Five-term sequence of a composite functor

Substitute $F=(-{)}^N$ and $T=(-{)}^Q$ into F1. The five terms become, in order, $H^1(Q,M^N)$, $H^1(G,M)$, $H^1(N,M{)}^Q$, $H^2(Q,M^N)$ and $H^2(G,M)$. F2 identifies the first and last maps with its bottom-cycle inflation, the second with restriction of invariant cocycles, and the middle with $d_2$ from $(0,1)$ to $(2,0)$. Hence every term and arrow agrees with F2, without importing a later cocycle-classification theorem.

To compute the specialization let $C_2=⟨s⟩$. Its trivial $\mathbb{Z}[C_2]$-module has the free resolution with augmentation and alternating differentials $d_1=s-1$, $d_2=1+s$, $d_3=s-1$, continuing periodically. Indeed for $a+bs$, the kernel of $s-1$ is $\mathbb{Z}(1+s)$ and the kernel of $1+s$ is $\mathbb{Z}(1-s)$; these are the preceding images. The augmentation kernel is also $\mathbb{Z}(1-s)$. Rank-one free terms are projective by lifting the image of their generator. Hom into trivial $\mathbb{Z}$ has successive differentials $0,2,0,2,\dots$. Thus F3 gives $H^1(C_2,\mathbb{Z})=0$ and $H^2(C_2,\mathbb{Z})=\mathbb{Z}\mathrm{/}2$.

For $Q=1$, invariants are identity and the positive cohomology is zero by exactness of any supplied resolution. Consequently step 1.1 has four zero non-initial terms before its final $\mathbb{Z}\mathrm{/}2$, as claimed. All maps in that displayed finite portion are zero, its transgression is zero and exactness holds at every required position. There is no final surjectivity. The rank-one periodic calculation itself is choice-free; only the common resolution-independent comparison convention is inherited.