Exact sequence out of commutative exact diagram I'm trying to get grip on the following commutative exact diagram:
 
I know where the maps come from and could verify the exactness and the other maps. (It is induced by the long exact sequence of Poitou-Tate, a local duality theorem and the definitions of the groups).
But now there is a lemma implying that we have an exact sequence
$$0 \rightarrow Ш^1(k_S,A') \rightarrow Ш^1(k_S,S \setminus T, A') \rightarrow coker^1(k_S,T,A)^\vee \rightarrow 0$$
I cannot see where the surjective map in this exact sequence comes from. I see, that for $s \in Ш^1(k_S,S \setminus T,A')$ we can map it to an element in the restricted product $\prod_S H^1(k_\mathfrak{p},A')$ by choice of an preimage. But from there I do not see any connection.
Does somebody has a hint how to proceed?
Thank you :-)
 A: It's not easy to explain a diagram chase in writing, but I'll try:
Let $x \in Ш^1(k_S, S \setminus T, A')$, map it to $y \in H^1(k_S | k, A')$ then to $z \in \prod_S H^1(k_\mathfrak{p}, A')$. The image of $z$ in $\prod_{S \setminus T} H^1(k_\mathfrak{p}, A')$ is zero by exactness of the first row, so actually $z \in \prod_T H^1(k_\mathfrak{p}, A')$. $z$ is mapped to $u$ in $\prod_T H^1(k_\mathfrak{p}, A')^\vee$. This $u$ is mapped to zero in $H^1(k_S|k, A)^\vee$, because of the exactness of the first row: $z$ came from $y \in H^1(k_S | k, A')$. Therefore $u \in \mathrm{coker}^1(k_S, T, A)^\vee$.
I'll leave you to check that this map is well-defined. It's surjective, because if $u$ is in the cokernel, you can map it to the "restricted product$^\vee$" and then to the "restricted product" (because of the iso), then it's easy enough to show that everything comes from an $x \in Ш^1(k_S, S \setminus T, A')$. Then the kernel is computed easily enough (it's very mechanical).
Hopefully this helps:

