Proof. There are finitely many possible conditions $\lambda $ , so $\mathcal {U}$ is finite.

Given any $x \in \text {Conf}_n(\Gamma ,\mathcal {P})$ , we may define a condition $\lambda $ as follows. Assign i to the open star whose central vertex is closest to $x_i$ ; if $x_i$ is at the midpoint of an edge between two vertices of degree $\geq 2$ , choose either open star for i and for all j for which $x_j = x_i$ . Let $P = \pi (x)$ . Define the total order on the indices assigned to $V_t$ by increasing distance from $v_t$ ; if multiple $x_j$ have the same distance from $v_t$ , choose any order for those indices that groups together indices of coincident points. By construction, $x \in U_{\lambda }$ .

We show that each $U_{\lambda }$ is open. For $x \in U_{\lambda }$ , define $\epsilon _0 = \min \{ d(x_i, x_j) \mid x_i \neq x_j\}$ , and define $\epsilon _t = \min \{ d(x_i,v_t) \mid i \not \sim _P \omega _t(1)\}$ for each $t \in \{1, \ldots s\}$ . That is, for each open star $V_t$ , $\epsilon _t$ is the minimum distance to its central vertex of points in the configuration which are not in the same part of P as the first index in the order $\omega _t$ . Now let $\epsilon = \min \{1,\epsilon _0, \epsilon _1, \ldots , \epsilon _s\}$ . We show the open ball $B(x,\epsilon /3)$ is contained in $U_{\lambda }$ . Let y be a configuration with $\delta (y,x) < \epsilon /3$ . Then, $d(y_i,x_i) < \epsilon /3$ for all $i \in N$ . We need to show that y conforms to $\lambda $ .

Since $d(y_i,x_i) < \epsilon /3 < 1$ , then $y_i$ and $x_i$ must lie on the same edge or on two edges that share a vertex. If they lie on the same edge, then $y_i$ lies in the same open star $\sigma (i)$ as $x_i$ . If they lie on two edges that share the central vertex of $\sigma (i)$ , then $y_i$ lies in $\sigma (i)$ . Assume that they lie on two adjacent edges that do not share the central vertex of $\sigma (i)$ . Then, they share a vertex with degree $\geq 2$ which is the central vertex $v_t$ of some open star $V_t \neq \sigma (i)$ . Since i is not assigned to $V_t$ by $\sigma $ , i cannot be in the same part of P as $\omega _t(1)$ , as P is a refinement of $\pi (\sigma )$ . Therefore, $d(x_i,v_t) \geq \epsilon _t$ . However, $d(y_i,x_i) \geq d(x_i,v_t)$ since the shortest part between $y_i$ and $x_i$ passes through $v_t$ . This contradicts $d(y_i,x_i) < \epsilon /3$ . Therefore, $y_i$ lies on $\sigma (i)$ .

Assume $y_i$ and $y_j$ lie on the same edge of $\Gamma $ and $\sigma (i) \neq \sigma (j)$ . Then, $x_i \neq x_j$ , so $d(x_i,x_j) \geq \epsilon $ and consequently $d(y_i,y_j)> \epsilon /3$ . If $x_i$ is not on this same edge, it must be on another edge of $V_t$ , so $d(y_i,v_t) < \epsilon /3$ . (If the two edges share both of their vertices and the shortest path between $x_i$ and $y_i$ were to pass through the other vertex, we would have $d(x_i,y_i)> \epsilon $ , which is a contradiction.) Meanwhile, $d(y_j,v_t)> 2\epsilon /3$ because $d(x_j,v_t) \geq \epsilon _k \geq \epsilon $ . Therefore, $y_i$ is strictly closer to the central vertex of $\sigma (i)$ than is $y_j$ . If $x_j$ is not on the same edge as $y_j$ and $y_i$ , then by a similar argument, $y_j$ must be closer to the other vertex of the edge than is $y_i$ , so therefore, $y_i$ is strictly closer to $v_t$ than is $y_j$ . The final case is that $x_i$ and $x_j$ are both on this edge, and since $\sigma (i) \neq \sigma (j)$ and x conforms to $\lambda $ , it must be the case that $d(x_i,v_t) < d(x_j,v_t)$ . Since they are on the same edge, $d(x_i,x_j) = d(x_j,v_t)-d(x_i,v_t) \geq \epsilon $ . We have $d(y_i,v_t)<d(y_i,x_i)+d(x_i,v_t) < \epsilon /3 + d(x_i,v_t)$ and $d(x_j,v_t)<d(y_j,v_t)+\epsilon /3$ , so $d(y_j,v_t)> d(x_j,v_t) - \epsilon /3 > d(x_i,v_t) +\epsilon /3 > d(y_i,v_t)$ . Thus, $y_i$ is strictly closer to the central vertex of $\sigma (i)$ than is $y_j$ .

Finally, for an open star $V_t$ , let $i < j$ with $\omega (i) \not \sim _P \omega (j)$ . Suppose $y_{\omega (i)}$ and $y_{\omega (j)}$ lie on the same edge. If $x_{\omega (i)}$ and $x_{\omega (j)}$ lie on this same edge, then by a similar argument as above, $d(x_{\omega (i)},v_t)<d(x_{\omega (j)},v_t)$ . We know $\omega (j) \not \sim _P \omega (1)$ . Therefore, $x_{\omega (j)}$ and $y_{\omega (j)}$ must lie on the same edge as the distance from $x_{\omega (j)}$ to any vertex of degree $\geq 2$ is at least $\epsilon $ , so $d(y_{\omega (j)},v_t)> 2\epsilon /3$ . If $x_{\omega (i)}$ is on a different edge from $y_{\omega (i)}$ , then $d(y_{\omega (i)},v_t) < \epsilon /3$ , which means $d(y_{\omega (i)},v_t) < d(y_{\omega (j)},v_t)$ .

Therefore, y conforms to $\lambda $ , so $U_{\lambda }$ is open.

Proof. Let A be a connected component of $U_{\Lambda }$ . Within A, the only points that are able to move between edges are Type 1. We construct a homotopy in two stages. First, we move all Type-1 points onto the central vertex of their designated open star via straight-line homotopy. This is possible because $i \sim _Q \omega ^{\lambda }(1)$ for all $\lambda \in \Lambda $ , so $x_i$ is not obstructed by any $x_j$ with $i \not \sim _Q j$ . Denote by B the resulting subspace of A. Within B, each Type-1 point must remain on its vertex, and no point can enter or leave any edge.

Consider an edge e of an open star $V_t$ and assume that at least one point of the configurations in B lies on e. There are two cases to consider.

In the first case, the other vertex of e has degree $1$ . Then, the only points on e are Type 2. Let m be the number of equivalence classes of Q that are represented by the Type-2 points lying on e. These equivalence classes are ordered by the distance of their closest representative to $v_t$ , since $\omega ^{\lambda }$ groups members of equivalence classes of $P^{\lambda }$ together in its total order. Let $y_e$ be the configuration of points on e where the points of the kth equivalence class of Q lie together at a distance of k/m away from $v_t$ .

In the second case, e connects $v_t$ to another central vertex $v_u$ . There may be points of Type 2 which are assigned to $V_t$ , points of Type 2 which are assigned to $V_u$ and points of Type 3 which lie in between. As before, the equivalence classes of Q represented by points on e can be ordered by the distance of their closest representative to $v_t$ . (Here, we use the facts that $d(x_i,v_t) = 1 - d(x_i,v_u)$ and equivalence classes represented by Type-2 points assigned to $V_u$ can be ordered by their distances from $v_u$ .) Let $m_1, m_2$ and $m_3$ denote the numbers of equivalence classes of Q represented by the points of Type 2 assigned to $V_t$ , Type 3 and Type 2 assigned to $V_u$ , respectively, and $m = m_1 + m_2 + m_3$ . Now define the configuration $y_e$ of points on e where the points of the kth equivalence class of Q lie together at a distance of ${k}/{(m+1)}$ away from $v_t$ . (Note that this will give the same configuration $y_e$ when defined for the open star $V_u$ , since the kth equivalence class from $v_t$ is the $(m-k+1)$ th equivalence class from $v_u$ .)

Now let y be the configuration of points that restricts to $y_e$ on each edge e of $\Gamma $ (and all Type-1 points lie on their designated central vertex). Use a straight-line homotopy to move the configurations of B to y. If $x_i$ and $x_j$ lie on the same edge with $i \not \sim _Q j$ with $d(x_i,v_t) < d(x_j,v_t)$ , then $d(y_i,v_t) < d(y_j,v_t)$ , so the straight-line homotopy will not create any collisions of points which are not in the same class of Q.

Proof. Since $\mathcal {D}_n(\Gamma ,\mathcal {P})$ is a regular cell complex, it is homeomorphic to the geometric realisation of $\Delta F$ . The $0$ -simplices of $\Delta F$ correspond to the elements of F (which are cells of $\mathcal {D}_n(\Gamma ,\mathcal {P})$ ). A k-simplex is a chain of $k + 1$ cells ordered by inclusion, and the points of the simplex are convex combinations of its vertices. A homeomorphic inclusion of $|\Delta F|$ into $\mathcal {D}_n(\Gamma ,\mathcal {P})$ is obtained by choosing a point in the interior of a cell $\tau $ for the $0$ -simplex corresponding to $\tau $ . This map is then defined linearly on the simplices.

For each cell $\tau = (\tau _1, \tau _2, \ldots , \tau _n) \in F_{\Lambda }$ , we show that $D_{\Lambda } \cap \tau $ is convex. Suppose $x, y \in D_{\Lambda } \cap \tau $ . Let $x_i, y_i \in \tau _i$ , so $z^t_i = (1-t)x_i + t y_i \in \tau _i$ for any $t \in [0,1]$ , so $z^t \in \tau $ . If $\tau _i$ and $\tau _j$ lie on the same edge e with vertex v of $\Gamma $ and $i \not \sim _Q j$ , then $d(x_i,v) < d(x_j,v)$ implies $d(y_i,v) < d(y_j,v)$ since both x and y conform to all $\lambda \in \Lambda $ . Therefore, $d(z_i,v) = (1-t)d(x_i,v) + t d(y_i,v) < (1-t)d(x_j,v)+t d(y_j,v) = d(z_j,v)$ . (If $i \not \sim _Q$ , and $\tau _i$ and $\tau _j$ lie on two edges $e_i$ and $e_j$ that meet at v with $\tau _i$ containing v, then if either $x_i$ or $y_i$ on v, then i is Type 1 and assigned to v under $\Lambda $ , and neither $x_j$ nor $y_j$ is allowed to be on v since $j \not \sim _Q i$ .) Therefore, for all $t \in [0,1]$ , $z^t_j$ must lie in the interior of $e_j$ and $z^t_i$ is either on $e_i$ or at v. Therefore, $z^t$ conforms to all $\lambda \in \Lambda $ .

Since $D_{\Lambda } \cap \tau $ is convex, it must lie in a connected component A of $U_{\Lambda }$ , so every Type-2 and Type-3 point is restricted to one edge of $\Gamma $ . Type-1 points might be able to move between edges within an open star. For each edge e of $\Gamma $ , let $m_e$ be the number of equivalence classes of Q whose points could lie on e in the connected component A of $U_{\Lambda }$ . These equivalence classes are ordered along e. (The Type-1 class assigned to the initial vertex of e first, then Type-2 classes assigned to the initial vertex, then, if the terminal vertex is also a central vertex of $\Gamma $ , Type-3 classes, and Type-2 and Type-1 classes assigned to the terminal vertex.)

For each $\tau $ , we will choose a point in the interior of $\tau $ that lies in $D_{\Lambda }$ . Each $\tau _i$ is either a node or a segment of $\Gamma ^{\prime }$ . If $\tau _i$ is an original vertex from $\Gamma $ of degree $\geq 2$ , then i must be of Type 1 in $\Lambda $ . For points $x_i$ of Types 2 and 3 under $\Lambda $ , $\tau _i$ must be a segment or an interior node on an edge e (possibly an original vertex of degree 1). We will now define a configuration $x^{\tau }$ in the interior of $\tau $ in $D_{\Lambda }$ .

• • If $\tau _i$ is a node, put $x_i^{\tau }$ at that node.

• • If $\tau _i$ is a segment on edge e and i belongs to the kth of the $m_e$ classes, put $x_i^{\tau }$ at ${k}/{(m_e +1)}$ along the length of $\tau _i$ from its initial node (the direction consistent with the direction chosen for e).

As defined, $x_i^{\tau }$ is contained in the interior of $\tau _i$ , so $x^{\tau } \in \tau $ . Also, if $x_i^{\tau } = x_j^{\tau }$ , then $i \sim _Q j$ , and the order of the equivalence classes along each edge of $\Gamma $ have been preserved, so $x^{\tau }$ conforms to all $\lambda \in \Lambda $ . Therefore, $x^{\tau } \in D_{\Lambda }$ .

We will first define a continuous map $f : D_{\Lambda } \rightarrow |\Delta F_{\Lambda }|$ and then create a straight-line homotopy from the identity map to f. For $x \in D_{\Lambda }$ , let $\tau $ be the minimal cell of $F_{\Lambda }$ containing x. For each $x_i$ , we will define a point $y_i \in \tau _i$ and then define $f(x) = y = (y_1, \ldots , y_n) \in |\Delta F_{\Lambda }|$ .

We note that, if $\tau \in F_{\Lambda }$ and $\tau _i$ is a segment from node a to node b, then the face of $\tau $ with $\tau _i$ replaced with a is only in $F_{\Lambda }$ if i is in the first equivalence class on $\tau _i$ (or Type 1 and assigned to a if a is a degree $\geq 2$ vertex of $\Gamma $ ). Likewise with b and the last equivalence class on $\tau _i$ . Furthermore, if i is in the first class on $\tau _i$ and j is in the last class on the preceding segment $\tau _j$ , then the face obtained by replacing both $\tau _i$ and $\tau _j$ with a is only in $F_{\Lambda }$ if $i \sim _Q j$ . Therefore, a configuration $y \in \tau $ is only in $|\Delta F_{\Lambda }|$ if each of its points $y_i$ either lies at $x_i^{\tau }$ or strictly between $x_i^{\tau }$ and a node of $\tau _i$ that i is free to occupy (in which case, i has ‘claimed’ that node, preventing any $y_j$ with $j \not \sim _Q i$ from also claiming it).

If $\tau _i$ is a node, then $x_i = x_i^{\tau }$ and we define $y_i = x_i$ .

Suppose $\tau _i$ is a segment with initial node a and terminal node b. If i is a member of an equivalence class of Q which is neither the first nor last with representatives in $\tau _i$ , then define $y_i = x_i^{\tau }$ .

We now suppose i is a member of the first of multiple equivalence classes of Q with representatives in $\tau _i$ .

In the case that a is an original vertex of $\Gamma $ with degree $\geq 2$ , if i is of Type 1 and assigned to a, then define $y_i$ to be the point closer to a out of $x_i$ and $x_i^{\tau }$ , and if i is not of that type, define $y_i = x_i^{\tau }$ .

If a is not an original vertex of $\Gamma $ , then the segment $\tau _i$ has a preceding segment on its edge. If other members of $[i]_Q$ are assigned to that preceding segment or if it contains no points at all in its interior, then define $y_i$ to be the point closer to a out of $x_i$ and $x_i^{\tau }$ .

Finally, suppose the last equivalence class with representatives assigned to that preceding segment is $[j]_Q$ , where $j \not \sim _Q i$ .

Let $d_i = \min \{ {d(x_i,a)}/{d(x_i^{\tau },a)},1 \}$ and let $d_{[i]} = \min \{d_k \mid k \in [i]_Q, \tau _k = \tau _i \}$ . Similarly, define $d_j$ and $d_{[j]}$ . Set $M = \max \{d_{[i]},d_{[j]}\}$ . If $d_i \geq d_{[j]}$ , set $y_i = x_i^{\tau }$ . If $d_i < 1 = d_{[j]}$ , set $y_i = x_i$ . Otherwise, we have $d_i < d_{[j]} = M < 1$ , and we set $y_i = ({d_i}/{M}) x_i^{\tau } + (1 - {d_i}/{M})a$ .

If i is a member of the last of multiple equivalence classes of Q in $\tau _i$ , $y_i$ can be defined symmetrically. (If $\tau _i$ is a segment whose terminal node b is a degree-1 vertex of $\Gamma $ , then set $y_i$ to be the point closer to b out of $x_i$ and $x_i^{\tau }$ .)

The last possibility is if all points in $\tau _i$ belong to the same equivalence class (so it is both the first and the last class on $\tau _i$ ). Then, if $x_i$ lies before $x_i^{\tau }$ , define $y_i$ according to the rules for the first equivalence class on a segment, and if it lies after, use the rules for the last equivalence class. If $x_i = x_i^{\tau }$ , both rules agree that $y_i = x_i^{\tau }$ .

We set $f(x) = y = (y_1, y_2, \ldots , y_n)$ . Then, f is continuous on each intersection $D_{\Lambda } \cap \tau $ , so f is continuous. Furthermore, f fixes each point of $|\Delta F_{\Lambda }|$ . The homotopy $H: D_{\Lambda } \times [0,1] \rightarrow D_{\Lambda }$ given by $H(x,t) = (1-t)x + t f(x)$ is therefore a deformation retraction of $D_{\Lambda }$ onto $|\Delta F_{\Lambda }|$ .

Proof. By the previous theorem, each connected component of $D_{\Lambda }$ has a connected component of $|\Delta F_{\Lambda }|$ as a deformation retract. We will show each connected component of $\Delta F_{\Lambda }$ is contractible.

Let J be a connected component of $\Delta F_{\Lambda }$ . For an edge e of $\Gamma $ , we will now define a Morse matching $M^1_e$ on the simplices of J. A k-simplex $\rho $ in J is a chain $\tau ^0 \subset \tau ^1 \subset \cdots \subset \tau ^k$ , where each $\tau ^j = (\tau ^j_1, \ldots \tau ^j_n) \in F_{\Lambda }$ . We will create pairs $\rho \subset \rho ^{\prime }$ , where $\rho ^{\prime }$ is a $(k+1)$ -simplex.

For $\rho = \tau ^0 \subset \cdots \subset \tau ^k$ , consider the maximal cell $\tau ^k$ . Let $C = C(\rho )$ be the last equivalence class with $\tau ^k_i \in Z_B$ for some $i \in C$ and $B \prec C$ . Let $\beta = \beta (\rho )$ be the last node or segment on e with $i \in C$ and $\tau ^k_i = \beta $ such that there exists an interior node $a \succ \tau ^k_i$ in $Z_C$ , which is unobstructed for i. Let $I = I(\rho )$ be the set of all $i \in C$ with $\tau ^k_i = \beta $ . Note that, since no class after C has members in any zone before its own, then if $Z_B$ is the first zone occupied by members of C, no zone after $Z_B$ up to and including $Z_C$ can contain any members of classes other than C. Since at least one member of C is in $Z_B$ , there is at least one node in $Z_C$ that is unobstructed for at least one $i \in C$ with $\tau _i^k$ lying before it.

If $\beta $ is a node $\alpha (0)$ , we define $\rho ^{\prime }$ by adding $\tau ^{k+1}$ to the end of the chain, where $\tau ^{k+1}_i = \alpha $ for all $i \in I$ and $\tau ^{k+1}_i = \tau ^k_i$ for all $i \notin I$ . This new cell $\tau ^{k+1}$ is still in $F_{\Lambda }$ because $\beta $ is the last cell with an unobstructed node in $Z_C$ after it, implying that $\alpha (1)$ must be unobstructed for $i \in I$ . Note that $C(\rho ^{\prime }) = C(\rho )$ , replacing $\alpha (0)$ with $\alpha $ cannot cause the class to leave a zone, and $\beta (\rho ^{\prime }) = \alpha $ . Similarly, if $\beta = \alpha $ is a segment and there exist $i \in C$ with $\tau ^{k}_i = \alpha (0)$ , we may define $\tau ^{k+1}_i = \alpha $ for such i, and $\tau ^{k+1}_i = \tau ^k_i$ for all other i, and add this $\tau ^{k+1}$ to the end of $\rho $ to produce $\rho ^{\prime }$ .

If $\beta = \alpha $ is a segment and there are no $i \in C$ with $\tau ^{k}_i = \alpha (0)$ , find the last $\tau ^j$ in $\rho $ such that $\tau ^j_i = \alpha (0)$ for some $i \in I$ . Define $\tau ^{\prime }$ by setting $\tau ^{\prime }_i = \alpha $ for all $i \in I$ with $\tau ^j_i = \alpha (0)$ and $\tau ^{\prime }_i = \tau ^j_i$ for all other i. We obtain $\rho ^{\prime }$ by inserting $\tau ^{\prime }$ in the chain immediately after $\tau ^j$ . (If $\tau ^{j+1} = \tau ^{\prime }$ already, then $\rho $ is paired by this rule with the $(k-1)$ -simplex obtained by removing $\tau ^{j+1}$ .) Note that $\tau ^{\prime }$ is in $F_{\Lambda }$ : $\tau ^{\prime } \subseteq \tau ^{j+1} \in F$ , so $\tau ^{\prime } \in F$ , and $\tau ^{\prime }$ has all the same cells as $\tau ^j$ except for a subset of one class C which changed from a node to an incident segment, so $\tau ^{\prime }$ still has nonempty intersection with $D_{\Lambda }$ .

If $\beta = \alpha $ is a segment and no $\tau ^j_i = \alpha (0)$ for any $i \in I$ , instead find the first $\tau ^j$ in $\rho $ with $\tau ^j_i = \alpha $ for some $i \in I$ . Define $\tau ^{\prime }$ by setting $\tau ^{\prime }_i = \alpha (1)$ for all $i \in I$ and $\tau ^{\prime }_i = \tau ^j_i$ for all other i. Then, $\rho ^{\prime }$ is obtained by inserting $\tau ^{\prime }$ immediately before $\tau ^j$ (or removing it if $\tau ^{j+1} = \tau ^{\prime }$ ). Note that $\tau ^{\prime } \in F_{\Lambda }$ : $\alpha (1) \subset \alpha $ , so $\tau ^{\prime } \in F$ , and since only cells for members of C are being changed from $\alpha $ to $\alpha (1)$ and C must be the last equivalence class on $\alpha $ , $\tau ^{\prime }$ has nonempty intersection with $D_{\Lambda }$ .

Note that if $\rho \subset \rho ^{\prime }$ is a pair, $C(\rho ) = C(\rho ^{\prime })$ and $I(\rho ) \subseteq I(\rho ^{\prime })$ . If $\beta (\rho ) = \alpha (0)$ , then $\beta (\rho ^{\prime }) = \alpha $ . Otherwise, if $\beta (\rho ) = \alpha $ , then $\beta (\rho ^{\prime }) = \alpha $ . In every case, $\beta (\rho ) \preceq \beta (\rho ^{\prime })$ .

Note also, the cell $\tau ^{\prime }$ advances some members of $C(\rho )$ one step along e from their position in $\tau ^j$ (replacing a node $\alpha (0)$ with its succeeding segment $\alpha $ , or replacing a segment $\alpha $ with its terminal node $\alpha (0)$ ), and all other coordinates remain in the same position.

Also, if $\tau ^1 \subset \tau ^2$ , then any $\tau ^1_i \subseteq \tau ^2_i$ .

Next, we consider how C may change for $\rho $ obtained by deleting the maximal cell $\tau ^{\max }(\psi )$ from a chain $\psi $ . Note that $\tau ^{\max }(\rho )$ is a face of $\tau ^{\max }(\psi )$ obtained by replacing some segments with nodes. The zone in which a given coordinate i is located can only change if $\tau ^{\max }(\psi )_i$ is a segment $\alpha $ , where two zones overlap and $\tau ^{\max }(\rho )_i$ is one of its nodes. That coordinate switches from lying in both zones to lying in only one. Therefore, if $\tau ^{\max }(\psi )_i$ does not lie in any zone prior to $Z_{[i]}$ , then neither does $\tau ^{\max }(\rho )_i$ . If $\tau ^{\max }(\psi )_i$ is the first segment of $Z_{[i]}$ and $\tau ^{\max }(\rho )_i$ is its terminal node, then i has exited the lower zone. Thus, $C(\rho ) \preceq C(\psi )$ .

Now assume $C(\rho ) = C(\psi ) = C$ . There are a few ways that $\beta $ and I could change. A segment in $Z_C$ appearing in $\tau ^{\max }(\psi )$ could be replaced in $\tau ^{\max }(\rho )$ by its terminal node, causing its initial node to become unobstructed for its preceding segment and node (appearing in $\tau ^{\max }(\rho )$ ), in which case $\beta (\rho )$ would be that preceding segment or node, $\beta (\rho ) \succ \beta (\psi )$ and $I(\rho ) \cap I(\psi ) = \emptyset $ . If $\beta (\psi ) = \alpha $ is a segment, all of its appearances in $\tau ^{\max }(\psi )$ could be replaced with $\alpha (0)$ in $\tau ^{\max }(\rho )$ , in which case, $\beta (\rho ) = \alpha (0)$ and $I(\psi ) \subseteq I(\rho )$ . If some of its appearances are replaced with $\alpha (1)$ in $\tau ^{\max }(\rho )$ , then $\beta (\rho ) \preceq \alpha (1)$ . Now, $\alpha (1)$ may or may not have unobstructed nodes of $Z_C$ ahead of it. If it does, $\beta (\rho ) = \alpha (1)$ and $I(\rho ) = I(\psi )$ . If it does not, $\beta (\rho ) \preceq \alpha (0)$ and there exists at least one coordinate $i \in I(\psi )$ such that $i \notin I(\rho )$ .

Note that if $\rho $ is obtained from $\psi $ by deleting its maximal cell, $I(\psi ) \not \subseteq I(\rho )$ implies $\tau ^{\max }(\rho )$ assigns a node a to a coordinate i for which $\tau ^{\max }(\psi )_i = \alpha $ with $\alpha (1) = a$ . If $\rho $ and $\psi $ have the same maximal cell, then $I(\psi ) = I(\rho )$ .

We now show $M^1_e$ is acyclic. Suppose we have a path

$$ \begin{align*} \rho_1 < \psi_1> \rho_2 < \psi_2> \cdots > \rho_{t-1} < \psi_{t-1}> \rho_t, \end{align*} $$

where $t \geq 3$ , and for $1 \leq r \leq t - 1$ , $\rho _r$ is paired with $\psi _r$ in the matching, and $\rho _{r+1}$ is a codimension-1 face of $\psi _r$ not equal to $\rho _r$ .

The chain $\psi _r$ is obtained from $\rho _r$ by inserting some $\tau ^{\prime }(r)$ based on the maximal cell $\tau ^{\max }(\rho _r)$ which determines the class $C(\rho _r)$ , cell $\beta (\rho _r)$ , and coordinate set $I(\rho _r)$ for the pairing. Here, $\rho _{r+1}$ is obtained by removing some cell other than $\tau ^{\prime }(r)$ from $\psi _r$ . Therefore, $\tau ^{\prime }(r+1) \neq \tau ^{\prime }(r)$ .

Suppose the path is a cycle, meaning $\rho _1 = \rho _t$ . Recall that $C(\rho _r) = C(\psi _r)$ and $C(\psi _r) \succeq C(\rho _{r+1})$ for all $1 \leq r \leq t-1$ . Therefore, since the path is a cycle, we must have $C(\psi _r) = C(\rho _{r+1})$ for all $1 \leq r \leq t-1$ .

Suppose $I(\psi _r) \not \subseteq I(\rho _{r+1})$ for some $1 \leq r \leq t - 1$ . Then, $\rho _{r+1}$ must be obtained by removing $\tau ^{\max }(\psi _r)$ from the chain, and there is a coordinate j such that $\tau ^{\max }(\rho _{r+1})_j$ is the terminal node $\alpha (1)$ of the segment $\alpha = \tau ^{\max }(\psi _r)_j$ . Therefore, $\tau _j = \alpha (1)$ for all cells $\tau $ in $\rho _{r+1}$ . The matching cannot introduce a preceding segment in a coordinate, so there is no way to regain $\tau ^{\max }(\psi _r)$ , which contradicts the path being a cycle. Therefore, we may assume $I(\psi _r) \subseteq I(\rho _{r+1})$ . We also know $I(\rho _r) \subseteq I(\psi _r)$ for all $1 \leq r \leq t-1$ . Since we have a cycle, we can conclude that $I(\rho _r) = I(\psi _r) = I$ for all $1 \leq r \leq t-1$ .

Now, for all $i \in I$ and $1 \leq r \leq t - 1$ , we must have $\beta (\rho _r) = \tau ^{\max }(\rho _r)_i$ and $\beta (\psi _r) = \tau ^ {\max }(\psi _r)_i$ . We have $\beta (\rho _r) \preceq \beta (\psi _r)$ , and $\beta (\psi _r)$ must be a segment $\alpha _r$ . If $\beta (\psi _r) \succ \beta (\rho _{r+1})$ , then $\rho _{r+1}$ is obtained by removing $\tau ^{\max }(\psi _r)$ and $\beta (\rho _{r+1}) = \tau ^{\max }(\rho _{r+1})_i = \alpha _r(0)$ . This forces the conclusion that $\tau ^{\max }(\psi _r) = \tau ^{\prime }(r)$ , so it could not have been removed from $\psi _r$ in a valid path. Therefore, $\beta (\psi _r) \preceq \beta (\rho _{r+1})$ . Since we have a cycle, this means $\beta $ must be the same segment $\alpha $ for all chains in the path.

We know $\tau ^{\prime }(1)$ is obtained by replacing $\alpha (0)$ with $\alpha $ if any $\tau ^j$ in $\rho _1$ includes $\alpha (0)$ in any of its coordinates, or by replacing $\alpha $ with $\alpha (1)$ otherwise. In the first case, $\rho _2$ must be obtained from $\psi _1$ by removing the predecessor to $\tau ^{\prime }(1)$ , a cell with $\alpha (0)$ in at least one coordinate i, where $\tau ^{\max }_i = \alpha $ , because any other removal would result in $\rho _2$ being paired with the lower dimensional chain obtained by removing $\tau ^{\prime }(1)$ . Such a cell cannot be regained later in the path because every cell in the chain has $\beta = \alpha $ , so any new cell will replace all $\alpha (0)$ of an existing cell with $\alpha $ or $\alpha $ with $\alpha (1)$ . Likewise, in the second case, $\rho _2$ must be obtained from $\psi _1$ by removing the successor to $\tau ^{\prime }(1)$ , a cell with $\alpha $ in at least one coordinate i, where $\tau ^{\prime }_i = \alpha $ . Such a cell cannot be regained later in the path because $\alpha (0)$ does not appear in any coordinate, so any new cell introduced by the matching will be replacing all $\alpha $ in an existing cell with $\alpha (1)$ . Therefore, the path cannot be a cycle.

Thus, $M^1_e$ is a Morse matching. The only simplices in J which are unpaired in $M^1_e$ are those chains $\rho = \tau ^1 \subset \cdots \subset \tau ^k$ , where no zone $Z_B$ on e contains any $\tau ^k_i$ with $[i] \succ B$ . These simplices form a subcomplex K: the order complex of cells $\tau \in J^0$ such that no $\tau _i$ in e lies in a zone $Z_B$ of e with $[i] \succ B$ .

We can define a similar Morse matching $M^2_e$ on K, where we consider the maximal cell $\tau ^k$ of a simplex $\rho $ . Let $C = C(\rho )$ be the first equivalence class such that $\tau ^k_i \in Z_B$ for some $i \in C$ and $C \prec B$ , and $\beta = \beta (\rho )$ be the first node or segment on e with $i \in C$ and $\tau ^k_i = \beta $ such that there exists an interior node $a \prec \tau ^k_i$ in $Z_C$ which is unobstructed for i, and let $I = I(\rho )$ be the set of all $i \in C$ with $\tau ^k_i = \beta $ . As before, no zone before $Z_B$ down to and including $Z_C$ can contain any member of a class other than C. From here, the matching $M^2_e$ may be defined similarly to $M^1_e$ , and it is similarly acyclic. Its critical simplices are those chains in K of cells $\tau $ for which no $\tau _i$ in e lies in a zone $Z_B$ with $[i] \prec B$ . However, since these chains are already in K, that means each $\tau _i$ on e lies in $Z_{[i]}$ . This forms a subcomplex L to which K now simplicially collapses.

The pairings $M^1_e$ and $M^2_e$ are determined by and only affect the $\tau _i$ in e. Therefore, if f is another edge of $\Gamma $ that contains points in $U_{\Lambda }$ , the matchings $M^1_f$ and $M^2_f$ induce matchings on the resulting subcomplex L to which J collapses. Collapsing via the two matchings for one edge of $\Gamma $ one at a time, we arrive at a subcomplex H consisting of chains of cells $\tau $ for which, if $\tau _i$ lies on an edge e, then $\tau _i$ lies in zone $Z_B$ if and only if $i \in B$ .

We will use $Z(i)$ to refer to the (closed) zone associated to i. If i belongs to a class B which is not Type 1, then $Z(i)$ is the interval from the first node to the last node of $Z_B$ on the edge containing $x_i$ in $U_{\Lambda }$ . If i belongs to a Type-1 class B, then $Z(i)$ is the union of the intervals from the central vertex to the last node of $Z_B$ on each edge of the open star.

Now, if $\tau \in H^0$ , each $\tau _i$ belongs to $Z(i)$ . Conversely, we will show that any cell $\tau $ such that each $\tau _i$ belongs to $Z(i)$ is in $H^0$ .

Since $Z(i) \cap Z(j) = \emptyset $ if $i \not \sim _Q j$ , if $\bigcap _{i \in C} \tau _i \neq \emptyset $ , then C must be a subset of a class of Q. Suppose there exists a partition P such that $\bigcap _{i \in C} \tau _i \neq \emptyset $ for all classes C of P. Then, P is a refinement of Q. Since $Q \notin \mathcal {P}$ , we have $P \notin \mathcal {P}$ , so $\tau $ is a cell of $\mathcal {D}_n(\Gamma ,\mathcal {P})$ . Since the zones are ordered in accordance with the orders of classes on each edge dictated by $\Lambda $ , $\tau $ has nonempty intersection with $U_{\Lambda }$ . Therefore, $\tau \in F_{\Lambda }$ , and thus it is in $H^0$ since it is unpaired in all of the Morse matchings.

It follows that H is the order complex of the face poset of the cubical complex $\{ (\tau _1, \ldots , \tau _n) \mid \tau _i \in Z(i)\}$ , which is contractible as the product of contractible zones.

Proof. We have seen that both $U_{\Lambda }$ and $D_{\Lambda }$ are disjoint unions of their connected components which are all contractible via deformation retraction. Each component of $D_{\Lambda }$ is contained in one component of $U_{\Lambda }$ . Within such a component of $U_{\Lambda }$ , there is a path between the two retract points. We therefore need to show a one-to-one correspondence between the connected components of $U_{\Lambda }$ and $D_{\Lambda }$ .

As described previously, the choice of $\Lambda $ determines which points are designated as Type 1, 2 and 3. Type-1 points are permitted to lie on their designated vertex, while Types 2 and 3 cannot move between edges. Furthermore, within each edge, points must remain ordered by equivalence class. We will therefore argue that the connected components of both spaces are in one-to-one correspondence with the possible assignments of Type-2 and Type-3 points to edges permissible by $\Lambda $ .

In the proof of Theorem 2.2, we showed that each connected component deformation retracts to a configuration where each point of Type 1 lies on its assigned vertex and points of Type 2 and Type 3 coincide with their equivalence class members on each edge, with the classes ordered in accordance with $\Lambda $ . Within $U_{\Lambda }$ , any two such configurations can be joined by a path, so there is only one connected component for each assignment.

In the proof of Theorem 3.3, we showed that each component deformation retracts to a cubical complex of ‘zones’. The zones are determined by the edge assignments, so there is only one connected component for each permissible assignment.

Proof. We have the spectral sequence associated to the open cover $\mathcal {U}$ of $\text {Conf}_n(\Gamma ,\mathcal {P})$ which converges to $H_*(\text {Conf}_n(\Gamma ,\mathcal {P}))$ , and we have the spectral sequence associated to the open cover $\{ D_{\lambda }\}$ of $\mathcal {D}_n(\Gamma , \mathcal {P})$ which converges to $H_*(\mathcal {D}_n(\Gamma , \mathcal {P}))$ . Since $D_{\Lambda }$ is homotopically equivalent to $U_{\Lambda }$ , for every set $\Lambda $ , by the inclusion map and the $d_1$ differentials are induced by inclusion, we have an isomorphism for each $p, q$ between the $E^1_{p,q}$ entries in the two spectral sequences commuting with $d_1$ . The two sequences are therefore equivalent, so $H_*(\text {Conf}_n(\Gamma ,\mathcal {P}))$ is isomorphic to $H_*(\mathcal {D}_n(\Gamma , \mathcal {P}))$ .