$G$ be a profinite group. I want to see that the profinite cohomology group $H^2_{cts}(G, \mathbb Q ) $ is $0$.
Using inflation maps we see profinite cohomology group as direct limit of cohomology groups $$ H^r _{cts}(G,\mathbb Q)= \varinjlim _H H^r(G/H,\mathbb Q ^H)$$
where $H$ runs through the open normal subgroups of $G$. These open normal subgroups have finite index as profinite groups are compact thus $G/H $ is finite. For $r>0$ $[G:H] H^r(G/H,\mathbb Q^H) =0$. So $ H^r _{cts}(G,\mathbb Q)$ is torsion.
$\mathbb Q$ is uniquely divisible. Hence so is $H^r(G/H,\mathbb Q ^H)$. Suppose for element $y$ of this group $ay=0$. But multiplication by $a$ is isomorphism so $y=0$.
Have I got it right ?