Relator expressions admit singular planar diagrams with controlled incidence

Write the supplied expression as ${\prod }_{j=1}^m{u}_j{r}_j^{{\epsilon }_j}{u}_j^{-1}$. Draw each whisker $u_j$ ending at a polygon reading $r_j^{{\epsilon }_j}$, in disjoint plane sectors attached at one base vertex, in the displayed product order. Empty relators may be omitted; one-letter relators are polygonal loops with one labelled occurrence and extra geometric bends, and two-letter relators are bigons with extra geometric bends if needed. The literal outer walk reads the product word. Every bounded graph region is its relator interior, with total face incidence $I\le Lm$. Label vertices by successive products starting at the identity. Each polygon closes consistently because its word represents the identity. The empty expression gives the one-vertex diagram.

Maintain these invariants during outer-word cancellation: every bounded graph region is occupied; each open face has a characteristic polygon injective on its interior; each occupied shore of an open graph edge is the image of exactly one characteristic side occurrence. Opposite shores may belong to the same face. Keep all attaching walks as occurrence lists and use distinct cyclic germs for the two ends of a loop. The initial diagram satisfies all these conditions.

First consider two distinct consecutive inverse-labelled outer edges forming a simple arc $A-B-C$ with $A\mathrm{\ne }C$. The empty exterior sector at $B$ and thin collars along the two edges give a polygonal disk $T$ meeting the carrier exactly in that arc; its remaining boundary arc joins $A$ to $C$ outside the carrier. Take a large enclosing rectangle $R$. Join an interior point of the third arc to $R$ through the unbounded region by a simple polygonal path disjoint from $T$ otherwise. Such a path exists: in an open connected plane region the points reachable by finite polygonal paths form an open-and-closed relative subset; loops of a finite path can be erased. A sufficiently narrow positive-width strip around it, widened at the start to the third arc and at the end to an interval of $R$, is disjoint from the carrier and from the interior of $T$. Finite vertex disks and strips from [F5] supply disjoint shores. Removing this notch from R leaves an embedded polygonal disk N containing T and the carrier. Set $K=\overline{N\setminus T}$, with closure in the plane. Its boundary replaces T's third arc by $A-B-C$; the relative interior of that third arc is omitted, so K is another embedded polygonal disk. Thus $K\cap T=A-B-C$ and $K\cup T=N$; these are actual planar disks, with no doubled slit boundary.

Set $T_0=\{(x,y):\mathrm{\mid }x\mathrm{\mid }\le y\le 1\}$ and $R_0=[-2,2]\times [-1,1]$. By [F5] map T to $T_0$, prescribing inverse edge parameters as $(-t,t)$ and $(t,t)$ for $0\le t\le 1$. Map K to the closed polygon $K_0=\overline{R_0\setminus T_0}$, whose boundary visits $(-2,-1),(2,-1),(2,1),(1,1),(0,0),(-1,1),(-2,1)$ in order. Its intersection with $T_0$ is exactly the two sloping sides. Use the same prescribed map on this shared arc and any compatible PL boundary homeomorphism on the remaining boundary. The two disk maps glue to a PL homeomorphism $N\to R_0$ with PL inverse after finite refinement. The carrier lies in $y\le 1$, and its exterior stays exterior: paths to the boundary of N transfer to paths to the boundary of $R_0$, and paths beyond either boundary miss the respective carrier.

In these coordinates define $Q(x,y)=(q(x,y),y)$. For $y<0$ let $q=x$; for $0\le y\le 1$ let $q=\mathrm{sgn}(x)\max (\mathrm{\mid }x\mathrm{\mid }-y,0)$; for $y>1$, use $q=(1-1\mathrm{/}y)x$ when $\mathrm{\mid }x\mathrm{\mid }\le y$ and $q=x-\mathrm{sgn}(x)$ otherwise. At $y=0$, $y=1$ and $\mathrm{\mid }x\mathrm{\mid }=y$ the formulas agree, so Q is continuous. Off $T_0$ every horizontal restriction is strictly increasing, with image the horizontal line minus $(0,y)$ when $0\le y\le 1$, and the whole line otherwise. Its inverse is $x=q+\mathrm{sgn}(q)y$ in the lower strip away from zero, the identity below zero, and the inverse of the two explicitly positive-slope pieces above one. These inverses agree on their seams. Thus Q is a homeomorphism off $T_0$ onto the complement of the collapsed vertical interval. It identifies exactly $(-t,t)$ with $(t,t)$ on the carrier, and is PL on its containing region $y\le 1$. The carrier remains finite polygonal; the complement is deformed, not fixed.

Compose each characteristic attaching map with Q. Open face interiors and all points off the folding arc remain embedded. The occupied shores formerly opposite the empty collar become opposite shores of the merged edge, so no occupied-side occurrence is duplicated on one shore. The entire labelled walk of every face is retained, possibly with more repeated edges or vertices. Following the exterior successor through the empty sector now skips exactly the inverse pair; all other exterior successor links are unchanged. The unbounded region remains connected, since a path entering the removed collar can pass around its third arc instead, in its disk chart. No empty bounded region is created. Hence all invariants of step 2.1 survive this fold.

Consider the remaining distinct-edge cases, in which a loop is involved or $A=C$. Parameterize the complete directed occurrences as $e:A\to B$ and $f:B\to C$, with $e$ labelled by $x$ and $f$ by $x^{-1}$. Mark the parameter values $1\mathrm{/}2$ by geometric points M,N only: they are not graph vertices, do not split an attaching-walk occurrence, and carry no labels. Apply the unlabelled PL collapse constructed in steps 3.1–5.1 first to the simple geometric arc $M-B-N$ and then to the images of the complementary subarcs $A-M$ and $N-C$. The intermediate image is used only to construct the composite quotient; it is not declared to be a labelled diagram, and the labelled-diagram conclusion of the preceding step is not invoked at that intermediate stage. If $A\mathrm{\ne }C$, the second arc is simple, and the composite identifies $e(t)$ with $f(1-t)$ for every $0\le t\le 1$. Only now compose the characteristic maps with the composite. Thus the final graph has one complete edge occurrence, not two labelled half-edges, and the preceding shore and exterior-successor argument applies to the complete matched parameter intervals. If $A=C$, the images of the two complementary subarcs form an embedded circle with vertices A and $P$, the common image of M,N; the subsequent circle-deletion argument completes the operation before any new diagram data are assigned. The same construction covers a closed bigon.

For that auxiliary circle, its two germs at P occur consecutively in the trace of the original exterior pair. The intervening exterior sector contains no other germ, so all other incident germs at P, including the image produced by the first auxiliary collapse, enter the bounded side of the circle. No edge can leave this side except through A: it cannot cross the embedded circle, and there is no outside germ at P. At A, the inside germs form one contiguous block in the cyclic order. Delete the circle and the carrier on its bounded side, retaining A. The outside graph is connected because the only attachment of the removed portion to it was A. In terms of the original occurrence list, this deletes the two complete occurrences $e,f$: their complementary subarcs form the circle and their other subarcs have image on its bounded side. Deleting the inside block at A joins the two outside sectors, while every other outside successor remains unchanged. Nothing strictly inside bordered the unbounded region.

Each open face lies wholly inside or wholly outside the circle, since it is connected and disjoint from its graph edges. An outside face cannot meet a circle subarc on its outside shore, which was exterior, or use P on its outside, whose sector was empty. It also cannot meet the auxiliary paired-subarc image, which lies on the bounded side by step 8.1. Thus every retained characteristic attaching walk avoids both complete deleted occurrences and all deleted vertices except possibly A, and is retained in full. No partial relator face is left. The deleted interiors join the exterior through the removed occurrence pair; other bounded regions remain their original faces. Opposite shores belonging to a single face cause no exception: its open interior lies on one side, so its entire characteristic face is retained or deleted. After this deletion the attaching maps are restricted only for wholly retained faces, so the final object again has labels only on complete graph edges.

If the cancelling occurrences are an edge and its own reverse, both shores are exterior and the empty successor corner at the middle endpoint implies that endpoint has no other incident germ. Delete this terminal spur. A loop cannot be this case: its bounded side would contain an occupied region by step 2.1, contrary to both shores being exterior. Spur deletion preserves all face walks and the literal exterior walk loses exactly the pair. Thus each cancellation operates on complete labelled occurrences: the simple-arc and $A\mathrm{\ne }C$ loop cases merge two whole edges, the $A=C$ case deletes both whole edges and only whole faces, and the spur case deletes one whole edge traversed twice. In all three cases the invariants of step 2.1 hold and the number of faces and total face incidence do not increase.

We prove contractibility for any resulting filled-region carrier by peeling faces. If a face remains, a generic polygonal path from a point of a face to infinity has a last transition from an occupied bounded graph region to the unbounded region. Its crossing edge has exactly one occupied shore, so appears exactly once in that face's characteristic boundary. It is a free open edge; every other boundary identification lies in the complementary characteristic boundary.

If that face has one labelled occurrence, its characteristic boundary minus the open free occurrence is one marked vertex v. Its disk meets the remainder of the carrier only at the image of v. Model the characteristic disk as a convex polygon with v on its boundary. The homotopy $H_t(z)=(1-t)z+tv$ contracts it to v fixing v, descends to the carrier and extends by the identity on the rest. If there are at least two labelled occurrences, their complement to the one open free side is a genuine closed arc in the abstract boundary, even if its image repeats vertices. By [F5] identify the characteristic disk with $T_0$, its free side with the top edge and the retained arc with the two sloping edges. The homotopy $H_t(x,y)=(x,(1-t)y+t\mathrm{\mid }x\mathrm{\mid })$ stays in $T_0$, retracts onto the retained arc, and fixes that arc pointwise. Consequently it respects all attaching identifications and extends by the identity to a deformation retraction of the carrier with that open face and free edge removed.

The emptied region now joins the exterior through the removed edge, while all other bounded regions remain filled. Repeat the retractions, decreasing the finite face count. At the end the connected graph has no cycle: a cycle contains an embedded polygonal circle, which would enclose an unfilled bounded region. A finite connected graph without cycles is a tree and contracts by deleting terminal edges successively (a longest simple path has terminal vertices of degree one). Thus the original carrier contracts to a point, including its monogons and all repeated retained boundary occurrences.

Each complete cancellation removes two literal outer letters, so finitely many such operations produce the reduced word of the product. By [F3] this is also the reduced word of w. Reverse a free reduction of w: for each insertion $x{x}^{-1}$ attach a new terminal edge labelled x at the specified outer occurrence in its exterior sector. Its endpoint receives the old endpoint label multiplied by x, and its two exterior traversals insert exactly that pair. This restores w literally without adding faces; adjoining a spur also preserves contractibility. Labels at folds agree because the two inverse traversals have equal endpoint group labels and matched intermediate letter parameters.

Every edge with a face incidence is counted at least once in I, hence there are at most I such edges. A thin edge cannot be on a cycle: a cycle contains a simple circle through it, and on the circle's bounded side some adjacent bounded graph region would have to occupy that shore, contrary to thinness. It is therefore a bridge, and both of its shores occur in the unique outer walk. There are at most $n\mathrm{/}2$ thin edges, since that walk has length n. Consequently $E\le I+n\mathrm{/}2\le Lm+n$. All claimed invariants, labelling and contractibility have been proved, independently of conjugator lengths.