Toric cycle classes are determined by two winding parities

The primal vertex-edge graph has the explicit spanning tree $T=\{h_{i,0}:0\le i<L-1\}\cup \{v_{i,j}:i\in {\mathbb{Z}}_L,0\le j<L-1\}$. The horizontal edges form a path through row zero, and each other row is attached to that path by one vertical column path; hence $T$ is connected and has no cycle. For each edge $e\mathrm{\notin }T$, add $e$ to the unique tree path between its endpoints. This gives a cycle ${\gamma }_e$ containing $e$ and no other edge outside $T$. These cycles are independent because the coefficient of each non-tree edge in a linear combination is its own coefficient. They span $Z_1$: subtract from any cycle the ${\gamma }_e$ corresponding to its non-tree edges; the remainder is an even-incidence chain supported on the tree, and a nonempty finite forest has a leaf of odd degree, so the remainder is zero. Thus $\dim Z_1=\mathrm{\mid }E\mathrm{\mid }-(\mathrm{\mid }V\mathrm{\mid }-1)=L^2+1$.

Every face boundary has even incidence at each vertex, so $B_1\subseteq Z_1$. By [F2], $\dim B_1=L^2-1$. For a cycle $c$, define $w_h^k(c)={\sum }_{j}c(h_{k,j})$ and $w_v^{\mathrm{\ell }}(c)={\sum }_{i}c(v_{i,\mathrm{\ell }})$, with all sums in ${\mathbb{F}}_2$. Summing the even-incidence equations over column $k$ shows $w_h^k=w_h^{k-1}$; summing over row $\mathrm{\ell }$ shows $w_v^{\mathrm{\ell }}=w_v^{\mathrm{\ell }-1}$. Thus the two parities $w_h,w_v$ are independent of the chosen cuts. Each face boundary crosses every such cut zero or two times, so both parities vanish on $B_1$.

The horizontal fundamental loop ${\gamma }_h={\sum }_i{h}_{i,0}$ has $(w_h,w_v)=(1,0)$, and the vertical fundamental loop ${\gamma }_v={\sum }_j{v}_{0,j}$ has $(w_h,w_v)=(0,1)$. Therefore no nonzero combination of these loops lies in $B_1$. A basis of $B_1$ together with ${\gamma }_h,{\gamma }_v$ is an independent family of $(L^2-1)+2=L^2+1$ cycles, which is the dimension of $Z_1$ from step 1.1. It is therefore a basis of $Z_1$, proving that the winding map identifies $Z_1\mathrm{/}{B}_1$ with ${\mathbb{F}}_2^2$. The dual square cellulation has the same coordinate graph and incidence counts, so the identical argument proves its two winding coordinates.