Let $G$ be a profinite group, $H \leq G$ open. It is known that thus $H$ is closed and has finite index in $G$.

Any $G-$module is an $H-$module and one can construct the restriction map as the extension of $H^0(G,A) = A^G \to A^H = H^0(H,A)$, using the universality of the cohomological functor.

The trace map works the other way: we have a natural map CoRes$:H^0(H,-) \Rightarrow H^0(G, -)$ given by $a \mapsto \sum ag_i$ where $g_i$ are representative of the right cosets of $H$ in $G$.

Clearly it follows that in dimension $0$, CoRes $\circ$ Res $ = [G:H]Id$.

In the book by Ribes he states that the reason this implies equality in all dimensions of cohomology is because of the following proposition:

If $\eta_0:H^0(G,-) \Rightarrow H^0(G,-)$ is an isomorphism, then its extension to the cohomology complex is also an isomorphism.


  1. I don't see why it follows from the above, does it?

  2. Doesn't it simply follow by the universality of the cohomological functor? The extensions of these natural transformations must equal. Am I missing something?


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