Each edge occurs twice among the square boundaries, so \(\prod _pU_p=1\). Conversely, choose a spanning tree in the connected dual graph, whose vertices are squares and whose edges cross their shared primal edges. Root this tree and set all primal edge signs to \(1\). Process dual leaves towards the root. At each leaf, its unique remaining dual-tree edge crosses a primal edge shared with its parent. Flip that primal sign exactly when needed to give the leaf its prescribed plaquette sign, then remove the leaf. This changes only the leaf and its parent among the unprocessed squares and changes no previously fixed square. At the last square, the global product constraint forces the prescribed root sign. This constructs every even-parity plaquette pattern.

The plaquette map is a homomorphism between finite abelian groups. Multiplication by one preimage identifies its kernel with any nonempty fiber. There are \(2^{P-1}\) target patterns, hence each fiber has \[\frac {2^{|E|}}{2^{P-1}}=2^{n+1}\] elements. For a connected graph, a gauge transformation fixes an abelian connection exactly when \(g_{h(e)}=g_{s(e)}\) on every edge, that is, when all vertex values are the same. The stabilizer has two elements, so every gauge orbit has \(2^{n-1}\) elements. The fiber therefore contains \(2^{n+1}/2^{n-1}=4\) orbits.

Take a horizontal fundamental cycle in one row and a vertical one in one column. Multiply by \(-1\) every horizontal edge crossing the periodic seam from \(x=N_1-1\) to \(x=0\). Each plaquette contains zero or two such edges, so its sign is unchanged, while the horizontal cycle holonomy flips and the vertical one does not. The analogous vertical seam operation flips just the other holonomy. The two operations give 4 possibilities for these gauge-invariant signs. They are distinct gauge orbits, and the count above shows that they exhaust the 4 orbits in each plaquette fiber.