My question is about the compatibility and consistency between two definitions of cohomology in two books.


I asked this question about 10 days ago [on MathSE](https://math.stackexchange.com/questions/4351968/compatibility-and-consistency-between-two-definitions-of-cohomology-in-two-books)
and I set a bounty on it, but I didn't receive an answer.


I was reading cohomology from Neukirch's book, and there he referenced Hall's book. The two approaches are almost the same (are they not?), and they should give us the same results (cohomology groups). But I can not see their compatibility and consistency, and I can not recover them from each other. My problem is that: **I can not propose the compatibility between these two books.**




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Hall's book:


**15.7. Cochains, Coboundaries, and Cohomology Groups.**


Given a double $\Omega$-modulo $A$ we define $C^n = C^n(A, \Omega)$ to be the additive group of all functions $f$ of $n$ variables which range independently over $\Omega$ and taking values in $A$, subject to the condition


$$ (15.7.1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
f(\xi\_1, \dots, \xi\_n) = 0,$$
whenever at least one of the $\xi\_i = 1$. The elements of $C\_n$ are called
*$n$-dimensional cochains*. $C^0 = A$ by definition and a zero dimensional
cochain is simply any element of A.
The coboundary operator $\delta$ maps each $C\_n$ into the next, $C\_{n+1}$ in
accordance with the rule


$$(\delta f)(\xi\_0, \xi\_1, \dots, \xi\_n)\\
=\xi\_0f(\xi\_1, \dots, \xi\_n) \\ 
+\sum\_{t=1}^{n} (-1)^t (\xi\_0, \xi\_1, \dots, \xi\_{t-2}, \xi\_{t-1}\xi\_{t}, \xi\_{t+1}, \dots, \xi\_{n}) \\ 
+f(\xi\_0\xi\_1, \dots, \xi\_{n-1})\xi\_n . \textit{ something that I cannot read}, 
$$


And let $Z^n=Z^n(A, \Omega)= \ker (C^n\longrightarrow C^{n+1})$, and let $B^n=B^n(A, \Omega)= Im (C^{n-1}\longrightarrow C^{n})$. The quotient group $Z^n/B^n$ is called the $n$-dimensional cohomology group of the double $\Omega$-module $A$. We write it $$H^n(A, \Omega) = Z^n/B^n.$$


In the part *something that I can not read*, **in the above formula**, probably it is written $(-1)^{n+1}$, but I am not sure.




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I have some confusions about coboundary operators and 1-cocycles and computing cohomology, and I can not find where the bugs are.


In Neukirch's Class field theory, page 13, does the last summand of $d\_q(\sigma\_1, \dots, \sigma\_q)$ written truly? Or should it be equal to


$$d\_q(\sigma\_1, \dots, \sigma\_q)= 
\sigma\_1(\sigma\_2, \dots, \sigma\_{q}) 
+ 
\sum\_{i=1}^{q-1} (-1)^i (\sigma\_1, \dots, \sigma\_{i-1}, \sigma\_{i}\sigma\_{i+1}, \sigma\_{i+2}, \dots, \sigma\_{q})
+ 
(-1)^q(\sigma\_1, \dots, \sigma\_{q-1})\sigma\_q ?$$


I've seen something closely similar to these relations in Hall's book (The theory of groups), but I think at least one of them should have a typo. I know that they are not exactly the same, but I think they are not compatible unless for instance we replace the definition of Neukirch as above.


Also I have ambiguity about cocycles and 1-cocycles. If the things in Neukirch's book are correct, then I expected the cocycle should be $f(\sigma\tau)=\sigma f(\tau)+f(\sigma)$, and I am confused with $\delta(f(\sigma, \tau))=\sigma f(\tau)-f(\sigma\tau)+f(\sigma)\tau$, and I can not see the compatibility.



Coboundary operators, 1-cocycles and computing cohomology
Let me just put my comments as an answer: (i) To compute the cohomology groups $H^n(G,A)$, for $A$ a (left) $G$-module and where
$$H^n(G,A)=\text{Ext}^n\_{{\mathbb Z}G}({\mathbb Z},A)$$
one needs a projective resolution of the $G$-module ${\mathbb Z}$, say
$$\cdots\stackrel{d\_3}\rightarrow P\_2\stackrel{d\_2}\rightarrow P\_1\stackrel{d\_0}\rightarrow P\_0\stackrel{\varepsilon}\rightarrow {\mathbb Z}\rightarrow 0$$
such that
$$H^n(G,A)=\text{Ext}^n\_{{\mathbb Z}G}(
{\mathbb Z},A)=\ker(d^\*\_{n+1})/\text{Im}(d\_n^\*).$$


(ii) One choice of projective resolution of ${\mathbb Z}$ is given by the free ${\mathbb Z}$-modules $P\_n$ with basis
the $(n+1)$-tuples $(g\_0,\ldots,g\_n)$ of elements of the group $G$ and where the action of $G$ is by (left) translation: $g(g\_0,\ldots,g\_n)=(gg\_0,\ldots,gg\_n)$. It follows that the $P\_n$ are free $G$-modules with basis the $(n+1)$-tuples of the form $(1,g\_1,\ldots,g\_n)$. The $G$-morphisms
$$d\_n:P\_n\rightarrow P\_{n-1}$$
are defined on the generators by the formula
$$d\_n(g\_0,\ldots,g\_n)=\sum\_{j=0}^n(-1)^n(g\_0,\ldots,\widehat{g\_j},\ldots,g\_n)$$
where $\widehat{g\_j}$ means that this term is omitted.
The morphism $\varepsilon$ sends a generator $(g\_0)$ to $1$. One shows that the $(P\_n,d\_n)$ form indeed an exact sequence and upon appliying the functor $\text{Hom}\_{G}(-,A)$ one obtains a complex whose cohomology groups are $H^n(G,A)$. All of these is pretty standard fare in any homological algebra textbook. The important point is now to observe that an $n$-cochain $f\in \text{Hom}\_G(P\_n,A)$ is determined by its values in the generators $(g\_0,\ldots,g\_n)$ of $P\_n$ and since the cochain is a $G$-module morphism (we say that it is $G$-covariant), then
$$f(g\_0,g\_1,\ldots,g\_n)=f(g\_0(g\_0^{-1}g\_1,\ldots,g\_0^{-1}g\_n))=g\_0f(1,g\_0^{-1}g\_1,\ldots,g\_0^{-1}g\_n)$$
i.e., $f$ is determined by its values on the generators of $P\_n$ of the form $(1,g\_1,\ldots,g\_n)$ and then one may also consider another projective resolution of ${\mathbb Z}$ given by the free $G$-modules $Q\_n$ with generators $[g\_1,\ldots,g\_n]$ (and where $Q\_0$ is the free $G$-module generated by a symbol $[\;]$. One then shows that $P\_n\simeq Q\_n$ as $G$-modules and defining
$$\delta\_n:Q\_n\rightarrow Q\_{n-1}$$
by
$$\delta\_n[g\_1,\ldots,g\_n]=g\_1[g\_2,\ldots,g\_n]+\sum\_{j=1}^{n-1}(-1)^j[g\_1,\ldots,g\_jg\_{j+1},\ldots,g\_n]+(-1)^n[g\_1,\ldots,g\_{n-1}]$$
the $(Q\_n,\delta\_n)$ form a free $G$-resolution of ${\mathbb Z}$ and there is an isomorphism of chain complexes $\varphi:(P\_n,d\_n)\rightarrow (Q\_n,\delta\_n)$, answering your question. Let just add some details: First, the morphisms $\varphi\_n:P\_n\rightarrow Q\_n$ are given by $\varphi\_n(g\_0,\ldots,g\_n)=g\_0[g\_0^{-1}g\_1,g\_1^{-1}g\_2,\ldots, g\_{n-1}^{-1}g\_n]$ and the corresponding $\psi\_n:Q\_n\rightarrow P\_n$ are $\psi\_n[g\_1,\ldots,g\_n]=(1,g\_1,g\_1g\_2,g\_1g\_2g\_3,\ldots, g\_1g\_2\cdots g\_n)$. A lenghty, but straightforward, computation shows that $\delta\_n=\varphi\_{n-1}\circ d\_n\circ \psi\_n$.


