2 Preliminaries

2.1 Ramification portraits associated to Thurston maps

Let $f:(S^2,P_f)\to (S^2,P_f)$ be a Thurston map of degree d. The ramification portrait of f is the weighted directed graph $\Gamma $ such that the vertex set $V(\Gamma )$ is the union of the set $C_f$ of critical points and the set $P_f$ of postcritical points, and for each vertex v, there is an edge from v to $f(v)$ with weight the local degree $\deg _f(v)$ of f at v. By the Riemann–Hurwitz formula,

$$ \begin{align*} \sum_{v\in C_f} (\deg_f(v)-1) = 2d-2. \end{align*} $$

Since f has degree d, at each vertex v, the sum of the weights of the incoming edges is at most d. Note that f is a topological polynomial if and only if there is a vertex v such that $f(v) = v$ and $\deg _f(v)=d$ .

3 Realizing a ramification portrait by an unobstructed map

In this section, we prove Theorem 1.1. We begin with an example to illustrate the construction. Consider the abstract ramification portrait $\Gamma $ that is shown as follows.

The proof defines an ordering of the finite postcritical vertices of $\Gamma $ . In this case, we use the ordering given by $a_1 < a_2 < a_3 < a_4 < a_5 < a_6$ . Following the terminology that will be defined in the proof, the ordered sets $(a_1,a_2,a_3)$ and $(a_4)$ are called type- $1$ chains and the ordered set $(a_5,a_6)$ is called a type- $2$ chain. (The first element of a type- $1$ chain is the image of a critical vertex that is not postcritical, and the first element of a type- $2$ chain is a periodic critical vertex.) The model subdivision complex $S_{\mathcal {R}}$ is shown in Figure 1 as a stereographic projection of $S^2$ to the plane. The 1-skeleton will always be a star graph with central vertex $\infty $ . The vertices of $\Gamma $ are identified with the vertices of $S_{\mathcal {R}}$ . The ordering of the finite postcritical vertices chosen above determines the counterclockwise ordering of the labels of the vertices in Figure 1. The tile type t is shown in Figure 2; $S_{\mathcal {R}}$ is the image of t under the characteristic map $\psi \colon \, t\to S_{\mathcal {R}}$ . The label of a vertex v of t is $\psi (v)$ ; if $\psi (v)\ne \infty $ , then v is called a finite vertex.

Figure 1

The model subdivision complex $S_{\mathcal {R}}$ .

Figure 2

The tile type t, which $\psi $ maps to $S_{\mathcal {R}}$ by identifying edges in pairwise fashion.

The following is an outline of the construction of the subdivision $\mathcal {R}(S_{\mathcal {R}})$ of $S_{\mathcal {R}}$ . We add edges to $S_{\mathcal {R}}$ in three stages.

In stage $1$ , we add edges to $S_{\mathcal {R}}$ that will ensure that the subdivision map cannot have any Levy cycles. Figure 3 shows the construction after the first stage for our example from the point of view of the tile type t. The tile t is a 12-gon. The label of every vertex v of t appears adjacent to v and outside of t. Every other vertex is labeled $\infty $ and the labels of the finite vertices are in the proper cyclic order. Every finite vertex v has an image label, which is $\sigma _{\mathcal {R}}(\psi (v))$ . It is drawn inside t. (Of course, this is abuse of notation, since we have not finished the construction yet and, hence, have not defined the subdivision map $\sigma _{\mathcal {R}}$ yet.)

Figure 3

The construction after stage $1$ .

In stage $2$ , we add edges to ensure that $\mathcal {R}(S_{\mathcal {R}})$ will have the correct number of subtiles and, hence, $\sigma _{\mathcal {R}}$ will have the correct degree. This is done in a way that ensures that the ramification portrait of $\sigma _{\mathcal {R}}$ will be isomorphic to $\Gamma $ . No further changes are made in the second stage for our example since there is already the correct number of subtiles.

In stage $3$ , we simply add stickers as needed in each subtile so that each subtile has the proper number of vertices and edges. (A sticker is an edge with a vertex of valence one, resembling a stick pin with a spherical head.) Figure 4 shows the result for our example.

Figure 4

The subdivision of the tile type after stage $3$ .

Figure 4 Long description

Starting at the top vertex, labeled a1 and infinity, the hexagon proceeds clockwise with vertices labeled a2, a3, a4, a5, and a6, each also paired with infinity at the corners. At each vertex, smaller points labeled a1 through a6 cluster near the main vertex. From the top left vertex, arcs radiate inward from a6 to a2, each connecting to its respective labeled point. At the bottom left, arcs from a6 and a5 extend inward to a5 and a4. At the bottom vertex, arcs from a5, a4, a3, a2, and a1 radiate from the main vertex a4. At the bottom right, arcs from a5, a6, a1, a2, and a3 cluster near a3. At the top right, arcs from a5, a6, a1, a2, and a3 cluster near a2. All arcs are gray, and the main hexagon outline is black and bold. The structure emphasizes the subdivision of each vertex into labeled subpoints and their interconnections.

All of this is done while ensuring that in every subtile, every other vertex is a vertex of t with label $\infty $ and that the image labels are in the proper cyclic order.

It is now straightforward to define the subdivision map $\sigma _{\mathcal {R}}$ so that its restriction to each open cell is a homeomorphism to an open cell and it takes each finite vertex to its image label. Figure 5 shows the subdivision $\mathcal {R}(S_{\mathcal {R}})$ .

Figure 5

The subdivision of the model subdivision complex.

Figure 5 Long description

Starting at the central node labeled infinity, six main branches radiate outward. Each branch is labeled at its endpoint: a sub 1, a sub 2, a sub 3, a sub 4, a sub 5, and a sub 6. Along each branch, intermediate nodes are labeled sequentially, such as a sub 2, a sub 3, a sub 4, a sub 5, and a sub 6, with each label corresponding to its position along the branch. Gray arcs connect nodes across branches, forming loops that link nodes with matching labels, such as a sub 1 to a sub 6, a sub 2 to a sub 5, and so forth. The structure illustrates the subdivision of the model subdivision complex, with spatial relationships defined by the arrangement and connectivity of labeled nodes.

The five new edges in Figure 3 ensure that if we extend the subtiling so that it combinatorially describes a finite subdivision rule, then the subdivision map cannot have a Levy cycle. This can be proven by contradiction as follows. Let $\gamma $ be an element of a Levy cycle. In the model subdivision complex, the new arc whose barycenter has label $a_4$ bounds an open disk D such that $D \cap P_f = \{a_3\}$ and its boundary contains $\infty $ . Lemma 3.1 implies, for every positive integer n, that $\gamma $ can be isotoped rel the postcritical set to be disjoint from all new edges of the nth subdivision $\mathcal {R}^{n}(S_{\mathcal {R}})$ . Since the disk D contains the single postcritical point $a_3$ and its boundary contains $\infty $ , the vertex $a_3$ cannot be in the connected component $D_{\gamma }$ of $S_{\mathcal {R}}\setminus \gamma $ which does not contain $\infty $ . In the next two subdivisions, there will be new edges enclosing the postcritical points $a_4$ , $a_2$ , and $a_1$ , so none of them can be in $D_{\gamma }$ . There is a new edge joining the vertex $a_5$ to $\infty $ , so $a_5$ cannot be in $D_{\gamma }$ . In the next subdivision, there will be a new edge joining $a_6$ to an $\infty $ , so $a_6$ cannot be in $D_{\gamma }$ . Hence, no finite postcritical point can be in $D_\gamma $ , so there are no Levy cycles and, hence, the subdivision map is equivalent to a rational map. This concludes our example.

We call an edge of a subdivision $\mathcal {R}^n(S_{\mathcal {R}})$ (of a finite subdivision rule $\mathcal {R}$ ) a new edge if it is not contained in an edge of $S_{\mathcal {R}}$ . The following lemma plays a crucial role in the proof of Theorem 1.1.

The ordering. Suppose for the recursive step that $A\not \subset \widetilde {A}$ , $i\in \{1,\dots ,n\}$ , and that we have already defined $a_j$ for $j\in \{1,\dots ,i-1\}$ . Suppose that there is a vertex in $P'\setminus \widetilde {A}$ that is not periodic under $\tau $ . Because $\tau (\widetilde {A})\subseteq \widetilde {A}$ , we can choose an element $v\in A\setminus \widetilde {A}$ such that v is not the image under $\tau $ of a postcritical vertex. Let $a_i = v$ and add v to $\widetilde {A}$ . If $\tau (v) \notin \widetilde {A}$ , we let $a_{i+1} = \tau (v)$ and add $\tau (v)$ to $\widetilde {A}$ . We continue until we reach an index j such that $\tau (a_j)$ is in $\widetilde {A}$ . At this point, we stop this iteration of the recursion. We define the ordered set $(a_i,\dots ,a_{j})$ to be the chain of each of its elements. We call it a type-1 chain. It begins with an element of A. The first element of the chain is $a_i$ and the last element of the chain is $a_{j}$ . The length of the chain is $j+1-i$ . After redefining i to be $j+1$ , we continue this recursive step as long as possible.

Once we can no longer continue this recursion, the elements of $P'$ which remain are exactly the elements of finite (attractor) cycles which are connected components of $\Gamma $ . To start the next recursion, we choose a critical vertex v in a remaining attractor cycle and let $a_i = v$ . Let k be the number of elements in the attractor cycle. For $1\le j < k$ , let $a_{i+j} = \tau ^{\circ j}(a_i)$ . As before, $a_i$ is the first element of the chain, $a_{i+k-1}$ is the last element of the chain, and the length of the chain is k. We call it a type-2 chain. We continue recursively to choose all of the points in the other attractor cycles. After doing this, the elements of $P'$ are $a_1,\dots ,a_n$ in order.

Figure 6

One or two chains of type $1$ .

Construction of $S_{\mathcal {R}}$ . We next construct the associated finite subdivision rule $\mathcal {R}$ . The $1$ -skeleton of the model subdivision complex $S_{\mathcal {R}}$ is a tree as in Figure 1. There is one central vertex. We identify $\infty \in \Gamma $ with this central vertex. There are n ‘stickers’ (a sticker is an edge of the graph with a vertex of valence one, like a stick pin with a spherical head) from $\infty $ to valence $1$ vertices $a_1,\dots ,a_n$ , in counterclockwise order. We identify $a_1,\dots ,a_n\in \Gamma $ with these valence $1$ vertices. The tile type t is a $(2n)$ -gon, which we think of as an n-gon with each edge bisected. The characteristic map $\psi :t\to S_{\mathcal {R}}$ maps the edge barycenters to the sticker heads and the other vertices to $\infty $ . The edge barycenters are called finite vertices and the others are called $\infty $ -vertices. More generally, a vertex of some subdivision of t which is not a vertex of t is called a finite vertex. Every vertex v of t is labeled by $\psi (v)$ . These vertex labels are placed outside t in Figures 6–10. We use clockwise order on $\partial t$ .

If k is an integer with $k\ge 2$ , a k-doodle is a graph with three vertices and k edges (none of them loops) such that one vertex (the central vertex) has valence k, one vertex (the head) has valence $1$ , and the third vertex (the foot) has valence $k-1$ . Note that a $2$ -doodle is a bisected arc.

We define a subdivision $\mathcal {R}(t)$ of t. We do this in three stages. We first define a subtiling of t into subtiles such that the finite vertices in each subtile are in the proper cyclic order. This means that there might be fewer than n of them, but they will have image labels, which are distinct elements of $\{a_1,\ldots ,a_n\}$ , and, when taken in clockwise order, their image labels have the same cyclic order as in $(a_1,\ldots ,a_n)$ . We then add arcs and k-doodles as determined by the critical vertices of $\Gamma $ so that we have d subtiles. Finally, we add stickers as necessary to get the subdivision $\mathcal {R}(t)$ . The tiles of the first stage will be defined so that the resulting subdivision map does not have Levy cycles. We do this by ensuring that there is an iterated subdivison $\mathcal {R}^n(t)$ of t such that for each finite vertex v of t except possibly one, either there is a new edge from v to an $\infty $ -vertex or there is an arc (made out of two or four new edges) from the $\infty $ -vertex before v to the $\infty $ -vertex after v.

As we construct $\mathcal {R}(t)$ , we will give image labels to its vertices. The image label of a vertex is the vertex it will map to under the analog of $\psi $ from $\mathcal {R}(t)$ to $S_{\mathcal {R}}$ . So, for a vertex v in t, the image label of v is defined to be $\tau (\psi (v))$ . We will keep track of the critical vertices that have already been accounted for during the construction of $\mathcal {R}(t)$ in a set $\widetilde {C}$ . For the beginning of the construction, we define $\widetilde {C} = \emptyset $ .

Figure 7

A chain of type $1$ followed by a chain of type $2$ .

Figure 8

A chain of type $2$ followed by a chain of type $1$ . Here, $i=n$ .

Figure 8 Long description

There are two panels. In both, a V-shaped chain has a central top vertex. The left branch is labeled a sub n, the right branch a sub 1. On the left panel, a sub n is repeated below the left branch, and tau of a sub 1 appears in gray below the right branch. The right panel repeats the main structure but adds a gray arrow from a sub n to a sub 1, labeled tau of a sub n, and a gray arrow from the top vertex to the right branch, labeled tau of a sub 1. All gray elements indicate mappings or transformations between nodes.

Figure 9

Two chains of type $2$ .

Figure 10

A single chain and it has type $1$ .

Figure 10 Long description

The left panel shows a V-shaped chain with three labeled points: a sub r minus 1 at the left vertex, a sub r at the right, and a gray arc from a sub r to tau of open parenthesis a sub r close parenthesis. The right panel shows a similar V-shaped chain with points a sub n at the left, a sub 1 at the right, and a sub r and a sub 1 below. Two gray arcs connect a sub 1 to tau of open parenthesis a sub 1 close parenthesis and a sub r to a sub 1.

Stage 1. Suppose $a_i$ is the last element of a chain and $a_j$ is the first element of the next chain (in cyclic order). So, either $1\le i < n$ and $j=i+1$ or $i=n$ and $j=1$ . The construction in stage 1 depends on the types of the chains which contain $a_i$ and $a_j$ . We consider various cases.

If $a_i$ and $a_{j}$ are in distinct chains of type $1$ or if they are in the same chain of type $1$ (there is only one chain) and $\tau (a_i)\ne a_j$ , then we add a k-doodle, with k being the degree of the critical vertex $c_{a_{j}}$ , with head the $\infty $ -vertex after $a_i$ and with foot the $\infty $ -vertex before $a_i$ . We give the central vertex of the k-doodle image label $a_{j}$ and add $c_{a_{j}}$ to $\widetilde {C}$ . See Figure 6, which, like Figures 7–10, is drawn with $k=2$ .

If $a_i$ is in a chain of type $1$ and $a_{j}$ is in a chain of type $2$ , then we add $k-1$ edges joining $a_j$ to the $\infty $ -vertex before $a_i$ (where $k = \deg (a_j)$ ) and add $a_{j}$ to $\widetilde {C}$ . See Figure 7.

Suppose $a_i$ is in a chain of type $2$ and $a_j$ is in a chain of type $1$ (this can only occur if $i=n$ and $j=1$ ). If $a_i$ is in a chain of length $1$ , then we do not do anything at this stage. If $a_i$ is in a chain of length greater than $1$ , then we add a k-doodle (with $k = \deg (c_{a_{j}})$ ) with head the $\infty $ -vertex after $a_i$ and with foot the $\infty $ -vertex before $a_i$ . We give the central vertex of the k-doodle image label $a_{j}$ and add $c_{a_{j}}$ to $\widetilde {C}$ . Figure 8 shows both possibilities.

If $a_i$ and $a_{j}$ are both in chains of type $2$ , and $a_i$ is in a chain of length $1$ , then to $a_{j}$ we add $k-1$ edges joining it to the $\infty $ -vertex before $a_{j}$ (where $k= \deg (a_j)$ ) and add $a_{j}$ to $\widetilde {C}$ . If $a_i$ and $a_{j}$ are both in chains of type $2$ , and $a_i$ is in a chain of length greater than $1$ , then to $a_{j}$ we add $k-1$ edges joining it to the $\infty $ -vertex before $a_{i}$ (where $k=\deg (a_j)$ ) and add $a_{j}$ to $\widetilde {C}$ . The two possibilities are shown in Figure 9.

Now, suppose that there is a single chain, it has type $1$ , and $\tau (a_n) = a_1$ . This is the only remaining case. The Riemann–Hurwitz condition implies that if there is just one finite critical value, then $\Gamma $ has only two critical vertices and their degrees both equal the degree of $\Gamma $ . Hence, the finite critical vertex is the only vertex of $\Gamma $ which $\tau $ maps to the finite critical value. This is impossible in the present case because $a_1\in A$ and $\tau (a_n)=a_1$ . So, either one of the $a_i$ elements is a critical vertex or one of the $a_i$ elements with $i>1$ is a critical value.

First, suppose that $r\in \{1,\dots ,n\}$ and $a_r$ is a critical vertex with degree k. We add $k-1$ arcs in t from $a_r$ to the $\infty $ -vertex before $a_r$ , and we add $a_r$ to $\widetilde {C}$ . See the left side of Figure 10.

If none of the $a_i$ elements is a critical vertex, then some $a_i$ with $i>1$ is a critical value. In this case, suppose $r\in \{2,\dots ,n\}$ such that $a_r$ is a critical value. Let $k_1 = \deg (c_{a_1})$ and let $k_r = \deg (c_{a_r})$ . We add a $k_r$ -doodle with head the $\infty $ -vertex after $a_n$ and with foot the $\infty $ -vertex before $a_n$ . We give its central vertex image label $a_r$ and we add $c_{a_r}$ to $\widetilde {C}$ . We then add a $k_1$ -doodle with head the $\infty $ -vertex after $a_n$ and with foot the $\infty $ -vertex before $a_n$ as indicated in Figure 10. We give its central vertex image label $a_1$ and we add $c_{a_1}$ to $\widetilde {C}$ . See the right side of Figure 10. This completes stage $1$ of the construction.

Verification of stage 1 consistency. We first verify that the stage 1 edge constructions can be made so that interiors of distinct edges are disjoint. This is clear unless one of the edges is as in Figure 7 or the right side of Figure 9. In both cases, the second chain $(a_j,\ldots )$ has type 2. Figures 7–9 show that a stage 1 edge can join the $\infty $ -vertex immediately preceding vertex $v_j$ to a vertex with image label in the chain $(a_j,\ldots )$ or an adjacent $\infty $ -vertex only in the situation of the left side of Figure 9. So, edges as in Figure 7 or the right side of Figure 9 can be constructed to meet other edges only in endpoints. Thus, the stage 1 edges can be constructed so that their interiors are disjoint.

We now look at what we have after stage $1$ . Every subtile except for the central one is either a $2$ -gon or a $4$ -gon, and so there are only one or two finite vertices. For a $2$ -gon, there is only one finite vertex and so its image label is in proper cyclic order.

For a $4$ -gon, there are two finite vertices, so their image labels are in proper cyclic order if they are distinct. Chains of type 2 consist of the vertices of connected components of $\Gamma $ which are cycles. Using this and the facts that elements of chains are distinct and distinct chains are disjoint, one verifies that the image labels of the 4-gons in Figures 7–10 are distinct. The only potential problem is if, in the notation of Figure 6, $\tau (a_i) = a_j$ . Suppose that this happens. Then, $a_i$ and $a_j$ are in different chains. Because $\tau (a_i) = a_j$ , and $a_i$ and $a_j$ are in different chains, $a_j$ is not periodic under $\tau $ . However, if there exists such a vertex when a chain is defined, then the first vertex of that chain must not be the image of a postcritical vertex. So, it is not possible that $\tau (a_i) = a_j$ . Hence, the two finite vertices of every 4-gon have different image labels.

Now, we verify that the same is true for the central subtile s. We first observe that the vertices of s alternate between finite postcritical vertices and $\infty $ -vertices, as required.

Note that if x is a vertex of t with label a, then the image label of x is $\tau (a)$ . So, considering only the vertices of s which are vertices of t, if they do not have image labels in the proper cyclic order, then the label of one of them is the last member of a chain. In this case, the left side of either Figures 8, 9, or 10 occurs. However, in these cases, it is not hard to see that our image labels are in proper cyclic order. So, the image labels of the vertices of s which are vertices of t are in the proper cyclic order.

It remains to interpolate the vertices of s which are not vertices of t. These occur in Figure 6, and the right sides of Figures 8 and 10. In the first two cases, the image label of the vertex in question is $a_j$ , and in the last case, this image label is $a_1$ . A short case analysis shows that proper cyclic order also occurs in these cases. We conclude that the image labels of the vertices of s are in proper cyclic order after stage 1.

Stage 2. For the second stage, we add subtiles (much as in stage 1) corresponding to the critical vertices in $C'$ that are not in $\widetilde {C}$ . We do this recursively. Each time we add subtiles because of an element c of $C' \setminus \widetilde {C}$ , we add this element to $\widetilde {C}$ . This guarantees that we visit every element of $C'$ exactly once, including the visits made in stage 1. Since $C'$ is finite, this process will terminate.

Suppose $c\in C' \setminus \widetilde {C}$ . Let $k = \deg (c)$ .

If $c\in P'$ , then we add $k-1$ edges from c to the $\infty $ -vertex of t before c and we add c to $\widetilde {C}$ . Image labels of finite vertices of all tiles remain in proper cyclic order.

Now, suppose $c\not \in P'$ , but that there is another element $c'\in C' \setminus \widetilde {C}$ with $c'\not \in P'$ and $\tau (c')\ne \tau (c)$ . Let $k' = \deg (c')$ . Let $i,j\in \{1,\dots ,n\}$ such that $\tau (c) = a_i$ and $\tau (c') = a_j$ . Choose a subtile s of t. If s does not contain a vertex with image label $a_i$ , then there is a unique $\infty $ -vertex in s such that we can add a k-doodle with head and foot this $\infty $ -vertex, and with image label $a_i$ , and still have the image labels be in cyclic order. We do this and we add c to $\widetilde {C}$ . Image labels of finite vertices of all tiles remain in proper cyclic order. Suppose s does contain a vertex v with image label $a_i$ . Then, we add a $k'$ -doodle to s with head the $\infty $ -vertex after v, with central vertex with image label $a_j$ , and with foot the $\infty $ -vertex before v. We then add a k-doodle to s with head the $\infty $ -vertex after v, with central vertex with image label $a_i$ , and with foot the $\infty $ -vertex before v. There are a new subtile in s with the same image labels (and in the same cyclic order) as for s, some $2$ -gons if $k>2$ or $k'> 2$ , and two $4$ -gons with a vertex labeled $a_i$ and a vertex labeled $a_j$ . So, we still have the image labels of the finite vertices of all of the subtiles in cyclic order. Finally, we add c and $c'$ to $\widetilde {C}$ .

To complete stage $2$ , we need to consider the case that all elements of $C' \setminus \widetilde {C}$ have the same image $a_i$ under $\tau $ . Let $r = \#( C' \setminus \widetilde {C} )$ and let m be the sum of the degrees of the elements of $C' \setminus \widetilde {C}$ . At every step of the construction thus far, the number of subtiles of t increases by $\deg (v)-1$ , where v is the vertex added to $\widetilde {C}$ . So, the number of subtiles of t created thus far is

$$ \begin{align*} 1+\Sigma_{v\in\widetilde{C}}(\mathrm{deg}(v)-1) & = 1+\Sigma_{v\in C'}(\mathrm{deg}(v)-1)-m+r\\ & = 1+d-1-m+r\\ & = (d-m)+r. \end{align*} $$

Because there remain m tiles to add with a vertex labeled $a_i$ and the total number of tiles at the end of this stage with a vertex labeled $a_i$ is at most d (the sum of the weights of the edges of $\Gamma $ with target vertex $a_i$ is at most d), there is currently a subtile that does not have a vertex with image label $a_i$ . There is a unique $\infty $ -vertex in this subtile such that we can add a k-doodle with head and foot this $\infty $ -vertex, and with image label $a_i$ , and still have the image labels be in cyclic order. We do this and we add c to $\widetilde {C}$ . This completes the recursive step, so we can continue the recursion until $C' = \widetilde {C}$ . This completes the second stage. At this point, there are d subtiles of t and in each subtile, the image labels of the finite vertices are in proper cyclic order.

Stage 3. Suppose s is a subtile of the construction after stage two and ${i\in \{1,\dots ,n\}}$ . If s does not have a vertex with image label $a_i$ , then there is a unique $\infty $ -vertex of s to which we can add a sticker whose other vertex has image label $a_i$ and still have the image labels of the finite vertices in cyclic order. We do this for every such s and i. This completes stage $3$ .

Completion of the construction of $S_{\mathcal {R}}$ . At this point, every subtile has n finite vertices and their image labels are in proper cyclic order. We define this to be the subdivision $\mathcal {R}(t)$ and we define its image under the characteristic map $t\to S_{\mathcal {R}}$ to be $\mathcal {R}(S_{\mathcal {R}})$ . It is straightforward to define a subdivision map that takes $\mathcal {R}(S_{\mathcal {R}})$ to $S_{\mathcal {R}}$ , which takes each open cell homeomorphically to an open cell, takes each $\infty $ -vertex to $\infty $ , and takes a vertex with image label $a_i$ to the vertex $a_i$ . This completes the definition of the finite subdivision rule $\mathcal {R}$ . It is clear from this construction that the ramification portrait of the subdivision map $\sigma _{\mathcal {R}}$ is isomorphic to $\Gamma $ .

Verification that $\sigma _{\mathcal {R}}$ has no Levy cycle. We prove by contradiction that $\sigma _{\mathcal {R}}$ cannot have a Levy cycle. Suppose $\{\delta _0,\dots ,\delta _{k-1},\delta _k=\delta _0\}$ is a Levy cycle for $\sigma _{\mathcal {R}}$ . Choose any $i\in \{1,\dots ,k\}$ and consider the component $D_i$ of $S^2 \setminus \delta _i$ that does not contain $\infty $ . Since $\delta _i$ is essential and is not peripheral, $D_i$ must contain at least two points of P. We will obtain a contradiction by showing that $D_i$ can contain at most one point of P. For this, consider a finite postcritical point $a_j$ of $\sigma _{\mathcal {R}}$ in $S_{\mathcal {R}}$ .

Suppose that there exists a positive integer m such that $\tau ^{\circ m}(a_j)$ is a critical vertex. Then, $a_j$ and $\infty $ are joined by a new edge in $\mathcal {R}^{m+1}(S_{\mathcal {R}})$ . By Lemma 3.1, we can isotope $\delta _i$ in $S^2\setminus P_f$ to be disjoint from this new edge, so $a_j$ and $\infty $ are in the same component of $S^2 \setminus \delta _i$ . Hence, $a_j\notin D_i$ . So, $a_j\notin D_i$ if either $a_j$ is in a type- $2$ chain or we are in the situation of the left half of Figure 10.

Now, suppose that $a_j$ is in a type- $1$ chain that is not followed by a type- $2$ chain. Let $a_r$ be the last element of the type- $1$ chain that contains $a_j$ . We are in the situation of either Figure 6 or the right half of Figure 10. So, there is a pair of new edges of $\mathcal {R}(S_{\mathcal {R}})$ that bounds an open disk D that contains $a_r$ and no other postcritical points. Hence, $a_r$ cannot be in the open disk $D_i$ since if so, Lemma 3.1 implies that we can isotope $D_i$ rel $P_f$ into D. Similarly, in $\mathcal {R}^{r-j+1}(S_{\mathcal {R}})$ , there is a pair of new edges that bounds an open disk that contains $a_j$ and no other postcritical points.

Finally, suppose $a_j$ is in a type- $1$ chain which is followed by a type- $2$ chain. This is the only remaining possibility. Let $a_r$ be the last element of the type- $1$ chain containing $a_j$ . Suppose that $j\ne r$ . Let u be the vertex of t with label $a_j$ . Let $t'$ be the tile of $\mathcal {R}^{r-j}(t)$ which contains u. Then, the label of u relative to $t'$ is $a_r$ , that is, the structure map of $t'$ from $t'$ to $S_{\mathcal {R}}$ maps u to $a_r$ . Let v be the finite vertex of $t'$ following u. The label of v relative to $t'$ is $a_{r+1}$ . Because (i) $j\ne r$ , (ii) $a_r$ is in a type-1 chain, and (iii) $a_{r+1}$ is in a type-2 chain, the definition of chains implies that v is not the finite vertex of t following u. So, the edge $e_1$ of $t'$ joining v and the $\infty $ -vertex of $t'$ following u must be a new edge. However, as in Figure 7, there is a new edge $e_2$ in $\mathcal {R}(t')$ joining v and the $\infty $ -vertex of $t'$ preceding u. As before, the two new edges in $\mathcal {R}^{r-j+1}(S_{\mathcal {R}})$ corresponding to $e_1$ and $e_2$ bound an open disk which contains $a_j$ and no other postcritical point. We have reduced to the case in which $j=r$ . Chains are defined so that at most one postcritical point has this property. So, the open disk $D_i$ can contain at most one postcritical point, which contradicts the assumption that $\delta _i$ is an element of a Levy cycle.

Since $\sigma _{\mathcal {R}}$ is a Thurston map whose ramification portrait is isomorphic to $\Gamma $ and $\sigma _{\mathcal {R}}$ has no Levy cycle, the proof of Theorem 1.1 is complete.

4 Completely unobstructed ramification portraits

5 Rose maps

To prove Theorem 1.9, we need a topological description for topological polynomials which may not be subdivision maps. We begin this section by discussing our approach to this. We define a rose to be the boundary of the union of finitely many closed topological disks in the 2-sphere which are disjoint except for having exactly one point in common. We view a rose as a graph with exactly one vertex. Its edges are called petals.

Let $S_1^2$ and $S_2^2$ be two copies of $S^2$ , and suppose that we have a finite branched covering map $g\colon \, S_1^2\to S_2^2$ whose critical values lie in a finite set $P\subseteq S_2^2$ . The restriction of g to $S_1^2\setminus g^{-1}(P)$ is a covering map from $S_1^2\setminus g^{-1}(P)$ onto $S_2^2\setminus P$ . In the context of covering maps, a straightforward thing to do in this situation is to use the fact that $S_2^2\setminus P$ is homotopic to a rose with $\#(P)-1$ petals—the fundamental group of $S_2^2\setminus P$ is a free group on $\#(P)-1$ generators. Let $R_2$ be a rose in $S_2^2\setminus P$ which is a spine and let $R_1=g^{-1}(R_2)$ . Because every connected component of $S_2^2\setminus R_2$ contains at most one branch value of g, every connected component of $S_1^2\setminus R_1$ is a disk, equivalently, $R_1$ is connected. The restriction of g to $R_1$ is a covering map onto $R_2$ and this restriction determines g up to homotopy.

With this in mind, we construct finite branched covering maps $g\colon \, S_1^2\to S_2^2$ as follows. Let $R_2\subseteq S_2^2$ be a rose with $n\ge 1$ petals. We orient $S_2^2$ and label n connected components of $S_2^2\setminus R_2$ each bounded by one petal with $1,\dotsc ,n$ in counterclockwise order. We label the remaining connected component of $S_2^2\setminus R_2$ with $\infty $ . Suppose that we have a finite connected graph $R_1\subseteq S_1^2$ whose vertices have small neighborhoods which look like small neighborhoods of the vertex of $R_2$ . Here is what this means. The connected components of $S_1^2\setminus R_1$ are labeled with $1,\dotsc ,n$ (duplications allowed) and $\infty $ . We choose a barycenter for every edge of $R_1$ and call the resulting edges half edges. These barycenters are not vertices of $R_1$ . Let v be a vertex of $R_1$ . Then, $2n$ half edges contain v. After orienting $S_1^2$ , they can be written as $\epsilon _1,\dotsc ,\epsilon _{2n}$ in counterclockwise order around v so that $\epsilon _{2i-1}$ and $\epsilon _{2i}$ are in the boundary of a connected component of $S^2_1\setminus R_1$ with label i for every $i\in \{1,\ldots ,n\}$ . Furthermore, $\epsilon _{2i}$ and $\epsilon _{2i+1}$ are in the boundary of a connected component of $S_1^2\setminus R_1$ with label $\infty $ for every $i\in \{1,\ldots ,n\}$ , where $\epsilon _{2n+1}=\epsilon _1$ . It is a straightforward matter, by mapping vertices, then half edges, and then disks, to construct a finite branched covering map $g\colon \, S_1^2\to S_2^2$ such that $g|_{R_1}$ is a covering map onto $R_2$ which maps vertices to vertices and edges to edges. This can be done so that g has at most one critical point in every connected component of $S_1^2\setminus R_1$ .

We define a rose map to be a map of pairs $g\colon \, (S_1^2,R_1)\to (S_2^2,R_2)$ , where $S_1^2$ and $S_2^2$ are two oriented copies of $S^2$ , $g\colon \, S_1^2\to S_2^2$ is an orientation-preserving finite branched covering map, $R_2\subseteq S_2^2$ is a rose, $R_1= g^{-1}(R_2)$ is a graph with pullback graph structure, and every connected component of $S_2^2\setminus R_2$ contains at most one critical value of g. The next lemma guarantees the existence of the rose maps that we will use for the proof of Theorem 1.9. We will precompose a rose map $g\colon \, S_1^2\to S_2^2$ with a homeomorphism $h\colon \, S_2^2\to S_1^2$ to obtain a desired topological polynomial $f=g\circ h$ . In the lemma, the connected components of $S_2^2\setminus R_2$ are labeled, which induces a labeling of the connected components of $S_1^2\setminus R_1$ , which induces a labeling of the vertices of the graph dual to $R_1$ .

Proof. We will construct $R_1$ rather explicitly.

To prepare for the construction of $R_1$ , let U be the set of critical vertices which $\tau $ maps to u and let W be the set of remaining finite critical vertices. So, $C_\Gamma =U\amalg W\amalg \{\infty \}$ . The Riemann–Hurwitz condition gives that

$$ \begin{align*} \sum_{w\in U}^{}(\deg_\tau(w)-1)+\sum_{w\in W}^{}(\deg_\tau(w)-1)=d-1, \end{align*} $$

where d is the degree of $\Gamma $ . So,

(5.1) $$ \begin{align} \sum_{w\in W}^{}(\deg_\tau(w)-1)=d-1+\#(U)-\sum_{w\in U}^{}\deg_\tau(w)\ge \#(U)-1. \end{align} $$

In particular, if W is empty, then $\#(U)=1$ . In this case, there is exactly one choice for $R_1$ up to isomorphism. Only two connected components of $S_1^2\setminus R_1$ are not monogons and one of these has label $\infty $ . Statement (1) is true in this case and statement (2) is true since $u=v$ . So, we henceforth assume that W is non-empty.

Let m be the maximum number of incoming edges from critical vertices at the critical values other than u. Then, it is possible to partition W into disjoint non-empty subsets $W_1,\dotsc ,W_m$ so that if $w,w'\in W_i$ for some i with $w\ne w'$ , then $\tau (w)\ne \tau (w')$ . We do this so that if $u\ne v$ , then there exists $w\in W_m$ such that $\tau (w)=v$ .

Now, we begin to construct $R_1$ . We enumerate the elements of $W_1$ and for every ${w\in W_1}$ , we construct a closed (two-dimensional) polygon $P_w\subseteq S_1^2$ with $\deg _\tau (w)$ sides whose interior has label $\tau (w)$ . These polygons are disjoint from each other, except that each has exactly one vertex in common with the next. We obtain a chain (not to be confused with the chains in §3) $C_1$ of polygons. We also construct such chains $C_2,\dotsc ,C_m$ for $W_2,\dotsc ,W_m$ so that the chains $C_1,\dotsc ,C_m$ are disjoint from each other and if $u\ne v$ , then the polygon with label v in $C_m$ is last.

The choices of u and m imply that U contains at least m elements. We choose $m-1$ distinct elements $u_1,\dotsc ,u_{m-1}\in U$ . For every $i\in \{1,\ldots ,m-1\}$ , we construct a polygon $P_{u_i}$ as before which is disjoint from the polygons already constructed and from the other $P_{u_j}$ polygons, except that one vertex of $P_{u_i}$ is a vertex of $C_i$ that is only in the last polygon of $C_i$ , and a different vertex of $P_{u_i}$ is a vertex of $C_{i+1}$ that is only in the first tile of $C_{i+1}$ . The polygons $P_{u_1},\dotsc ,P_{u_{m-1}}$ join the chains $C_1,\dotsc ,C_m$ to form a single chain C.

We intend to also construct similar polygons $P_v$ for the remaining elements v of U. We intend to construct each of them in one of two ways. One way to construct $P_v$ is to choose a vertex of C contained in only one polygon and to construct $P_v$ so that it meets the polygons already constructed exactly in this vertex. Here is another way to construct $P_v$ . Choose $i\in \{1,\ldots ,m\}$ , and two consecutive polygons P and $P'$ in $C_i$ . Let x be the vertex common to P and $P'$ . We modify P and $P'$ slightly near x, pulling them apart so that they become disjoint. We then construct $P_v$ so that it contains both of the new vertices in P and $P'$ while being otherwise disjoint from the polygons already constructed.

The only obstacle to performing the constructions described in the previous paragraph is that C might not contain enough vertices to accommodate all the elements of U. However, it is not difficult to see that the number of elements of U that can be accommodated, including $u_1,\dotsc ,u_{m-1}$ , is

$$ \begin{align*} 1+\sum_{w\in W}(\deg_\tau(w)-1). \end{align*} $$

Thus, line equation (5.1) shows that C does indeed have enough vertices to accommodate all the elements of U. So, we construct a polygon $P_v$ as described in the previous paragraph for every $v\in U\setminus \{u_1,\dotsc ,u_{m-1}\}$ .

Because U contains at least m elements, this can be done so that

(5.2) $$ \begin{align} \text{the first of these polygons meets exactly one polygon in } C, \text{the first polygon in } C_1.\quad \end{align} $$

Furthermore, if u has an incoming edge from a non-critical vertex, then the inequality in line equation (5.1) is strict, so we may also construct these polygons so that

(5.3) $$ \begin{align} &\text{if } u \text{ has an incoming edge from a non-critical vertex, then the last polygon in }\nonumber\\ & C_m \text{ contains a vertex not in any of these polygons.} \end{align} $$

Every vertex in the complex constructed thus far is contained in either one or two polygons. If there are two, then their labels are different. Hence, it is possible to add monogons with labeled interiors at each of these vertices so that the cyclic order of labels about each vertex agrees with the cyclic order of the labels of $R_2$ . We have $R_1$ . The discussion preceding the lemma describes how to construct a rose map $g\colon \, (S_1^2,R_1) \to (S_2^2,R_2)$ from this information. Line equation (5.2) implies statement (1) of the lemma and line equation (5.3) implies statement (2).

6 Obstructed ramification portraits

Proof of Theorem 1.9

We first prove the theorem in cases (i) and (ii). Assume that $\Gamma $ satisfies either condition (i) or (ii). We will use the following example to illustrate various constructions in the proof. Consider the abstract polynomial ramification portrait that is shown as follows.

It satisfies the conditions of case (i).

We prepare to apply Lemma 5.1. Suppose that $\Gamma $ has n finite postcritical vertices $v_0,\dotsc ,v_n$ with $v_n=v_0$ such that $\tau (v_i)=v_{i+1}$ for every $i\in \{0,\ldots ,n-1\}$ . Of course, if $\Gamma $ satisfies condition (i), then $n=p^k$ . Choose a finite critical value u of $\Gamma $ with the maximum number of incoming edges from critical vertices. We redefine $v_0,\dotsc ,v_n$ if necessary so that $v_0=u$ without changing the assumptions.

In particular, if $\Gamma $ satisfies condition (i), then there exists a critical value ${v_i\in \{v_0,\dotsc ,v_{p^k-1}\}}$ such that i is not a multiple of $p^{k-1}$ . We set $v=v_i$ . In our example, we take $u=v_0$ and $v=v_2$ .

To define v in the case of condition (ii), suppose that $\Gamma $ satisfies condition (ii). Because u is in a non-attractor cycle, $\tau $ maps some vertex which is not critical to u. It follows that the inequality in line equation (5.1) is strict and so $\Gamma $ has more than one finite critical value. Let v be any finite critical value other than u.

We next define a positive integer m. If $\Gamma $ satisfies condition (i), then we set $m=p^{k-1}$ . Suppose that $\Gamma $ satisfies condition (ii). Let i be the index such that $v=v_i$ . Because n is not a prime power, it is the product of two relatively prime proper divisors. Because they are relatively prime, if both of these two proper divisors divide i, then n divides i, which is not true. Hence, some positive proper divisor of n does not divide i. Let m be such a positive proper divisor of n.

So, in either case (i) or (ii), m is a positive proper divisor of n such that $v=v_i$ and m does not divide i.

Lemma 5.1 implies that there exists a rose map $g\colon \, (S_1^2,R_1)\to (S_2^2,R_2)$ realizing the branch data of $\Gamma $ such that n connected components of $S_2^2\setminus R_2$ each bounded by a petal of $R_2$ are labeled $v_0,\dotsc ,v_{n-1}$ in counterclockwise order, the remaining connected component is labeled $\infty $ , and $R_1$ has a dual graph $R_1^*$ for which:

1. (1) the boundary of one connected component $C_u$ of $S_1^2\setminus R_1^*$ contains exactly two critical points: one vertex with label u and one vertex with label $\infty $ ;

2. (2) the boundary of one connected component $C_v$ of $S_1^2\setminus R_1^*$ contains exactly two critical points: one vertex with label v and one vertex with label $\infty $ .

By modifying $R_1^*$ if necessary, we may assume that the restriction of g to both $C_u$ and $C_v$ is injective. We identify every postcritical vertex w of $\Gamma $ with a point in the connected component of $S_2^2\setminus R_2$ with label w. These serve as the vertices of a graph $R_2^*$ dual to $R_2$ . Their g-pullbacks serve as the vertices of $R_1^*$ . Figure 11 depicts important features of the rose map g for our example.

Figure 11

Rose map $g:S_{1}^2\to S_{2}^2$ .

Figure 11 Long description

At the center is a node labeled infinity, marked with a red dot. Two large petal-shaped clusters, S squared sub 1 and S squared sub 2, are positioned to the left and right, each containing nine petals labeled v sub 0 through v sub 8. Each petal has a black dot at its tip. The left cluster, S squared sub 1, is labeled at the bottom left, and the right cluster, S squared sub 2, is labeled at the bottom right. Both clusters are connected to the central infinity node by curved lines. Above the central node, a third cluster is shown, also with nine petals labeled v sub 0 through v sub 8, and is labeled S squared sub 2 at the lower right. Dotted lines and ellipsis indicate continuation or repetition between the two main clusters. The spatial arrangement emphasizes the central infinity node as a hub connecting multiple petal graphs.

Now, we choose m disjoint closed topological disks $D_i\subseteq S_2^2\setminus \{\infty \}$ such that: (i) $v_j$ is in the interior of $D_i$ if $j\equiv i\text { (mod }m)$ for $i\in \{0,\dotsc ,m-1\}$ and $j\in \{0,\dotsc ,n-1\}$ ; (ii) $D_0$ is in the open disk $g(C_v)$ ; and (iii) for $i\in \{1,\dotsc ,m-1\}$ , $D_i$ is in the open disk $g(C_u)$ . The restriction of g to $C_v$ is a homeomorphism, so there exists a unique lift $\widetilde {D}_0$ of $D_0$ to $C_v$ . We denote the lift of $v_j$ to $C_v$ by $\widetilde {v}_j$ for every index $j\equiv 0\text { (mod } m)$ . There likewise exist unique lifts $\widetilde {D}_i$ of $D_i$ to $C_u$ for every $i\in \{1,\dotsc ,m-1\}$ . We denote the lift of $v_j$ to $C_u$ by $\widetilde {v}_j$ for every index $j\not \equiv 0\text { (mod }m)$ . Figure 12 shows all of these points and disks for our example.

Figure 12

Rose map $g:S_{1}^2\to S_{2}^2$ .

Figure 12 Long description

Starting at the left, the first petal-shaped graph is centered at a black dot with eight radiating edges labeled v sub 1 through v sub 8, each ending in a black dot. Gray paths loop around subsets of these vertices, forming nested contours. The region is labeled S super 2 sub 1 and C sub u. A thin curve connects this cluster to a central infinity symbol, then continues to a mirrored petal-shaped graph on the right, labeled S super 2 sub 2 and C sub v, with vertices v sub 0 through v sub 8. Gray paths again highlight connections among vertices. At the lower right, a smaller petal-shaped graph repeats the structure with vertices v sub 1 through v sub 6, gray paths, and a central black dot. All graphs are drawn with bold black lines for petals and dots for vertices, with gray shading indicating specific paths or cycles.

We next construct an orientation-preserving homeomorphism $h\colon \, S_2^2\to S_1^2$ such that: (i) $h(v_j)=\widetilde {v}_{j+1}$ for $j\in \{0,\dotsc ,n-1\}$ ; (ii) $h(\infty )=g^{-1}(\infty )$ ; and (iii) $h(D_i)=\widetilde {D}_{i+1}$ for $i\in \{0,\dotsc ,m-1\}$ , where $D_m=D_0$ . See Figure 13.

Finally, define $f:=g\circ h$ . The map f is a Thurston map whose ramification portrait is isomorphic to $\Gamma $ and the boundaries of the disks $D_i$ form a degenerate Levy cycle. This proves Theorem 1.9 in cases (i) and (ii).

Now, suppose that $\Gamma $ satisfies condition (iii). Let C be a non-attractor cycle of length $\ell \ge 2$ that does not contain all of the finite critical values of $\Gamma $ . Let $v_1,\dotsc ,v_n$ be the finite postcritical vertices of $\Gamma $ indexed so that $v_1,\dotsc ,v_\ell $ are the vertices of C, $\tau (v_\ell )=v_1$ , and $\tau (v_i)=v_{i+1}$ for $i\in \{1,\ldots ,\ell -1\}$ . By Lemma 5.1, there exists a critical value $v\notin C$ and a rose map $g\colon \, (S_1^2, R_1)\to (S_2^2,R_2)$ realizing the branch data of $\Gamma $ such that n connected components of $S_2^2\setminus R_2$ each bounded by a petal of $R_2$ are labeled $v_1,\dotsc ,v_n$ in counterclockwise order, the remaining connected component is labeled $\infty $ , and the boundary of one connected component $C_v$ of $S_1\setminus R_1^*$ contains exactly two critical points: one vertex with label v and one vertex with label $\infty $ . We may assume that the restriction of g to $C_v$ is injective. As in cases (i) and (ii), we identify every postcritical vertex w of $\Gamma $ with a point in the connected component of $S_2^2\setminus R_2$ with label w.

Let D be a closed topological disk in the open disk $g(C_v)$ whose interior contains $v_1,\dotsc ,v_\ell $ , but such that D contains no other postcritical vertex of $\Gamma $ . Let $\widetilde {D}$ be the lift of D to $C_v$ . Let $\widetilde {v}_i$ be the lift of $v_i$ to $C_v$ for $i\in \{1,\ldots ,\ell \}$ . Now, construct an orientation-preserving homeomorphism $h\colon \, S_2^2\to S_1^2$ such that: (i) $h(v_i)=\widetilde {v}_{i+1}$ for $i\in \{1,\ldots ,\ell -1\}$ ; (ii) $h(v_\ell )=v_1$ ; (iii) $h(\infty )=g^{-1}(\infty )$ ; (iv) $h(D)=\widetilde {D}$ ; and (v) the ramification portrait of $f:=g\circ h$ is isomorphic to $\Gamma $ . Since $\partial D$ is by itself a degenerate Levy cycle for f, this proves Theorem 1.9 in case (iii).

Finally, we consider case (iv), where $\Gamma $ has at least two non-attractor cycles of length one. Let a and b denote the vertices of two non-attractor cycles of $\Gamma $ of length $1$ . Define the abstract ramification portrait $\Gamma '$ as follows: $V(\Gamma ')=V(\Gamma )$ , $\tau '(a)=b$ , $\tau '(b)=a$ , and $\tau '(x)=\tau (x)$ for all $x\in V(\Gamma ')\setminus \{a,b\}$ . So, $\Gamma '$ satisfies condition (iii). The proof for case (iii) shows that there is a Thurston map f realizing $\Gamma '$ that has a degenerate Levy cycle consisting of a single curve which bounds a disk D such that $D\cap P_f = \{a,b\}$ . Postcompose f with an orientation-preserving homeomorphism $h:S^{2}\to S^{2}$ such that h is the identity in the complement of D, $h(b)=a$ , and $h(a)=b$ . Then, $h\circ f$ is an obstructed Thurston map whose ramification portrait is isomorphic to $\Gamma $ .

Figure 13

The obstructed Thurston map $f:S^{2}_{2}\to S^{2}_{2}$ .

Figure 13 Long description

Starting at the top left, the first panel shows a graph with five labeled vertices v tilde 4, v tilde 2, v tilde 1, v tilde 5, v tilde 7, and v tilde 8 connected by curved edges, labeled S super 2 sub 1. The top right panel shows a three-pronged structure with vertices v tilde 0, v tilde 3, and v tilde 6. An arrow labeled h points from the top left to the bottom left panel. The bottom left panel, labeled S super 2 sub 2, displays a more complex graph with vertices v 0 through v 8, each connected by multiple curved and straight edges. The bottom right panel, also labeled S super 2 sub 2, shows a similar structure but with different edge arrangements. An arrow labeled g points from the top right to the bottom right panel, and an arrow labeled f points from the bottom left to the bottom right panel. All vertices and edges are clearly labeled, and the mappings illustrate the transformation of the graph structures between surfaces.

7 Classification of completely unobstructed ramification portraits