Toric incidence maps have exactly one relation each

Define $S:{\mathbb{F}}_2^V\to {\mathbb{F}}_2^E$ by sending a coordinate vector at a vertex to its star-support vector and extending linearly. The coefficient of an edge joining vertices $q,q^{\mathrm{\prime }}$ in $S(a)$ is $a_q+a_{q^{\mathrm{\prime }}}$. Thus $\ker S$ consists exactly of assignments constant along every edge. By [F1] the vertex graph is connected, so $\ker S$ is the one-dimensional space of constant assignments. Since $\dim {\mathbb{F}}_2^V=L^2$, [F2] gives $\dim \mathrm{im}S=L^2-1$. The all-vertex sum is zero because each edge has two endpoints, so this is the only star relation.

Define $T:{\mathbb{F}}_2^F\to {\mathbb{F}}_2^E$ by sending a face coordinate vector to that face's boundary and extending linearly. An edge separating faces $f,f^{\mathrm{\prime }}$ has coefficient $b_f+b_{f^{\mathrm{\prime }}}$ in $T(b)$. Thus $\ker T$ consists exactly of assignments constant along adjacent faces. By [F1] the face-adjacency graph is connected, so $\ker T$ is the one-dimensional space of constant assignments. Since $\dim {\mathbb{F}}_2^F=L^2$, [F2] gives $\dim \mathrm{im}T=L^2-1$. The all-face sum is zero because each edge borders two faces, so this is the only face-boundary relation. The dual face-adjacency graph has the same square-grid incidence, proving its incidence rank is also $L^2-1$.