Deriving an exact couple once

Since $k^1=0$, the differential $j^1{k}^1$ is zero and $E^2=E^1$ by the identity cycle quotient. For $D_{p,-p}^2$ take the image of $D_{p-1,1-p}^1\to D_{p,-p}^1$. It is zero for $p\le 0$, the subgroup $\{0,2\}$ for $p=1$, and the whole $\mathbb{Z}\mathrm{/}4$ for $p\ge 2$. These are the displayed terms.

The restricted $i^2$ at $p=1$ is the inclusion $\{0,2\}\hookrightarrow \mathbb{Z}\mathrm{/}4$ and at $p\ge 2$ is identity; at smaller indices it has zero source. For $a\in D_{1,-1}^2=\{0,2\}$ its preimage under the old inclusion is the same element of $D_{0,0}^1$, so $j^2(a)=a\in E_{0,0}^2$. At $p=2$ the old $i^1$ is identity on $\mathbb{Z}\mathrm{/}4$, so $j^2(a)$ is its parity class in $E_{1,-1}^2$. Every other $j^2$ has zero target, and $k^2=0$ by its defining formula. Thus the $j$ index changes by $(-1,1)$ rather than remaining degree zero.

At $D^2$ before $j^2$, when $p=1$ both the incoming $i^2$ image and the kernel are zero; when $p=2$ both are $\{0,2\}$; when $p\ge 3$ both are the whole group; and when $p\le 0$ both are zero. Every $j^2$ onto a nonzero $E^2$ term is surjective, so its image equals $\ker k^2$. All $i^2$ maps are injective, so their kernels are zero, exactly the incoming $k^2$ images. Off-diagonal terms are zero. This verifies all exactness claims of [F3] directly, with the transition indices now one and two. The derivation used only literal subgroup inclusions and quotient maps; no chosen section or AC occurs.