 Solution

Let $\hat{\zeta }\in H^2(\hat{\Gamma },M)$ lie in the kernel of ${\iota }^{\ast }$, and represent it by a profinite extensionIts pullback to $\Gamma$ is split, so there is a homomorphism $s:\Gamma \to E$ satisfying $\pi s=\iota$. By part b, $s$ extends uniquely to a continuous homomorphism $\hat{s}:\hat{\Gamma }\to E$. The continuous maps $\pi \hat{s}$ and the identity of $\hat{\Gamma }$ agree on the dense image of $\Gamma$, hence agree everywhere. Thus $\hat{s}$ is a section of $\pi$, the extension splits, and $\hat{\zeta }=0$. Therefore