Prove
Lemma 13 ([Reduction rules]). In either minimum diagram the following…Theorem 7.1 ([Geometric conclusion]). There are absolute constants C_*,C_ such…Proposition
An arc of length greater than two is freely reduced. Otherwise an inverse pair with an unmarked bivalent midpoint can be erased, identifying its two outer vertices. Those outer vertices are distinct in the diagram: the interior of a maximal arc has no repetitions, and a loop arc of length greater than two has this property for each such pair. The two-edge path is therefore a tree, and its contraction preserves the number of faces and the plane embedding. Its outer group values agree; its bivalent middle vertex has no other incidences. On every assigned source walk foldedness identifies the inverse pair as a backtrack. The contraction therefore preserves all assignments and decreases the number of edges, which is impossible. If an arc has a generator-side incidence it has length one. Otherwise an arc incident with a bounded face has two source assignments. Its other incidence cannot be on the same bounded face, since an edge with the same face on both sides is a bridge, excluded by Lemma 13 ([Reduction rules]). If the arc were at least L_i long on either compared assignment, its freely reduced word would determine its projected nonbacktracking path in Y . Both projections would be this same path. The two lifts of that path then differ by a single deck transformation, so their source incidences are aligned. Lemma 13 ([Reduction rules]) forbids that alignment. Arcs of length at most two already satisfy the claimed inequalities. If a face had fewer than twenty arc incidences, select its longest arc, of length a , and let i be its initial layer in the face assignment. Its perimeter would be at most 19a<20L_i=2000H_i/_2d . By, every layer j on it has H_j≥H_i/2 for large ℓ . The face word is not freely trivial. Cancel backtracks in its closed source walk, and then cancel around its cyclic ends. Foldedness makes these actual source cancellations; the remaining walk is nonempty and cyclically nonbacktracking. Its length is at most the displayed perimeter, while C_*H_j/_2d≥5000H_i/_2d . This contradicts . Let v be the number of vertices of the compressed diagram. It is connected and satisfies b=v+J-1 by the plane Euler formula, also for loops, parallel edges and bridges. Every unmarked vertex has degree at least three; a graph consisting of one marked vertex and no edges also satisfies the resulting inequality. Hence 2b≥3(v-m) , which gives the first inequality of . Each arc has two face incidences in total, counting the exterior, so the twenty-incidence bound gives its second inequality. Combining them gives b≤(30/7)(m-1)≤6m whenever edges occur; the same final bound holds for the trivial diagram.