The exact couple of a two step filtration

Homology of each piece equals its degree-zero group and vanishes in all other degrees. The successive quotient at $p=0$ is $\{0,2\}$, isomorphic to $\mathbb{Z}\mathrm{/}2$ by $[a{]}_2\mapsto [2a{]}_4$. At $p=1$ it has cosets $\{0,2\},\{1,3\}$, identified with $\mathbb{Z}\mathrm{/}2$ by parity. All other graded quotients are zero. This proves every displayed $D^1$ and $E^1$ term, including the infinite constant $D^1$ tail.

The $i$ arrow from $D_{0,0}^1$ to $D_{1,-1}^1$ is the inclusion of $\{0,2\}$; at every $p\ge 1$ it is identity into the next $\mathbb{Z}\mathrm{/}4$. At $p<0$ it is the map from zero. The $j$ arrows are the stated identity and parity quotient at $p=0,1$, and zero to zero targets elsewhere. Every $k$ lowers total degree to minus one, where the $D^1$ terms vanish, so $k=0$. These maps have the exact degrees in [F3].

Check exactness at $D^1$ before $j$: at $p=0$, the incoming $i$ image and $\ker j$ are zero; at $p=1$, both are $\{0,2\}$; at $p\ge 2$, both are all of $\mathbb{Z}\mathrm{/}4$; at $p<0$ both are zero. At every $E^1$ term $j$ is onto, so $\mathrm{im}j=\ker k=E^1$. At $D^1$ before $i$, each $i$ is injective, so $\ker i=0=\mathrm{im}k$. All off-diagonal terms give zero equalities. Thus all vertices are explicitly exact. The $d^1=jk$ differentials are zero, and the nonzero $(1,-1)$ term prevents a first-quadrant interpretation. No AC or representative section is used.