Computations of cup products in Serre's Local Fields I have been reading the appendix in Serre's Local fields, to do with explicit computations of cup products (pg 176), but I'm stuck on one bit of lemma 4. It goes as follows
Let B be a $G$-module, $u: G \times G \rightarrow B$ a 2-cocycle and $\bar{u} \in H^2(G,B)$. Then for all $s \in G$ with $\bar{s} \in \widehat{H}^{-2}(G,\mathbb{Z})$. Then $$\bar{s} \cup \bar{u} = \bar{a}, \qquad \text{with  } a=\sum_{t \in G} u(t,s)$$
Now to prove it he begins by taking a sequence $0 \rightarrow B \rightarrow B' \rightarrow B'' \rightarrow 0$ with $B'$ induced. Now since $H^2(G,B')=0$, there is a 1-cochain $f': G \rightarrow B$ such that $$u(x,y)=xf'(y)-f'(xy) + f(x), \qquad x,y \in G.$$
Then composing $f'$ with $B' \rightarrow B''$ we get a 1-cocycle $f'':G \rightarrow B$, with $d(\bar{f''})=\bar{u}$.
From which he deduces that $$\bar{s} \cup \bar{u} = \bar{s} \cup d(\bar{f''})=d(\bar{s} \cup \bar{f''})$$
Now by the previous lemma we know $\bar{s} \cup \bar{f''}= \bar{f''(s)}_0$ (here the subscript 0 denotes its class in $H^{-1}(G,B''))$. Now this is where I'm stuck he then says that $d(\bar{s} \cup \bar{f''})=d( \overline{f''(s)}_0)= \overline{N(f'(s))}^0$ (where the superscript 0 denotes its class in $\widehat{H}^0$), my question is why does the $f''$ change to $f'$? I cant see why this is so, he also does something similar in lemma 2 of this same appendix, but this is the lemma I need to use.
Thank you
 A: The critical point with your problem is to understand how the connecting homomorphism is computed. Let $P$ be a complete resolution . Then Tate cohomology is the cohomology of the cocomplex $Hom_G(P,-)$. Let 
$$0 \to B \xrightarrow{i}B' \xrightarrow{\rho}B''\to 0$$
be exact. We have a commutative diagramm with exact rows: 
$$\begin{array}{ccccc}
Hom_G(P_{-1},B) & \overset{i_\ast}{\hookrightarrow} & Hom_G(P_{-1},B') & \overset{\rho_\ast}{\twoheadrightarrow} & Hom_G(P_{-1},B'') \newline 
\delta \downarrow & & \downarrow\delta' & & \downarrow\delta'' \newline 
Hom_G(P_0,B) & \overset{i_\ast}{\hookrightarrow} & Hom_G(P_0,B') & \overset{\rho_\ast}{\twoheadrightarrow} & Hom_G(P_0,B'') 
\end{array}\tag{1}$$
By using $P_0=\mathbb{Z}G$ and $P_{-1}=Hom_G(P_0,\mathbb{Z}G)$, diagram $(1)$ becomes 
$$\begin{array}{ccccc}
B_G & \overset{i}{\hookrightarrow} & B_G^' & \overset{\rho}{\twoheadrightarrow} &  B_G^''\newline 
\delta \downarrow & & \downarrow\delta' & & \downarrow\delta'' \newline 
B^G & \overset{i_\ast}{\hookrightarrow} & B^{'G} & \overset{\rho}{\twoheadrightarrow} & B^{''G} 
\end{array}$$
where $B_G = B/C$ with $C := \langle (g-1)b\mid g\in G,b\in B\rangle$ are the coinvariants and 
$\delta=N$ is multiplication with the norm of $G$ (this is shown at the beginning of VI,§ 4 in Brown's book).  
Now $\overline{f''(s)} := f''(s) + C^'' =\rho(f'(s)) + C^''\in B_G^''$ represents a cohomology class. Hence $f'(s) + C^' \in B^'_G$ is a lift. By definition, $d(\overline{f''(s)}_0)$ is represented by $\delta'(f'(s) + C^')=N\cdot f'(s)$. Therefore, it suffices to show $Nf'(s)=a$: 
$$Nf'(s)=\sum_{g \in G}gf'(s)=\sum_g f'(s) - \sum_g f'(gs) + \sum_g f'(g) = \sum_g u(g,s)=a.$$
Feel free to ask if you have any questions to my computations.
