Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 151 / 5 / c

Choose $\hat{\zeta }\in H^2(\hat{\Gamma },M)$ with ${\iota }^{\ast }(\hat{\zeta })=\zeta$, and letbe its extension. By the given fact, $E$ is a profinite group, hence is residually finite. The extension $E_{\zeta }^{\prime }$ over $\Gamma$ is the pullback of $E$ along the injective map $\iota$. Part 5(a)(iii) embeds $E_{\zeta }^{\prime }$ into $E$. Since every subgroup of a residually finite group is residually finite, so is $E_{\zeta }^{\prime }$.

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