# CoRes $\circ$ Res $= [G:H]Id$ on the cohomology groups of a profinite group

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.

However,

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?