I am reading about 'Cup products' from the book 'Algebraic Number Theory' by Cassels and Frohlich. The following theorem appears on page 105.
Theorem. $G$ is a finite group then there exists one and only one family of morphisms $$\hat H^p(G,A ) \otimes \hat H^q(G,B ) \rightarrow \hat H^{p+q}(G,A \otimes B ) $$ defined for integers $p, q$ and $G$ modules $A,B$ such that
*the usual properties of cup products are listed
The second of these listed properties is
(ii) For $p=q=0 $ they (the family of morphism) are induced by natural product $$A^G \otimes B^G= (A \otimes B)^G $$
I suppose the natural map is $a\otimes b \mapsto a \otimes b$. While this map works for cohomology groups, when we are dealing with Tate cohomology groups $\hat H^0(G,A)=A^G/Nm_G A $. So the cup product morphism is $$A^G/Nm_G A \otimes B^G/Nm_G B \rightarrow (A\otimes B)^G/Nm_G(A \otimes B)$$ The natural induced map should be, to my mind, $\overline a \otimes \overline b \mapsto \overline{ a\otimes b }$.
But I don't see how this map is well defined? I know that in general the map $A/A' \otimes B/B' \rightarrow A \otimes B/(A'\otimes B')$ is not well defined.
Suppose for example $a' \in Nm_GA, b' \in Nm_G B $ then is $\overline {a+a'\otimes b+b'} = \overline {a \otimes b} $.
I am stuck at $ \overline {a+a'\otimes b+b'}=\overline {a \otimes b} + \overline {a \otimes b'} + \overline {a' \otimes b}$
Feel free to give any reference and any help is appreciated.
To show that the map$\newcommand\Nm{\mathrm{Nm}}$ $$\frac{A^G}{\Nm_GA}\otimes\frac{B^G}{\Nm_GB}\to\frac{(A\otimes B)^G}{\Nm_G(A\otimes B)}$$ is well-defined, we need it to be induced from a bilinear map $$\frac{A^G}{\Nm_GA}\times\frac{B^G}{\Nm_GB}\to\frac{(A\otimes B)^G}{\Nm_G(A\otimes B)}.$$ In other words, we need the map $$A^G\times B^G\ni(a,b)\longmapsto\overline{a\otimes b}\in\frac{(A\otimes B)^G}{\Nm_G(A\otimes B)},$$ which is visibly bilinear, to vanish when $a\in\Nm_GA$ or $b\in\Nm_GB$ (so it descends to a bilinear map on the product of the quotient spaces). First suppose $$a=\Nm_Gx=\sum_{g\in G}g\cdot x$$ for some $x\in A$. Then, for any $b\in B^G$, $$(a,b)\mapsto\overline{\sum_{g\in G}(g\cdot x)\otimes b}=\overline{\sum_{g\in G}(g\cdot x)\otimes(g\cdot b)}=\overline{\Nm_G(x\otimes b)}=0\in\frac{(A\otimes B)^G}{\Nm_G(A\otimes B)},$$ as desired. One similarly shows that for any $a\in A^G$ and $b=\Nm_Gy$ ($y\in B$), $(a,b)\mapsto0$. Thus, the map $\overline a\otimes\overline b\mapsto\overline{a\otimes b}$ you wrote down is well-defined.