Finite square-lattice cycles are boundaries

For each row $j$, sum the cycle equations over $i\in \mathbb{Z}$. The horizontal terms cancel in pairs, giving $S_j=S_{j-1}$, where $S_j:={\sum }_i{V}_{i,j}$. Finite support gives $S_j=0$ for all sufficiently large $\mathrm{\mid }j\mathrm{\mid }$, so $S_j=0$ for every $j$. Define a face coefficient $F_{i,j}:={\sum }_{k>i}{V}_{k,j}$ in ${\mathbb{F}}_2$. For fixed $j$, this coefficient is zero for $i$ sufficiently large by finite support and for $i$ sufficiently negative because then it equals $S_j=0$. It is also zero for all but finitely many rows. Hence $F$ is a finitely supported face chain.

The vertical edge $v_{i,j}$ belongs to the boundaries of $f_{i-1,j}$ and $f_{i,j}$, so its coefficient in $\mathrm{\partial }F$ is $F_{i-1,j}+F_{i,j}=V_{i,j}$. Thus $\mathrm{\partial }F$ and the given chain agree on every vertical edge.

The horizontal edge $h_{i,j}$ belongs to the boundaries of $f_{i,j-1}$ and $f_{i,j}$. Its coefficient in $\mathrm{\partial }F$ is $F_{i,j-1}+F_{i,j}={\sum }_{k>i}(V_{k,j-1}+V_{k,j})$. The cycle equation at $(k,j)$ replaces each summand by $H_{k,j}+H_{k-1,j}$; summing over $k>i$ telescopes to $H_{i,j}$ because the sequence is finitely supported. Thus $\mathrm{\partial }F$ and the given chain agree on every horizontal edge, proving $\mathrm{\partial }F=(H,V)$. Translating the dual square lattice by $(1\mathrm{/}2,1\mathrm{/}2)$ identifies it with the same integer grid, so the identical construction proves the dual assertion.