Proof. This is [Reference Dahmani, Hagen and Sisto19, Corollary 3.6] (which applies to DT_K in view of [Reference Dahmani, Hagen and Sisto19, Proposition 5.1]).

Proof. This is essentially [Reference Behrstock, Hagen, Martin and Sisto8, Proposition 8.31], whose proof only uses [Reference Dahmani, Hagen and Sisto19, Corollary 3.6] and therefore works for any deep enough subgroup. Said proposition is stated for $m_0=3 \text{Gen}(S)+2p(S)-3$, but the only reason for that particular threshold in [Reference Behrstock, Hagen, Martin and Sisto8] is that this is the maximal number of sides of generalised m-gons considered in that paper, but the proof works for any fixed m ₀. Finally, [Reference Behrstock, Hagen, Martin and Sisto8, Proposition 8.31] as stated deals only with the case $b\geqslant5$, but with similar tools it is easy to prove the uniqueness of the orbit of lifts also for edges and triangles in the Farey complex. This shall be done in Lemmas 2.8 and 2.9, whose proofs are prototypical of many arguments throughout the paper.

Proof. By definition of the quotient graph, there exist two adjacent vertices $x,y\in \mathcal{C}_4$ whose projections are $\overline x$ and $\overline y$, respectively. Therefore, there exists at least one lift. Now let $\{x',y'\}$ be another lift of $\{\overline x, \overline y\}$. Up to the action of $\mathcal{N}$ we can assume that $y=y'$. Therefore, we have a path $\{x,y,x'\}$ inside $\mathcal{C}_4$, and there is an element $g\in \mathcal{N}$ such that $g(x)=x'$.

If g is the identity we are done, otherwise let $(s,\gamma_s)$ as in Proposition 2.3. If $d_\mathcal{C}(x,s)\leqslant 1$, so that γ_s fixes x, we can apply γ_s to both edges. Now we have a path $\{\gamma_s(x),\gamma_s(y),\gamma_s(x')\}$, and $x=\gamma_s(x)$ is sent to $\gamma_s(x')$ by the element $\gamma_s g$. Since $\alpha(\gamma_s g) \lt \alpha(g)$ we can proceed by induction on the complexity $\alpha(g)$.

Otherwise $d_s(x,x') \gt \Theta$. We claim that $d_\mathcal{C}(y,s)\leqslant 1$. If this is not the case then $\pi_s(y)$ is well-defined, and by triangle inequality for annular projections either $d_s(x,y) \gt \Theta/2$ or $d_s(x',y) \gt \Theta/2$. Without loss of generality, we can assume to be in the first case. If we choose $\Theta \geqslant 2B$, where B is the constant from the bounded geodesic image Theorem 2.4, we get that every geodesic from x to y (that is, the edge between these vertices) should pass through the star of s, which is a contradiction. Therefore, γ_s must fix y, and if we replace xʹ with $\gamma_s(x')$ we can again reduce the complexity of g, while preserving the fact that the two edges share an endpoint.

At the end of the inductive argument we have that $x=x'$, thus proving uniqueness of the orbit of lifts of an edge.

Proof. First we show that $\overline\Delta$ has a lift. Let $\{x,y,z,x'\}$ a lift of the triangle, which we see as a closed path $\{\overline x, \overline y,\overline z,\overline x\}$. Such a lift exists by how edges in the quotient graph are defined, but it is possibly open. Let $g\in \mathcal{N}$ be an element mapping x to xʹ, and suppose that g is not the identity. Then let $(s,\gamma_s)$ as in Proposition 2.3. If $d(x,s)\leqslant 1$, we can apply γ_s to the whole path, and proceed by induction on the complexity $\alpha(g)$.

Otherwise we claim that either $d_{\mathcal{C}}(s,y)\leqslant1$ or $d_{\mathcal{C}}(s,z)\leqslant1$. If none of this happens then $\pi_s(y)$ and $\pi_s(z)$ are both defined, and by triangle inequality we can assume, without loss of generality, that $d_s(x,y) \gt \Theta/3$. But then again if we choose $\Theta\geqslant 3B$, where B is the constant from the Bounded Geodesic Image Theorem 2.4, we have a contradiction. Hence γ_s fixes a cut point between y and z, and we can apply γ_s “beyond” this point. Then we get a new lift of the path, whose endpoints are x and $\gamma_s(x')=\gamma_s g(x)$, and again we can proceed by induction.

At the end of the inductive procedure we will have glued x to xʹ, thus we will have a closed lift of the triangle.

Now let $\Delta=\{x,y,z\}$ and $\Delta'=\{x',y',z'\}$ be two lifts inside $\mathcal{C}_4$ of the same triangle $\overline\Delta=\{\overline x,\overline y,\overline z\}$. By Lemma 2.8 we know that the edge $\overline y,\overline z$ admits a unique $\mathcal{N}$-orbit of lifts, therefore we can assume that $y=y'$ and $z=z'$. Moreover let $g\in\mathcal{N}$ be an element mapping x to xʹ.

If g is the identity we are done, otherwise let $(s,\gamma_s)$ as in Proposition 2.3. If $d(x,s)\leqslant 1$ then γ_s fixes x, and we can apply γ_s to both triangles and proceed by induction on the complexity $\alpha(g)$. Otherwise $d_s(x,x') \gt \Theta$, and with the same argument as in Lemma 2.8 we have that γ_s fixes both y and z. But then if we replace $\Delta'$ with $\gamma_s(\Delta')$ we can again reduce the complexity of g, while preserving the fact that the two triangles share an edge.

At the end of the inductive argument we have that $\Delta=\Delta'$, thus proving uniqueness of the orbit of lifts of a triangle.

Proof. This is the “moreover” part of [Reference Dahmani, Hagen and Sisto19, Proposition 4.3], whose proof only uses [Reference Dahmani, Hagen and Sisto19, Corollary 3.6].

Proof. By Theorem 2.7, whenever $\mathcal{N}$ is deep enough every special pentagon $\overline {\mathfrak{P}}$ as in Definition 2.15 admits a lift $\mathfrak{P}$, which remains isometrically embedded since the projection map is 1-Lipschitz. More precisely, if $\overline x,\overline y\in \overline{\Gamma}$ and $x,y\in \Gamma$ are their lift then

\begin{equation*}d_{\mathcal{C}^1}(x,y)\geqslant d_{\mathcal{C}^1/DT_K}(\overline x, \overline y)=d_{\overline{\mathfrak{P}}}(\overline x, \overline y)=d_{\mathfrak{P}}(x, y).\end{equation*}

Proof. P cuts out a sphere S ₄ with four punctures, and by Remark 2.13 having special intersection implies that the intersection number is 2. Thus we want to show that, if $\alpha,\alpha'$ are both adjacent to β in the Farey complex and D is one of the two disks bounded by β, then $\alpha'=H_D^k(\alpha)$ for some $k\in \mathbb{Z}$. By properties of the Farey complex there exists a sequence of triangles $T_1,\ldots, T_k$ such that each triangle contains β, $\alpha\in T_1$, $\alpha'\in T_k$ and every two consecutive triangles T_i and $T_{i-1}$ share and edge containing β. Thus it suffices to prove that if $\alpha,\alpha',\beta$ are the vertices of a triangle then $\alpha'=H^{\pm1}_D(\alpha)$. This is true, since the two curves $H^{\pm1}_D(\alpha)$ are adjacent to both α and β, and α and β belong to exactly two triangles.

Proof. First we show that β bounds a twice-punctured disk. To see this, notice that, by construction of the graph $X_{b-1}$, if two curves $x,y\in X_{b-1}$ are at distance 1 then there exists a curve $z\in X_{b-1}$ which is at distance 2 from both. This fact still holds if we replace distances in $X_{b-1}$ with distances in the curve graph, since $X_{b-1}$ is isometrically embedded, and it translates to the fact that whenever two curves in $X_{b-1}$ are disjoint there is a third curve which intersects both. Hence the union of all curves of $X_{b-1}$ must lie on the same connected component of $S_b\setminus\beta$, which is a disk that we call Sʹ. But then Sʹ contains b − 4 pairwise disjoint curves (other than the boundary curve), and therefore must contain at least b − 2 punctures. Thus the other connected component of $S_b\setminus\beta$ must be a disk with at most 2 punctures, and since β is an essential curve it cannot bound disks with less than two punctures.

Now, since β bounds a twice-punctured disk it is easy to find some $X_b'$ which contains β as a minimal curve. Let $X_{b-1}'=\text{Lk}_{X_b'}(\beta)$, and consider the $S_{b-1}$ obtained by shrinking β to a puncture B. By finite rigidity of $X_{b-1}$ there exists a mapping class $f\in MCG(S_{b-1})$ mapping $X_{b-1}'$ to $X_{b-1}$, which we may choose to be the identity in a neighbourhood of B, up to rotations of $S_{b-1}$ and isotopy. Therefore, f extends to a mapping class $F\in MCG(S_b)$ (for example, by setting F to be the identity in a neighbourhood of the twice-punctured disk surrounded by β), and now $F(X_b')$ is a copy of X_b that completes $\beta\star X_{b-1}$.

For the “moreover” part, choose a copy of X_b that completes $\beta\star X_{b-1}$ and let $\alpha'$ be the curve in this copy that corresponds to α (i.e., the curve that has special intersection with β with respect to the same P). By Lemma 2.17 and the discussion in Remark 2.18 there is a suitable power $H_\beta^k$ of the half twist around β that maps α to $\alpha'$. Since H_β fixes $\beta\star X_{b-1}$ we may apply this mapping class to obtain the desired copy of X_b.

Proof. Let $R\subset X_{b-1}$ be a simplex of codimension two such that $\alpha,z\in\text{Lk}(R)$, and complete it to a codimension one simplex P by adding some curve $x\in \text{Lk}(\alpha_i)\cap X_{b-1}$ that intersects α. For example, x and R can be chosen as in Figure 6. We need to show that if some curve zʹ lies in $\text{Lk} (P)\cap \text{Lk}(\alpha)$ then it is unique. First notice that zʹ must be an essential curve of the five-punctured sphere S ₅ that R cuts out (that is, it cannot surround only one puncture of S ₅, or it would coincide with some curve in R). Moreover, we know that α and x are both essential curves of S ₅ and they have intersection number 2, so we are in the situation depicted in Figure 7. Now, α and x together cut S ₅ in three once-punctured disk and a twice-punctured disk. Since zʹ is essential in S ₅ it must be the boundary of the twice-punctured disk, and thus it is unique.

Figure 7.

The five-punctured sphere that R cuts out.

Proof. The proof is very similar to that of [Reference Dahmani, Hagen and Sisto19, Lemma 4.4]. Suppose by contradiction that there is a shorter path from $\overline{x}$ to $\overline{y}$ inside $\text{Lk}(\overline\Delta)$. Lift this shorter path as a geodesic segment $[y, x']\subseteq \text{Lk}(\Delta)$, which can be done by Lemma 2.7. There exists $\gamma \in \mathcal{N}$ such that $\gamma x = x'$. We proceed by induction on the complexity of γ, to prove that $d_{\text{Lk}(\Delta)}(y, x') = d_{\text{Lk}(\Delta)}(y, x)$ (thus contradicting that $d_{\text{Lk}(\overline\Delta)}(\overline x,\overline y) \lt d_{\text{Lk}(\Delta)}(x, y)$). If γ = 1 we are done, otherwise let $(s, \gamma_s)$ be as in 2.3. If $d_{\mathcal{C}}(x,s)\leqslant1$ then $\gamma_s x = x$, and we can apply γ_s to both geodesics and to Δ and conclude by the induction hypothesis. Otherwise $d_s(x, \gamma x) \gt \Theta$, and arguing as in Lemma 2.8 we see that γ_s must fix Δ pointwise. Moreover, either $d_\mathcal{C}(y,s)\leqslant1$ or $d_s(y,x')\geqslant d_s(x,x')-d_s(x,y)$ must be large, since $d_s(x, y)$ is assumed to be small. In both cases there must be some $s' \in [y, x']$ in the star of s. Thus, one can change the lift of $[y, x]$ as $[y, \gamma_s x']$, while keeping it an isometric lift inside the link of Δ. One concludes by induction hypothesis, which applies to $\gamma_s\gamma$.

Proof. Since V is finite, we only need to show that for every $x,z\in \mathcal{C}$ there is some constant $M(x,z)$ such that $\sup_{s\in \mathcal{C}} d_s(x,z) \lt M(x,z)$, because then we can choose $\Theta \gt M=\max_{x,z\in V} M(x,z)$ to ensure that the hypothesis of Lemma 3.3 applies. One way to see this is to complete $x,z$ to complete clean markings $\mu,\nu$, in the sense of [Reference Masur and Minsky26, Section 2.5]. By the Distance formula [Reference Masur and Minsky26, Theorem 6.12] there exists a constant $M'(S)$ such that the sum $\sum_{s\in \mathcal{C},\,d_s(\mu,\nu) \gt M'} d_s(\mu,\nu)$ is bounded above in terms of the distance between µ and ν in the marking graph. In particular the sum is finite, and since every term is greater than a constant there must be a finite number of terms. Moreover, for every $s\in \mathcal{C}$ for which the quantity $d_s(x,y)$ is defined (in particular, not a base curve of µ or ν), let $x'\in \mu$ and $y'\in \nu$ be curves which realise $d_s(\mu,\nu)$. Then by triangle inequality

\begin{equation*}d_s(x,y)\leqslant d_s(x,x')+d_s(x',y')+d_s(y',y)=d_s(x,x')+d_s(\mu,\nu)+d_s(y',y)\end{equation*}

and since annular projections are 2-Lipschitz (see [Reference Masur and Minsky26, Lemma 2.3]) we have that

\begin{equation*}d_s(x,y)\leqslant d_s(\mu,\nu)+4\end{equation*}

Thus it suffices to choose $M(x,z) \gt \max_{s\in \mathcal{C},\,d_s(\mu,\nu) \gt M'} d_s(\mu,\nu)+4$.

Proof. Fix a copy X_b and let $M=\max_{x,z\in Y_b} M(x,z)$, where $M(x,z)$ is defined as in Corollary 3.4. We are left to prove that, if $X_b'$ is another copy of the rigid set and $Y_b'$ is the corresponding union of half twists, then M works also for $Y_b'$. Take an extended mapping class f that maps X_b to $X_b'$, which exists by Theorem 2.19. In particular f maps minimal curves to minimal curves. Therefore,

\begin{align*}Y_b'&=X_b'\bigcup_{\beta'\in X_b'\,minimal} H_{\beta'}^{\pm1}(X_b')=f(X_b)\bigcup_{\beta\in X_b\,minimal} H_{f(\beta)}^{\pm1}\circ f(X_b)\\ &=f(X_b)\bigcup_{\beta\in X_b\,minimal} f\circ H_{\beta}^{\pm1}(X_b)=f(Y_b)\end{align*}

Now the conclusion follows, since for every $x,z\in Y_b'$ and $s\in \mathcal{C}$ we have that

\begin{equation*}d_s(x,z)=d_{f^{-1}(s)}(f^{-1}(x),f^{-1}(z))\leqslant M.\end{equation*}

Proof. Just combine Lemma 3.6 and Corollary 3.4.

Proof. For every curve $x\in \mathcal{C}$ and every pseudo-Anosov mapping class $g\in MCG(S)$ there exists $\Theta \gt 0$ such that $\sup_{s\in\mathcal{C},n\in\mathbb{Z}}d_s(x, g^n(x)) \lt \Theta$. This follows from an argument in the proof of [Reference Dahmani, Hagen and Sisto19, Theorem 2.1] which just uses the Bounded geodesic image Theorem 2.4 and the fact that for every $x,z\in\mathcal{C}$ then $\sup_{s\in\mathcal{C}}d_s(x,z) \lt \infty$, which we proved in Corollary 3.4. Hence, if $\mathcal{N}$ is deep enough with respect to Θ, by Lemma 3.3 the projection is an isometry on the orbit $\{g^n(x)\}_{n\in\mathbb{Z}}$, which is infinite since g is pseudo-Anosov.

Proof. Recall that x and y fill Σ if and only if $d_{\text{Lk}(\Delta)}(x,y)\geqslant3$. If x and y satisfy the hypothesis of Lemma 3.3 then the distance between $\overline x$ and $\overline y$ in $\text{Lk}(\overline\Delta)$ remains at least three. But then every pair of lifts must be at distance at least three, since the projection map is 1-Lipschitz.

Proof. In all cases we will use that the projection is 1-Lipschitz, since the action of H over $\mathcal{C}$ is simplicial.

1. (1) Two adjacent curves in the Farey complex induce different puncture separations, therefore their projections must be at distance 1 since they cannot coincide.

2. (2) If α and $\alpha'$ are disjoint then they must induce different puncture separations, otherwise they would bound an unpunctured annulus and therefore they would be isotopic. Then their projections can only be at distance 1, for the same reasons as before.

3. (3) Suppose that α and $\alpha'$ intersect and induce non-nested puncture separations. Firstly, α and $\alpha'$ cannot have the same projection, since they induce different puncture separations. Furthermore, if by contradiction $\pi(\alpha)$ and $\pi(\alpha')$ are the endpoints of an edge then we may lift it, since the action is simplicial. In other words there exists some $g\in H$ such that α is disjoint from $g(\alpha')$. But then α and $g(\alpha')$ must induce nested puncture separations, and so must do α and $\alpha'$ since Dehn twists preserve puncture separations by Remark 3.11.

Proof of existence

We proceed by induction, using the case b = 5 as the base case since X ₅ is a pentagon, which lifts by Theorem 2.7. Assume that $b\geqslant6$ and that every copy of $X_{b-1}$ has a lift inside $\mathcal{C}_{b-1}$, and let $\overline{X}_b$ be a copy of X_b inside $\mathcal{C}_b/\mathcal{N}$. Let $\overline\beta\in \overline{X}_b$ be a minimal vertex, and let $\overline{X}_{b-1}:=\text{Lk}_{\overline{X}_b}(\overline\beta)$. In order to apply the inductive hypothesis we need to show that $\text{Lk}_{\mathcal{C}_b}(\beta)$ is isomorphic to $\mathcal{C}_{b-1}$ and that its projection is isomorphic to $\mathcal{C}_{b-1}/\mathcal{N}(S_{b-1})$. We start with the first assertion.

Now, Corollary 3.2 shows that the projection of $\text{Lk}_{\mathcal{C}_b}(\beta)$ is isomorphic to $\mathcal{C}_{b-1}/\mathcal{N}(S_{b-1})$. Therefore, we can use the inductive hypothesis to lift $\overline{X}_{b-1}$ to some $X_{b-1}$ inside $\text{Lk}_{\mathcal{C}_b}(\beta)$. Now the key point of the following construction will be this uniqueness lemma:

Proof. Since both x and xʹ have special intersection with y, with respect to the same facet P, by Lemma 2.17 there exists $k\in \mathbb{Z}$ such that, in the four-holed sphere S ₄ that P cuts out, $x'=H_D(x)$, where D is one of the disks bounded by y. Now we claim that $k=2m$ must be even, and therefore $H^{2m}_D=T_y^m$ extends to the whole surface S. To see this first notice that since x and xʹ are in the same $\mathcal{N}$-orbit they must induce the same puncture separation by Remark 3.11. Moreover x and y intersect twice, thus they separate the punctures of S into four sets $\mathcal{A}^\pm,\mathcal{B}^\pm$, each of which corresponds to one of the punctures of S ₄. Suppose that x induces the separation $\{\mathcal{A}^+\cup\mathcal{B}^+,\mathcal{A}^-\cup\mathcal{B}^-\}$ while y induces the separation $\{\mathcal{A}^+\cup\mathcal{A}^-,\mathcal{B}^+\cup\mathcal{B}^-\}$, and that D contains the punctures corresponding to $\mathcal{A}^\pm$. Then H_D swaps the two punctures corresponding to $\mathcal{A}^\pm$, which shows that $H_D^k(x)$ induces the same puncture separation as x if and only if k is even.

Now let $\gamma\in \mathcal{N}$ such that $x'=\gamma(x)$. If γ = 1 then we are done. Otherwise let $(s,\gamma_s)$ be as in Proposition 2.3. If $d_{\mathcal{C}}(x,s)\leqslant1$ then we may apply γ_s to the whole data and proceed by induction on the complexity of γ. Otherwise $d_s(x,x')\geqslant \Theta$. Now we claim that any other $z\in S\cup\{y\}$ is in the star of s, so that we can replace xʹ with $\gamma_s(x')$ and proceed by induction. If this is not the case then $\pi_s(z)$ is well-defined, and we have that

\begin{align*}d_s(x,x')&\leqslant d_s(x,z)+d_s(x',z)=d_s(x,z)+d_s(T^m_y(x),T^m_y(z))\\ &=d_s(x,z)+d_{T_y^{-m}(s)}(x,z)\end{align*}

where we used that z belongs to the star of y and is therefore fixed by T_y. But now

\begin{equation*}\Theta\leqslant d_s(x,x')\leqslant d_s(x,z)+d_{T_y^{-m}(s)}(x,z)\leqslant 2\max_{p,q\in X_{b},\, s\in\mathcal{C}} d_s(p,q)\leqslant 2M \end{equation*}

where M is the constant from Lemma 3.6, which does not depend on Θ. This is a contradiction if $\mathcal{N}$ is deep enough.

Now, in order to complete our lift of $\overline X_b$ we still need to lift the vertices $\overline{\alpha}_1,\ldots,\overline\alpha_{b-3}\in \overline X_b$ outside the link of $\overline{\beta}$. For every $i=1,\ldots, b-3$ we can find vertices $\overline x, \overline y\in \overline X_{b-1}$ and a codimension two simplex $\overline R\subset \overline X_{b-1}$ such that $\overline{\beta},\, \overline{\alpha}_1,\, \overline{\alpha}_i,\, \overline x,\, \overline y$ and $\overline R$ span a special pentagon $\overline{\mathfrak{P}}$. For example, one could look at Figure 6, replace z with α_i and then consider the corresponding vertices inside $\overline X_{b}$.

Lift $\overline{\mathfrak{P}}$ to a special pentagon $\mathfrak{P}'$, let $Q'=\beta'\star y'\star R'$ be the lift of the simplex $\overline Q=\overline \beta\star \overline y\star \overline R$ inside $\mathfrak{P}'$, and let $Q=\beta\star y\star R$ be the corresponding simplex inside $\beta\star X_{b-1}$. Since there is a unique $\mathcal{N}$-orbit of lifts of $\overline Q$ there exists $g\in \mathcal{N}$ mapping Qʹ to Q. Thus we can apply g to $\mathfrak{P}'$ and assume that $Q=Q'$. Moreover, let $x'\in \mathfrak{P}'$ be the lift of $\overline x$, and x the corresponding lift inside $X_{b-1}$. If we apply Lemma 4.4 to $x,x',y$ and $S=\beta\star R$ we find some element $g\in \mathcal{N}$ that maps xʹ to x and fixes Q. Thus, we can apply g to $\mathfrak{P}'$ and assume that also $x=x'$.

Now, let $G_i=\mathfrak{P}'\cup X_{b-1}$. Let $\alpha_1^i\in \mathfrak{P}'$ be the lift of $\overline{\alpha}_1$ inside the special pentagon. Lemma 2.21 provides a copy $X_b^i$ which extends $\beta\star X_{b-1}$ and contains $\alpha_1^i$. By Lemma 2.22, $X_b^i$ also contains α_i, therefore $G_i\subset X_b^i$. Now, consider the copies $X_b^2$ and $X_b^j$, for $3\leqslant j\leqslant b-3$. These copies coincide on $\beta\star X_{b-1}$, and by construction α ₁² and $\alpha_1^j$ project to the same element $\overline{\alpha}_1$. Therefore, if we apply Lemma 4.4 with $x=\alpha_1^2$, $x'=\alpha_1^j$, $y=\beta$ and $S=X_{b-1}$, we find an element $g\in \mathcal{N}$ that maps α ₁² to $\alpha_1^j$ and fixes $\beta\star X_{b-1}$. Then we conclude that $g(X_{b}^2)=X_b^j$ by Lemma 2.22, which implies that the projection of $X_b^2$ coincides with $\overline{X}_b$ also on $\overline{\alpha_j}$. Since we may do this for every $3\leqslant j\leqslant b-3$ we see that $X_b^2$ is the lift we were looking for.

Proof. Proof of uniqueness

Again we proceed by induction. First we discuss the base case b = 5.

Proof. The proof is just a sequence of iterations of Lemma 4.4. More precisely, let $\{\overline{\gamma}_i\}_{i=1,\ldots,5}$, be the vertices of $\overline X_5$, and let γ_i and $\gamma_i'$ be lifts that form two pentagons X ₅ and $X_5'$. Up to applying some element of $\mathcal{N}$ to $X_5'$ we may assume that $\gamma_1=\gamma_1'$ and $\gamma_2=\gamma_2'$, because the edge $\overline\gamma_1,\overline\gamma_2$, which is a simplex of dimension 1, admits a unique orbit of lifts by Lemma 2.7. Now, if we apply Lemma 4.4 with $x=\gamma_3$, $x'=\gamma_3'$, $y=\gamma_1$ and $S=P=\gamma_2$ we find some element $g\in\mathcal{N}$ mapping γ ₃ to $\gamma_3'$ and fixing $\gamma_1,\gamma_2$, so we can apply g to X ₅ and assume that $\gamma_3=\gamma_3'$. Moreover, if we repeat the argument with $x=\gamma_4$, $x'=\gamma_4'$, $y=\gamma_2$, $P=\gamma_3$ and $S=\{\gamma_1,\,\gamma_3\}$ we may also assume that $\gamma_4=\gamma_4'$. Now X ₅ and $X_5'$ coincide on four curves, and therefore they must coincide because of Lemma 2.24.

Proof. Let $G=\beta\star X_{b-1}$ and $G'=\beta'\star X_{b-1}'$ be two lifts of $\overline G$. Up to an element of $\mathcal{N}$ we may assume that $\beta=\beta'$. Now $X_{b-1}$ and $X_{b-1}'$ are in $\text{Lk}_{C_b}(\beta)\cong\mathcal{C}_{b-1}$, where again we see $S_{b-1}$ as the surface obtained by shrinking β to a puncture. Moreover, as already noticed the projection of $\text{Lk}_{\mathcal{C}_b}(\beta)$ is isomorphic to $\mathcal{C}_{b-1}/\mathcal{N}(S_{b-1})$, thus by induction there is an element $h\in \mathcal{N}(S_{b-1})$ that maps $X_{b-1}$ to $X_{b-1}'$. If we extend h to be the identity on the twice-punctured disk bounded by β we get a mapping class $H\in \mathcal{N}(S_b)$ that maps G to Gʹ, as required.

Proof. Item (1) is true by construction of the Farey complex. Regarding Item (2) fix a pair $\alpha,\,\beta$ of adjacent lifts of $\overline\alpha,\,\overline\beta$. Any triangle $\overline{T}=\{\overline{\alpha},\,\overline\beta,\,\overline\gamma\}$ has a lift $\{\alpha',\,\beta',\,\gamma'\}$, and by uniqueness of the orbit of lifts of an edge (which is a particular case of Theorem 2.7) we may assume that $\alpha=\alpha'$ and $\beta=\beta'$. Therefore, $\overline T$ is the projection of one of the two triangles containing $\alpha,\,\beta$.

Proof of Proposition 5.3

Let $T\subseteq\mathcal{C}_4$ be a triangle, and let $\overline{T}$ be its projection, which is still an isometrically embedded triangle by Item (1) of Lemma 3.13. Let $\widetilde{T}$ be a lift of $\overline{\phi}(\overline T)$, which exists by Theorem 2.7. By Theorem 5.1 there is a mapping class $h\in MCG^\pm(S_4)$ mapping T to $\widetilde{T}$, which means that $\overline{\phi}$ coincides with the induced map $\overline{h}$ on $\overline{T}$. In other words, we showed that the following diagram commutes on $T\subset\mathcal{C}_4$:

Our next goal is to show that, if the diagram commutes on some triangle T, then it commutes on any triangle Tʹ that shares an edge with T. If we prove the claim we are done, since any triangle in the Farey complex maps to a fixed triangle via a finite sequence of reflections along sides. Let $\alpha,\,\beta\in T$ be the vertices of the common edge. Notice that h, which is a graph automorphism, must map Tʹ to the triangle $\widetilde{T}$ that contains $h(\alpha)$ and $h(\beta)$ and is not $\widetilde{T}$, and similarly $\overline\phi$ must map $\pi(T')$ to $\pi(\widetilde{T}')$ since there are no other triangles containing $\pi(\alpha)$ and $\pi(\beta)$. Therefore Diagram (5.1) commutes also on Tʹ, as required.

Proof. Choose a copy $\overline T\subset \mathcal{C}/\mathcal{N}$ of the rigid set, and let T be one of its lifts. Then g(T) is also a lift of $\overline T$, and by Lemma 2.9 there is an element $h\in \mathcal{N}$ mapping T to g(T). Hence $h^{-1}\circ g$ is the identity on T, and therefore belongs to $\mathcal{K}$ by the “moreover” part of Theorem 5.1.

Proof. The first bullet is clearly true, so we focus on the second. Let α be a curve in X_b that intersects β, and let P be a facet that both α and β complete, which cuts out a four-holed sphere S ₄. Now, clearly $H_\beta(\alpha)$ has special intersection with both α and β with respect to P (the curves γ and δ are easy to find in both cases); therefore, it suffices to notice that there are only two curves on S ₄ with intersection 2 with both α and β, since in the curve graph of S ₄ the edge with endpoints α and β belongs to only two triangles.

Proof. Consider X_b inside $\mathcal{C}$ and let $\overline{X}_b$ be its projection, which is still a copy of X_b by Theorem 3.7. Moreover, let $\widetilde{X_b}$ be a lift of $\overline{\phi}(\overline{X}_b)$. By Theorem 2.19 there is an extended mapping class $h\in MCG^\pm(S_b)$ mapping X_b to $\widetilde{X_b}$, which means that the following diagram commutes on $X_b\in\mathcal{C}$:

Now, choose a minimal vertex $\beta\in X_b$, and let r be the reflection fixing X_b. We know that, if we fix a curve $\alpha_1\in X_b$ that intersects β, h must map $H_{\beta}(\alpha_1)$ to either $H_{h(\beta)}(h(\alpha_1))$ or $H_{h(\beta)}^{-1}(h(\alpha_1))$, since these are the only two curves that satisfy the graph-theoretic characterisation of the half twist, Theorem 5.6. Section 5.3 shows that the same argument works in the quotient, hence there are only two possibilities for $\overline\phi\circ H_{\overline\beta}(\overline{\alpha}_1)$. Up to replacing h with $h\circ r$ we may therefore assume that Diagram 5.2 commutes also on $H_{\beta}(\alpha_1)$, and hence on the whole $H_{\beta}(X_b)$, which by Lemma 2.22 is uniquely determined by $H_{\beta}(\alpha_1)$ (and the copy of $X_{b-1}$ which is the link of β in X_b; this is fixed by H_β).

We now claim that, if the diagram commutes on $X_b\cup H_\beta(X_b)$ and ζ is a minimal curve in X_b that intersects β then the diagram commutes also on $H_{\zeta}(X_b)$. In fact $H_\zeta^{-1}(\beta)=H_\beta(\zeta)$, as shown in Figure 11, therefore we already know that the diagram commutes on $H_\zeta^{-1}(\beta)$ (and therefore on $H_\zeta(\beta)$ which is the only other curve that satisfies the same characterisation as $H_\zeta^{-1}(\beta)$). Therefore, the diagram commutes also on the whole $H_{\zeta}(X_b)$, which is uniquely determined by $H_{\zeta}(\beta)$.

Figure 11.

The two minimal curves from the proof of theorem 5.12 and their respective half twists.

We may repeat the previous argument to show that Diagram 5.2 commutes on X_b and on any half twist around one of its minimal curves. Now it suffices to prove that the orbit of X_b by successive half twists covers $\mathcal{C}$. That is, we are left to prove:

Claim: Given two copies X_b and $X_b'$, there is a sequence $X_b=X^0, X^1, \ldots, X^k=X_b'$ of copies of X_b such that, for every $i=0,\ldots,k-1$, $X^{i+1}=H_\theta^{\pm}(X^i)$ for some minimal curve $\theta\in X^i$.

Proof of Claim

To show this let h be a (orientation-preserving) mapping class that maps X_b to $X_b'$. Now, if $\theta_1,\ldots,\theta_b$ are the minimal curves in X_b we know that $H_{\theta_1},\ldots,H_{\theta_b}$ generate the mapping class group. Therefore, let $h=H_{\theta_{i_1}}^{\pm}\ldots H_{\theta_{i_k}}^{\pm}$, for some possibly repeated indices $i_1,\ldots,i_k\in\{1,\ldots,b\}$. Now we proceed by induction on k. If k = 1 we are done. Otherwise notice that, by the properties of half Dehn twists,

\begin{equation*}h=H^{\pm}_{\theta_{i_1}}\ldots H^{\pm}_{\theta_{i_k}}=H^{\pm}_{H^{\pm}_{\theta_{i_1}}(\theta_{i_2})}\ldots H^{\pm}_{H^{\pm}_{\theta_{i_1}}(\theta_{i_k})}H^{\pm}_{\theta_{i_1}}\end{equation*}

Now we have expressed h in terms of $H_{\theta_{i_1}}^{\pm}$ and k − 1 half twists around some minimal curves of $X^1:=H_{\theta_{i_1}}^{\pm}(X_b)$. Therefore, we conclude by the inductive hypothesis.

This concludes the proof of Theorem 5.12.

Proof. Choose a copy of $\overline{X}_b\subseteq \mathcal{C}/\mathcal{N}$, and let X_b be one of its lifts. Then $g(X_b)$ is also a lift of $\overline{X}_b$, and by the uniqueness part of Theorem 4.1 there is an element $h\in \mathcal{N}$ mapping X_b to $g(X_b)$. Therefore, $h^{-1}\circ g$ is the identity on X_b, and by the “moreover” part of Theorem 2.19 we must have that $h^{-1}\circ g\in\left\langle r\right\rangle$. Now, $h^{-1} g$ induces the identity map on $\mathcal{C}/\mathcal{N}$, as g does so by hypothesis and h is an element of $\mathcal{N}$. Therefore, in order to prove that $h^{-1}g$ must be the identity, we must prove that r does not induce the identity on $\mathcal{C}/\mathcal{N}$. If this is not the case then, for any minimal curve $\beta\in X_b$, we have that $H_\beta(X_b)$ and $r\circ H_\beta(X_b)=r\circ H_\beta\circ r(X_b)=H_\beta^{-1}(X_b)$ coincide in the quotient (here we used that $r(X_b)=X_b$ and the properties of a half Dehn twist under conjugation). Therefore, by uniqueness of the orbit of lifts, there is an element $k\in \mathcal{N}$ which, for every curve $\gamma\in X_b$, maps $H_\beta^{-1}(\gamma)=r\circ H_\beta(\gamma)$ to $H_\beta(\gamma)$. By the “moreover” part of Theorem 2.19 there is only one orientation-preserving mapping class with this property, and therefore we must have that $T_\beta=k\in \mathcal{N}$ since the Dehn twist T_β clearly maps $H_\beta^{-1}(\gamma)$ to $H_\beta(\gamma)$. But then since $\mathcal{N}$ is a normal subgroup it must contain all conjugates of T_β, that is, all Dehn twists around curves that bound twice-punctured disks. These mapping classes generate the pure mapping class group $PMCG(S_b)$, which is the finite index subgroup of $MCG(S_b)$ consisting of elements that fix each puncture individually. But this subgroup is not deep enough if we choose a large enough constant Θ (for example, since the quotient $\mathcal{C}/PMCG$ is finite, which contradicts Corollary 3.8).

Proof. For the existence part let X_b and $X_b'$ be lifts of $\overline{X}_b$ and $\phi(\overline{X}_b)$, respectively. Then by finite rigidity there is a mapping class $h\in MCG$ mapping X_b to $X_b'$, and the induced map $\overline{h}\in MCG/\mathcal{N}$ maps $\overline{X}_b$ to $\phi(\overline{X}_b)$.

For the uniqueness part, if h and hʹ both induce maps that map $\overline{X}_b$ to $\phi(\overline{X}_b)$ then $h(X_b)$ and $h'(X_b)$ are both lifts of $\phi(\overline{X}_b)$, hence by uniqueness of the orbit of lifts there exists a $g\in \mathcal{N}$ such that $g\circ h|_{X_b}=h'|_{X_b}$. But then $g\circ h$ and hʹ must differ by an element of $\text{Pstab}(X_b)$.