I have been reading Cassels-Frohlich, and I have a question about a fact that is cited in a proof about the cohomology of pro-finite groups (the exact palace is the proof of Prop 4 of section 2.8 of Chapter V).
Let $G$ be a pro-finite group, $H$ a closed normmal subgroup, and $A$ a discrete $G$-module.
Near the beginning of the proof, it is cited that $H^1(H,A)=0$ is equivalent to for all open normal subgroups $U$ of $G$, $H^1(HU/U,A^U)=0$. This does not make sense to me. $H^1(H,A)$ is the limit of the $H^1(H/H\cap U,A^{H\cap U})$. The second isomorphism theorem gives $HU/U\cong H/H\cap U$, but besides this, I do not see how to relate $A^U$ and $A^{H\cap U}$.
From the above assertion it seems plausible that $H^q(H,A)=\varinjlim_U H^q(HU/U,A^U)$ where the limit is taken over open normal subgroups of $G$, but looking at $q=0$ this seems false, as $H^0(HU/U,A^U)=A^{HU}$ and $H^0(H/H\cap U,A^{H\cap U})$ is $A^H$. So does the statement at the beginning of this paragraph hold for $q=1$, or does the statement being used in Cassel-Frohlich follow from some other method.
Thank you for any help
I have figured an answer to my question.
In particular, I claim that $\varinjlim_U H^q(HU/U,A^U)\cong \varinjlim_U H^q(H/H\cap U,A^{H\cap U})$.
To do this consider the isomorphism $\phi_U :HU/U\to H_{H\cap U}$ of groups and the inclusion $\psi_U:A^U\to A^{H\cap U}$. Consider $\psi$, the directed limit of the maps induced by $\psi_U$ on cohomology. I claim $\psi$ is an isomorphism. First assume that $x\mapsto 0$. Then we have some $U$ such that $x\in H^q(HU/U,A^U)$ and its image in $H^q(H/H\cap U,A^{H\cap U})$. $x$ is represented by some $f:(HU/U)^q\to A^U$. If it is a coboundary inside of $H^q(H/H\cap U,A^{H\cap U})$, there exists some $f':(H/H\cap U)^{q-1}\to A^{U\cap H}$ such that $df'=f$. However, note that $f'$ takes finitely many values, so by making $U$ smaller we find $V$ so that $A^V$ contains all the finitely many values, we see that $x=0$ in $H^q(HU/ U,A^{U})$ and in fact $x=0$ in $\varinjlim_U H^q(HU/U,A^U)$.
We must also prove surjectivity. Let $x\in H^q(H/H\cap U,A^{H\cap U})$, then $x$ is represented by some $f:(H/H\cap U)^q\to A^{U\cap H}$. So by making $U$ smaller, we find $V$ such that the image of $f$ is in $A^V$, and we get that $x$ is in the image of $H^q(HV/V,A^{H})$. This we have bijectivity.