Suppose that we have a morphism between profinite groups $f: G_{1}\rightarrow G_{2}$ such that $f^{\ast}:H_{cont}^{\ast}(G_{2},A)\rightarrow H_{cont}^{\ast}(G_{1},A) $ is an isomorphism for any finite trivial $G_{i}$-module $A$. What can we say in general about $f$ ? and in particular when $G_{i}=\mathrm{Gal}(\overline{k_{i}}|k_{i})$, $i=1,2$.
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1$\begingroup$ By "trivial $G_i$-module" you mean the module has the trivial action, right? And presumably $f$ is continuous? So it looks to me like this says no more and no less than the statement that the map from the abelianisation of $G_1$ to the abelianisation of $G_2$ induced by $f$ is an isomorphism. By the way, how will continuous cohomology differ from profinite cohomology if $A$ is finite? Maybe they're the same? Or have I missed a subtlety? $\endgroup$– zntCommented Dec 20, 2016 at 16:02
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$\begingroup$ @znt Yes I do mean trivial action, and f is continuous with respect to the profinite topology. I do not know what is the profinite cohomology. I suspect that it is the same notion, no ? $\endgroup$– Muhammed AliCommented Dec 20, 2016 at 16:34
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$\begingroup$ By profinite cohomology I meant the direct limit of group cohomology $H^*(G/N,A^N)$ as $N$ runs through the normal subgroups -- the usual way one defines cohomology of a profinite group acting discretely (stabilisers are open) on an abelian group. It's an exact functor (unlike continuous cohomology, for which there are traps). $\endgroup$– zntCommented Dec 20, 2016 at 20:29
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$\begingroup$ @znt that is fine! we are talking about the same thing in this case I guess. $\endgroup$– Muhammed AliCommented Dec 20, 2016 at 20:59
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Here is one recent result in this direction, taken from I. Efrat and J. Minac, Galois groups and cohomological functors, Trans. of the AMS, http://www.ams.org/journals/tran/0000-000-00/S0002-9947-2016-06724-0/home.html :
Let $q=p^s$ be a prime power. Suppose that $G_i=\mathrm{Gal}(\overline{K_i}/K_i)$ for fields $K_i$, both containing a root of unity of order $q$, and take the trivial $G_i$-module $A=\mathbb{Z}/q\mathbb{Z}$ ($i=1,2$). For a profinite group $G$ consider the closed subgroups $G^{(2)}=G^q[G,G]$ and $G^{(3)}=(G^{(2)})^q[G^{(2)},G]$ (resp., $G^{(3)}=(G^{(2)})^{2q}[G^{(2)},G]$) when $p>2$ (resp., $p=2$). Then the map $f^*$ is an isomorphism if and only if $f$ induces an isomorphism $G_1/G_1^{(3)}\cong G_2/G_2^{(3)}$ of profinite groups.
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$\begingroup$ Thank you very much! I have finally found the full version! very impressive and in the same time it is great that the question was solved! $\endgroup$ Commented Dec 21, 2016 at 14:47