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

Choose a normalized set-theoretic section $s:H\to E$ of $\pi$, so $s(1)=1$. Define the extension cocycleAssociativity of $s(h)s(k)s(l)$ gives, in multiplicative notation for $M$,which is exactly the two-cocycle identity. Thus $\varphi$ represents the class of the group extension in $H^2(H,M)$.

The fiber product of groupsis a group under componentwise multiplication. The maps $m\mapsto (m,1)$ and $(e,g)\mapsto g$ give an exact sequenceThe section $g\mapsto (s(f(g)),g)$ has extension cocycleTherefore this pullback extension represents $f^{\ast }([\varphi ])\in H^2(G,M)$.

Assume $f$ is injective. If two elements $(e,g),(e^{\prime },g^{\prime })\in E^{\prime }$ have the same image under the first projection $E^{\prime }\to E$, then $e=e^{\prime }$. Applying $\pi$ gives $f(g)=f(g^{\prime })$, so injectivity of $f$ gives $g=g^{\prime }$. Hence the natural map $E^{\prime }\to E$ is injective.

For every open normal subgroup $U◃G$, the compositehas finite image and therefore factors uniquely through the profinite completion $\hat{\Gamma }$. These factor maps are compatible as $U$ varies. The universal property of an inverse limit consequently produces a continuous homomorphismwith $\hat{f}\iota =f$. It is unique because $\iota (\Gamma )$ is dense in $\hat{\Gamma }$ and two continuous maps into the Hausdorff group $G$ that agree on a dense subset agree everywhere. This is the universal property of profinite completion.

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