Lawrence–Krammer–Bigelow Representations and Linearity

Interpret transverse intersections and their number in the interiors; common fixed boundary endpoints are retained and not counted. The graph/face construction of [F2] puts the disjoint family $\mathcal{N}$ in polygonal coordinates: its union with the outer circle divides the disk into finitely many Jordan faces, prescribe the boundary and crosscut maps, and extend over the faces by [F1], correcting the finitely many marked points in their own faces. This is a fixed change of coordinates, not a moving family. In those coordinates apply [F2]'s relative smooth-representative construction to $T$, keeping $P$ and the outer boundary fixed. Its finite normal strips permit polygonal interpolation; finitely many small vertex perturbations give finitely many transverse interior intersections with the fixed polygonal crosscuts. At a common boundary endpoint separate the two germs by the supported half-disk shear and straighten them as in [F2]; their endpoint remains fixed. These are ambient isotopies of $T$, not simultaneous motions of the reference crosscuts. Thus the set of finite attainable total intersection counts is a nonempty subset of $\mathbb{N}$, and its least member is attained. This proves (i). If a disjoint representative exists the least count is zero, and every least-count representative has that count; conversely a zero-count representative supplies the required disjointness under the stated shared-boundary-endpoint convention. This proves (ii), with common endpoints treated consistently.

We record the additional extraction needed for avoidance and for the consumers: if $T$ can individually be isotoped off each of a disjoint family of crosscuts and meets that family, there is a clean puncture-free bigon reducing its total count. For a crosscut $M$ met by $T$, take the supplied relative-endpoint isotopy to a representative disjoint from $M$. Compact-square continuity permits terminal strips missing $M$ when the endpoints differ; if there is a common boundary endpoint use the normalized half-disk sector construction of [F2]. Perturb the compact remainder piecewise transversely. The inverse image of $M$ consists of finitely many arcs and possibly circles. Circles need not be removed and contribute no boundary endpoints. When the endpoint sets are disjoint, the terminal strips miss $M$. If the initial and final counts are $I,J$, and $A,B,C$ count components with respectively two initial, two final, and one of each boundary endpoint, then $I=2A+C$ and $J=2B+C$. For a shared fixed boundary endpoint, [F2] normalizes the terminal half-disk angles and makes their encounters with the fixed crosscut germ occur at finitely many parameter times. Ignore the constant endpoint edge itself in the inverse image; incident components can then have one end at such a time and one on an initial or final side. Let $E_0,E_1$ count these two kinds. The correct counts are $I=2A+C+E_0$ and $J=2B+C+E_1$; components with both ends on endpoint edges and closed components contribute neither. In the disjoining case $J=0<I$, so either $A>0$ or $E_0>0$. In the first case a returning component gives a nullhomotopic loop consisting of a $T$ subpath and an $M$ subpath. In the second it gives a parameter sector with one fixed boundary endpoint: its two sides map along $T$ and $M$ and its endpoint edge maps constantly to that actual boundary point. The normalized half-disk sector construction of [F2] therefore applies. No half-bigon between two distinct fixed boundary points is used.

Lift that loop to the simply connected cover of the punctured disk. This is the cut-disk tree cover used in [F2]; compact pieces lie in a finite subtree enlarged by the corner stars and bounded puncture-collar rectangles, hence in a disk chart where Jordan separation applies. The lifts of the proper arcs are locally finite, and there are finitely many transverse crossings in that compact region. The returning pair of lifted subpaths contains a lifted bigon: erase repeated traversal segments and choose the first return bounding a disk between the two embedded lifted lines; an outermost entering segment gives a smaller such disk. Reduce across ALL lifts of $T$ and ALL crosscuts until none enters its interior. A different crosscut cannot cross the crosscut side, so any entering component has both contacts on the $T$ side and cuts off a smaller bigon; a lifted $T$ component similarly returns to the other side. The same finite reduction handles further lifts of $M$. Boundary projection is injective: an identified point of different sides gives another transverse lifted crossing on a side, one branch entering the disk, contrary to the reduction. Same-side identifications are excluded by embeddedness. Consequently distinct deck translates of the boundary are disjoint. If their disk interiors overlap, Jordan nesting gives a deck map or its inverse carrying the compact disk into itself, contrary to Brouwer and the fixed-point-free deck action. Thus the entire disk projects injectively to an ordinary compact disk missing $P$. At a common boundary endpoint the same argument is in a half-disk chart with that endpoint an actual covering point; no ideal puncture tip is used. This is the projection justification, rather than an inference that a face of a null-homotopy domain embeds in the surface.

A clean bigon push is supported in a slightly enlarged disk, not literally only in the original closed bigon. Its moving crosscut is prescribed to pass to a small pushoff of the opposite side, fixing the support boundary; [F1] extends the prescriptions over both Jordan faces, and [F2]'s Alexander move realizes them. At a boundary corner use a half-disk and fix its boundary edge. Choose the enlargement so it misses all other crosscuts and all punctures. Two interior crossings disappear (one for a fixed endpoint sector), and no new crossing appears. Starting with an arc individually disjoinable from each crosscut, these ambient isotopies preserve each such individual isotopy class. Step 2.1 therefore gives another clean bigon whenever any intersections remain. Strict decrease of the finite total count terminates with simultaneous disjointness. This proves the additional individual-to-simultaneous disjoining assertion used by the kernel argument.

For (iii), each component of $C$ is proper by the original Given convention. A clean bigon from step 3.1 cannot meet $C$: a component meeting its interior cannot cross either boundary side, since $C$ is disjoint from both $T$ and the crosscuts; it cannot remain wholly inside, since its endpoints are on $P\cup \mathrm{\partial }D$, whereas the ordinary bigon interior contains no such point. In an endpoint-sector disk its only boundary endpoint is on $T$, hence is also excluded by $C\cap T=\mathrm{\varnothing }$. Compactness then allows the enlarged support of step 4.1 to miss $C$. After each push $T$ remains disjoint from $C$, so the same argument applies at the next stage. The finite composite disjoins $T$ while its moving part avoids $C$, proving (iii). AC is used in the general planar graph/face and smoothing constructions; the proper-end convention is essential here. A floating unmarked interior-ended obstacle could lie wholly inside a bigon and would not satisfy that convention.

Work first in the ordered configuration space, a subset of $D\times D$. Denote its finitely many distance functions by $d_0=\mathrm{\mid }{z}_1-z_2\mathrm{\mid }$ and $d_{ki}=\mathrm{\mid }{z}_k-p_i\mathrm{\mid }$. Their minimum is positive and locally Lipschitz. Choose ${\epsilon }_0$ so small that $12{\epsilon }_0$ is less than the minimum distance between distinct punctures, $12{\epsilon }_0<{\min }_i(1-\mathrm{\mid }{p}_i\mathrm{\mid })$, and $4{\epsilon }_0<1$; omit the first bound if there is only one puncture. A prescribed base configuration can additionally be kept outside ${\nu }_{2{\epsilon }_0}$ by decreasing ${\epsilon }_0$. Such a choice uses only minima of finite positive lists.

At a point with $f\le 3{\epsilon }_0$, draw the graph whose vertices are the two mobile points and the fixed punctures and whose edges are precisely the distances equal to $f$. No component contains two punctures: a path between them would use at most the two mobile vertices, hence have length at most $3f\le 9{\epsilon }_0$, contradicting step 1.1. In a component containing a puncture $p$, set the velocity of each mobile vertex to $z_k-p$ and leave the puncture fixed. Every active distance in this component has positive derivative equal to that distance, since the whole component is dilated about $p$. All moved vertices are within $2f\le 6{\epsilon }_0$ of $p$, so these velocities can be used in a neighbourhood disjoint from the disk boundary. Two disjoint puncture components use their respective dilations simultaneously. Isolated mobile vertices have zero velocity.

The remaining possible active component consists of the two mobile vertices without a puncture. Put $d=z_1-z_2$, $V_1=(I-z_1{z}_1^T)d$, and $V_2=-(I-z_2{z}_2^T)d$, using real coordinates in ${\mathbb{R}}^2$. This vector field is tangent to each disk boundary factor, and the derivative of $\mathrm{\mid }d{\mathrm{\mid }}^2$ is $2d^T(I-z_1{z}_1^T)d+2d^T(I-z_2{z}_2^T)d>0$. Each summand is nonnegative on the disk and is positive if that point is interior. If both points are on the boundary, equality would require $d$ parallel to both boundary normals; distinct such points are antipodal, with distance $2$, excluded by $f\le 3{\epsilon }_0<1$. Thus this candidate also increases every active distance. The cases in steps 2.1 and 3.1 exhaust the graph possibilities, including ties between a collision and a puncture distance and two separate puncture distances.

Fix $0<\delta <\epsilon <{\epsilon }_0$. The band $B=\{\delta \mathrm{/}4\le f\le 3\epsilon \}\subset D^2$ is compact and avoids all punctures and collisions. Each candidate of steps 2.1 and 3.1 is smooth near the point at which it was selected and strictly increases every active distance there. By continuity, the same holds in a neighbourhood: distances inactive at that point have a positive gap from the minimum, so none can become active on a sufficiently small neighbourhood unless its derivative was already positive. Cover $B$ by finitely many such neighbourhoods. For the puncture candidates restrict these neighbourhoods so that every moved coordinate is interior; for the collision candidate tangency holds identically. Take Lipschitz weights subordinate to this finite cover by the distance-to-complement construction, shrinking supports using the maximum-distance threshold as necessary, and normalize their sum. Their weighted sum is locally Lipschitz, tangent to every boundary factor, and strictly increases every active distance. Average it with its coordinate-interchanged translate to make it invariant under mobile interchange; positivity and tangency survive the averaging. Extend it to a neighbourhood of $B$ with a Lipschitz cutoff equal to one on the smaller band used by the trajectories. All these operations concern a finite compact band.

Write $V$ for this field. The set of pairs $(x,j)$ with $x\in B$ and $d_j(x)=f(x)$ is compact. The continuous quantities $D{d}_j(x)V(x)$ are positive on it, so have a positive common lower bound $c$. Along the flow of $-V$, the derivative of the minimum, at every time at which it is differentiable, is the derivative of one of its active distances and is at most $-c$. This also follows directly from the one-sided derivative of a finite minimum. The minimum is Lipschitz along the flow and hence its integrated decrease is at least $c$ per unit time while the trajectory stays in $B$. The field preserves each disk boundary factor: on a boundary factor it is tangent, and uniqueness of solutions prevents an interior trajectory from crossing that factor. Thus these trajectories are valid configurations and preserve $\mathrm{\partial }C$.

The required flow needs no additional existence assumption. On a compact neighbourhood with Lipschitz constant $L$ and bound $M$, the operator $u(t)\mapsto x+{\int }_0^t(-V)(u(s))ds$ is a contraction on the closed sup-norm ball of paths for time $h$ with $Lh<1$ and $Mh$ smaller than its radius. Starting with the constant path, its iterates have geometrically bounded consecutive differences, so converge uniformly to the unique integral solution. The same estimates give continuous dependence on the initial point. Repeat on finitely many compact-band time intervals as needed; a solution cannot cease to exist while staying in the band. This proves the local flow and the extension needed here. For an initial configuration with $\delta \mathrm{/}2<f\le 2\epsilon$, step 5.1 shows that the first hitting time $T(x)$ of $f=\delta \mathrm{/}2$ is finite, bounded by $(2\epsilon -\delta \mathrm{/}2)\mathrm{/}c$, and continuous in $x$; the strict decrease and continuous dependence give the last assertion by bracketing the hitting time on either side. Set $T=0$ at $f\le \delta \mathrm{/}2$.

Choose a Lipschitz function $\eta$ equal to one for $f\le \epsilon$ and zero for $f\ge 2\epsilon$. For $f<2\epsilon$ flow for time $s\eta (f(x))T(x)$, $0\le s\le 1$, and fix configurations with $f\ge 2\epsilon$ or $f\le \delta \mathrm{/}2$. The bounded hitting times ensure continuity at the cutoff. This gives an interchange-invariant homotopy $R_s:C\to C$ from the identity to $r=R_1$. It never increases $f$ where it moves a point, preserves $\mathrm{\partial }C$, and sends ${\nu }_{\epsilon }$ into $\{f\le \delta \mathrm{/}2\}\subset {\nu }_{\delta }$. It preserves ${\nu }_{\delta }$ throughout as well. Therefore $r$ is a map of pairs in the reverse direction of each identity inclusion in the statement, and $R_s$ gives the two inverse homotopies, as homotopies of the respective pairs. For the boundary-union pairs, a boundary point with larger $f$ remains a boundary point; no cutoff across $f=\epsilon$ is being used on that boundary.

The homotopy fixes the chosen base configuration. Lift it starting at the identity of the covering; uniqueness of homotopy lifting in evenly covered neighbourhoods implies that the lift commutes with every deck transformation. Each lifted map preserves the preimages of the end neighbourhoods and the boundary exactly when its base map does. Thus step 7.1 proves deck-equivariant homotopy equivalences of both lifted pairs. It follows directly on singular relative chains, using the prism homotopy, that the induced maps are isomorphisms.

For $\gamma <\delta <\epsilon$ the natural inclusions satisfy $i_{\gamma ,\epsilon }=i_{\delta ,\epsilon }{i}_{\gamma ,\delta }$, so their unique inverses satisfy $k_{\delta ,\gamma }{k}_{\epsilon ,\delta }=k_{\epsilon ,\gamma }$, and likewise for the boundary-union groups. The inverse is independent of every vector-field or cutoff choice because it is the inverse of a specified canonical homomorphism. The direct limit over decreasing sufficiently small radii therefore has compatible canonical isomorphisms from every one of these groups, and its universal property identifies it with any of them. The same conclusion holds after passage to a smaller cofinal interval of radii. This establishes both the stabilization and the directed convention claimed.

Regard $z=x+iy$ with $x,y\in {\mathbb{R}}^2$. Let $\mathcal{A}$ be the real lines $x_1=p_i$, $x_2=p_i$, and $x_1=x_2$. For a line with affine equation $\mathrm{\ell }(x)=0$ and linear part ${\mathrm{\ell }}_0$, its complexification contains $x+iy$ exactly when $\mathrm{\ell }(x)={\mathrm{\ell }}_0(y)=0$. Thus membership in $M$ is a condition on the lines containing the real part and the directions of the imaginary part. Choose a square $R=(-L,L{)}^2$ containing all real arrangement vertices, and a strictly increasing continuous function $h:\mathbb{R}\to (-L,L)$ fixing an interval containing every $p_i$. For example, keep $h(u)=u$ on $[-L\mathrm{/}2,L\mathrm{/}2]$ after taking $L$ large enough, and on each tail use the increasing exponential interpolation to the endpoint $\pm L$. Applying $(1-s)u+sh(u)$ to both real coordinates and leaving imaginary coordinates unchanged preserves every equality $x_k=p_i$ and the equality/order of $x_1,x_2$. It therefore gives an equivariant homotopy equivalence between $M$ and its subspace with real part in $R$; the same homotopy restricts to that subspace.

Here is the finite good-cover argument needed for this particular cover. For a finite open cover $U_{\alpha }$ of a metric space $X$, let $d_{\alpha }(x)=d(x,X\setminus U_{\alpha })$, using the constant function $1$ if the complement is empty, and put $D(x)={\max }_{\alpha }{d}_{\alpha }(x)>0$. Normalize the nonnegative functions $\max (0,d_{\alpha }-D\mathrm{/}2)$ to get a partition ${\lambda }_{\alpha }$. Its support is locally contained in $U_{\alpha }$, since a point in the closure of its nonzero set satisfies $d_{\alpha }\ge D\mathrm{/}2>0$. Form its thick nerve by gluing $U_{\sigma }\times {\mathrm{\Delta }}^{\sigma }$ for all nonempty intersections $U_{\sigma }={\bigcap }_{\alpha \in \sigma }{U}_{\alpha }$, using the face inclusions. Projection to $X$ has the continuous section $x\mapsto (x,({\lambda }_{\alpha }(x)))$; local containment of supports justifies continuity in this gluing topology. Straight interpolation in the simplex coordinates contracts its fibres onto this section, because the union of the two supports still consists of sets containing $x$. Projection to the ordinary nerve is also a homotopy equivalence when all $U_{\sigma }$ are contractible. To see this, filter by simplex dimension: each step attaches the products $U_{\sigma }\times {\mathrm{\Delta }}^{\sigma }$ along $U_{\sigma }\times \mathrm{\partial }{\mathrm{\Delta }}^{\sigma }$, and projection on both products is a homotopy equivalence. The boundary product is a cofibration: a collar of the boundary of the finite simplex supplies its homotopy extension explicitly, after taking a product with $U_{\sigma }$. Consequently the pushout comparison preserves a homotopy equivalence. Indeed replace an attachment by its double mapping cylinder, extend the collar across it, and use the contractions of the factors on the two ends; the resulting homotopies descend to the pushout. Induction over the finitely many simplices proves the assertion. This proof uses finite choices only.

Cellulate the closed square by the arrangement lines and its boundary, and barycentrically subdivide this finite convex polyhedral cellulation. Choose a point $v_F$ in the relative interior of each face $F$, respecting coordinate interchange; actual barycentres do this. An interior face means one not contained in the artificial square boundary, and corresponds to exactly one real arrangement facet. For each interior $F$, let $S_F$ be its open vertex star in this subdivision, intersected with $R$. These stars cover $R$: a point in the relative interior of a face has a positive weight at an interior face vertex in its barycentric simplex. A nonempty intersection $S_{F_0}\cap \cdots \cap S_{F_k}$ occurs exactly when the $F_j$ form a strict inclusion chain. It contracts to the centroid of their vertices by interpolating barycentric weights to those of that centroid; all their weights stay positive, and the centroid lies in $R$. In particular $k\le 2$. A real point in $S_F$ belongs to a cofacet of $F$, so every arrangement line containing that point contains $F$.

For every $F$, partition imaginary space by the linear hyperplanes parallel to the arrangement lines containing $F$. Let $K_{F,C}$ be an open chamber of this local arrangement; for a two-dimensional $F$ there are no such lines and the chamber is all of ${\mathbb{R}}^2$. Put $U_{F,C}=S_F+i{K}_{F,C}$. Step 2.1 and the membership criterion of step 1.1 show that this is an open subset of $M$ with real part in $R$. These finitely many sets cover that space: at a real point in facet $G$, use a star whose positive top face is $G$ and the local chamber containing its imaginary part. Intersections have contractible real factor by step 2.1 and convex imaginary factor, the intersection of finitely many strict linear half-spaces. A nonempty intersection is therefore contractible, and its indices form a face chain of length at most three. Coordinate interchange permutes the cover and its intersections.

Apply step 1.2 to step 3.1. The nerve has dimension at most two. Its barycentric simplices are exactly chains of facets with compatible local chamber labels. Every local chamber at a vertex is a sector between two or three incident lines; at an edge it is one of the two sides; at a real chamber there is a single label. Thus the nerve is the barycentric subdivision of the following regular complex: one vertex for each real chamber, two directed edges across each real edge, and one disk for every pair consisting of an arrangement vertex and an incident real chamber. The disk boundary follows the two shortest directed paths around that vertex from this chamber to the opposite chamber, one on each side. This follows directly by grouping the face-chain triangles with the same vertex-sector label; each group is the cone on the cyclic sequence of incident edges between those opposite chambers. This is the two-dimensional Salvetti cell description, here obtained from the explicit cover.

All constructions are equivariant. In step 1.2 use the Euclidean metric, so the partition is equivariant. No nerve simplex is setwise fixed by $\tau$: a simplex has at most one facet of each dimension, so a fixed simplex would have fixed labels at each dimension; a real chamber cannot be fixed since it lies wholly on one side of the diagonal, and a fixed facet on the diagonal has its two sides or its incident sectors interchanged. Choose the finite contractions and extension data once for each orbit of simplices, and use their translates on the other members. The inductive pushout comparison of step 1.2 then provides equivariant inverse maps and equivariant homotopies. In particular these descend to homotopy equivalences of the quotients; an ordinary nonequivariant homotopy equivalence is not being used to justify this descent.

Label quotient chamber vertices by $P_{ij}$, $1\le i\le j\le n+1$, according to the two puncture intervals containing the coordinates, with the diagonal splitting a same-interval square into two exchanged chambers. There are diagonal loops $c_i$, pairs of directed edges $a_{ij},{\bar{a}}_{ij}$ between $P_{ij}$ and $P_{i,j+1}$, and $b_{ij},{\bar{b}}_{ij}$ between $P_{i+1,j+1}$ and $P_{i,j+1}$. Read the disk boundaries of step 4.1 at a double intersection and a triple intersection. At a double intersection $i<j$ they are $(b_{i,j-1}{a}_{ij})(a_{i+1,j}{b}_{ij}{)}^{-1}$, $({\bar{a}}_{i+1,j}{b}_{i,j-1})(b_{ij}{\bar{a}}_{ij}{)}^{-1}$, $(a_{ij}{\bar{b}}_{ij})({\bar{b}}_{i,j-1}{a}_{i+1,j}{)}^{-1}$, and $({\bar{b}}_{ij}{\bar{a}}_{i+1,j})({\bar{a}}_{ij}{\bar{b}}_{i,j-1}{)}^{-1}$. At a triple intersection they are $(a_{ii}{\bar{b}}_{ii}{c}_{i+1})(c_i{a}_{ii}{\bar{b}}_{ii}{)}^{-1}$, $({\bar{b}}_{ii}{c}_{i+1}{b}_{ii})({\bar{a}}_{ii}{c}_i{a}_{ii}{)}^{-1}$, and $(c_{i+1}{b}_{ii}{\bar{a}}_{ii})(b_{ii}{\bar{a}}_{ii}{c}_i{)}^{-1}$. This exhausts the two types of vertices of this arrangement, and is obtained by following the four or six sectors cyclically.

The fourth double-intersection disks and their barred edges form the ordinary staircase grid complex on the $P_{ij}$. It is contractible: in the realization as a square grid below a staircase, move horizontally to its leftmost column and then vertically to its bottom vertex; the horizontal sections are intervals containing that column. The same description includes $n=1$, when it is an interval. Collapsing this subcomplex to a point is a homotopy equivalence, because it is a finite CW subcomplex and its contraction extends across collars of the attached cells. The second and third double-intersection disks now have boundaries $b_{i,j-1}{b}_{ij}^{-1}$ and $a_{ij}{a}_{i+1,j}^{-1}$. Collapse these bigons successively to identify each row of $b$-edges with $b_i$ and each column of $a$-edges with $a_j$. Their remaining boundaries are precisely the four words in the statement, with one vertex and no cells above dimension two.

For a homotopy equivalence carrying a specified normal subgroup of the fundamental group to the corresponding subgroup, lift it and an inverse after fixing basepoint lifts. Their compositions lift the identity homotopies; uniqueness of a lifted path, checked in successive evenly covered neighbourhoods, gives lifted homotopies between the compositions and the identity. The lifts commute with every deck transformation by the same uniqueness. Thus the corresponding covers are deck-equivariantly homotopy equivalent, and their ordinary absolute singular homology groups are identified. Apply this to the homotopy equivalences above and use [F1] for the unordered interpretation.

For the unit closed disk choose $r_0<1$ larger than the modulus of every puncture. A strictly increasing radial map $f:[0,1]\to [0,1)$ fixing $[0,r_0]$ gives an injection of the disk into its interior fixing all punctures. Interpolating its radial function with the identity remains injective, fixes the punctures, and takes interior points to interior points. Applying it to both coordinates therefore proves that the interior inclusion is a homotopy equivalence of punctured configurations, also after coordinate interchange. An increasing radial homeomorphism $[0,\mathrm{\infty })\to [0,1)$ fixing $[0,r_0]$ identifies the punctured plane with the punctured open disk, fixing the punctures. Both operations preserve the puncture and mutual winding characters, since their homotopies keep pairs distinct and avoid punctures. The lifting argument of step 8.1 therefore applies to these identifications too.

Choose the directed edges in [F1] to make a positively oriented puncture meridian on returning along its barred edge. The barred grid is contractible, so its lift can be fixed consistently with every barred edge having displacement $1$. The resulting $a_i$ and $b_i$ loops move one mobile point counterclockwise once around $p_i$, with the other outside that small meridian disk. Thus their exponents are $(a,b)=(1,0)$ and their displacement is $q$. Each $c_i$ crosses the diagonal between two real chambers exchanged by coordinate interchange; in the unordered quotient it exchanges the mobile points counterclockwise in their common puncture interval, enclosing no puncture. Its exponents are $(0,1)$ and its displacement is $t$. Paths in the barred grid provide the basepoint paths, and changing those paths by a homotopy has no effect on these displacements. Both $q$ and $t$ occur, so the deck group is ${\mathbb{Z}}^2$.

Lift every cell after fixing one basepoint lift. For a word, a positive edge contributes its prefix displacement times that edge, and an inverse edge contributes minus the displacement after traversing that inverse edge times the positive edge. Apply this to the four attaching words of [F1]. The word $b_i{a}_j{b}_i^{-1}{a}_j^{-1}$ contributes $b_i+q{a}_j-q{b}_i-a_j$. The word $a_i{c}_{i+1}{a}_i^{-1}{c}_i^{-1}$ contributes $a_i+q{c}_{i+1}-t{a}_i-c_i$. The word $c_{i+1}{b}_i{a}_i^{-1}{c}_i^{-1}$ contributes $c_{i+1}+t{b}_i-t{a}_i-c_i$. Finally $c_{i+1}{b}_i{c}_i^{-1}{b}_i^{-1}$ contributes $c_{i+1}+t{b}_i-q{c}_i-b_i$. These are exactly the displayed differential formulas. The lifted model has no 3-cells, so [F1] and [F2] identify $H_2(\tilde{C};\mathbb{Z})$ with $\ker d\subset C_2$. These are absolute chains; no end neighbourhood or relative quotient has entered the construction.

Over $K$, let $W$ be the span of all $B_{ir}$. Suppose ${\sum }_i(r_i{B}_{i1}+s_i{B}_{i2}+u_i{B}_{i3})$ has zero boundary. Its $a_i$ coefficient gives $s_i=(1-t)t^{-1}{r}_i$, and its $b_i$ coefficient then gives $u_i=r_i$. The remaining boundary is $(q+t^{-1}){\sum }_i{r}_i(c_{i+1}-c_i)$. Since $q+t^{-1}\mathrm{\ne }0$, its $c_1$ coefficient forces $r_1=0$, its $c_2$ coefficient then forces $r_2=0$, and successive coefficients force every $r_i=0$. Hence also $s_i=u_i=0$, and $d{\mathrm{\mid }}_W$ is injective. Projection onto the $\left(\genfrac{}{}{0px}{}{n}{2}\right)$ $A$-coordinates is consequently injective on $\ker (d\otimes K)$, giving an upper bound $\left(\genfrac{}{}{0px}{}{n}{2}\right)$ for its dimension.

Put $S=(t-1)(qt+1)$ and define $V_{ib}=-qt{B}_{i1}+q(t-1)B_{i2}+B_{i3}$, $V_{ia}=B_{i1}+q(t-1)B_{i2}-qt{B}_{i3}$, and $V_{i0}=-t{B}_{i1}+(t-1)B_{i2}-t{B}_{i3}$. Substitution gives $d{V}_{ib}=S{b}_i-(q-1)(qt+1)c_{i+1}$, $d{V}_{ia}=-S{a}_i+(q-1)(qt+1)c_i$, and $d{V}_{i0}=(qt+1)(c_i-c_{i+1})$. Therefore the integral chains $E_{ij}=S{A}_{ij}+(q-1)V_{ib}+(q-1)V_{ja}+{\sum }_{i<k<j}(q-1{)}^2{V}_{k0}$ are cycles: the $a_j,b_i$ terms cancel the boundary of $S{A}_{ij}$, and the $c$ terms telescope. Their $A$-coordinates are $S$ in coordinate $(i,j)$ and zero elsewhere. Since $S\mathrm{\ne }0$, they are independent over $K$. Together with step 3.1 this proves that the field kernel has dimension $\left(\genfrac{}{}{0px}{}{n}{2}\right)$ and basis $\{E_{ij}\}$. For $n=1$ there are no such chains, and step 3.1 says the full kernel is zero.

The ring $\mathrm{\Lambda }$ is a domain: it is the localization of the polynomial domain $\mathbb{Z}[q,t]$ by monomials. Its finite free module $C_2$ is torsion-free, and so is its submodule $\ker d$. If an element of $\ker d$ maps to zero after localization, a nonzero denominator annihilates it; torsion-freeness makes it zero. Thus the localization map on absolute $H_2$ is injective. Moreover every field cycle becomes an integral cycle after multiplying by a common nonzero denominator of its finitely many cellular coordinates. Hence localization of $\ker d$ is exactly $\ker (d\otimes K)$, not just a subspace thereof. Step 4.1 now gives the asserted dimension. Integral spanning by the $E_{ij}$ was never used or inferred.

Construct compatible models directly. For ordered configurations write $z=x+iy$, with $x,y\in {\mathbb{R}}^2$, and use the real arrangement $x_1=p_i$, $x_2=p_i$, $x_1=x_2$. A complexified line excludes $x+iy$ exactly when its affine equation vanishes at $x$ and its linear part vanishes at $y$. Choose $L$ with all punctures in $(-L,L)$ and an increasing map $h:\mathbb{R}\to (-L,L)$ fixing an interval containing them. Applying $(1-s)u+sh(u)$ to both real coordinates preserves their order and all equalities with punctures, and leaves imaginary coordinates unchanged. It gives a homotopy equivalence to the subspace with real part in $R=(-L,L{)}^2$, and restricts to the smaller configuration, whose compressed real domain is $R^{<}=(-L,p_n{)}^2$. Cellulate the closed square by the arrangement and its boundary, choose the barycentres $v_F$ of its faces, and barycentrically subdivide. For every face not contained in the square boundary let $S_F$ be its open vertex star intersected with $R$. These stars cover $R$; intersections occur exactly for face chains and contract by barycentric interpolation to the centroid of the specified vertices. A point in $S_F$ lies in a cofacet of $F$, so every arrangement line through that point contains $F$. Call $F$ retained when its relative interior lies in $R^{<}$. Its star lies entirely in $R^{<}$: every coface stays on the left side of each last-puncture line, and the positive weight at $v_F$ makes both inequalities strict. The retained stars cover $R^{<}$, and their intersection contractions stay there.

For each $F$ and each open chamber $Q$ of the linear arrangement parallel to the lines containing $F$, put $U_{F,Q}=S_F+iQ$. The membership criterion of step 1.1 shows these are open sets of the ordered configuration space. They cover it: at a real point in facet $F$, its imaginary part avoids precisely that local arrangement. Each nonempty intersection has a contractible star-intersection factor and a convex imaginary factor. Its labels form a face chain of length at most three. For retained $F$, neither last-puncture line contains $F$, so the local imaginary arrangement is exactly the smaller one. The retained $U_{F,Q}$ therefore give a good cover of the smaller space by the same sets as in the larger cover. Their nerve is the subcomplex on the retained labels.

Here the good-cover comparison can be made compatible without choosing compatible inverse homotopies. For a finite open cover of a metric space $X$, form the thick nerve from $U_{\sigma }\times {\mathrm{\Delta }}^{\sigma }$ over its nonempty intersections, with the face identifications. Projection to $X$ is a homotopy equivalence: normalize $\max (0,d(x,X\setminus U_{\alpha })-D(x)\mathrm{/}2)$, where $D(x)$ is the maximum of these distances, to obtain a partition with supports locally contained in the covering sets; its graph is a section, and straight interpolation of simplex coordinates gives the inverse homotopy. Projection to the ordinary nerve is also a homotopy equivalence: filter by simplex dimension and compare the attachments $U_{\sigma }\times \mathrm{\partial }{\mathrm{\Delta }}^{\sigma }\subset U_{\sigma }\times {\mathrm{\Delta }}^{\sigma }$ with the simplex attachments. The projections on these products are homotopy equivalences since $U_{\sigma }$ is contractible; collars of simplex boundaries give cofibrations, so the pushout comparison, equivalently its double-mapping-cylinder comparison, preserves homotopy equivalences at each finite stage. Both projection squares commute with inclusion for step 2.1's covers. Coordinate interchange preserves these covers. No nerve simplex is setwise fixed: it has at most one label of each face dimension, a fixed diagonal facet has its local chambers exchanged, and a two-dimensional facet lies on one side of the diagonal. Thus choose the finite contraction and extension data orbit by orbit; the comparisons are equivariant and descend to the unordered quotients. This identifies the actual half-plane inclusion with the nerve-subcomplex inclusion on absolute homology.