\to M$ by a function functions of sets $c:G^i\to M$, is not a trick: it is just the observation that $G$-linearity implies that, since $\mathbb{Z} G^{i+1}=\mathbb ZG\otimes\mathbb Z{G^i}$ as a left $G$-module, there is a natural isomorphism $$\hom_G(\mathbb ZG^{i+1},M)=\hom_G(\mathbb ZG\otimes\mathbb Z^{G^i},M)=\hom_{\mathbb Z}(\mathbb Z G^i,M)$$ by standard properties of the tensor product. Now $\mathbb ZG^i$ is free abelian on the set $G^i$, so there is a natural isomorphisms isomorphism $$\hom_{\mathbb Z}(\mathbb Z G^i,M)\cong\hom_{\mathrm{Set}}(G^i,M).$$ The composition of theses maps gives the identification between $f$'s and $c$. c$'s. You can write them down explicitelyexplicitly, and compute their inverses: in both cases a one on the left" is involved and this is simply a reflection that we are using left modules. On clunkynessclunkiness The form you call 'clunky' has the following origin. The usual bar resolution for a group$G$has in degree$n$the$\mathbb ZG$-module$P_n=\mathbb ZG^{n+1}$. We can think of it as a free left$G$-module generated by the set of symbols symbols$[g_1,\dots,g_n]$. The differential of the complex is then left$G$-linear and for example, $$d[g_1,g_2,g_3] = g_1[g_2,g_3]-[g_1g_2,g_3]+[g_1,g_2g_3]-[g_1,g_2],$$ and this plainly involves the multiplication of the group. This is the so-called inhomogeneous description. We can also think of$P_n$in another, homogenous, way: define$(g_0,\dots,g_n)=g_0[g_0^{-1}g_1,\dots,g_0^{-1}g)n]$.$(g_0,\dots,g_n)=g_0[g_0^{-1}g_1,\dots,g_0^{-1}g_n].$$The set of all such symbols is now a basis of P_n over \mathbb Z, with the left action of G given by$$g\cdot(g_0,\dots,g_n)=(gg_0,\dots,gg_n),$$which is a bit more annoying that before, but . The good thing is that now the boundary is given by$$d(g_0,\dots,g_n)=\sum_{i=0}^n(-1)^i(g_0,\dots,\hat g_i,\dots g_n)$$g_n),$$ which does not involve the product in $G$ at all , and, in fact, is precisely the same formula for as the one giving the boundary in a simplicial complex.

The clunkyness clunkiness you refer to is nothing but the translation formula between these two descriptions of the complex, as $$[g_1,\dots,g_n]=(1,g_1,g_1g_2,g_1g_2g_3,\dots,g_1\cdots g_n).$$

The original description of the complex is the inhomogeneous one, because it was found topologically. The two can be seen as a description of the classifying space for $G$ either by labelling paths in the category by the objects you go through, or by the arrows you follow, or something...

4 typo

The "trick" you are referring to, of replacing $G$-linear morphisms $f:\mathbb G^{i+1}\to ZG^{i+1}\to M$ by a function of sets $c:G^i\to M$, is not a trick: it is just the observation that $G$-linearity implies that, since $\mathbb{Z} G^{i+1}=\mathbb ZG\otimes\mathbb Z{G^i}$ as a left $G$-module, there is a natural isomorphism $$\hom_G(\mathbb ZG^{i+1},M)=\hom_G(\mathbb ZG\otimes\mathbb Z^{G^i},M)=\hom_{\mathbb Z}(\mathbb Z G^i,M)$$ by standard properties of the tensor product. Now $\mathbb ZG^i$ is free abelian on the set $G^i$, so there is a natural isomorphisms $$\hom_{\mathbb Z}(\mathbb Z G^i,M)\cong\hom_{\mathrm{Set}}(G^i,M).$$ The composition of theses maps gives the identification between $f$'s and $c$. You can write them down explicitely, and compute their inverses: in both cases a one on the left" is involved and this is simply a reflection that we are using left modules.

On clunkyness

The form you call 'clunky' has the following origin. The usual bar resolution for a group $G$ has in degree $n$ the $\mathbb ZG$-module $P_n=\mathbb ZG^{n+1}$. We can think of it as a free left $G$-module generated by the set of symbols symbols $[g_1,\dots,g_n]$. The differential of the complex is then left $G$-linear and for example, $$d[g_1,g_2,g_3] = g_1[g_2,g_3]-[g_1g_2,g_3]+[g_1,g_2g_3]-[g_1,g_2],$$ and this plainly involves the multiplication of the group. This is the so-called inhomogeneous description.

We can also think of $P_n$ in another, homogenous, way: define $(g_0,\dots,g_n)=g_0[g_0^{-1}g_1,\dots,g_0^{-1}g)n]$. The set of all such symbols is now a basis of $P_n$ over $\mathbb Z$, with the left action of $G$ given by $$g\cdot(g_0,\dots,g_n)=(gg_0,\dots,gg_n),$$ which is a bit more annoying that before, but now the boundary is given by $$d(g_0,\dots,g_n)=\sum_{i=0}^n(-1)^i(g_0,\dots,\hat g_i,\dots g_n)$$ which does not involve the product in $G$ at all, and in fact is the same formula for the boundary in a simplicial complex.

The clunkyness you refer to is nothing but the translation formula between these two descriptions of the complex, as $$[g_1,\dots,g_n]=(1,g_1,g_1g_2,g_1g_2g_3,\dots,g_1\cdots g_n).$$

The original description of the complex is the inhomogeneous one, because it was found topologically. The two can be seen as a description of the classifying space for $G$ either by labelling paths in the category by the objects you go through, or by the arrows you follow.

3 More; deleted 4 characters in body

On clunkyness

The form you call 'clunky' has the following origin. The usual bar resolution for a group $G$ has in degree $n$ the $\mathbb ZG$-module $P_n=\mathbb ZG^{n+1}$. We can think of it as a free left $G$-module generated by the set of symbols symbols $[g_1,\dots,g_n]$. The differential of the complex is then left $G$-linear and for example, $$d[g_1,g_2,g_3] = g_1[g_2,g_3]-[g_1g_2,g_3]+[g_1,g_2g_3]-[g_1,g_2],$$and this plainly involves the multiplication of the group. This is the so-called inhomogeneous description.

We can also think of $P_n$ in another, homogenous, way: define $(g_0,\dots,g_n)=g_0[g_0^{-1}g_1,\dots,g_0^{-1}g)n]$. The set of all such symbols is now a basis of $P_n$ over $\mathbb Z$, with the left action of $G$ given by $$g\cdot(g_0,\dots,g_n)=(gg_0,\dots,gg_n),$$ which is a bit more annoying that before, but now the boundary is given by $$d(g_0,\dots,g_n)=\sum_{i=0}^n(-1)^i(g_0,\dots,\hat g_i,\dots g_n)$$ which does not involve the product in $G$ at all, and in fact is the same formula for the boundary in a simplicial complex.

The clunkyness you refer to is nothing but the translation formula between these two descriptions of the complex, as $$[g_1,\dots,g_n]=(1,g_1,g_1g_2,g_1g_2g_3,\dots,g_1\cdots g_n).$$

The original description of the complex is the inhomogeneous one, because it was found topologically. The two can be seen as a description of the classifying space for $G$ either by labelling paths in the category by the objects you go through, or by the arrows you follow.

2 added 71 characters in body
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