I've been reading the book 'Cohomology of Number Fields' for years. But I couldn't check the commutativity of the diagram
on page 126 until now. So I ask for help.
The diagram is induced by taking cup-product with $t$ on a double complex, so there must be some commutative diagram, since $tr$ is also an edge morphism.
More explicitly, the edge morphism is defined as following
$Z^{p,0}_{2}/(B^{p,0}_{1}+Z^{p+1,-1}_{1}) \rightarrow Z^{p,0}_{2}/(B^{p,0}_{m-1}+Z^{p+1,-1}_{1}) \simeq Z^{p,0}_{m}/(B^{p,0}_{m-1}+Z^{p+1,-1}_{m-1})=E^{p,0}_{m}$ for sufficiently large $m$.
In this view, the lower arrow is induced by $C^{n-j}(G/U, H^{0}(U, \tilde{X}^{n}(G,A)^{\ast}) \rightarrow C^{n}(G/U, H^{0}(U, \tilde{X}^{n}(G,\mathbb{Z})^{\ast})$.
And when we define the edge morphism, we are using the Lemma 2.2.4, hence the $E^{p}$ is represented by elements of $C^{0}(G/U, H^{0}(U, \tilde{X}^{j}(G,A)^{*})$ i.e. (0,p)-component.
Eventually I could see that the lower arrow can be induced by $C^{0}(G/U, H^{0}(U, \tilde{X}^{j}(G,A)^{\ast}) \rightarrow C^{j}(G/U, H^{0}(U, \tilde{X}^{j}(G,\mathbb{Z})^{\ast})$
But I couldn't send the element of $C^{j}(G/U, H^{0}(U, \tilde{X}^{j}(G,\mathbb{Z})^{\ast})$ to the (0,n) component. The image of $tr$ is represented by (0,n)-component.
So I ask for any kind of help.
And I want to ask how I can understand the dualizing modules, Tate spectral sequence intuitively. Any kind of references are welcome.
Thank you.
