Proof. Initially use every point of \(S^h\) as a mark, independently choosing a uniform direction for the line through it. The number of directions is \[J=\frac{q^h-1}{q-1}.\] For a fixed \(z\in P\), each mark other than \(z\) gives an incidence with probability \(1/J\). Its degree is therefore binomial, with mean \(\mu_z\sim cq\) uniformly in \(z\) and variance at most \(\mu_z\). For large \(q\), \(\mu_z\ge8r^2\), so the second-moment inequality gives \[\Pr\bigl(|I(z)|<4r^2\bigr) \le \frac{\mu_z}{(\mu_z-4r^2)^2}=O(q^{-1}).\] The expected number of low-degree points is \(O(m/q)\). By Markov’s inequality, with probability tending to one there are at most \(m q^{-1/2}\) such points.

For \(2\le s\le6\), count simple cycles with \(s\) point vertices and \(s\) line labels. There are at most \(m^s\) ordered choices of the points. Consecutive points force a geometric line, on which there are at most \(q\) possible marks. For distinct line labels, the prescribed directions occur with probability \(J^{-s}\) by independence. Choices whose mark is a forbidden punctured endpoint can only increase this upper bound. The expected number of these cycles is consequently at most \[m^s q^s J^{-s}\le q^{2s},\] since \(J\ge q^{h-1}\). This argument also allows different labels to specify the same geometric line: their random directions remain independent. Summing over \(s\) shows that, with probability tending to one, the total number of cycles of length at most \(12\) is at most \(q^{13}\).

Choose an outcome satisfying both bounds. Delete every label appearing in one of those cycles, at most \(6q^{13}\) labels, and let \(E\) contain the old low-degree points and all points of the discarded lines. Deleting labels creates no cycle, and outside \(E\) it changes no degree. Moreover, \[|E|\le m q^{-1/2}+6q^{14}=o(m),\qquad N=|S|^h-O(q^{13})=(c+o(1))m.\] As \(cr\ge100\), these estimates give (2) for large enough \(q\): eventually \(N/m\ge c/2\) and \(r|E|/m\le1\), whereas \(cr/8\ge12.5>3\). ◻

Proof. At \(O\), each block gives a complete bipartite link on its \(L\)- and \(R\)-directions. For fixed \(u,v,j,k\), the equations \[\alpha_b(i)=j-v,\qquad \beta_b(i)=k-u\] specify exactly one \(i\in I_b\). Thus there is one link edge per pair of directions, and every nonempty cyclically reduced closed walk has at least four edges. Its angular length is at least \(12\pi/5\).

Project the link at \(V\) to the incidence graph of Lemma 4 by sending an \(x_{iuv}\)-direction to \(i\) and an \(L_{b,u,j}\)-direction to \(p(b)\). This map is locally injective on edges. At \(x_{iuv}\) there is at most one incident triangle over any point \(z\), because \(i\) lies in at most one block at \(z\). At \(L_{b,u,j}\), fixing \(i\) determines at most one \(v\in[r]\) from \(j=\alpha_b(i)+v\), since \(s_b\ge r\). The link at \(W\) has the same property using \(t_b\ge r\) and the \(R\)-directions. Every cyclically reduced link walk therefore has a reduced image and at least \(14\) edges, of angular length at least \(14\pi/5\).

A reduced path between distinct \(x\)-directions with the same \(i\) projects to a positive closed walk reduced at every internal vertex. It need not be reduced at its closing seam, but it still contains a simple cycle: choose a shortest positive subwalk with equal endpoints. A subwalk of length two would be an internal reversal, and one of length one is impossible in this graph. The path therefore has at least \(14\) edges. A shortest link path is reduced, proving the distance assertion. ◻

Proof. Let \(d=\mathop{\mathrm{rank}}_{\mathbb R}M\); the assertion is immediate for \(d=0\). A nonzero \(d\times d\) minor \(B\) has \[1\le|\det B|\le r^{d/2},\] by Hadamard’s determinant inequality, obtained by orthogonalizing its rows. If \(d'=\mathop{\mathrm{rank}}_{\mathbb F_q}B\), choose \(d'\) rows spanning its row space modulo \(q\). Subtract integer lifts of their linear combinations from the other rows. These determinant-preserving operations make \(d-d'\) rows divisible by \(q\), whence \[q^{d-d'}\mid\det B,\qquad q^{d-d'}\le r^{d/2}.\] Since \(q>r\), we obtain \(d'\ge d/2\), as required. ◻

Proof. At a point \(x\), cut the incident cells into sectors centered at \(x\) and choose \(\rho>0\) smaller than the distances to their nonincident faces. The resulting truncated star consists of finitely many sectors and, possibly, isolated radial intervals, glued along radial sides. Its radial coordinate \(r\) agrees across these gluings, and changes by at most the length of a polygonal path. In particular, a path from \(x\) leaving the star has length at least \(\rho\). For a point of radius \(r<\rho\), the radial segment has length \(r\); any shorter competitor stays in the star and has length at least \(r\). Hence its distance from \(x\) is exactly \(r\).

This description proves positivity of the distance and identifies the small metric neighborhoods with the usual star neighborhoods. The metric therefore induces the complex topology. It also supplies contracting neighborhoods, by radial contraction. Finiteness of the complex gives compactness. Parameterize a sequence of polygonal paths whose lengths decrease to the distance between two fixed points proportionally to arclength on \([0,1]\). Their speeds are uniformly bounded. Compactness and equicontinuity give a uniformly convergent subsequence: this follows, for example, by a diagonal choice on a countable dense subset of \([0,1]\) and then the common modulus of continuity. For every fixed partition, the sum of distances between successive points passes to the limit. Taking the supremum over partitions shows that the limiting path has length at most the limiting lengths. Since every path has length at least the distance of its endpoints, this is a shortest path.

Consider now a sufficiently short unit-speed minimizing arc starting at \(x\). Its image lies in the truncated star, and the preceding radial formula gives \(r(t)=t\). At positive radius, its direction varies continuously in the finite graph of directions of the star. If that direction were nonconstant, it would traverse an interval in the interior of one of the graph’s edges: a continuous map with image in the finite set of junctions and isolated directions is constant. Choose a subarc inside the corresponding open sector whose endpoint directions differ. The sector is locally Euclidean, and the length of this subarc is at least the Euclidean distance of its endpoints, which is strictly larger than the difference of their radii. This contradicts \(r(t)=t\) and unit speed. Thus the arc is radial and straight in one sector or radial edge. Apply the argument on both sides of each parameter value of a local geodesic. A finite subcover of its compact parameter interval gives finitely many straight pieces. For a closed curve use the parameter circle, so this also includes its seam.

Finally, an angular path of length less than \(\pi\) between the incoming backward direction and the outgoing direction yields a shorter chord. Indeed, take two nearby points at equal distance from the turn, unfold the consecutive sectors traversed by that angular path, and join the points by the straight chord in the sector of total angle less than \(\pi\). Its length is less than the sum of the two radial lengths. The chord maps back to a path in the complex; the unfolded sector need not be embedded there. This contradicts local minimality. ◻

Proof. We form a labelled picture from a disk filling, minimize its number of nodes, erase components disjoint from the boundary, and take the planar dual, retaining the prescribed boundary word, including spurs. Start with a continuous filling of the path by a disk, with its boundary mapped linearly along each prescribed edge traversal. Mark all boundary breakpoints. Barycentric coordinates for the vertices of the genuine simplicial complex \(S\), extended by zero outside their stars, are continuous. Triangulate the disk sufficiently finely, respecting the marked boundary points, that each coordinate varies by less than \(1/4\) on each domain triangle. Label a domain vertex by a target vertex of largest coordinate in its image. This coordinate is at least \(1/3\), because \(S\) has dimension at most two. At every point of a domain triangle all its chosen labels therefore have positive coordinate, since \(1/3-1/4>0\). They belong to one target simplex. Extending the labels linearly gives a simplicial map, allowing collapsed edges and triangles.

Along each prescribed boundary edge the only possible labels are its two endpoints. The boundary parametrization is monotone along that edge, so the largest-coordinate label changes from the initial endpoint to the terminal endpoint exactly once; either choice at a midpoint tie has this property. Consequently the noncollapsed boundary steps are exactly the prescribed edge word, in its prescribed order. The same argument on an interval or circle shows that every loop based at a vertex has a based edge-loop representative: the original map and its approximation interpolate inside common target simplices, fixing the basepoint.

For the simplicial filling, draw the inverse images of ties for the largest target coordinate. Inside a domain triangle mapped onto a target triangle they are three arcs meeting at a trivalent node. Inside a triangle mapped to an edge they form one separating arc, and inside a triangle mapped to a vertex they are absent. Suppress all bends of valence two. We obtain a finite polygonal picture: its regions are labelled by target vertices; an arc separates distinct labels that span a target edge; and the three labels at every interior trivalent node span a target triangle. Its boundary endpoints are leaves, one for each letter of the prescribed edge word. Components disjoint from the boundary, including circles, are allowed.

Among all labelled pictures with these local rules and these boundary data, choose one with the fewest trivalent nodes. No arc can join two nodes labelled by the same target triangle. To see this, let the labels on its two sides be \(a,b\), and let \(d\) be the third vertex of that triangle. The selected arc and its endpoints form an embedded interval. A sufficiently thin regular neighborhood is a disk meeting the four other incident half-arcs in four distinct boundary slots (Figure 2). The two slots on the \(a\)-side both separate \(a\) from \(d\); those on the \(b\)-side both separate \(b\) from \(d\). Erase the two nodes and the intervening arc and join each of these two pairs along its side of the neighborhood, with label \(d\) between the new arcs. The new arcs do not cross, and every boundary label of the neighborhood is unchanged. This gives a picture with two fewer nodes.

The argument is unaffected if some of the four half-arcs reconnect outside the neighborhood: their slots remain distinct and their outside continuations retain the same labels. In particular, an additional arc between the nodes merely makes a lens outside the chosen neighborhood. An arc cannot join a node to itself, since the three incident arcs there have three different target-edge labels.

Next erase every component of the picture that misses the boundary. Here the existence of consistent region labels deserves attention. The retained picture together with the boundary circle is a connected polygonal graph \(H\). Each of its inside regions is a disk interior whose boundary is a walk, possibly with repeated vertices and edges. One way to see this is to build \(H\) from the boundary circle: first attach tree edges reaching all its other vertices, which slit disk regions, and then add the remaining arcs, each of which splits a disk region into two. Cutting apart repeated boundary incidences gives a polygon for each region, with its interior embedded in the original disk.

Take a thin regular neighborhood of \(H\) missing all erased components, which are finitely many compact sets disjoint from \(H\). For each inside region, one connected curve in the boundary of this neighborhood follows its entire boundary walk on the region side. In particular it follows the two sides of a bridge separately, and follows every incidence at a repeated vertex. This curve crosses no arc of the old picture and so lies in one old labelled region. All labels prescribed along the boundary incidences of the retained region are therefore equal. Extend that label over the region. Erasing these components cannot increase the number of nodes, so minimality is preserved.

Now construct its dual. Put a vertex in each region, and in that region’s polygon join this vertex to the midpoint of each incident arc side by radii disjoint except at the vertex. Join the radii across the arcs to obtain dual edges. Around each trivalent node the three consecutive pairs of radii enclose a triangular region. Its three dual vertices are distinct, since their region labels are the three distinct vertices of a target triangle. The dual graph is connected, as can be seen by taking paths between regions transverse to the picture arcs.

An Euler count verifies that these triangles fill all its bounded faces. Let \(E,N,L\) denote the numbers of retained picture arcs, trivalent nodes, and boundary leaves. The boundary word is nonempty, so \(L>0\). Adjoining the circle subdivided at the leaves gives a connected planar graph with \(N+L\) vertices and \(E+L\) edges. If \(F\) is its number of inside regions, Euler’s formula gives \[(N+L)-(E+L)+(F+1)=2, \qquad F=E-N+1.\] The dual graph has \(F\) vertices and \(E\) edges, hence exactly \(E-F+1=N\) bounded complementary faces. The \(N\) disjoint triangular regions already constructed exhaust them. Filling these triangles gives the required planar diagram with no holes. The labels define its map to \(S\); assign each edge and triangle the metric of its image.

At a boundary leaf of the picture, its dual edge has an exterior incidence. The interval between two successive leaves identifies the corresponding sector at a region vertex. Following the original boundary circle thus follows the dual exterior walk in exactly the prescribed order. In particular, a picture arc with both endpoints on the boundary produces an edge with no incident triangle, encountered twice. If there are no nodes, the dual has \(E+1\) vertices and \(E\) edges and is a tree; its spurs are still part of the exterior walk.

At an interior dual vertex, an immediate reversal of two successive link edges would mean that the corresponding adjacent triangles map to the same target triangle. Their picture nodes are joined by the arc dual to their shared edge, contradicting minimality. This includes the last and first corners in the cyclic order. Finally, every simple edge circuit in the diagram bounds its filled planar disk. Any closed edge walk splits at repeated vertices into such circuits and cancellable backtracks, so every edge loop is nullhomotopic in the diagram. ◻

Proof. First work in a simplicial subdivision and suppose such a path were nullhomotopic. Apply Lemma 8. At an interior vertex the summand in (15) is nonpositive. At a vertex with \(n_y=1\), its consecutive interior corners supply a link path between the two boundary directions, of length \(\theta_y\). The prescribed turn therefore implies \(\theta_y\ge\pi\), so this summand too is nonpositive. The only possibility with no intervening corner is a tip of a naked edge: its two directions coincide, giving turn distance zero, and is excluded by the hypothesis. If \(n_y\ge2\), then \((2-n_y)\pi-\theta_y\le0\) without any further condition. Thus every summand is nonpositive, contradicting (15).

For a path in the original complex, pass to the barycentric subdivision. Its old turn distances are unchanged. At a new point along an original edge, the straight-through directions have link distance \(\pi\) when the edge belongs to a triangle, and infinite distance otherwise. The subdivided path therefore satisfies the same hypothesis. ◻

Proof. We first justify a shortest-loop argument using only the local metric description. The contracting star neighborhoods and compactness give \(\epsilon>0\) such that every ball of radius \(5\epsilon\) is contained in a neighborhood contractible in \(K\). Indeed, take a finite cover by contracting stars with Lebesgue number \(a>0\), and choose \(\epsilon<a/20\), so each such ball has diameter less than \(a\). Every loop of length less than \(5\epsilon\) is contained in such a ball about one of its points and is nullhomotopic.

Uniformly \(\epsilon\)-close loops are freely homotopic. To verify this, partition their common parameter circle so that on each interval the first loop stays within \(\epsilon\) of its starting point. Join the corresponding partition points by shortest paths, of length less than \(\epsilon\), using Lemma 7 and the same connector at the cyclic seam. On each interval the second loop stays within \(2\epsilon\) of that same starting point, by pointwise closeness; no speed bound on the second loop is needed. The two subarcs and the two connectors form a quadrilateral loop contained in a ball of radius \(3\epsilon\). It contracts in the chosen larger neighborhood. Fill each such quadrilateral and glue along the shared connectors to obtain an annular homotopy.

Let \(g\ne1\) in \(\pi_1(K)\) and consider the free homotopy class of a loop representing \(g\). It does not contain a constant loop. It has rectifiable representatives by the edge approximation used in Lemma 8, and their lengths have infimum \(\ell\ge5\epsilon\). Parameterize a minimizing sequence proportionally to arclength on the parameter circle. Its speeds are bounded, so the compactness argument of Lemma 7 supplies a uniformly convergent subsequence. The limit lies in the same free homotopy class by the preceding closeness argument. Lower semicontinuity of length shows that its length is \(\ell\).

Parameterize this minimizer by arclength. It is a local geodesic, including across the cyclic seam. Otherwise some subarc of length less than \(\epsilon\) could be replaced by a shorter shortest path with the same endpoints. Both paths lie within \(\epsilon\) of the initial endpoint, so the loop formed by them contracts in the chosen neighborhood. The replacement preserves the free homotopy class and shortens its representative, a contradiction.

By Lemma 7, this closed local geodesic has finitely many straight pieces. Insert them into a finite compatible Euclidean subdivision of the barycentric triangulation. First mark all curve-piece endpoints on its edges, including edges belonging to no triangle. In each triangle, insert their finitely many distinct supporting lines and mark their endpoints, intersections, and ends of overlaps. The lines cut the triangle into convex polygons. On each shared edge, collect all its marks, including those supplied by its incident triangles, and impose this finite set on every incident triangle. Cone each convex polygon from an interior point to the consecutive marked points of its boundary. This introduces no further boundary marks, so it gives a compatible, genuine finite Euclidean triangulation containing the curve in its edges. The subdivision preserves the angular links and their lower bounds, as described at the start of this section. Every turn of the curve has link distance at least \(\pi\) by local geodesicity.

If \(g\) had finite order, a positive repetition of this curve would be nullhomotopic: free homotopy changes its based element only by conjugacy. The repeated curve is a nonempty closed edge path in the new triangulation, and all its cyclic turns, including its repeated seams, still have distance at least \(\pi\). This contradicts Proposition 9. ◻

Proof. We isolate a simple geodesic polygon, use angle excess and the fixed barycentric mesh to bound distance to its boundary, and then use a level arc to bound distance from one side to the others.

Fix a point \(p\) of one geodesic side. If \(p\) belongs to either other side, there is nothing to prove. Otherwise let \((a,b)\) be the component of this side outside the union of the other two sides which contains \(p\). The endpoints exist since the endpoints of the entire side lie on that union. There is a simple path \(Q\) from \(a\) to \(b\) on the other sides consisting of at most two geodesic subarcs. If one side contains both \(a\) and \(b\), use its subarc. In the remaining case write the other sides as \(\alpha,\beta\), with \(a\in\alpha\), \(b\in\beta\), and common corner \(c\). Follow \(\alpha[a,c]\) to its first meeting \(w\) with the compact arc \(\beta[b,c]\), and use \(\alpha[a,w]\cup\beta[w,b]\). The first-meeting condition makes the two interiors disjoint, irrespective of any later intersections of \(\alpha\) and \(\beta\). The interior of \([a,b]\) misses \(Q\) by its definition.

Consequently \([a,b]\cup Q\) is a simple closed curve with at most three geodesic sides. It bounds a disk \(\Omega\subset\Delta\): an omitted point inside this curve would belong to a bounded complementary component of the planar diagram, contrary to its having no holes. By Lemma 7, the sides have finitely many straight pieces. Insert these pieces in the diagram, subdivide at their intersections with its edges, and triangulate the resulting pieces of \(\Omega\).

Let \(n\) count the diagram vertices in \(\Omega^\circ\) mapping to original vertices of \(K\). At every such vertex all its diagram sectors are present in \(\Omega\): a planar neighborhood of an interior point of the Jordan disk is contained in that disk. Its angle therefore has the full excess of at least \(\eta\). Other interior vertices have angle at least \(2\pi\); newly inserted ones have angle exactly \(2\pi\). At a boundary vertex other than the at most three corners, the angle on the inside is at least \(\pi\), since a smaller angle would give a local shortcut to the geodesic side. Each corner has boundary deficit at most \(\pi\). Applying (15) to this disk gives \[2\pi\le 3\pi-n\eta,\qquad n\le\frac{\pi}{\eta}.\]

We first bound the distance to the polygon boundary. All distances here and below are \(d_\Delta\), even when minimizing routes leave \(\Omega\). Set \(f(x)=d_\Delta(x,\partial\Omega)\). For an interior point \(x\), choose a closed triangle of the fixed barycentric diagram containing \(x\), before the auxiliary subdivision along the polygon boundary. Join \(x\) inside it to its vertex mapping to an original vertex of \(K\); the segment has length less than \(M\). Either this segment first meets \(\partial\Omega\), in which case \(f(x)<M\), or it ends at one of the \(n\) vertices just counted. Thus every point with \(f(x)>M\) lies in an ambient ball of radius \(M\) about one of those vertices. Since \(f\) is \(1\)-Lipschitz, each such ball contributes values in an interval of length \(2M\). The continuous function \(f\) on the connected compact disk takes every value from \(0\) to its maximum \(F\). If \(F>M\), the interval \((M,F]\) is therefore covered by \(n\) intervals of total length at most \(2Mn\). Hence \[ d_\Delta(x,\partial\Omega)\le M+2Mn\le A \quad (x\in\Omega). \tag{16}\] The segments used in this argument are actual paths in \(\Delta\); allowing additional routes outside \(\Omega\) only decreases the distances.

Suppose now that \(R=d_\Delta(p,Q)>10(A+1)\). On a sufficiently fine Euclidean subdivision of \(\Omega\), interpolate the values of \(d_\Delta(p,\cdot)\) linearly at the vertices to obtain a continuous piecewise linear function \(h\) with \[|h(x)-d_\Delta(p,x)|<1.\] Include \(p\) and all side breakpoints among the subdivision vertices. On \([a,b]\) the interpolant agrees exactly with distance to \(p\), since that side is geodesic and its distance function is linear on each side of \(p\). On \(Q\) we have \(h\ge R-1\). Choose \(t\in(R/3,2R/3)\) different from all vertex values of \(h\). Because \(a,b\in Q\), both distances \(d_\Delta(p,a)\) and \(d_\Delta(p,b)\) are at least \(R\). Thus the level \(h=t\) has exactly two boundary endpoints, both on \([a,b]\), one on each side of \(p\), and no endpoint on \(Q\).

Inside each small triangle the level consists of a segment or is empty; segments join in pairs across interior edges. A component with a boundary endpoint must therefore be an arc, and the two boundary endpoints must belong to the same component. Call that arc \(\gamma\). For \(y\in\gamma\), \[\frac R3-1<d_\Delta(p,y)<\frac{2R}3+1, \qquad d_\Delta(y,Q)\ge R-d_\Delta(p,y)>\frac R3-1>A.\] By (16), a nearest boundary point to \(y\) is therefore on \([a,b]\) and is at distance at most \(A\). At the two endpoints of \(\gamma\), the values of \(d_\Delta(a,\cdot)\) lie on opposite sides of \(d_\Delta(a,p)\). The intermediate value theorem supplies \(y\in\gamma\) such that \(d_\Delta(a,y)=d_\Delta(a,p)\). Choose \(z\in[a,b]\) with \(d_\Delta(y,z)\le A\). Figure 3 records the topological configuration. Since \([a,b]\) is geodesic, \[d_\Delta(z,p) =\bigl|d_\Delta(a,z)-d_\Delta(a,p)\bigr| =\bigl|d_\Delta(a,z)-d_\Delta(a,y)\bigr|\le A.\] It follows that \(d_\Delta(p,y)\le2A\), whereas \(d_\Delta(p,y)>R/3-1>2A\). This contradiction proves \(d_\Delta(p,Q)\le\delta\). Since \(Q\) lies on the other two sides, it proves the lemma. ◻

Proof. A point has a nearest point on \(J\), because a geodesic segment is compact. We first prove a projection estimate. Suppose \(R_0>4\delta\), \(d(x,J),d(y,J)\ge R_0\), and \(d(x,y)\le R_0/2\). Choose nearest points \(x',y'\in J\). Every point \(z\in[x',y']\subset J\) is within \(2\delta\) of one of the other three sides of the geodesic quadrilateral \(x,x',y',y\); a diagonal reduces this assertion to two thin triangles. It cannot be within \(2\delta\) of \(w\in[x,y]\), for then \[R_0\le d(x,J)\le d(x,z) \le d(x,w)+2\delta\le R_0/2+2\delta<R_0.\] If instead \(d(z,w)\le2\delta\) for \(w\in[x,x']\), nearest-point minimality gives \[d(x,x')\le d(x,z)\le d(x,w)+2\delta, \qquad d(w,x')\le2\delta,\] so \(d(z,x')\le4\delta\). The analogous bound holds near \([y,y']\). Thus every point of \([x',y']\) is within \(4\delta\) of one of its endpoints. Applying this to its midpoint proves \[ d(x',y')\le8\delta. \tag{18}\]

Choose, for example, \(R_0=1+32\lambda\delta\); then \(R_0>4\delta\) and \(16\lambda\delta/R_0<1/2\). Every component of an excursion of \(\sigma\) beyond distance \(R_0\) from \(J\) has both endpoints at distance \(R_0\), because the endpoints of the whole path belong to \(J\). Let \(H\) be the length of its closure. Partition it by arclength into at most \(1+2H/R_0\) pieces of length at most \(R_0/2\). All partition points are at distance at least \(R_0\) from \(J\), so (18), summed over consecutive points, bounds the distance between the projections of the excursion endpoints by \(8\delta(1+2H/R_0)\). Applying (17) to the excursion gives \[ H\le\lambda(2R_0+8\delta) +\frac{16\lambda\delta}{R_0}H+C. \tag{19}\] Consequently \(H\le2[\lambda(2R_0+8\delta)+C]\). Every point of the excursion is within arclength \(H/2\) of one of its endpoints. Hence the whole path is within distance \[D=R_0+\lambda(2R_0+8\delta)+C\] of \(J\), including points outside the excursions.

For the reverse inclusion, let \(s\) be the initial endpoint and fix \(q\in J\). By continuity along \(\sigma\) there is a point \(x\in\sigma\) with \(d(s,x)=d(s,q)\). A nearest point \(x'\in J\) satisfies \(d(x,x')\le D\), whence \[d(q,x')=\bigl|d(s,q)-d(s,x')\bigr| =\bigl|d(s,x)-d(s,x')\bigr|\le D.\] Therefore \(d(q,x)\le2D\). This uses the distance coordinate along the fixed geodesic \(J\), and requires no continuity of nearest-point choices. Finally replace each side of a path triangle by a geodesic. A point on one path is within \(D\) of its geodesic, then within \(\delta\) of another geodesic, and then within \(2D\) of the corresponding path. ◻

Proof. First compare the metric of a diagram with its graph metric. There is a constant \(B_1>0\), depending only on the fixed triangulation of \(K\), such that for diagram vertices \(u,v\), \[ d_{\Delta^{(1)},\mathrm{edges}}(u,v)\le B_1d_\Delta(u,v). \tag{20}\] Indeed a geodesic from \(u\) to \(v\) has finitely many straight pieces. Replace each passage through a triangle interior by a route on its boundary. For points on sides meeting at angle \(\omega\), the route through their common corner has length at most \(1/\sin(\omega/2)\) times their Euclidean distance: for radial lengths \(a,b\), the law of cosines gives \(|x-y|^2\ge(a+b)^2\sin^2(\omega/2)\). For points on the same side use the side segment. Finitely many model angles give one length factor \(c\ge1\). Partial edge traversals in the resulting graph path can be combined, deleting reversals, to give a walk of full edges with no greater length. If \(\ell_*>0\) is the minimum model edge length, that walk uses at most \(c\,d_\Delta(u,v)/\ell_*\) edges. This proves (20), also for diagrams with no faces.

Fix a vertex \(o\in K\) and a spanning tree in its graph. Let \(\tau_v\) be the tree path from \(o\) to \(v\). For every oriented edge \(e:u\to v\), take the element represented by \(\tau_u e\tau_v^{-1}\), and let \(S\) be the finite symmetric set of its nonidentity values. These elements generate \(G\): every based edge loop factors into them by cancellation of consecutive tree paths, and every loop has an edge-loop representative. If \(S\) is empty then \(G\) is trivial and the assertion is immediate. Otherwise choose one such based edge loop for each \(s\in S\), with the reverse loop for \(s^{-1}\), and choose \(B_2>0\) bounding all their metric lengths.

Consider a geodesic triangle with vertices in the Cayley graph \(\operatorname{Cay}(G,S)\). Expand its successive letters into the chosen loops in \(K\). The empty triangle is already thin. Otherwise their concatenation is a nonempty nullhomotopic edge path and has a reduced diagram. Mark each transition between letters on its outside boundary walk. Thus each of the three boundary paths is partitioned into letter loops of length at most \(B_2\); its marks correspond to the successive group vertices on its Cayley side.

For any two marks \(u,v\) with group labels \(g_u,g_v\), \[ d_S(g_u,g_v)\le d_{\Delta^{(1)},\mathrm{edges}}(u,v) \le B_1d_\Delta(u,v). \tag{21}\] To justify the first inequality, any diagram edge path from \(u\) to \(v\) has the same group displacement as the corresponding outside walk: the two paths differ by an edge loop contractible in the diagram. All marks map to \(o\), and inserting the tree paths at successive target vertices makes the displacement a product with at most one letter of \(S\) per edge. This argument also covers pinched boundary walks. In particular, two marks which are the same diagram vertex have the same group label. Distinct positions on a nonconstant geodesic Cayley side therefore cannot become the same marked diagram vertex.

Between two marks on one side, the number of letters is precisely their group distance, since that Cayley side is geodesic. Its expanded subarc consequently has length at most \(B_2d_S(g_u,g_v)\le B_1B_2d_\Delta(u,v)\). For an arbitrary parameter subarc from \(x\) to \(y\), extend to enclosing marks \(u,v\), adding at most \(B_2\) at either end. With \[\lambda=\max\{1,B_1B_2\},\qquad C=2\lambda B_2,\] we obtain \[\operatorname{length}(x\text{ to }y) \le\lambda d_\Delta(u,v) \le\lambda d_\Delta(x,y)+2\lambda B_2.\] This bound holds even when a letter loop retraces edges or the two subarc endpoints coincide in the diagram.

Lemmas 11 and 12 now give uniform thinness \(T=3D+\delta\) for the expanded path triangle. A mark on one side is within \(T\) of some point on another side, hence within \(T+B_2\) of a mark on that side. By (21), every group vertex of a Cayley triangle is within \[H_0=B_1(T+B_2)\] of a group vertex on another side. Every other point on a side is within distance \(1\) of a side vertex. Thus every geodesic triangle whose corners are group vertices is \(H\)-thin, where \(H=H_0+1\).

For completeness, this gives uniform thinness also when triangle corners lie inside edges. A geodesic polygon with \(n\ge3\) vertex corners is \((n-2)H\)-thin: draw geodesic diagonals from an endpoint of the side under consideration and successively apply triangle thinness. In an arbitrary geodesic triangle, keep on each side the segment between its first and last graph vertices. If the side contains no vertex, it is contained in one edge and has length less than \(1\); replace it by a constant segment at an endpoint of that edge. Each kept or constant segment lies within distance \(1\) of its original side, and every discarded initial or final part has length at most \(1\). Join consecutive replacement segments by geodesics between their vertex endpoints. Those endpoints are each within \(1\) of the common original corner, so every joining segment has length at most \(2\). This yields a six-sided vertex geodesic polygon, allowing constant sides and repeated corners. The diagonal argument applies unchanged to these degeneracies and gives thinness at most \(4H\). Each point of a joining side is within \(1\) of its nearer endpoint, hence within \(2\) of the corresponding original corner. That corner belongs to an original side other than any fixed side under consideration. The other replacement segments either lie on their original sides or are within distance \(1\) of them. A point on a kept segment is consequently within \(4H+2\) of another original side. Discarded tails are within \(1\) of a shared corner; if an original side has no vertices, every one of its points is within \(1\) of its corners. Thus \(4H+2\) is a uniform thinness constant for all geodesic triangles in the Cayley graph. This is word-hyperbolicity. ◻