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I'm reading the book "Cohomology of Number fields" by Neukirch(Proposition 1.4.3) in which I am not unable to follow one statement:

Let $0 \rightarrow A^{'} \rightarrow A \rightarrow A^{''} \rightarrow 0$ and $0 \rightarrow C^{'} \rightarrow C \rightarrow C^{''} \rightarrow 0$ be exact sequences of $G$ modules. A pairing of $G$ modules is bilinear map $$\theta:U \times V \rightarrow W$$ such that $\theta(\sigma u ,\sigma v)=\sigma \theta( u , v)$ for all $\sigma \in G$ and where $U$ and $V$ are $G$ modules. Let $B$ be another $G$ module.

What do we mean by the statement " Suppose we are given a pairing $\psi: A\times B \rightarrow C $ which induces pairing $ A^{'}\times B \rightarrow C^{'}$ and $ A^{''}\times B \rightarrow C^{''}. $ " Whether here he is assuming $\psi(A^{'}\times B)= 0?$

math
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    Given a pairing $\psi$ as above, we have an immediate restriction of it, $\psi_1:A'\times B\to C$, and we moreover require=assume that the image is in $C'$. For the second case, let us assume $\pi_{C''}\psi(A'\times B)=0$. In this case, the computation of $\pi_{C''}\psi(a'',b):=\pi_{C''}\psi(a,b)$ for an $a''\in A''$ does not depend on a lift $a\in A$ of $a''$, since two lifts differ by an $a\in A'\subseteq B$ (let us suppose that the monomorphism $A'\to A$ is in fact an inclusion). This leads to a map $\psi:A''\times B\to C''$. Here $\pi_{C''}:C\to C''$ is the projection from the 2. seq. – dan_fulea Jan 17 '20 at 22:05
  • why you need the condition $A^{'} \subset B. $ @dan_fulea – math Jan 20 '20 at 14:39
  • If we do not have the inclusion $A'\subseteq B$, then i have to make the "complicated proposition" ... the two lifts differ by an image of an element $a\in A'$ via the (mono)morphism $A'\to B$... It is just a matter of being exact / pedant. – dan_fulea Jan 20 '20 at 19:56
  • My point is even without assuming the condition $A^{'}\subset B$ still we can show that $\psi:A^{''}\times B\rightarrow C^{''}$ is well defined. We just need the condition $\pi_{C^{''}}\psi(A^{'}\times B)=0.$@dan_fulea – math Jan 21 '20 at 00:00
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    The assumption $A'\subseteq B$ is not so important, we may assume this or not. (Assuming it makes propositions and thinking simpler. Else we have to mention the morphism $A'\to A$ all the time, a morphism that was not denoted by a letter in the OP, and one that may be replaced by the inclusion of its image in $A$. For the other side, i had to write that $\pi_{C''}$, the quotient morphism on this side does matter, and it was easier to put a letter on it.) – dan_fulea Jan 21 '20 at 12:05

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