Five-term homology sequence of a group extension

Supply the resolution convention needed by [F3] as follows. For each relevant group $K$ and left $\mathbb{Z}K$-module $M$, take the free module on the underlying set of $M$, augment $[m]\mapsto m$, and repeat on each successive kernel. Each augmentation is onto its kernel target, so the resulting complex is an exact free, hence projective, resolution, specified uniformly in $M$. By [A1] and [L1], DC also holds. Apply [F2] to [F1]. Since $H_0(N;\mathbb{Z})=\mathbb{Z}$ with trivial $Q$-action, its terms are $E_{2,0}^2=H_2(Q;\mathbb{Z})$, $E_{0,1}^2=H_0(Q;H_1(N;\mathbb{Z}))$, $H_1=H_1(G;\mathbb{Z})$, and $E_{1,0}^2=H_1(Q;\mathbb{Z})$. This gives an exact sequence with the coinvariant group in the middle.

By [F3], $H_1(N;\mathbb{Z})=N\mathrm{/}[N,N]$ under the map sending $n$ to its class. The $Q$-coinvariants impose the relations $[gn{g}^{-1}]=[n]$ for every lift $g\in G$ and $n\in N$. Their subgroup in $N\mathrm{/}[N,N]$ is exactly $[G,N]\mathrm{/}[N,N]$, because $[N,N]\subseteq [G,N]$. Therefore $H_0(Q;H_1(N;\mathbb{Z}))\cong N\mathrm{/}[G,N]$.

Substituting the identification of step 1.2 into step 1.1 gives the displayed sequence. To check naturality, take a homomorphism of extensions. The bar classifying-space maps give a commuting square from $BG\to BQ$ to $B{G}^{\mathrm{\prime }}\to B{Q}^{\mathrm{\prime }}$; mapping-path replacement gives a map of their fibrations. Its fiber map induces the specified homomorphism $N\to N^{\mathrm{\prime }}$ on fundamental groups. Under the fiber identifications with $BN,B{N}^{\mathrm{\prime }}$ in [F1], [F5] identifies its homology map with that induced by $N\to N^{\mathrm{\prime }}$. By [F4], the map of spectral sequences commutes with $d^2$, all edge maps, and the abutment filtration. The five-term splice in [F2] uses only these maps and the natural subquotient identifications, so its squares commute. Finally, step 1.2 sends each $n$ to its class in the coinvariants, hence commutes with the homomorphism on $N$ and its conjugation action. This proves naturality of the displayed sequence, including the transgression.