######  Proposition 8.2.

For $\mu_{0}$-almost every $x\in\mathbb{O}$, the fiber
$\pi^{-1}(x)\subset\Lambda$ is trivial. Consequently, $F|_{\Lambda}$ has a
unique invariant probability measure $\mu:=\pi^{*}\mu_{0}$.

###### Proof.

Let $\mu$ be an ergodic probability measure on $\Lambda$. Note that $\mu$
cannot have two negative Lyapunov exponents since it is not supported on a
sink. By Proposition 3.24, for any $\kappa>0$, there exists $L\geq 1$ such
that $\mu(\hat{\Lambda}^{P}_{L,\varepsilon})>1-\kappa$ (recall that if
$p\in\hat{\Lambda}^{P}_{L,\varepsilon}$, then $p$ is Pesin
$(L,\varepsilon)$-regular, and $W^{ss}_{\Omega}(p)$ is proper in $\Omega$).

Let $p\in\hat{\Lambda}^{P}_{L,\varepsilon}$. For $n\geq 0$, let
$\mathcal{B}_{p}^{n}$ be the connected component of $\mathcal{B}_{p}\cap
F^{n}(\Omega)$ containing $p$, where $\mathcal{B}_{p}$ is a regular box at
$p$. Suppose
$q^{n}_{r},q^{n}_{l}\in\hat{\Lambda}^{P}_{L,\varepsilon}\cap\mathcal{B}_{p}^{n}$
are points located to the right and left of
$W^{ss}_{\Omega}(p_{0})\cap\bar{\mathcal{B}}_{p}^{n}$ respectively. Then we
can form a rectangle
$\mathcal{B}_{p}^{n}(q^{n}_{r},q^{n}_{l})\subset\mathcal{B}^{n}_{p}$ enclosed
by $W^{ss}_{\Omega}(q^{n}_{r})$ and $W^{ss}_{\Omega}(q^{n}_{l})$. If
$q^{n}_{r}$ and $q^{n}_{l}$ are sufficiently close to $p$, then by Proposition
7.2, we have

$\operatorname{diam}(\mathcal{B}_{p}^{n}(q^{n}_{r},q^{n}_{l}))=O(\lambda^{(1-\varepsilon)n}).$

Let $Z_{p}$ be the connected component of $\Lambda$ containing $p$. If
$q^{n}_{r},q^{n}_{l}\notin Z_{p}$ for all $n$, then
$Z_{p}\subset\mathcal{B}_{p}^{n}(q^{n}_{r},q^{n}_{l})$. This implies that
$Z_{p}=\\{p\\}$.

Let $q\in\hat{\Lambda}_{L,\varepsilon}$, and let $Z_{q}$ be the connected
component of $\Lambda$ that contains $q$. We claim that if $q$ cannot be
approached by points in $\hat{\Lambda}_{L,\varepsilon}\setminus Z_{q}$ from
both sides of $W^{ss}_{\Omega}(q)$, then there exists a rectangle
$R_{q}\subset\mathcal{B}_{q}$ containing $q$ such that
$R_{q}\cap\hat{\Lambda}_{L,\varepsilon}\subset Z_{q}$. The set of all such
components $Z_{q}$ is countable, and the result follows.

If $q$ is isolated in $\hat{\Lambda}^{P}_{L,\varepsilon}$, or can only be
approached from within $\hat{\Lambda}^{P}_{L,\varepsilon}$ by points contained
in $Z_{q}$, then clearly the claim holds. If $q$ can only be approached by
points in $\hat{\Lambda}^{P}_{L,\varepsilon}\setminus Z_{q}$ from one side of
$W^{ss}_{\Omega}(q)$, then we can form a rectangle $R_{q}$ so that one of its
sides is contained in $W^{ss}_{\Omega}(q)$, and
$R_{q}\cap\hat{\Lambda}_{L,\varepsilon}\subset W^{ss}_{\Omega}(q)$. Since
$W^{ss}_{\Omega}(q)\cap\hat{\Lambda}_{L,\varepsilon}\subset Z_{q}$, the claim
again holds. ∎

######  Proposition 8.3.

The unique invariant measure $\mu$ for $F|_{\Lambda}$ is not hyperbolic. That
is, the Lyapunov exponents of $F|_{\Lambda}$ must be $0$ and $\log\lambda$ for
some $\lambda\in(0,1)$.

###### Proof.

Suppose $\mu$ is hyperbolic. Then it has a period $p\geq 1$. Up to take an
iterate, one can assume that $p=1$. Then for almost every points
$x,y\in\Lambda$, there exists a homoclinic point $z$ such that

$\operatorname{dist}(f^{n}(x),f^{n}(z))\to 0\hskip 14.22636pt\text{and}\hskip
14.22636pt\operatorname{dist}(f^{-n}(y),f^{-n}(z))\to 0\hskip
14.22636pt\text{as}\hskip 14.22636ptn\to\infty.$

This is a contradiction, since one can choose $x,y$ to belong to different
connected components of $\Lambda$. ∎

##  9\. Regular Unicriticality

Let $F$ be the mildly dissipative, infinitely renormalizable diffeomorphism
considered in Section 8. Let $\eta,\varepsilon,\delta\in(0,1)$ be small
constants such that $\bar{\eta}<\varepsilon$; ${\bar{\varepsilon}}<\delta$ and
${\bar{\delta}}<1.$ Suppose that $F$ is
$(\delta,{\bar{\varepsilon}})$-regularly unicritical. Let
$\\{c_{m}\\}_{m\in\mathbb{Z}}\subset\Lambda$ be a regular critical orbit with
$c_{1}\in\mathcal{D}^{n}$ for all $n\in\mathbb{N}.$ Denote
$\mathcal{D}^{n}_{i}:=F^{i-1}(\mathcal{D}^{n})$ for $i\in\mathbb{N}.$

By Propositions 8.2, 4.1 and 5.7; and Remark 1.6, we can replace $F$ by
$F^{R_{N}}|_{\mathcal{D}^{N}}$ for some sufficiently large $N\in\mathbb{N}$
such that the following properties hold:

  1. i)

$F$ is $\eta$-homogeneous on $\Lambda\cap\mathcal{D}^{N}$;

  2. ii)

$c_{1}\in\mathcal{D}^{n}_{1}$ is an $\varepsilon$-regular critical value for
$F$; and

  3. iii)

$F$ is $(\delta,{\bar{\varepsilon}})$-regular unicritical on
$\Lambda\cap\mathcal{D}^{n}_{1}$.

Let

$\\{\Phi_{c_{m}}:(\mathcal{U}_{c_{m}},c_{m})\to(U_{c_{m}},0)\\}_{m\in\mathbb{Z}}$

be a uniformization of the orbit of the $c_{1}$ given in Theorem 5.8. Let
$t_{0}\in(0,1)$ be a sufficiently small uniform constant so that

$\mathbb{D}_{c_{-i}}(t_{0}\lambda^{{\bar{\varepsilon}}i})\subset\mathcal{U}_{c_{-i}}\hskip
14.22636pt\text{for}\hskip 14.22636pti\geq 0.$

Let $L_{0}:=L(t_{0})\geq 1$ be the regularity factor given in Definition 1.3
ii).

By Proposition 3.2, $c_{0}$ is infinitely backward
$(O(1),{\bar{\varepsilon}})_{v}$-regular along

$E^{v}_{c_{0}}:=D_{0}\Phi_{c_{0}}^{-1}(E^{gv}_{0}).$

######  Proposition 9.1.

Let $t\in(0,t_{0})$. After enlarging $L(t)\geq 1$ by a uniform amount if
necessary, the following condition holds. If

$c_{-n}\notin\bigcup_{i=0}^{n-1}\mathbb{D}_{c_{-i}}({t}\lambda^{{\bar{\varepsilon}}i})$

for some $n\in\mathbb{N}$, then $c_{-n}$ is $n$-times forward
$(L(t),\delta)_{v}$-regular along $E^{v}_{c_{-n}}$.

###### Proof.

By the regular unicriticality condition, $c_{-n}$ is $n$-times forward
$(L(t),\delta)_{v}$-regular along some direction
$\tilde{E}^{v}_{c_{-n}}\in\mathbb{P}^{2}_{c_{-n}}$. Let $0\leq j\leq n$ be the
largest integer such that $-j$ is a Pliss moment in the backward orbit of
$c_{0}$. This means that $c_{-j}$ is $(\infty,j)$-times
$(K,{\bar{\varepsilon}})_{v}$-regular along $E_{c_{-j}}^{v}$ for some uniform
constant $K\geq 1$. By Lemma 4.3, we have $j>(1-{\bar{\varepsilon}})n.$

It follows from Proposition 3.4 that $c_{-n}$ is $n$-times forward
$(K\lambda^{-{\bar{\varepsilon}}n},{\bar{\varepsilon}})_{v}$-regular along
$E_{c_{-k}}^{v}$. Using Proposition 3.11, we see that

$\measuredangle(E^{v}_{c_{-n}},\tilde{E}_{c_{-n}}^{v})<\overline{L(t)}\lambda^{-{\bar{\varepsilon}}n}\lambda^{(1-\delta)n}.$

If

$\measuredangle(E^{v}_{c_{-n}},\tilde{E}_{c_{-n}}^{v})<k\lambda^{(1+\eta){\bar{\varepsilon}}n}$

for some uniformly small constant $k>0$, then

$\measuredangle(E^{v}_{c_{-i}},\tilde{E}_{c_{-i}}^{v})<k\hskip
14.22636pt\text{for all}\hskip 14.22636ptj\leq i\leq n.$

Thus, if $n>C\log_{\lambda}L(t)$ for some uniform constant $C\geq 1$, then we
see that $c_{-n}$ is $n$-times forward $(\overline{L(t)},\delta)_{v}$-regular
along $E^{v}_{c_{-n}}$. ∎

###  9.1. Critical discs

Let $\varepsilon_{i}$ with $i\in\\{0,1,2\\}$ be small constants with
$\varepsilon_{0}:=\varepsilon$ and
${\bar{\varepsilon}}_{i}<\varepsilon_{i+1}<\underline{\delta}$ for $0\leq i<2$
such that

$\operatorname{diam}(\mathcal{U}_{c_{m}})>{t_{0}}\lambda^{\varepsilon_{1}|m|}.$

For ${t}\in(0,{t_{0}})$ and $m\in\mathbb{Z}$, the $({t},\varepsilon_{2})$-disc
at $c_{m}$ is defined as

$\mathbb{D}^{t}_{c_{m}}:=\mathbb{D}_{c_{m}}({t}\lambda^{\varepsilon_{2}|m|})\setminus\\{c_{m}\\}\subset\mathcal{U}_{c_{m}}\setminus\\{c_{m}\\}.$

######  Lemma 9.2.

For $i\geq 1$, we have
$F(\mathbb{D}^{t}_{c_{-i}})\subset\mathbb{D}^{t}_{c_{-i+1}}$.

###### Proof.

This follows immediately from the $\eta$-homogeneity of $F$. ∎

Let $p\in\Omega$. If $p\in\mathbb{D}^{t}_{c_{-n}}$ for some $n\geq 0$, let
$e_{n}(p)>0$ be the smallest integer such that
$F^{-e_{n}(p)}(p)\notin\mathbb{D}^{t}_{c_{-n-e_{n}(p)}},$ and let
$0\leq\operatorname{d}(p)\leq\infty$ be the largest integer such that
$p\in\mathbb{D}^{t}_{c_{-\operatorname{d}(p)}}.$ If, on the other hand,
$p\notin\mathbb{D}^{t}_{c_{-n}}$ for all $n\geq 0$, then denote
$\operatorname{d}(p)=\varnothing$.

######  Lemma 9.3.

Let $p\in\mathbb{D}^{t}_{c_{-n}}$. Then

$e_{n}(p)<{\bar{\varepsilon}}_{2}^{-1}\log_{\lambda}\left(\operatorname{dist}(p,c_{-n})\right).$

###### Proof.

By $\eta$-homogeneity, we have

$\operatorname{dist}(F^{-i}(p),c_{-n-i})>\lambda^{\eta
i}\operatorname{dist}(p,c_{-n}).$

Since

$\operatorname{radius}(\mathbb{D}^{t}_{c_{-n-i}}):={t}\lambda^{\varepsilon_{2}(n+i)},$

the claim follows. ∎

######  Lemma 9.4.

If $c_{-n}\in\mathbb{D}^{t}_{c_{0}}$ for some $n>0$, then

$\operatorname{dist}(c_{-n},c_{0})>\left(\operatorname{radius}(\mathbb{D}^{t}_{c_{-n}})\right)^{{\bar{\varepsilon}}_{1}}.$

###### Proof.

By Proposition 5.10, we have

$\lambda^{{\bar{\varepsilon}}_{1}n}<\operatorname{dist}(c_{-n},c_{0}).$

The claim now follows from the fact that

$\left(\operatorname{radius}(\mathbb{D}^{t}_{c_{-n}})\right)^{{\bar{\varepsilon}}_{1}}={t}^{{\bar{\varepsilon}}_{1}}\lambda^{\varepsilon_{2}{\bar{\varepsilon}}_{1}n}<\lambda^{{\bar{\varepsilon}}_{1}n}.$

∎

######  Lemma 9.5.

Let $p_{0}\in\mathbb{D}^{t}_{c_{0}}$. If $p_{-n}\in\mathbb{D}^{t}_{c_{0}}$ for
some $0<n<e_{0}(p_{0})$, then

$\operatorname{dist}(p_{-n},c_{0})>\left(\operatorname{dist}(p_{0},c_{0})\right)^{{\bar{\varepsilon}}_{1}}.$

###### Proof.

Note that by Lemma 9.4

$\displaystyle\operatorname{dist}(p_{-n},c_{0})$
$\displaystyle>\operatorname{dist}(c_{-n},c_{0})-\operatorname{dist}(p_{-n},c_{-n})$
$\displaystyle>\left(\operatorname{radius}(\mathbb{D}^{t}_{c_{-n}})\right)^{{\bar{\varepsilon}}_{1}}-\operatorname{radius}(\mathbb{D}^{t}_{c_{-n}})$
$\displaystyle\asymp\lambda^{\varepsilon_{2}{\bar{\varepsilon}}_{1}n}.$

By $\eta$-homogeneity, we have

$\operatorname{dist}(p_{0},c_{0})<\lambda^{-\eta
n}\operatorname{dist}(p_{-n},c_{-n})<\lambda^{-\eta
n}\operatorname{radius}(\mathbb{D}^{t}_{c_{-n}})\asymp\lambda^{(\varepsilon_{2}-\eta)n}.$

The result follows. ∎

######  Proposition 9.6.

Let $p_{0}\in\mathbb{D}^{t}_{c_{0}}$. If $\operatorname{d}(p_{0})<\infty$,
then there exists a smallest integer $0\leq\operatorname{e}(p)<\infty$ such
that $d\left(F^{-\operatorname{e}(p)}(p)\right)=\varnothing$. Moreover,
$\operatorname{e}(p_{0})<(1+{\bar{\varepsilon}}_{1})e_{0}(p_{\operatorname{d}(p_{0})})$.

###### Proof.

Denote $q_{0}:=p_{\operatorname{d}(p_{0})}\in\mathbb{D}^{t}_{c_{0}}.$ Note
that $\operatorname{d}(q_{0})$ may be greater than $0$.

Consider a sequence $\\{n_{i}\\}_{i=0}^{j}$ defined as follows. Let
$n_{0}:=0$. For $i\geq 0$, if
$\operatorname{e}(q_{-n_{i}})>e_{0}(q_{-n_{i}}),$ then let

$n_{i+1}^{\prime}:=n_{i}+e_{0}(q_{-n_{i}}),$

and let $n_{i}=n_{i}^{0}<n_{i}^{1}<\ldots<n_{i}^{k}\leq n_{i}^{\prime}$ be a
sequence of moments such that $q_{-n_{i}^{l}}\in\mathbb{D}^{t}_{c_{0}}$ for
$0\leq l\leq k.$ Define $n_{i}=n_{i}^{l}$, where $0\leq l\leq k$ is the
integer such that $n_{i}^{l}+e_{0}(q_{-n_{i}^{l}})$ is maximized.

We claim that $n_{i+1}>n_{i}$. Otherwise, we have
$q_{-n}\not\in\mathbb{D}^{t}_{c_{0}}$ for $n_{i}<n\leq n_{i+1}^{\prime},$ and
hence
$\operatorname{d}(q_{-n_{i+1}^{\prime}})>\operatorname{e}_{0}(q_{-n_{i}}).$
Denote

$b_{i}:=n_{i+1}^{\prime}-\operatorname{d}(q_{-n_{i+1}^{\prime}}).$

Then by Lemma 9.2, we have $q_{-b_{i}}\in\mathbb{D}^{t}_{c_{0}}$ and

$b_{i}+\operatorname{e}_{0}(q_{-b_{i}})>n_{i+1}^{\prime}=n_{i}+\operatorname{e}_{0}(q_{-n_{i}}).$

If $b_{i}>n_{k-1}\geq 0$ for some $k\leq i$, then this contradicts the
maximality of $n_{k}+\operatorname{e}_{0}(-n_{k})$. If $b_{i}<0$, then

$\operatorname{d}(q_{-n_{i+1}^{\prime}})=n_{i+1}^{\prime}-b_{i}>n_{i+1}^{\prime}.$

Hence,

$\operatorname{d}(p_{0})=\operatorname{d}(q_{-\operatorname{d}(p_{0})})\geq\operatorname{d}(q_{-n_{i+1}^{\prime}})-n_{i+1}^{\prime}+\operatorname{d}(p_{0})>\operatorname{d}(p_{0}),$

which is a contradiction. Thus, the claim holds.

Let $l_{i}:=\operatorname{dist}(c_{-n_{i}},c_{0}).$ By Lemma 9.5, we have

$l_{i+1}>l_{i}^{{\bar{\varepsilon}}_{1}}>l_{0}^{{\bar{\varepsilon}}_{1}^{i+1}}.$

This immediately implies that $j<\infty$.

Denote $m_{i+1}:=n_{i+1}-n_{i}.$ By $\eta$-homogeneity, we have

${t}\lambda^{\varepsilon_{2}m_{i}}=\operatorname{diam}(\mathbb{D}^{t}_{c_{-m_{i}}})>\operatorname{dist}(c_{-n_{i}},c_{-m_{i}})>\lambda^{\eta
m_{i}}l_{i-1}.$

So

$m_{i+1}<{\bar{\varepsilon}}_{2}^{-1}\log_{\lambda}(l_{i})<{\bar{\varepsilon}}_{2}^{-1}{\bar{\varepsilon}}_{1}^{i}\log_{\lambda}(l_{0}).$

Let $m_{j+1}:=\operatorname{e}_{0}(q_{-n_{j}}).$ Then by Lemma 9.3,

$m_{j+1}<{\bar{\varepsilon}}_{2}^{-1}l_{j}<{\bar{\varepsilon}}_{2}^{-1}{\bar{\varepsilon}}_{1}^{j}\log_{\lambda}(l_{0}).$

Finally, observe that by Proposition 5.10, we have

$\operatorname{e}_{0}(q_{0})>n_{1}>{\bar{\varepsilon}}_{1}^{-1}\log_{\lambda}(l_{0}).$

Consequently,

$\operatorname{e}(p_{0})\leq\operatorname{e}(q_{0})=\sum_{i=0}^{j+1}m_{i}<(1+{\bar{\varepsilon}}_{1})\operatorname{e}_{0}(q_{0})$

as claimed. ∎

###  9.2. Pinching at the critical point

######  Theorem 9.7.

We have

$\Lambda\cap\mathbb{D}_{c_{0}}(1/\overline{L_{0}})\subset\mathcal{T}_{c_{0}}^{1/{\bar{\delta}}},$

where $\mathcal{T}_{c_{0}}^{1/{\bar{\delta}}}\subset\mathcal{U}_{c_{0}}$ is a
$1/{\bar{\delta}}$-pinched regular neighborhood at $c_{0}$.

###### Proof.

Let $p_{0}\in\Lambda\cap\mathbb{D}_{c_{0}}(1/\overline{L_{0}})$. Write
$z_{0}=(x_{0},y_{0}):=\Phi_{c_{0}}(p_{0}).$ Suppose towards a contradiction
that $p_{0}\notin\mathcal{T}_{c_{0}}^{1/{\bar{\delta}}}$. For
$m\in\mathbb{Z}$, consider the $({t_{0}},{\bar{\varepsilon}})$-disc
$\mathbb{D}_{c_{m}}^{t_{0}}$ at $c_{m}$. By Theorem 3.6 and Proposition 3.9,
we have

$|y_{-i}|>k\lambda^{-(1-{\bar{\varepsilon}})i}|y_{0}|>k\lambda^{-(1-{\bar{\varepsilon}})i}|x_{0}|^{1/{\bar{\delta}}}\hskip
14.22636pt\text{for}\hskip 14.22636pt0\leq i<\operatorname{e}_{0}(p_{0})$

for some uniform constant $k>0$. If

$k\lambda^{-(1-{\bar{\varepsilon}})i}|x_{0}|^{1/{\bar{\delta}}}<{t_{0}}\lambda^{{\bar{\varepsilon}}i}$

then it follows by straightforward computation that

$i<{\bar{\delta}}^{-1}\log_{\lambda}|x_{0}|.$ (9.1)

First, suppose that $\operatorname{d}(p_{0})=0$. By Proposition 9.6 and (9.1),
we have

$N:=\operatorname{e}(p_{0})<(1+{\bar{\varepsilon}})\operatorname{e}_{0}(p_{0})<{\bar{\delta}}^{-1}\log_{\lambda}|x_{0}|<\infty.$
(9.2)

We also conclude from

$\lambda^{-(1-{\bar{\varepsilon}})i}\overline{L_{0}}^{-1}<k\lambda^{-(1-{\bar{\varepsilon}})i}|y_{0}|<{t_{0}}\lambda^{{\bar{\varepsilon}}i}$

that

$N>-\log_{\lambda}\overline{L_{0}}.$ (9.3)

If $p_{0}=c_{-n}$ for some $n\in\mathbb{N}$, then for $l\in\mathbb{Z}$, denote
$\tilde{E}_{p_{l}}^{v}:=E^{v}_{c_{-n+l}}.$ Otherwise, denote
$\tilde{E}_{p_{l}}^{v}:=E^{s}_{p_{l}}.$ In the former case,
$p_{-N}=c_{-(n+N)}$ is $(n+N)$-times forward $(L_{0},\delta)_{v}$-regular
along $\tilde{E}_{p_{-N}}^{v}$ by Proposition 9.1. Consequently, by
Proposition 3.4, we see that $p_{1}=c_{-n+1}$ is $(n-1)$-times forward
$(L_{0}\lambda^{-{\bar{\delta}}N},\delta)_{v}$-regular along
$\tilde{E}_{p_{-n+1}}^{v}$. Similarly, in the latter case, $p_{-N}$ is
infinitely forward $(L_{0},\delta)_{v}$-regular along
$\tilde{E}_{p_{-N}}^{v}$, and thus, $p_{1}$ is infinitely forward
$(L_{0}\lambda^{-{\bar{\delta}}N},\delta)_{v}$-regular along
$\tilde{E}_{p_{-n+1}}^{v}$.

Let $M\in\mathbb{N}$ be the largest integer such that
$p_{1+i}\in\mathcal{U}_{c_{1+i}}$ for $0\leq i<M.$ Then by Theorem 3.6 and
Proposition 3.9, we have
$\lambda^{{\bar{\varepsilon}}M}<|x_{M}|<\lambda^{-{\bar{\varepsilon}}M}|x_{1}|.$
We also have

$|y_{0}|<\lambda^{(1-{\bar{\varepsilon}})N}.$ (9.4)

If $|y_{0}|>|x_{0}|$, then $|x_{1}|\asymp|y_{0}|$. Hence,

$\lambda^{{\bar{\varepsilon}}M}<\lambda^{-{\bar{\varepsilon}}M}|y_{0}|<\lambda^{(1-{\bar{\varepsilon}})N}.$

It follows that $N<{\bar{\varepsilon}}M$. If $|y_{0}|<|x_{0}|$, then
$|x_{1}|=O(|x_{0}|)$. So

$\lambda^{{\bar{\varepsilon}}M}<\lambda^{-{\bar{\varepsilon}}M}|x_{0}|<\lambda^{-{\bar{\varepsilon}}M}|y_{0}|^{{\bar{\delta}}}<\lambda^{{\bar{\delta}}N}.$

Hence, we again have $N<{\bar{\varepsilon}}M$.

Let

$E^{v}_{p_{1}}:=D_{z_{1}}\Phi_{c_{1}}^{-1}(E^{gv}_{z_{1}}).\hskip
14.22636pt\text{and}\hskip
14.22636pt\theta_{1}:=\left|\measuredangle(\tilde{E}_{p_{1}}^{v},E^{v}_{p_{1}})\right|.$

By Proposition 3.11, we have

$\theta_{1}<L_{0}\lambda^{-{\bar{\delta}}N}\lambda^{(1-{\bar{\delta}})M}<L_{0}\lambda^{{\bar{\varepsilon}}^{-1}N}<|y_{0}|^{1/{\bar{\varepsilon}}}.$

Moreover, by Theorem 5.8, we have
$\theta_{0}<O\left(|x_{0}|+\theta_{1}\right).$ If $|y_{0}|<|x_{0}|$, then
$\theta_{0}<O\left(|x_{0}|\right).$ This, (9.2), (9.3) and Proposition 6.2
contradict the $N$-times forward $(L_{0},\delta)_{v}$-regularity of $p_{-N}$
along $\tilde{E}^{v}_{p_{-N}}$.

If $|y_{0}|>|x_{0}|$, then $\theta_{0}<O\left(|y_{0}|\right).$ By Theorem 3.6
and Proposition 3.9, we have

$\theta_{-i}<K\lambda^{-(1+{\bar{\varepsilon}})i}\theta_{0}<{\bar{K}}\lambda^{-(1+{\bar{\varepsilon}})i}|y_{0}|.$
(9.5)

Let $j\geq 0$ be the smallest integer less than $\operatorname{e}_{0}(p_{0})$
such that $\theta_{-j}>\lambda^{{\bar{\varepsilon}}j}\nu.$ for some small
$\nu>0$. By (9.5), we have

$j>(1-{\bar{\varepsilon}})(\log_{\lambda}|y_{0}|+\log_{\lambda}({\bar{K}})).$

Thus, by (9.4), we see that $j>kN$ for some uniform constant $k>0$. Finally,
by Proposition 3.9 and (9.3), we have

$\|DF^{-N}|_{\tilde{E}^{v}_{p_{0}}}\|<\lambda^{-{\bar{\varepsilon}}j}\lambda^{-(1+\eta)(N-j)}<\lambda^{-(1-k+{\bar{\varepsilon}})N}=\lambda^{(k-{\bar{\delta}})N}\lambda^{-(1-\delta)N}<L_{0}^{-1}\lambda^{-(1-\delta)N}.$

This contradicts the $N$-times forward $(L_{0},\delta)_{v}$-regularity of
$p_{-N}$ along $\tilde{E}^{v}_{p_{-N}}$.

Finally, consider the case when
$p_{0}\in\mathbb{D}_{c_{0}}(1/\overline{L_{0}})$, and
$\operatorname{d}(p_{0})=n>0$. This means that
$c_{-n}\in\mathbb{D}^{t_{0}}_{c_{0}}$. Write
$(x_{0},y_{0}):=\Phi_{c_{0}}(c_{-n}).$ It follows by Proposition 5.10 that

$\operatorname{radius}(D^{t_{0}}_{c_{-n}})<|x_{0}|^{1/{\bar{\varepsilon}}}.$

Thus, by slightly increasing ${\bar{\delta}}$ a uniform amount if necessary,
we obtain $p_{0}\in
D^{t_{0}}_{c_{-n}}\subset\mathcal{T}_{c_{0}}^{1/{\bar{\delta}}}.$ ∎

###  9.3. Refinement of regular unicriticality

Let ${t_{1}}:=1/\overline{L_{0}}<\underline{t_{0}}$ be a sufficiently small
uniform constant such that so that

$\mathbb{D}^{t_{1}}_{c_{m}}\subset\mathcal{U}_{c_{m}}\setminus\\{c_{m}\\}\hskip
14.22636pt\text{for}\hskip 14.22636ptm\in\mathbb{Z}\hskip
14.22636pt\text{and}\hskip
14.22636pt\Lambda\cap\mathbb{D}^{t_{1}}_{c_{0}}\subset\mathcal{T}_{c_{0}}^{1/\underline{\delta}}.$

Let ${L_{1}}:=L({t_{1}})\geq 1$ be the regularity factor given in Definition
1.3 ii). By increasing ${L_{1}}$ if necessary, we may assume that
${L_{1}}^{-\varepsilon}<{t_{1}}.$

Denote

$\delta>\delta^{\prime}:=(1-{\bar{\delta}})\delta>\underline{\delta}.$

For ${t}\in(0,{t_{0}}]$ and $n\geq 0$, let $\mathcal{T}_{c_{-n}}({t})$ and
$\mathcal{T}_{c_{1+n}}({t})$ be the $1/\underline{\delta}$-pinched,
$\delta^{\prime}$-truncated critical tunnel and valuable crescent at $c_{-n}$
and $c_{1+n}$ respectively of length ${t}$ (see Subsection 6.1).

Denote

$\hat{E}^{v}_{c_{-n}}:=D_{0}\Phi_{c_{-n}}^{-1}(E^{gv}_{0})=DF^{-n}(E^{v}_{c_{0}})$

and

$\hat{E}^{h}_{c_{1+n}}:=D_{0}\Phi_{c_{1+n}}^{-1}(E^{gh}_{0})=DF^{1+n}(E^{v}_{c_{0}}).$

Let $p\in\Lambda$. For $L\geq 1$, we say that $p$ is forward
$(L,\delta)$-regular (without specifying time or direction) if either

  * •

$p=c_{-n}$ for some $n\geq 0$, and $p$ is $n$-times forward
$(L,\delta)_{v}$-regular along $\hat{E}^{v}_{c_{-n}}$; or

  * •

$p\neq c_{-n}$ for all $n\geq 0$, and $p$ is infinitely forward
$(L,\delta)_{v}$-regular along $\hat{E}^{v}_{p}:=E^{s}_{p}$.

Similarly, we say that $p$ is backward $(L,\delta)$-regular if either

  * •

$p=c_{n}$ for some $n\geq 1$, and $p$ is $n$-times backward
$(L,\delta)_{h}$-regular along $\hat{E}^{h}_{c_{n}}$; or

  * •

$p\neq c_{n}$ for all $n\geq 1$, and $p$ is infinitely backward
$(L,\delta)_{h}$-regular along $\hat{E}^{h}_{p}:=E^{c}_{p}$.

######  Theorem 9.8 (Geometric criterion for regularity).

For ${t}\in(0,{L_{1}}^{-1}]$, let $L:=1/\underline{t}$. Then for all
$p\in\Lambda$, the following statements hold.

  1. a)

If $\displaystyle p\notin\bigcup_{i=0}^{\infty}\mathcal{T}_{c_{-i}}({t}),$
then $p$ is forward $(L,\delta)$-regular. If $p$ is also contained in
$\mathcal{T}_{c_{1}}({t})$, then $p$ is forward $(1,\delta)$-regular.

  2. b)

If $\displaystyle p\notin\bigcup_{i=0}^{\infty}\mathcal{T}_{c_{1+i}}({t}),$
then $p$ is backward $(L,\delta)$-regular. If $p$ is also contained in
$\mathcal{T}_{c_{0}}({t})$, then $p$ is backward $(1,\delta)$-regular.

  3. c)

If $\displaystyle
p\notin\bigcup_{i=-\infty}^{\infty}\left(\mathcal{T}_{c_{i}}({t})\cup\\{c_{i}\\}\right)$
then $p$ is Pesin $(L,\delta)$-regular.

Let $\varepsilon_{i}$ with $i\in\\{0,1,2\\}$ be constants satisfying
$\varepsilon_{0}:=\varepsilon$ and
${\bar{\varepsilon}}_{i}<\varepsilon_{i+1}<\underline{\delta}$ for $0\leq
i<2,$ so that we have

$\operatorname{diam}(\mathcal{U}_{c_{m}})>{t_{0}}\lambda^{\varepsilon_{1}|m|}\hskip
14.22636pt\text{and}\hskip
14.22636pt\operatorname{diam}(\mathbb{D}^{t_{1}}_{c_{m}}):={t_{1}}\lambda^{\varepsilon_{2}|m|}\hskip
14.22636pt\text{for all}\hskip 14.22636ptm\in\mathbb{Z}.$

Let $p_{0}\in\Omega$. If $p_{0}\in\mathbb{D}^{{t_{1}}}_{c_{-n}}$ for some
$n\geq 0$, let $e_{n}(p_{0})>0$ be the smallest integer such that
$p_{-e_{n}(p_{0})}\notin\mathbb{D}^{{t_{1}}}_{c_{-n-e_{n}(p_{0})}},$ and let
$0\leq\operatorname{d}(p_{0})\leq\infty$ be the largest integer such that
$p_{0}\in\mathbb{D}^{{t_{1}}}_{c_{-\operatorname{d}(p)}}.$ In this case,
$0\leq\operatorname{e}(p_{0})<\infty$ is the integer given by Proposition 9.6
(taking ${t}={t_{1}}$). If, on the other hand,
$p_{0}\notin\mathbb{D}^{t_{1}}_{c_{-n}}$ for all $n\geq 0$, then denote
$\operatorname{d}(p_{0})=\varnothing$.

If $p_{0}\in\mathbb{D}^{2{t_{1}}}_{c_{-n}}$ for some $n\geq 0$, let
$\hat{d}(p_{0}),\hat{e}_{n}(p_{0})\geq 0$ be the smallest integers so that
$p_{0}\in\mathbb{D}^{2{t_{1}}}_{c_{-\hat{d}(p_{0})}}$ and
$p_{-\hat{e}_{n}(p_{0})}\notin\mathbb{D}^{2{t_{1}}}_{c_{-n-\hat{e}_{n}(p_{0})}}.$
Otherwise, denote $\hat{d}(p_{0})=\varnothing$.

For ${t}\in(0,{t_{1}}]$, define the ${t}$-critical depth
$\operatorname{cd}_{t}(p_{0})$ and the ${t}$-valuable depth
$\operatorname{vd}_{t}(p_{0})$ of $p_{0}$ as in Subsection 6.1. Denote

$\mathcal{T}_{c_{m}}:=\mathcal{T}_{c_{m}}({t_{1}})\hskip
14.22636pt\text{for}\hskip 14.22636ptm\in\mathbb{Z};$

and

$\operatorname{cd}(p_{0}):=\operatorname{cd}_{{t_{1}}}(p_{0})\hskip
14.22636pt\text{and}\hskip
14.22636pt\operatorname{vd}(p_{0}):=\operatorname{vd}_{{t_{1}}}(p_{0}).$

######  Lemma 9.9.

Let $p_{-n}\in\mathbb{D}^{2{t_{1}}}_{c_{-n}}$ for some $n\geq 0$. Then one of
the following statements hold.

  1. i)

We have $\operatorname{cd}(p_{-n})\in\\{\varnothing,\infty\\}$, and
$\operatorname{cd}(p_{1})=\operatorname{cd}(p_{-n}).$

  2. ii)

We have $\operatorname{cd}(p_{-n})=-n$, and
$\operatorname{cd}(p_{1})=\varnothing.$

  3. iii)

We have ${\bar{\varepsilon}}_{1}^{-1}n<\operatorname{cd}(p_{-n})<\infty$, and
$\operatorname{cd}(p_{1})=\operatorname{cd}(p_{-n})-n-1.$

###### Proof.

Suppose that $p_{1}\in\mathcal{T}_{c_{-k}}$ for some $k\geq 0$. Then by
Proposition 5.10, we have $k>{\bar{\varepsilon}}_{1}^{-1}n$. Observe that, for
some uniform constant $K\geq 1$, we have

$\displaystyle\operatorname{dist}(c_{-k+1},c_{0})$
$\displaystyle<K\operatorname{dist}(c_{-k},p_{1})+\operatorname{dist}(p_{0},c_{0})$
$\displaystyle<K\operatorname{dist}(c_{-k},p_{1})+\lambda^{-\eta
n}\operatorname{dist}(p_{-n},c_{-n})$
$\displaystyle<\lambda^{{\bar{\delta}}k}+2{t_{1}}\lambda^{(\varepsilon_{2}-\eta)n}$
$\displaystyle<\lambda^{{\bar{\varepsilon}}_{2}n}.$

Thus, by Lemma 5.11, we have

$p_{-n}\in F^{-n-1}(\mathcal{T}_{-k})=\mathcal{T}_{c_{-k-n-1}}.$

∎

######  Lemma 9.10.

Let $p\in\Omega$. If $\hat{d}(p)\neq\varnothing$, then
$\operatorname{d}(c_{-\hat{d}(p)})=\varnothing$.

###### Proof.

Denote $\hat{d}:=\hat{d}(p)$. Suppose
$c_{-\hat{d}}\in\mathbb{D}^{t_{1}}_{c_{-n}}$ for some $n\geq 0$. By Lemma 9.9,
we see that $\hat{d}>{\bar{\varepsilon}}_{1}^{-1}n$. Thus,

$\displaystyle\operatorname{dist}(p,c_{-n})$
$\displaystyle<\operatorname{dist}(p,c_{-\hat{d}})+\operatorname{dist}(c_{-\hat{d}},c_{-n})$
$\displaystyle<2{t_{1}}\lambda^{\varepsilon_{2}\hat{d}}+{t_{1}}\lambda^{\varepsilon_{2}n}$
$\displaystyle<2{t_{1}}\lambda^{\varepsilon_{2}n}.$

So $p\in\mathbb{D}^{2{t_{1}}}_{c_{-n}}$, which is a contradiction. ∎

Let $\tilde{N}_{1}:=-\delta^{-1}\log_{\lambda}\overline{L_{1}},$ and let
$N_{1}$ be the first Pliss moment larger than $\tilde{N}_{1}$ in the forward
orbit of $c_{1}$. By Proposition 4.2, we have $\tilde{N}_{1}\leq
N_{1}<(1+{\bar{\varepsilon}}_{1})\tilde{N}_{1}.$ Denote

${t_{2}}:=\operatorname{radius}(U_{c_{1}}^{{\bar{\varepsilon}}_{1}N_{1}})\asymp\lambda^{{\bar{\varepsilon}}_{1}N_{1}}={L_{1}}^{-{\bar{\varepsilon}}_{1}}<{t_{1}},$

and let

$\mathbf{T}_{c_{n}}:=F^{n-1}(\mathcal{T}_{c_{1}}({t_{2}}))\hskip
14.22636pt\text{for}\hskip 14.22636pt0\leq n\leq N_{1}.$

######  Lemma 9.11.

Let $p_{1}\in\mathcal{T}_{c_{1}}\setminus\mathbf{T}_{c_{1}}$. If
$\operatorname{d}(p_{1})=\varnothing$, then

$\hat{e}(p_{1})<-{\bar{\varepsilon}}_{1}\log_{\lambda}{L_{1}}.$

###### Proof.

Letting $t=2{t_{1}}$ and applying Lemma 9.3, we have

$\hat{e}_{0}(p_{0})<{\bar{\varepsilon}}_{2}^{-1}\log_{\lambda}{t_{2}}=-{\bar{\varepsilon}}_{1}\log_{\lambda}{L_{1}}.$

The result now follows from Proposition 9.6. ∎

######  Lemma 9.12.

Let $p_{N_{1}}\in\mathbf{T}_{c_{N_{1}}}$ be $n$-times forward
$({\overline{L_{1}}},\delta)_{v}$-regular along
$\tilde{E}^{v}_{p_{N_{1}}}\in\mathbb{P}^{2}_{p_{N_{1}}}$ with
$n>-\log_{\lambda}{\overline{L_{1}}}.$ Then $p_{1}$ is $(n+N_{1})$-times
forward $(1,\delta)_{v}$-regular along $\tilde{E}^{v}_{p_{1}}$. Moreover,
$\tilde{E}^{v}_{p_{1}}$ is sufficiently stable-aligned in
$\mathcal{U}_{c_{1}}$.

###### Proof.

Since $p_{1}\in\mathcal{U}_{c_{1}}^{{\bar{\varepsilon}}_{1}N_{1}}$, we see
that $p_{i}\in\mathcal{U}_{c_{i}}$ for $1\leq i\leq M$ with

$M>\varepsilon_{1}^{-1}({\bar{\varepsilon}}_{1}N_{1})>{\bar{\varepsilon}}_{1}^{-1}N_{1}.$

By Proposition 3.11, we have for some uniform constant $K\geq 1$:

$\measuredangle(\tilde{E}^{v}_{p_{N_{1}}},E^{v}_{p_{N_{1}}})<K\lambda^{-\varepsilon_{1}N_{1}}{\overline{L_{1}}}\lambda^{(1-\delta)n}<K{L_{1}}^{{\bar{\varepsilon}}_{1}}{\overline{L_{1}}}\lambda^{(1-\delta)n}<{\overline{L_{1}}}^{-1}<\lambda^{{\bar{\varepsilon}}_{1}N_{1}}.$

Thus, $\tilde{E}^{v}_{p_{N_{1}}}$ is sufficiently vertical in
$\mathcal{U}_{c_{N_{1}}}$.

By Proposition 3.9, we conclude that $p_{1}$ is $N_{1}$-times forward
$(1,{\bar{\varepsilon}}_{1})_{v}$-regular along $\tilde{E}^{v}_{p_{1}}$. Since

$\lambda^{(1-{\bar{\varepsilon}}_{1})N_{1}}{\overline{L_{1}}}\lambda^{(1-\delta)i}={\overline{L_{1}}}\lambda^{(\delta-{\bar{\varepsilon}}_{1})N_{1}}\lambda^{(1-\delta)(N_{1}+i)}<\lambda^{(1-\delta)(N_{1}+i)}\hskip
14.22636pt\text{for}\hskip 14.22636pti\geq 0,$

the result follows. ∎

######  Lemma 9.13.

Let $p_{1}\in\mathbf{T}_{c_{1}}$. Then one of the following statements hold.

  1. i)

We have $\operatorname{cd}(p_{1})\in\\{\varnothing,\infty\\}$, and
$\operatorname{cd}(p_{N_{1}})=\operatorname{cd}(p_{1}).$

  2. ii)

We have ${\bar{\varepsilon}}_{1}^{-1}N_{1}<\operatorname{cd}(p_{1})<\infty$,
and $\operatorname{cd}(p_{N_{1}})=\operatorname{cd}(p_{1})-N_{1}+1.$

###### Proof.

Since $N_{1}$ is a Pliss moment, and

$\operatorname{dist}(c_{0},c_{m-N_{1}})<\lambda^{{\bar{\varepsilon}}_{1}N_{1}},$

it follows from Lemma 5.11 that either $\operatorname{cd}(p_{1})=\varnothing$
or $\operatorname{cd}(p_{1})>{\bar{\varepsilon}}_{1}^{-1}N_{1}.$

Suppose $p_{N_{1}}\in\mathcal{T}_{c_{-n}}$ for some
$n>{\bar{\varepsilon}}_{1}^{-1}N_{1}$. Then by Proposition 6.8, we have

$\operatorname{dist}(c_{-n-N_{1}},c_{0})<\operatorname{diam}(\mathcal{T}_{c_{-n-N_{1}}})+\operatorname{diam}(\mathbf{T}_{c_{1}})<\lambda^{\delta(n+N_{1})}+\lambda^{{\bar{\varepsilon}}_{1}N_{1}}<\lambda^{{\bar{\varepsilon}}_{1}N_{1}}.$

Thus, by Lemma 5.11, the result follows. ∎

######  Lemma 9.14.

Let $p_{1}\in\Lambda\cap\mathcal{T}_{c_{1}}$ be a point such that
$\operatorname{cd}(p_{1})=\varnothing$, and let
$\tilde{E}^{v}_{p_{1}}\in\mathbb{P}^{2}_{p_{1}}$ be sufficiently stable-
aligned in $\mathcal{U}_{c_{1}}$. Then for all $n\geq 1$ such that
$d_{\mathcal{T}}(p_{-n})=\varnothing$, the point $p_{-n}$ is $n$-times forward
$({L_{1}}^{1+{\bar{\varepsilon}}},\delta)_{v}$-regular along
$\tilde{E}^{v}_{p_{-n}}$.

###### Proof.

Let $-j_{0}<0$ be the tunnel escape moment of $p_{0}$, and
$\hat{e}_{0}:=\hat{e}_{0}(p_{0})$. By Proposition 6.13 and Lemma 9.10, we see
that for $j_{0}\leq i\leq\hat{e}_{0}$ such that $\hat{d}(p_{-i})=i$, the point
$p_{-i}$ is $i$-times $(O({L_{1}}),\delta)_{v}$-regular along
$\tilde{E}^{v}_{p_{-i}}$.

Let $m_{1}>j_{0}$ be the smallest integer such that
$p_{-m_{1}}\in\mathcal{T}_{c_{0}}$. We claim that
$\hat{d}(p_{-m_{1}+1})=m_{1}+1$. Otherwise, there exists
$0<m_{1}^{\prime}<j_{0}$ such that

$\hat{d}(p_{-m_{1}^{\prime}-i})=-i\hskip 14.22636pt\text{for}\hskip
14.22636pt0\leq i\leq m_{1}.$

Then by Lemma 9.9, we must have

$j_{0}>\hat{e}_{0}(p_{-m_{1}^{\prime}})>m_{1},$

which is a contradiction.

If $p_{-m_{1}}\notin\mathbf{T}_{c_{0}}$, then by Lemma 9.11, we see that for

$0\leq i\leq l\leq\hat{e}_{1}:=\hat{e}(p_{-m_{1}}),$

we have

$\|DF^{i}|_{\tilde{E}^{v}_{p_{-m_{1}-l}}}\|<\lambda^{-\eta
i}=\lambda^{-(1-\delta+\eta)i}\lambda^{(1-\delta)i}\leq{L_{1}}^{{\bar{\varepsilon}}_{1}}\lambda^{(1-\delta)i}.$

Moreover, by Proposition 6.13 and Lemma 9.10, the point $p_{-m_{1}+1}$ is
$(m_{1}-1)$-times $(O({L_{1}}),\delta)_{v}$-regular along
$\tilde{E}^{v}_{p_{-m_{1}+1}}$. Hence, $p_{-m_{1}-l}$ is $(m_{1}+l)$-times
$({L_{1}}^{1+{\bar{\varepsilon}}_{1}},\delta)_{v}$-regular along
$\tilde{E}^{v}_{p_{-m_{1}-l}}$.

Suppose instead that $p_{-m_{1}}\in\mathbf{T}_{c_{0}}$, so that
$p_{-m_{1}+N_{1}}\in\mathbf{T}_{c_{N_{1}}}$. By Proposition 6.13, Lemma 9.10
and Lemma 9.13, we see that $p_{-m_{1}+N_{1}}$ is $(m_{1}-N_{1})$-times
$(O({L_{1}}),\delta)_{v}$-regular along $\tilde{E}^{v}_{p_{-m_{1}+N_{1}}}$. By
Lemma 9.12, the point $p_{-m_{1}+1}$ is $(m_{1}-1)$-times
$(1,\delta)_{v}$-regular along $\tilde{E}^{v}_{p_{-m_{1}+1}}$, and in
particular, is sufficiently stable-aligned in $\mathcal{U}_{c_{1}}$.

Proceeding by induction, the result follows. ∎

######  Lemma 9.15.

If

$p\in\Lambda\setminus\bigcup_{i\geq
0}^{\infty}\mathcal{T}_{c_{-i}}({L_{1}}^{-{\bar{\varepsilon}}}),$

then $p$ is forward $({L_{1}}^{1+{\bar{\varepsilon}}},\delta)$-regular. If $p$
is also contained in
$\mathbf{T}_{c_{1}}=\mathcal{T}_{c_{1}}({L_{1}}^{-{\bar{\varepsilon}}})$, then
$p$ is forward $(1,\delta)$-regular.

###### Proof.

Let $p_{0}\in\Lambda\cap\mathcal{T}_{c_{0}}$. If
$p_{0}\notin\mathbf{T}_{c_{0}}$, then by Lemma 9.11, we see that for

$0\leq i\leq l\leq\hat{e}(p_{0}),$

we have

$\|DF^{i}|_{\tilde{E}^{v}_{p_{-l}}}\|<\lambda^{-\eta
i}=\lambda^{-(1-\delta+\eta)i}\lambda^{(1-\delta)i}\leq{L_{1}}^{{\bar{\varepsilon}}_{1}}\lambda^{(1-\delta)i}.$

If $\operatorname{d}(p_{0})=\varnothing$, then this implies that $p_{-l}$ is
forward $({L_{1}}^{1+{\bar{\varepsilon}}_{1}},\delta)$-regular. Otherwise,
$d_{1}:=\operatorname{d}(p_{0})>0$, and by Lemma 9.5, we have

$\operatorname{dist}(p_{d_{1}},c_{0})<\operatorname{dist}(p_{0},c_{0})^{1/{\bar{\varepsilon}}_{1}}.$

Proceeding by induction, we obtain a sequence $0=:d_{0}<d_{1}<\ldots<d_{M}$
with $M\geq 0$ such that $d_{i+1}:=\operatorname{d}(p_{d_{i}})$, and either
$\operatorname{d}(p_{d_{M}})=\varnothing$, or
$p_{d_{M}}\in\mathbf{T}_{c_{0}}$. In the former case, we can argue as before
to conclude that $p_{d_{M}-l}$ is forward
$({L_{1}}^{1+{\bar{\varepsilon}}_{1}},\delta)$-regular for all $0\leq
l\leq\hat{e}(p_{d_{M}})$.

It remains to prove the result for $p_{0}\in\mathbf{T}_{c_{0}}$ satisfying
$\operatorname{cd}(p_{0})<\infty$. By Lemma 9.9 ii) and Lemma 9.13, we have

$\operatorname{cd}(p_{d_{\mathcal{T}}(p_{0})+N_{1}})=\varnothing.$

Thus, by replacing $p_{N_{1}}$ with $p_{\operatorname{cd}(p_{N_{1}})+N_{1}}$,
we may assume without loss of generality that
$\operatorname{cd}(p_{N_{1}})=\varnothing$.

If there exists $M\geq 0$ such that $p_{M}\in\mathbf{T}_{c_{N_{1}}}$ and
$\operatorname{d}(p_{M})=\varnothing$, then the result follows by Lemma 9.12
and Lemma 9.14. Suppose this is not the case. Then there exists an infinite
sequence $0\leq M_{1}<\ldots$ such that

$p_{M_{i}}\in\mathbf{T}_{c_{N_{1}}},\hskip
14.22636pt\operatorname{cd}(p_{M_{i}})=\varnothing\hskip
14.22636pt\text{and}\hskip
14.22636pt\operatorname{d}(p_{M_{i}})=M_{i+1}-M_{i}-N_{i}.$

Let $E^{i}_{p_{M_{i}}}:=E^{v}_{p_{M_{i}}}$, so that

$D\Phi_{c_{N_{1}}}\left(E^{i}_{p_{M_{i}}}\right)=E^{gv}_{\Phi_{c_{N_{1}}}(p_{M_{i}})}.$

By Lemma 9.12 and Lemma 9.14, $p_{0}$ is $M_{i}$-times forward
$({L_{1}}^{1+{\bar{\varepsilon}}},\delta)_{v}$-regular along $E^{i}_{p_{0}}$.
Hence, by Proposition 3.11, $E^{i}_{p_{0}}$ converge exponentially fast as
$i\to\infty$ to $E^{s}_{p_{0}}$, along which $p_{0}$ is forward
$({L_{1}}^{1+{\bar{\varepsilon}}},\delta)_{v}$-regular. ∎

Let $p_{0}\in\mathcal{T}_{c_{0}}$. Write $(x_{0},y_{0}):=\Phi_{c_{0}}(p_{0}).$
Denote

$\tilde{l}:=(1+{\bar{\varepsilon}}_{1})\log_{\lambda}|x_{0}|.$

By Lemma 4.2, there exists a Pliss moment $-l<0$ in the backward orbit of
$c_{0}$ such that $\tilde{l}\leq l<(1+{\bar{\varepsilon}}_{1})\tilde{l}.$ Such
a moment is referred to as a direction aligning moment for $p_{0}$.

######  Lemma 9.16.

Let $p_{0}\in\mathcal{T}_{c_{0}}$, and let $-l<0$ be a direction aligning
moment for $p_{0}$. If $\tilde{E}^{v}_{p_{1}}\in\mathbb{P}^{2}_{p_{1}}$ is
sufficiently stable-aligned in $\mathcal{U}_{c_{1}}$, then
$\tilde{E}^{v}_{p_{-n}}$ is sufficiently vertical in $\mathcal{U}_{c_{-n}}$
for $l\leq n<\hat{e}_{0}(p_{0})$. If, additionally, we have

$p_{-n}\in\mathbf{T}_{c_{1}}\hskip 14.22636pt\text{and}\hskip
14.22636pt\hat{d}(p_{-n})=n,$

then $\tilde{E}^{v}_{p_{-n}}$ is sufficiently stable-aligned in
$\mathcal{U}_{c_{1}}$.

###### Proof.

The first claim follows immediately from Proposition 6.13 ii).

Write $(x_{1},y_{1}):=\Phi_{c_{1}}(p_{-n+1})$ and
$(\tilde{x}_{1},\tilde{y}_{1}):=\Phi_{c_{1}}(c_{-n+1}).$ By Lemma 9.4, we have

$|x_{1}-\tilde{x}_{1}|^{{\bar{\varepsilon}}_{1}}<|\tilde{x}_{1}|.$ (9.6)

Consequently, $c_{-n+1}\in\mathcal{U}_{c_{1}}^{{\bar{\varepsilon}}_{1}N_{1}}$.
Moreover, by Lemma 9.10 and Lemma 9.13, we have
$\operatorname{d}(c_{-m+N_{1}})=\varnothing$. Arguing as in the proof of Lemma
9.12, we see that $c_{-n+1}$ is forward $(1,\delta)$-regular, and hence,
$\hat{E}^{v}_{c_{-n+1}}$ is sufficiently stable-aligned in
$\mathcal{U}_{c_{1}}$. Since $\tilde{E}^{v}_{p_{-n+1}}$ is sufficiently
vertical in $\mathcal{U}_{c_{-n+1}}$, (9.6) implies that
$\tilde{E}^{v}_{p_{-n+1}}$ is also sufficiently stable-aligned in
$\mathcal{U}_{c_{1}}$. ∎

######  Lemma 9.17.

Let $p_{0}\in\mathcal{T}_{c_{0}}$, and let $-l<0$ be a direction aligning
moment for $p_{0}$. If $p_{-n}\in\mathcal{T}_{c_{0}}$ for some $1\leq
n\leq\hat{e}_{0}(p_{0})$, then either $n>l$ or $l>\hat{e}_{0}(p_{-n})$.

###### Proof.

The result follows from a similar argument as the one used in the proof of
Lemma 9.9. ∎

###### Proof of Theorem 9.8 a).

Let ${t}\in(0,{L_{1}}^{-1}]$. Consider $\mathbf{p}\in\Lambda$ such that

$M:=\operatorname{cd}(\mathbf{p})\geq 0\hskip 14.22636pt\text{and}\hskip
14.22636pt\operatorname{cd}_{t}(\mathbf{p})=\varnothing.$

Then we have

$p_{1}:=F^{M+1}(\mathbf{p})\in\mathbf{T}_{c_{1}}\setminus\mathcal{T}_{c_{1}}({t}),$

and by Lemma 9.9 and Lemma 9.13,

$\operatorname{cd}(p_{1})=\operatorname{cd}(p_{N_{1}})=\varnothing=\operatorname{cd}_{t}(p_{1}).$

Thus, it follows from Lemma 9.12 that $p_{1}$ is forward
$(1,\delta)_{v}$-regular along $\hat{E}^{v}_{p_{1}}$, and
$\hat{E}^{v}_{p_{1}}$ is sufficiently stable-aligned in $\mathcal{U}_{c_{1}}$.

Let $-l_{0}<0$ be a direction aligning moment for $p_{0}$. By Lemma 9.16, for
$l_{0}\leq n\leq\hat{e}_{0}(p_{0}),$ the direction $\hat{E}^{v}_{p_{-n}}$ is
sufficiently vertical in $\mathcal{U}_{c_{-n}}$. Denote the vertical direction
at $p_{-n}$ in $\mathcal{U}_{c_{-n}}$ by $\tilde{E}^{0}_{p_{-n}}$, so that

$D\Phi_{c_{-n}}(\tilde{E}^{0}_{p_{-n}})=E^{gv}_{\Phi_{c_{-n}}(p_{-n})}.$

Let $m_{0}:=0$, and for $i\geq 1$, let

$m_{i-1}+l_{i-1}<m_{i}<m_{i-1}+\hat{e}_{0}(p_{-m_{i-1}})$

be an integer such that

$p_{-m_{i}}\in\mathbf{T}_{c_{0}}\setminus\mathcal{T}_{c_{0}}({t})\hskip
14.22636pt\text{and}\hskip 14.22636pt\hat{d}(p_{-m_{i}})=-m_{i}+m_{i-1}.$

Write $z^{i}=(x^{i},y^{i}):=\Phi_{c_{0}}(p_{-m_{i}}).$ Then we have by Lemma
9.5,

$|x^{i}|>|x^{i-1}|^{{\bar{\varepsilon}}_{1}}.$ (9.7)

Let

$-l_{i}=-(1+{\bar{\varepsilon}}_{1})\log_{\lambda}|x^{i}|,$

be a direction aligning moment for $p_{-m_{i}}$. By Lemma 9.16,
$\hat{E}^{v}_{p_{-m_{i}+1}}$ is sufficiently stable-aligned in
$\mathcal{U}_{c_{1}}$, and for $l_{i}\leq n\leq\hat{e}_{0}(p_{-m_{i}}),$ the
direction $\hat{E}^{v}_{p_{-m_{i}-n}}$ is sufficiently vertical in
$\mathcal{U}_{c_{-n}}$.

We claim that the initially given point $p_{-M}=\mathbf{p}$ is forward
$(L,\delta)$ regular along $\hat{E}^{v}_{p_{-M}}$ with $L=1/\underline{{t}}$.
Let

$0:=m_{0}<m_{1}<\ldots<m_{I}\leq M$

with $I\geq 0$ be a maximal sequence of returns to
$\mathbf{T}_{c_{0}}\setminus\mathcal{T}_{c_{0}}({t})$ as specified above.

For $0\leq i<I$, denote $n_{i}:=m_{i+1}-m_{i}.$ Since for $1\leq n\leq
n_{i}-l_{i}$, the direction $\hat{E}^{v}_{p_{-m_{i+1}+n}}$ is sufficiently
vertical in $\mathcal{U}_{c_{-n_{i}+n}}$, and $c_{-n_{i}}$ is forward
$(1,\delta)$-regular, we conclude that $p_{-m_{i+1}}$ is $(n_{i}-l_{i})$-times
forward $(1,\delta)$-regular. It follows that $p_{-m_{i+1}}$ is $n_{i}$-times
forward $(L_{i},\delta)$-regular, where

$L_{i}<\lambda^{-(1-\delta-\eta)l_{i}}.$

Denote $n_{I}:=M-m_{I}.$ If $n_{I}>l_{I}$, then for $0\leq n\leq n_{I}-l_{I}$,
the direction $\hat{E}^{v}_{p_{-m_{i+1}+n}}$ is sufficiently vertical in
$\mathcal{U}_{c_{-n_{i}+n}}$. Moreover, since $\hat{d}(p_{-M})=n_{i},$ the
point $c_{-n_{i}}$ is forward $({L_{1}},\delta)$-regular. Thus, $p_{-M}$ is
$(n_{I}-l_{i})$-times forward $({L_{1}},\delta)$-regular. It follows that
(whether $n_{I}>l_{I}$ or not) $p_{-M}$ is $n_{I}$-times forward
$({L_{1}}L_{I},\delta)$-regular, where
$L_{I}<\lambda^{-(1-\delta-\eta)l_{I}}.$

Lastly, observe that by (9.7), we have $l_{i}<{\bar{\varepsilon}}_{1}l_{i-1}$
for $1\leq i\leq I,$ and hence

$\sum_{i=0}^{I}l_{i}<Kl_{0}<-K\log_{\lambda}|x^{0}|<K\log_{\lambda}{t}.$ (9.8)

Concatenating the above regularities, we conclude that $p_{-M}$ is forward
$({L_{1}}L^{\prime},\delta)_{v}$-regular along $\tilde{E}^{v}_{p_{-M}}$, where

${L_{1}}L^{\prime}<{t}^{-1}\lambda^{-(1-\delta-\eta)\sum_{i=0}^{I}l_{i}}<1/\underline{{t}}.$

∎

Let $N\geq 0$ and $K\geq 1$. We say that a point $p_{0}\in\Omega$ is $N$-times
consistently forward $(K,\delta)_{v}$-regular along
$\tilde{E}^{v}_{p_{0}}\in\mathbb{P}^{2}_{p_{0}}$ if, for all $0\leq i\leq N$,
the point $p_{i}$ is $(N-i)$-times forward $(K,\delta)_{v}$-regular along
$\tilde{E}^{v}_{p_{i}}$.

######  Lemma 9.18.

If for some $N\geq 0$ and $K\geq 1$, a point $p_{0}\in\Omega$ is $N$-times
consistently forward $(K,\delta)_{v}$-regular along
$\tilde{E}^{v}_{p_{0}}\in\mathbb{P}^{2}_{p_{0}}$, then $p_{i}$ is
$(i,N-i)$-times $(K,\delta)_{v}$-regular along $\tilde{E}^{v}_{p_{i}}$.

######  Lemma 9.19.

Let $p_{0}\in\Omega$ be $M$-times backward $(K,\delta)_{v}$-regular along
$\tilde{E}^{v}_{p_{0}}\in\mathbb{P}^{2}_{p_{0}}$ for some $M\geq 0$ and $K\geq
1$. If $\tilde{E}^{h}_{p_{-M}}\in\mathbb{P}^{2}_{p_{-M}}$ is a direction
satisfying

$\measuredangle(\tilde{E}^{h}_{p_{-M}},\tilde{E}^{v}_{p_{-M}})>k$

for some uniform constant $k>0$, then $p_{0}$ is $M$-times backward
$({\bar{K}},\delta)_{h}$-regular along $\tilde{E}^{h}_{p_{0}}$. Moreover, we
have

$\measuredangle(\tilde{E}^{h}_{p_{0}},\tilde{E}^{v}_{p_{0}})>{\bar{K}}^{-1}.$

Let $\delta_{1}=\underline{\delta}$, so that $\mathcal{T}_{c_{0}}$ is a
$1/\underline{\delta_{1}}$-pinched neighborhood. Denote

$\delta_{1}^{\prime}:=(1-{\bar{\delta}}_{1})\delta_{1}.$

For ${t}\in(0,{L_{1}}^{-1}]$ and $n\geq 0$, let
${\hat{\mathcal{T}}}_{c_{-n}}({t})$ be the $n$-times $1/\delta_{1}$-pinched,
$\delta_{1}^{\prime}$-truncated tunnel of length ${t}$. Numbers $-l<-j<0$ are
the slow and fast tunnel escape moment for $p_{0}\in\mathcal{T}_{c_{0}}({t})$
respectively if $l$ and $j$ are the smallest integers such that

$p_{-l}\notin{\hat{\mathcal{T}}}_{c_{-l}}({t})\hskip
14.22636pt\text{and}\hskip 14.22636ptp_{-j}\notin\mathcal{T}_{c_{-j}}({t}).$

If $p\in{\hat{\mathcal{T}}}_{c_{-n}}({t})$ for some $n\geq 0$, then denote by

$d:=d_{{\hat{\mathcal{T}}}_{-}({t})}(p)\in\mathbb{N}\cup\\{0,\infty\\}$

the largest integer such that $p\in{\hat{\mathcal{T}}}_{c_{-d}}({t})$.
Otherwise, denote

$d_{{\hat{\mathcal{T}}}_{-}({t})}(p)=\varnothing.$

######  Lemma 9.20.

Let ${t}\in(0,{L_{1}}^{-1}]$ and $L:=1/\underline{{t}}$. For
$p_{0}\in\Lambda\cap\mathcal{T}_{c_{0}}({t})$, let $-l<-j<0$ be the slow and
fast tunnel escape moment for $p_{0}$. Suppose
$\tilde{E}^{v}_{p_{0}},\tilde{E}^{h}_{p_{0}}\in\mathbb{P}^{2}_{p_{0}}$ are
directions such that $\tilde{E}^{v}_{p_{1}}$ is sufficiently stable-aligned in
$\mathcal{U}_{c_{1}}$; the point $p_{-l}$ is $M$-times backward
$({\bar{L}},\delta)_{h}$-regular along $\tilde{E}^{h}_{p_{-l}}$ for some
$M\geq 0$; and

$\measuredangle(\tilde{E}^{v}_{p_{-l}},\tilde{E}^{h}_{p_{-l}})>{\bar{L}}^{-1}.$

Then for $n\in\\{0,j\\}$, the point $p_{-n}$ is $(M+l-n)$-times backward
$(1,\delta)_{h}$-regular along $\tilde{E}^{h}_{p_{0}}$. Moreover,

$\measuredangle(\tilde{E}^{h}_{p_{-j}},\tilde{E}^{v}_{p_{-j}})>k$

for some uniform constant $k>0$, and $\tilde{E}^{h}_{p_{0}}$ is sufficiently
center-aligned in $\mathcal{U}_{c_{0}}$.

###### Proof.

Write $z_{0}=(x_{0},y_{0}):=\Phi_{c_{0}}(p_{0})$;
$E^{v}_{p_{0}}:=D_{z_{0}}\Phi_{c_{0}}^{-1}(E^{gv}_{z_{0}})$ and
$\tilde{E}^{h}_{z_{0}}:=D\Phi_{c_{0}}(\tilde{E}^{h}_{p_{0}}).$ Then

$l=(1+{\bar{\delta}})\delta_{1}^{-1}\log_{\lambda}|x_{0}|\hskip
14.22636pt\text{and}\hskip
14.22636ptj<\underline{\delta}^{-1}\log_{\lambda}|x_{0}|.$

By Proposition 6.13 and Proposition 3.11, we have

$\measuredangle(\tilde{E}^{v}_{p_{-l}},E^{v}_{p_{-l}})<\lambda^{(1-{\bar{\delta}})l}<|x_{0}|^{(1-{\bar{\delta}})/\delta_{1}}<{\bar{L}}^{-(1-{\bar{\delta}})/\delta_{1}}\ll{\bar{L}}^{-1}$
(9.9)

Hence,

$\measuredangle(\tilde{E}^{h}_{p_{-l}},E^{v}_{p_{-l}})>{\bar{L}}^{-1}.$

Denote

$\theta^{v/h}_{0}:=\measuredangle(\tilde{E}^{h}_{z_{0}},E^{gv/gh}_{z_{0}}).$

For $0\leq i\leq l$, if $\theta^{v}_{-i}<k$ for some uniform constant $k>0$,
then by Theorem 3.6 and Proposition 3.9, we have

$\theta^{v}_{-i+1}>\lambda^{-(1-{\bar{\varepsilon}})}\theta^{v}_{-i}.$ (9.10)

Similarly, if $\theta^{h}_{-i}<k$, then

$\theta^{h}_{-i+1}>\lambda^{1-{\bar{\varepsilon}}}\theta^{h}_{-i}.$ (9.11)

Let $I_{1},I_{2}$ be the smallest integers such that $\theta^{v}_{-l+I_{1}}>k$
and $\theta^{h}_{-l+I_{2}}<k.$ Inequalities (9.10) and (9.11) imply

$I_{1}<-\log_{\lambda}{\bar{L}}<K\log_{\lambda}|x_{0}|\hskip
14.22636pt\text{and}\hskip 14.22636ptI_{2}<I_{1}+K$

for some uniform constant $K\geq 1$.

Let $N>0$ be the largest integer such that
$\theta_{-N}^{h}<\lambda^{{\bar{\varepsilon}}(l-N)}.$ By a straightforward
computation, we obtain

$N>l-I_{2}-{\bar{\varepsilon}}\log_{\lambda}|x_{0}|>\delta_{1}^{-1}\log_{\lambda}|x_{0}|.$

Consequently,

$l-N<{\bar{\delta}}_{1}N\hskip 14.22636pt\text{and}\hskip
14.22636ptN-j>(1-\underline{\delta}^{-1}\delta_{1})N=(1-{\bar{\delta}}_{1})N.$

Note that for $0\leq i\leq N$, the direction $\tilde{E}^{h}_{p_{-i}}$ is
sufficiently horizontal in $\mathcal{U}_{c_{-i}}$.

By Proposition 3.9, $p_{-n}$ with $n\in\\{0,j\\}$ is $(N-n)$-times backward
$(1,{\bar{\varepsilon}})_{h}$-regular along $\tilde{E}^{h}_{p_{-n}}$. For
$0\leq i<l-N$, we have

$\displaystyle\|D_{p_{-j}}F^{-(N+i-j)}\|$
$\displaystyle<\lambda^{-{\bar{\varepsilon}}(N-j)}\lambda^{-(1+\eta)(l-N)}$
$\displaystyle<\lambda^{-{\bar{\delta}}_{1}N}$
$\displaystyle<\lambda^{-{\bar{\delta}}_{1}N}\lambda^{\delta(N+i-j)}\lambda^{-\delta(N+i-j)}$
$\displaystyle<\lambda^{\delta(1-{\bar{\delta}}_{1})N}\lambda^{-\delta(N+i-j)}$
$\displaystyle<{t}^{1/{\bar{\delta}}_{1}}\lambda^{-\delta(N+i-j)}$
$\displaystyle<{\bar{L}}^{-1}\lambda^{-\delta(N+i-j)}.$

Concatenating with the $M$-times backward $({\bar{L}},\delta)_{h}$-regularity
of $p_{-l}$ along $\tilde{E}^{h}_{p_{-l}}$, we see that the point $p_{-n}$
with $n\in\\{0,j\\}$ is $(M+l-n)$-times backward $(1,\delta)_{h}$-regular
along $\tilde{E}^{h}_{p_{-n}}$.

The other claims follow immediately from Lemma 9.19 and Proposition 6.12. ∎

###### Proof of Theorem 9.8 b).

For ${t}\in(0,{L_{1}}^{-1}]$, let $L:=1/\underline{{t}}$ be the forward
regularity factor given in part a). Let $p_{0}\in\Lambda$ be a point
satisfying $\operatorname{vd}_{t}(p_{0})=\varnothing.$ We prove that $p_{0}$
is backward $({\bar{L}},\delta)$-regular.

First, suppose $\operatorname{cd}_{t}(p_{0})=\varnothing,$ so that $p_{0}$ is
forward $(L,\delta)_{v}$-regular along $\hat{E}^{v}_{p_{0}}$. Let $N\geq 0$ be
an integer such that $\operatorname{cd}_{t}(p_{-N})=\varnothing;$ and let
$\tilde{E}^{N}_{p_{-N}}\in\mathbb{P}^{2}_{p_{-N}}$ be a direction satisfying

$\measuredangle(\tilde{E}^{N}_{p_{-N}},\hat{E}^{v}_{p_{-N}})>k$

for some uniform constant $k>0$. We claim that $p_{0}$ is $N$-times backward
$({\bar{L}},\delta)_{h}$-regular along $\tilde{E}^{N}_{p_{0}}$.

Let $0\leq m_{1}<\ldots<m_{I}<N$ for some $I\geq 0$ be the sequence of all
integers satisfying

$p_{-m_{i}}\in\mathcal{T}_{c_{1}}({t})\hskip 14.22636pt\text{and}\hskip
14.22636pt\operatorname{cd}_{t}(p_{-m_{i}})=\varnothing.$

Let $-l_{i}<-j_{i}<0$ be the slow and fast tunnel escape moment for
$p_{-m_{i}-1}$, and let $n_{i}>1$ be the crescent escape moment for
$p_{-m_{i}}$. By Proposition 6.9, we have

$\operatorname{cd}_{t}(p_{-m_{i}-l_{i}})=\operatorname{cd}_{t}(p_{-m_{i}-j_{i}})=\operatorname{cd}_{t}(p_{-m_{i}+n_{i}})=\varnothing,$

and

$m_{i}+l_{i}<m_{i+1}-n_{i+1}.$

Hence,

$\operatorname{cd}_{t}(p_{-m})=\varnothing\hskip 14.22636pt\text{for}\hskip
14.22636ptm_{i}+j_{i}\leq k\leq m_{i+1}-n_{i+1}.$

By Lemma 9.18, $p_{-m_{I}-l_{I}}$ is $(N-m_{I}-l_{I})$-times backward
$(L,\delta)_{v}$-regular along $\hat{E}^{v}_{-m_{I}-l_{I}}$. Applying Lemma
9.19, Lemma 9.20 and Proposition 6.14, we see that $p_{-m_{I}+n_{I}}$ is
$(N-m_{I}+n_{I})$-times backward $(O(1),\delta)_{h}$-regular along
$\tilde{E}^{N}_{p_{-m_{I}+n_{I}}}$, and

$\measuredangle(\tilde{E}^{N}_{p_{-m_{I}+n_{I}}},\hat{E}^{v}_{p_{-m_{I}+n_{I}}})>k.$

Proceeding by induction, we conclude that $p_{-m_{1}+n_{1}}$ is
$(N-m_{1}+n_{1})$-times backward $(O(1),\delta)_{h}$-regular along
$\tilde{E}^{N}_{p_{-m_{1}+n_{1}}}$, and

$\measuredangle(\tilde{E}^{N}_{p_{-m_{1}+n_{1}}},\hat{E}^{v}_{p_{-m_{1}+n_{1}}})>k.$

By Lemma 9.18 and Lemma 9.19, $p_{0}$ is $(m_{1}-n_{1})$-times backward
$({\bar{L}},\delta)_{h}$-regular along $\tilde{E}^{N}_{p_{0}}$. Concatenating
the regularities, it follows that $p_{0}$ is $N$-times backward
$({\bar{L}},\delta)_{h}$-regular along $\tilde{E}^{N}_{p_{0}}$.

Now, suppose

$0\leq M:=\operatorname{cd}_{t}(p_{0})<\infty.$

Denote $m_{0}:=M$. Let $-l_{0}<-j_{0}<0$ be the slow and fast tunnel escape
time for $p_{m_{0}}\in\mathcal{T}_{c_{0}}({t})$. Then

$\operatorname{cd}_{t}(p_{m_{0}+1})=\operatorname{cd}_{t}(p_{m_{0}-j_{0}})=\operatorname{cd}_{t}(p_{m_{0}-l_{0}})=\varnothing.$

Observe that $j_{0}>m_{0}$. Let $m_{1}$ be the largest integer less than
$m_{0}$ such that $p_{m_{1}}\in\mathcal{T}_{c_{0}}({t})$, and if
$-l_{1}<-j_{1}<0$ are the slow and fast tunnel escape time for $p_{m_{1}}$,
then $j_{1}>m_{1}$. Clearly, we have

$p_{m_{1}}\in\mathcal{T}_{c_{m_{1}-m_{0}}}({t})\hskip
14.22636pt\text{and}\hskip
14.22636pt\operatorname{cd}_{t}(c_{m_{1}-m_{0}})=\varnothing.$

Moreover, by Proposition 6.9, we have

$0>m_{1}-l_{1}>m_{0}-j_{0}.$

Proceeding by induction, we obtain the maximal sequence
$M=:m_{0}>m_{1}>\ldots>m_{I}>0$ with $I\geq 0$, such that for $0<i\leq I$, we
have

$p_{m_{i}}\in\mathcal{T}_{c_{m_{i}-m_{i-1}}}({t})\hskip
14.22636pt\text{and}\hskip
14.22636pt\operatorname{cd}_{t}(c_{m_{i}-m_{i-1}+1})=\varnothing;$

and if $-l_{i}<-j_{i}<0$ are the slow and fast tunnel escape time for
$p_{m_{i}}$, then

$j_{i}>m_{i}\hskip 14.22636pt\text{and}\hskip
14.22636pt0>m_{i}-l_{i}>m_{i-1}-j_{i-1}.$

Let $\hat{E}^{i}_{p_{m_{i}}}$ be the vertical direction at $p_{m_{i}}$ in
$\mathcal{U}_{c_{0}}$, so that

$D\Phi_{c_{0}}(\hat{E}^{i}_{p_{m_{i}}})=E^{gv}_{\Phi_{c_{0}}(p_{m_{i}})}.$

Then by the assumption on the critical depth of $c_{m_{i}-m_{i-1}+1}$, the
direction $\hat{E}^{i-1}_{p_{m_{i}+1}}$ is sufficiently stable-aligned in
$\mathcal{U}_{c_{1}}$. Hence, by Lemma 6.13, $p_{m_{i}-l_{i}}$ is
$l_{i}$-times forward $(1,\delta)_{v}$-regular along
$\hat{E}^{i-1}_{p_{m_{i}-l_{i}}}$.

Let $N\geq l_{0}-m_{0}$ be an integer such that
$\operatorname{cd}_{t}(p_{-N})=\varnothing;$ and let
$\tilde{E}^{N}_{p_{-N}}\in\mathbb{P}^{2}_{p_{-N}}$ be a direction satisfying

$\measuredangle(\tilde{E}^{N}_{p_{-N}},\hat{E}^{v}_{p_{-N}})>k$

for some uniform constant $k>0$. We claim that $p_{0}$ is $N$-times backward
$({\bar{L}},\delta)_{h}$-regular along $\tilde{E}^{N}_{p_{0}}$.

Denote

$\hat{E}^{-1}_{p_{0}}:=\hat{E}^{v}_{p_{0}}\hskip 14.22636pt\text{and}\hskip
14.22636ptk_{0}:=N-(l_{0}-m_{0});$

and

$k_{i}:=(j_{i-1}-m_{i-1})-(l_{i}-m_{i})>0\hskip 14.22636pt\text{for}\hskip
14.22636pt1\leq i\leq I.$

Suppose that for $0\leq i\leq I$, the point $p_{m_{i}-l_{i}}$ is $k_{i}$-times
backward $({\bar{L}},\delta)_{h}$-regular along
$\tilde{E}^{N}_{p_{m_{i}-l_{i}}}$, and

$\measuredangle(\tilde{E}^{N}_{p_{m_{i}-l_{i}}},\hat{E}^{i-1}_{p_{m_{i}-l_{i}}})>{\bar{L}}^{-1}.$

For $i=0$, this was proved in the previous case.

Observe that

$p_{m_{i-1}-n}\in\mathcal{T}_{c_{-n}}({t})\hskip 14.22636pt\text{for}\hskip
14.22636ptn<j_{i-1};$

and

$\operatorname{cd}_{t}(c_{-m_{i-1}+m_{i}-n})=\varnothing\hskip
14.22636pt\text{for}\hskip 14.22636ptn\in\\{-1,j_{i},l_{i}\\}.$

Hence, $\tilde{E}^{i-1}_{p_{m_{i}+1}}$ is sufficiently stable-aligned in
$\mathcal{U}_{c_{1}}$, and $p_{m_{i}-l_{i}}$ is $l_{i}$-times forward
$(1,\delta)_{v}$-regular along $\tilde{E}^{i-1}_{p_{m_{i}+1}}$. By Lemma 9.20,
$p_{m_{i}-j_{i}}$ is $(k_{i}+l_{i}-j_{i})$-times backward
$(1,\delta)_{h}$-regular along $\tilde{E}^{N}_{p_{m_{i}-j_{i}}}$, and

$\measuredangle(\tilde{E}^{N}_{p_{m_{i}-j_{i}}},\hat{E}^{i}_{p_{m_{i}-j_{i}}})>k.$

Since

$\operatorname{cd}_{t}(c_{-m_{i}+m_{i+1}-l_{i+1}})=\varnothing,$

it follows from the previous case that $p_{m_{i+1}-l_{i+1}}$ is
$k_{i+1}$-times backward $({\bar{L}},\delta)_{h}$-regular along
$\tilde{E}^{N}_{p_{m_{i+1}-l_{i+1}}}$, and

$\measuredangle(\tilde{E}^{N}_{p_{m_{i+1}-l_{i+1}}},\hat{E}^{i}_{p_{m_{i+1}-l_{i+1}}})>{\bar{L}}^{-1}.$

By induction, we see that $p_{m_{I}-l_{j}}$ is $(N-(j_{I}-m_{I}))$-times
backward $(1,\delta)_{h}$-regular along $\tilde{E}^{N}_{p_{m_{I}-j_{I}}}$, and

$\measuredangle(\tilde{E}^{N}_{p_{m_{I}-j_{I}}},\hat{E}^{I}_{p_{m_{I}-j_{I}}})>k.$

Since $\operatorname{cd}_{t}(c_{-m_{I}})=\varnothing,$ by applying the
previous case again, we conclude that $p_{0}$ is $N$-times backward
$({\bar{L}},\delta)_{h}$-regular along $\tilde{E}^{N}_{p_{m_{I}-j_{I}}}$.

Lastly, suppose that $\operatorname{cd}_{t}(p_{0})=\infty.$ Let $D\geq 0$ be
an integer such that $p_{0}\in\mathcal{T}_{c_{-D}}({t})$. Note that
$d_{{\hat{\mathcal{T}}}_{-}({t})}(c_{-D})\neq\infty.$ Moreover, since
$\operatorname{vd}_{t}(p_{0})=\varnothing,$ we may assume by Proposition 3.9
that $\operatorname{vd}_{t}(c_{-D})=\varnothing.$

Let $N>0$ be an integer such that $-(D+N)<0$ is the slow tunnel escape moment
for $p_{D}\in\mathcal{T}_{c_{0}}$. Observe that

$\operatorname{dist}(p_{D},c_{0})<\lambda^{-\eta
D}\operatorname{dist}(p_{0},c_{-D})<\lambda^{(1+{\bar{\delta}})\delta D}$

for $D$ sufficiently large by Lemma 6.8. Consequently,

$N>(1-{\bar{\delta}})\delta_{1}^{-1}\delta D>{\bar{\delta}}_{1}^{-1}D.$

Let $\hat{E}^{v}_{c_{0}}\in\mathbb{P}^{2}_{c_{0}}$ be the vertical direction
at $c_{0}$ in $\mathcal{U}_{c_{0}}$, so that

$D\Phi_{c_{0}}(\hat{E}^{v}_{c_{0}})=E^{gv}_{0}.$

Let $\tilde{E}^{N}_{c_{-(D+N)}}\in\mathbb{P}^{2}_{c_{-(D+N)}}$ be a direction
satisfying

$\measuredangle(\tilde{E}^{N}_{c_{-(D+N)}},\hat{E}^{v}_{c_{-(D+N)}})>k.$

By the previous two cases, we see that $c_{-D}$ is $N$-times backward
$({\bar{L}},\delta)_{h}$-regular along $\tilde{E}^{N}_{c_{-D}}$. By
Proposition 3.9, there exists $\tilde{E}^{N}_{p_{0}}\in\mathbb{P}^{2}_{p_{0}}$
such that $p_{0}$ is $N$-times backward $({\bar{L}},\delta)_{h}$-regular along
$\tilde{E}^{N}_{p_{0}}$.

We proved that for $N$ arbitrarily large, there exists
$\tilde{E}^{N}_{p_{0}}\in\mathbb{P}^{2}_{p_{0}}$ such that $p_{0}$ is
$N$-times backward $({\bar{L}},\delta)_{h}$-regular along
$\tilde{E}^{N}_{p_{0}}$. By Proposition 3.12, we conclude that as
$N\to\infty$, the direction $\tilde{E}^{N}_{p_{0}}$ converges exponentially
fast to $E^{c}_{p_{0}}$, along which $p_{0}$ is infinitely backward
$({\bar{L}},\delta)_{h}$-regular. ∎

###### Proof of Theorem 9.8 c).

For ${t}\in(0,{L_{1}}^{-1}]$, let $L:=1/\underline{{t}}$. Let
$\mathbf{p}\in\Lambda$ be a point satisfying

$\operatorname{vd}_{t}(\mathbf{p})=\operatorname{cd}_{t}(\mathbf{p})=\varnothing.$

By Theorem 9.8 a), $\mathbf{p}$ is forward $(L,\delta)_{v}$-regular along
$\hat{E}^{v}_{\mathbf{p}}$. We show that $\mathbf{p}$ is also backward
$({\bar{L}},\delta)_{v}$-regular along $\hat{E}^{v}_{\mathbf{p}}$.

Let $-m<0$ be the first moment in the backward orbit of $\mathbf{p}$ such that

$p_{0}:=F^{-m}(\mathbf{p})\in\mathcal{T}_{c_{0}}({t}).$

Write $z_{0}=(x_{0},y_{0}):=\Phi_{c_{0}}(p_{0}).$ Let $-l<-j<0$ be the slow
and fast tunnel escape moment of $p_{0}$ respectively, and let $n>0$ be the
crescent escape moment of $p_{1}$. Recall that

$l>\underline{\delta}^{-1}\log_{\lambda}|x_{0}|>\underline{\delta}^{-1}\log_{\lambda}L\hskip
14.22636pt\text{and}\hskip
14.22636ptj,n>\delta^{-1}\log_{\lambda}|x_{0}|>-k\delta^{-1}\log_{\lambda}L$

for some uniform constant $k>0$. Observe that
$\operatorname{cd}_{t}(p_{i})=\varnothing$ for $1\leq i\leq m.$ Moreover,

$\operatorname{cd}_{t}(p_{i})=\operatorname{vd}_{t}(p_{i})=\varnothing\hskip
14.22636pt\text{for}\hskip 14.22636pti\in\\{-j,-l\\}.$

By Theorem 9.8 a) and Lemma 9.18, $p_{m}$ is $m$-times backward
$(L,\delta)_{v}$-regular along $\hat{E}^{v}_{p_{m}}$. Arguing as in the proof
of Theorem 9.8 b), we see that $p_{0}$ is $l$-times backward
$({\bar{L}},\delta)_{v}$-regular along $\hat{E}^{v}_{p_{0}}$.

Let $m_{1}>1$ be a Pliss moment in the forward orbit of $c_{1}$ so that

$Kn\leq m_{1}<(1+{\bar{\varepsilon}})Kn,$

where $K\geq 1$ is a sufficiently large uniform constant. If $n<m<m_{1}$, then
arguing as in the proof of Lemma 9.12, we see that $\hat{E}^{v}_{p_{m}}$ is
sufficiently vertical in $\mathcal{U}_{c_{m}}$. Since $c_{m}$ is $m$-times
backward $(\lambda^{-{\bar{\varepsilon}}m},\varepsilon)_{v}$-regular along
$E^{v}_{c_{m}}$, we have

$\|DF^{-m}|_{\hat{E}^{v}_{p_{m}}}\|>\lambda^{{\bar{\varepsilon}}m}\lambda^{-(1-\varepsilon)m}>\lambda^{-(1-{\bar{\varepsilon}})m}.$

If $m>m_{1}$, then again, $\hat{E}^{v}_{p_{m_{1}}}$ is sufficiently vertical
in $\mathcal{U}_{c_{m_{1}}}$, and we have

$\|DF^{-m_{1}}|_{\hat{E}^{v}_{p_{m_{1}}}}\|>\lambda^{-(1-{\bar{\varepsilon}})m_{1}}>\lambda^{-(\delta-{\bar{\varepsilon}})Kn}\lambda^{-(1-\delta)m_{1}}>{\bar{L}}\lambda^{-(1-\delta)m_{1}}.$

By Proposition 6.12, $\hat{E}^{v}_{p_{1}}$ is sufficiently stable-aligned in
$\mathcal{U}_{c_{1}}$, and hence, $\hat{E}^{v}_{p_{-j}}$ is sufficiently
vertical in $\mathcal{U}_{c_{-j}}$. Thus,

$\|DF^{-(l-j)}|_{\hat{E}^{v}_{p_{-j}}}\|>\lambda^{-(1-{\bar{\varepsilon}})(l-j)}>\lambda^{-(\delta-{\bar{\varepsilon}})(l-j)}\lambda^{-(1-\delta)(l-j)}>L^{\underline{\delta}^{-1}}\lambda^{-(1-\delta)(l-j)}.$

Concatenating the above estimates, we conclude that $\mathbf{p}=p_{m}$ is
$(m+l)$-times backward $({\bar{L}},\delta)_{v}$-regular along
$\hat{E}^{v}_{\mathbf{p}}$. Moreover, since

$\|D_{\mathbf{p}}F^{-(m+l)}|_{\hat{E}^{v}_{\mathbf{p}}}\|>\lambda^{-(1-\delta)(m+l)},$

we see that if $p_{-(m+l)}$ is $M$-times backward
$({\bar{L}},\delta)_{v}$-regular along $\hat{E}^{v}_{p_{-(m+l)}}$, then
$\mathbf{p}$ is $(m+l+M)$-times backward $({\bar{L}},\delta)_{v}$-regular
along $\hat{E}^{v}_{\mathbf{p}}$. The result follows by induction. ∎

###  9.4. Irregular points

######  Proposition 9.21.

Let $\Delta_{c_{0}}^{c/fc}$ and $\Delta_{c_{1}}^{v}$ be the critical/full
critical dust of $c_{0}$ and the valuable dust of $c_{1}$ respectively. For
$K\geq 1$, let $\Lambda_{K}^{+/-/P}$ be the set of all infinitely
forward/backward/Pesin $(K,\delta)$-regular points in $\Lambda$. Then we have

$\Lambda\setminus\left(\bigcup_{K\geq
1}\Lambda_{K}^{+}\right)\subset\Delta_{c_{0}}^{c}\hskip
14.22636pt\text{and}\hskip 14.22636pt\Lambda\setminus\left(\bigcup_{K\geq
1}\Lambda_{K}^{-}\right)\subset\Delta^{v}_{c_{1}}.$

Moreover,

$\Lambda\setminus\left(\bigcup_{K\geq
1}\Lambda_{K}^{P}\right)\subset\Delta_{c_{0}}^{fc}\cup\\{c_{m}\\}_{m\in\mathbb{Z}}.$

###### Proof.

We prove the last containment, as the other two can be proved by a similar and
simpler argument.

If

$p_{0}\notin\Delta_{c_{0}}^{fc}:=\Delta_{c_{0}}^{c}\cup\Delta^{v}_{c_{1}},$

then $\operatorname{cd}(p_{0})\neq\infty$ and
$\operatorname{vd}(p_{0})\neq\infty.$ By replacing $p_{0}$ by
$p_{-\operatorname{vd}(p_{0})-1}$ if necessary, we can assume that
$\operatorname{vd}(p_{0})=\varnothing.$ If
$d:=\operatorname{cd}(p_{0})\neq\varnothing,$ let $-j<0$ be the tunnel escape
moment for $p_{d}$. Then by Proposition 6.9, we see that for $q_{0}:=p_{d-j},$
we have $\operatorname{cd}(q_{0})=\operatorname{vd}(q_{0})=\varnothing.$ Thus,
$p_{0}$ is Pesin regular by Theorem 9.8 c). ∎

######  Proposition 9.22.

If $F$ is regularly unicritical on $\Lambda$, then there exists a unique
regular critical orbit $\\{c_{m}\\}_{m\in\mathbb{Z}}\subset\Lambda$.

###### Proof.

Let $q_{1}\in\Lambda\setminus\\{c_{1}\\}$ be a generalized critical value.
Since $q_{1}$ cannot be Pesin regular, $q_{1}\in\Delta_{c_{0}}^{fc}$ by
Proposition 9.21. For concreteness, assume $q_{1}\in\Delta_{c_{0}}^{c}$. Then
$q_{1}\in\mathcal{T}_{c_{-N}}$ for $N$ arbitrarily large. Let $-l<0$ be the
slow tunnel escape moment for $q_{1+N}$. Then we have
$l>{\bar{\delta}}_{1}^{-1}N.$ Since $c_{-N}$ is $(l-N,N)$-times
$(\lambda^{-{\bar{\varepsilon}}N},\varepsilon)_{v}$-regular along
$\hat{E}^{v}_{c_{-N}}$, we contradict Proposition 5.4. ∎

##  10\. Total Ordering Inside a Unicritical Cover

Let $F$ be the mildly dissipative, infinitely renormalizable and regularity
unicritical diffeomorphism considered in Section 9.

In Theorem 9.8, choose

$t=\mathbf{t}:={L_{1}}^{-1}\hskip 14.22636pt\text{and}\hskip
14.22636ptL:=1/\underline{\mathbf{t}}.$

Lastly, denote

$\mathcal{T}_{c_{m}}:=\mathcal{T}_{c_{m}}(\mathbf{t})\hskip
14.22636pt\text{for}\hskip 14.22636ptm\in\mathbb{Z},$

and

$\operatorname{cd}:=\operatorname{cd}_{\mathbf{t}}\hskip
14.22636pt\text{and}\hskip
14.22636pt\operatorname{vd}:=\operatorname{vd}_{\mathbf{t}},$

where the $t$-critical depth $\operatorname{cd}_{t}$ and the $t$-valuable
depth $\operatorname{vd}_{t}$ are defined as in Subsection 6.1.

###  10.1. Outer boundaries of tunnels and crescents

Let

$\delta^{\prime}:=(1-{\bar{\delta}})\delta.$

For $n\geq 0$ and $i\in\\{0,1\\}$, denote

$U_{c_{i}}^{\delta^{\prime}n}(a):=\mathbb{B}(a\lambda^{\delta^{\prime}n}l_{c_{i}},l_{c_{i}}):=(-a\lambda^{\delta^{\prime}n}l_{c_{i}},a\lambda^{\delta^{\prime}n}l_{c_{i}})\times(-l_{c_{i}},l_{c_{i}})\subset\mathbb{R}^{2},$

where $l_{c_{i}}>0$ is the regular radius at $c_{i}$. For $a>0$, let

$T_{c_{0}}(a):=\\{(x,y)\in
U_{c_{0}}\;|\;|x|<a\mathbf{t}\;\text{and}\;|y|<|x|^{1/\underline{\delta}}\\}$

and

$\bar{\partial}T_{c_{0}}(a):=\\{(x,y)\in
U_{c_{0}}\;|\;|x|<a\mathbf{t}\;\text{and}\;|y|=|x|^{1/\underline{\delta}}\\}.$

Let $j_{n}\geq 0$ be the largest integer such that $j_{n}\leq n$ and $-j_{n}$
is a Pliss moment in the backward orbit of $c_{0}$. Define

$\mathcal{T}_{c_{-n}}(a):=F^{-n}\circ\Phi_{c_{0}}^{-1}(T_{c_{0}}(a)\cap
U_{c_{0}}^{\delta^{\prime}j_{n}}(a))$ (10.1)

and

$\bar{\partial}\mathcal{T}_{c_{-n}}(a):=F^{-n}\circ\Phi_{c_{0}}^{-1}(\bar{\partial}T_{c_{0}}(a)\cap
U_{c_{0}}^{\delta^{\prime}j_{n}}(a)).$

Note that $\mathcal{T}_{c_{-n}}=\mathcal{T}_{c_{-n}}(1)$. The outer boundary
of $\mathcal{T}_{c_{-n}}$ is defined as

$\bar{\partial}\mathcal{T}_{c_{-n}}:=\bar{\partial}\mathcal{T}_{c_{-n}}(2).$

A Jordan arc $\Gamma:(0,1)\to\mathcal{T}_{c_{-n}}(2)$ is said to be vertically
proper in $\mathcal{T}_{c_{-n}}$ if $\Gamma$ extends continuously to $[0,1]$,
and $\Gamma(0)$ and $\Gamma(1)$ are in distinct components of
$\bar{\partial}\mathcal{T}_{c_{-n}}\setminus\\{c_{-n}\\}$. More generally, a
Jordan arc $\Gamma$ in $\Omega$ is said to be vertically proper in
$\mathcal{T}_{c_{-n}}$ if $\Gamma\cap\mathcal{T}_{c_{-n}}(2)$ is.

For $a>0$, let

$T_{c_{1}}(a):=F_{c_{0}}(T_{c_{0}}(a))\hskip 14.22636pt\text{and}\hskip
14.22636pt\bar{\partial}T_{c_{1}}(a):=F_{c_{0}}(\bar{\partial}T_{c_{0}}(a)),$

where $F_{c_{0}}:=\Phi_{c_{1}}\circ F\circ\Phi_{c_{0}}^{-1}.$ For $n\geq 1$,
let $i_{n}\geq 1$ be the largest integer such that $i_{n}\leq n$ and $i_{n}$
is a Pliss moment in the forward orbit of $c_{1}$. Define

$\mathcal{T}_{c_{n}}(a):=F^{n-1}\circ\Phi_{c_{1}}^{-1}(T_{c_{1}}(a)\cap
U_{c_{1}}^{\delta^{\prime}i_{n}}(a))$ (10.2)

and

$\bar{\partial}\mathcal{T}_{c_{n}}(a):=F^{n-1}\circ\Phi_{c_{1}}^{-1}(\bar{\partial}T_{c_{1}}(a)\cap
U_{c_{1}}^{\delta^{\prime}i_{n}}(a)).$

Note that $\mathcal{T}_{c_{n}}=\mathcal{T}_{c_{n}}(1)$. The outer boundary of
$\mathcal{T}_{c_{n}}$ is defined as

$\bar{\partial}\mathcal{T}_{c_{n}}:=\bar{\partial}\mathcal{T}_{c_{n}}(2).$

Let

$\mathbb{L}^{h,+}_{0}:=\\{(x,0)\;|\;0<x<l_{c_{1}}\\}$

and

$W^{h,+}_{c_{n}}:=F^{n-1}\circ\Phi_{c_{1}}^{-1}(\mathbb{L}^{h,+}_{0}\cap
U_{c_{1}}^{\delta^{\prime}i_{n}}(2)).$

Denote

$\tilde{\mathcal{U}}_{c_{1}}^{i_{n}}(2):=\mathcal{U}_{c_{1}}^{\delta^{\prime}i_{n}}(2)\cap\mathcal{U}_{c_{1}}^{0}(2\mathbf{t}^{2}).$

The crescent gap $\mathcal{T}^{g}_{c_{n}}$ at $c_{n}$ is defined to be the
connected component of
$F^{n-1}(\tilde{\mathcal{U}}_{c_{1}}^{\delta^{\prime}i_{n}}(2))\setminus\mathcal{T}_{c_{n}}(2)$
that contain $W^{h,+}_{c_{n}}$. The filled valuable crescent at $c_{n}$ is
defined as

$\mathcal{T}^{f}_{c_{n}}:=\mathcal{T}_{c_{n}}^{2}\cup\mathcal{T}^{g}_{c_{n}}.$

A Jordan arc $\Gamma:(0,1)\to\mathcal{T}^{f}_{c_{n}}$ is said to be vertically
proper in $\mathcal{T}_{c_{n}}$ if $\Gamma$ is a concatenation of three curves
$\Gamma_{1}:(0,a_{1})\to\mathcal{T}_{c_{n}}(2)$;
$\Gamma_{2}:[a_{1},a_{2}]\to\mathcal{T}^{g}_{c_{n}}$ and
$\Gamma_{3}:(a_{2},1)\to\mathcal{T}_{c_{n}}(2)$ such that $\Gamma_{1}$ and
$\Gamma_{3}$ extends continuously to $[0,a_{1}]$ and $[a_{2},1]$ respectively;
and for $i\in\\{1,2,3\\}$, the endpoints of $\Gamma_{i}$ are in distinct
components of $\bar{\partial}\mathcal{T}_{c_{n}}\setminus\\{c_{n}\\}$. More
generally, a Jordan arc $\Gamma$ in $\Omega$ is said to be vertically proper
in $\mathcal{T}_{c_{n}}$ if $\Gamma\cap\mathcal{T}_{c_{n}}(2)$ is.

######  Lemma 10.1.

If a $C^{1}$-curve $\Gamma$ is sufficiently stable-aligned in
$\mathcal{U}_{c_{1}}$, and

$\Gamma\cap\mathcal{T}_{c_{1}}^{f}\neq\varnothing,$

then $\Gamma$ is vertically proper in $\mathcal{T}_{c_{1}}$.

Let

$\hat{\delta}:=(1+{\bar{\delta}})\delta.$

For $a>0$, let

$\hat{T}_{c_{0}}(a):=\\{(x,y)\in
U_{c_{0}}\;|\;|x|<a\mathbf{t}\;\text{and}\;|y|<|x|^{1/\hat{\delta}}\\}\supset
T_{c_{0}}(a).$

For $n\geq 0$, define ${\hat{\mathcal{T}}}_{c_{-n}}(a)$ by replacing
$T_{c_{0}}(a)$ in (10.1) with $\hat{T}_{c_{0}}(a)$. Let
$\hat{\mathcal{U}}_{c_{-n}}$ be a regular neighborhood at $c_{-n}$ associated
with a linearization of its $(\infty,n)$-time orbit with vertical direction
$\hat{E}^{v}_{c_{-n}}$. This means, in particular, that whenever $-n$ is a
Pliss moment in the backward orbit of $c_{0}$, we have
$\operatorname{radius}(\hat{\mathcal{U}}_{c_{-n}})\asymp 1.$ The set

${\hat{\mathcal{T}}}_{c_{-n}}:={\hat{\mathcal{T}}}_{c_{-n}}(1)\cap\hat{\mathcal{U}}_{c_{-n}}\supset\mathcal{T}_{c_{-n}}$
(10.3)

is referred to as a thickened critical tunnel at $c_{-n}$. A thickened
valuable crescent ${\hat{\mathcal{T}}}_{c_{1+n}}$ at $c_{1+n}$ is defined by
replacing $T_{c_{1}}(a)$ in (10.2) with

$\hat{T}_{c_{1}}(a):=\Phi_{c_{1}}\circ
F\circ\Phi_{c_{0}}^{-1}(\hat{T}_{c_{0}}(a)),$

and letting

${\hat{\mathcal{T}}}_{c_{1+n}}:={\hat{\mathcal{T}}}_{c_{1+n}}(1)\supset\mathcal{T}_{c_{1+n}}.$
(10.4)

######  Lemma 10.2.

Let $0\leq m<n$. If $c_{-n}\in\mathcal{T}_{c_{-m}}$, then
${\hat{\mathcal{T}}}_{c_{-n}}\subset{\hat{\mathcal{T}}}_{c_{-m}}.$ If
$c_{n}\in\mathcal{T}_{c_{-m}}$, then
$\mathcal{T}_{c_{n}}\subset{\hat{\mathcal{T}}}_{c_{-m}}.$

###### Proof.

By Proposition 6.9, we have $n>{\bar{\varepsilon}}^{-1}m.$ Write
$(x_{0},y_{0}):=\Phi_{c_{0}}(c_{-n+m}).$ Since

$\operatorname{radius}(\mathcal{U}_{c_{-n+m}})>\lambda^{{\bar{\varepsilon}}(n-m)},$

we have

$n>{\bar{\varepsilon}}^{-1}\log_{\lambda}|x_{0}|$

by Proposition 5.9. Hence, by Proposition 6.8

$\operatorname{diam}({\hat{\mathcal{T}}}_{c_{-n+m}})<\lambda^{\delta
n}<|x_{0}|^{1/{\bar{\varepsilon}}}.$

Since

$\mathbb{D}_{c_{-n+m}}(k|x_{0}|^{1/\hat{\delta}})\subset{\hat{\mathcal{T}}}_{c_{0}},$

for some uniform constant $k>0$, the first claim follows.

The proof of the second claim is similar, and will be omitted. ∎

###  10.2. Disjoint vertical manifolds

Let $p\in\Lambda\setminus\\{c_{-n}\\}_{n=0}^{\infty}$. If
$p\notin\Delta^{c}_{c_{0}}$ (where $\Delta^{c}_{c_{0}}$ is the critical dust
of $c_{0}$), then by Proposition 9.21, $p$ has a well-defined local strong-
stable manifold $W^{ss}_{\operatorname{loc}}(p)\subset\mathcal{U}_{p}$. If
$\operatorname{cd}(p)=\varnothing$, define

$\hat{W}^{v}(p):=W^{ss}_{\operatorname{loc}}(p).$

If $\operatorname{cd}(p)=d\in[0,\infty)$, define

$\hat{W}^{v}(p):=W^{ss}_{\operatorname{loc}}(p)\cap{\hat{\mathcal{T}}}_{c_{-d}}.$

For $n\geq 0$, we have $c_{-n}\notin\Delta^{c}_{c_{0}}$ by Proposition 6.9.
Let $\hat{\mathcal{U}}_{c_{-n}}$ be a regular neighborhood at $c_{-n}$
associated with a linearization of its $(\infty,n)$-time orbit with vertical
direction $\hat{E}^{v}_{c_{-n}}$. For $q\in\mathcal{T}_{c_{-n}}$, define
$W^{v}_{c_{-n}}(q)$ to be the local vertical manifold at $q$ in
$\hat{\mathcal{U}}_{p_{0}}$. Moreover, let

$W^{v}_{c_{-n}}(\mathcal{T}_{c_{-n}}):=\bigcup_{q\in\mathcal{T}_{c_{-n}}}W^{v}_{c_{-n}}(q)$

(which is not a curve but an open set foliated by vertical curves). If
$\operatorname{cd}(c_{-n})=\varnothing$, define

$\hat{W}^{v}(q):=W^{v}_{c_{-n}}(q)\hskip 14.22636pt\text{for}\hskip
14.22636ptq\in\mathcal{T}_{c_{-n}}$

and

$\hat{W}^{v}(\mathcal{T}_{c_{-n}}):=W^{v}_{c_{-n}}(\mathcal{T}_{c_{-n}}).$

If $\operatorname{cd}(c_{-n})=d\in[0,\infty)$, define

$\hat{W}^{v}(q):=W^{v}_{c_{-n}}(q)\cap{\hat{\mathcal{T}}}_{c_{-d}}\hskip
14.22636pt\text{for}\hskip 14.22636ptq\in\mathcal{T}_{c_{-n}}$

and

$\hat{W}^{v}(\mathcal{T}_{c_{-n}}):=W^{v}_{c_{-n}}(\mathcal{T}_{c_{-n}})\cap{\hat{\mathcal{T}}}_{c_{-d}}.$

######  Proposition 10.3.

Let $p,q\in\Lambda\setminus\Delta^{c}_{c_{0}}$. If $p$ and $q$ are not
contained in the same connected component of $\Lambda$, then
$\hat{W}^{v}(p)\cap\hat{W}^{v}(q)=\varnothing.$

###### Proof.

The result is obvious unless either $p$ or $q$ is equal to $c_{-n}$ for some
$n\geq 0$. For concreteness, suppose it is the former. Let
$\hat{\mathcal{U}}_{c_{-n}}$ be a regular neighborhood at $c_{-n}$ associated
with a linearization of its $(\infty,n)$-time orbit with vertical direction
$\hat{E}^{v}_{c_{-n}}$. Then
$F^{n}(\hat{\mathcal{U}}_{c_{-n}}^{{\bar{\delta}}n})\subset\mathcal{U}_{c_{0}}.$

If

$\hat{W}^{v}(q)\cap\hat{W}^{v}(c_{-n})\neq\varnothing,$

then we have

$F^{n}(\hat{W}^{v}(q))\subset\hat{W}^{v}(F^{n}(q))\subset{\hat{\mathcal{T}}}_{c_{0}}.$

This contradicts the fact that
${\hat{\mathcal{T}}}_{c_{0}}\cap\hat{W}^{v}(c_{0})=\varnothing.$ ∎

###  10.3. Unicritical cover

For $p\in\Lambda^{P}_{L,\delta}$, let $\bar{\mathcal{B}}_{p}$ be an enclosed
regular box at $p$ with outer boundary $\bar{\partial}\bar{\mathcal{B}}_{p}$
(see Section 7). The collection

$\mathcal{C}:=\\{\bar{\mathcal{B}}_{p}\\}_{p\in\Lambda^{P}_{L,\delta}}\cup\\{\mathcal{T}_{c_{m}}\\}_{m\in\mathbb{Z}}$

is a cover of $\Lambda$ by (pinched) neighborhoods. We refer to $\mathcal{C}$
as a unicritical cover of $\Lambda$.

######  Proposition 10.4.

By slightly adjusting the sizes of the elements in $\mathcal{C}$ if necessary,
the following properties hold for all $Q\in\mathcal{C}$.

  1. i)

If for some $p\in\Lambda\setminus\Delta^{c}_{c_{0}}$, we have
$\hat{W}^{v}(p)\cap Q\neq\varnothing$, then $p\in Q$.

  2. ii)

If for some $n\geq 0$, we have $\hat{W}^{v}(\mathcal{T}_{c_{-n}})\cap
Q\neq\varnothing$, then $\mathcal{T}_{c_{-n}}\subset Q$.

###### Proof.

Let $\\{-j_{i}\\}_{i=0}^{\infty}$ be the decreasing sequence of Pliss moments
in the backward orbit of $c_{0}$. For $i\geq 0$, let
$a_{i}\in(\lambda^{\underline{\delta}},1]$. Consider a sequence
$\\{{\check{\mathcal{T}}}_{c_{-n}}\\}_{n=0}^{\infty}$ of slightly smaller
critical tunnels that satisfy

$\mathcal{T}_{c_{-j_{i}}}(a_{i})\subset{\check{\mathcal{T}}}_{c_{-j_{i}}}\subset\mathcal{T}_{c_{-j_{i}}}$

and

${\check{\mathcal{T}}}_{c_{-j_{i}-l}}=F^{-l}({\check{\mathcal{T}}}_{c_{-j_{i}}})\hskip
14.22636pt\text{for}\hskip 14.22636pt0\leq l<j_{i+1}-j_{i}.$

Observe that all previous results about critical tunnels still hold if we
replace $\\{\mathcal{T}_{c_{-n}}\\}_{n=0}^{\infty}$ by
$\\{{\check{\mathcal{T}}}_{c_{-n}}\\}_{n=0}^{\infty}$.

Let $p_{1}\in\Lambda\cap\mathcal{T}_{c_{1}}$ with
$\operatorname{cd}(p_{1})=\varnothing$. Write
$(x_{0},y_{0}):=\Phi_{c_{0}}(p_{0}).$ By Theorem 9.8 and Proposition 6.12,
$\hat{W}^{v}(p_{1})$ is sufficiently stable-aligned in $\mathcal{U}_{c_{1}}$.
If $p_{0}=c_{-m}$ for some $m>0$, then by Lemma 9.5, we have
$\operatorname{diam}(\mathcal{T}_{c_{-m}})<|x_{0}|^{1/{\bar{\varepsilon}}}.$

If $p_{0}=c_{-m}$, let $q_{0}\in\mathcal{T}_{c_{-m}}$ and
$\tilde{W}^{v}(q_{0}):=W^{v}_{c_{-m}}(q_{0}).$ Otherwise, let $q_{0}=p_{0}$.
Let $\tilde{E}^{v}_{q_{0}}\in\mathbb{P}^{2}_{q_{0}}$ be the direction tangent
to $\tilde{W}^{v}(q_{0})$. Then $\tilde{E}^{v}_{q_{1}}$ and
$\tilde{W}^{v}(q_{1})$ are sufficiently stable-aligned in
$\mathcal{U}_{c_{1}}$. Let $E^{v/h}_{q_{0}}$ be the vertical/horizontal
direction at $q_{0}$ in $\mathcal{U}_{c_{0}}$. Then

$\measuredangle(\tilde{E}^{v}_{q_{0}},E^{h}_{q_{0}})>k|x_{0}|$

for some uniform constant $k>0$.

Suppose $q_{-j_{i}}\notin\mathcal{T}_{c_{-j_{i}}}$ and
$\tilde{W}^{v}(q_{-j_{i}})\cap\mathcal{T}_{c_{-j_{i}}}\neq\varnothing.$ By
Proposition 6.13, $q_{-j_{i}}$ is $j_{i}$-times forward
$(1,\delta)_{v}$-regular along $\tilde{E}^{v}_{q_{-j_{i}}}$. By Proposition
3.11, we see that $\tilde{W}^{v}(q_{-j_{i}})$ is
$\lambda^{(1-{\bar{\delta}})j_{i}}$-vertical in $\mathcal{U}_{c_{-j_{i}}}$. On
the other hand,

$|x_{-j_{i}}|>\lambda^{{\bar{\varepsilon}}j_{i}}|x_{0}|>\lambda^{(1-{\bar{\delta}})\delta
j_{i}}.$

It follows that

$\tilde{W}^{v}_{q_{-j_{i}}}\cap\mathcal{T}_{c_{-j_{i}}}(1-\lambda^{(1-{\bar{\delta}})j_{i}})=\varnothing.$

Thus, we may choose ${\check{\mathcal{T}}}_{c_{-j_{i}}}$ such that

$\mathcal{T}_{c_{-j_{i}}}\supset{\check{\mathcal{T}}}_{c_{-j_{i}}}\supset\mathcal{T}_{c_{-j_{i}}}(1-\lambda^{(1-{\bar{\delta}})j_{i}})\supset\mathcal{T}_{c_{-j_{i}}}(\lambda^{\underline{\delta}}),$

and
$\tilde{W}^{v}_{q_{-j_{i}}}\cap{\check{\mathcal{T}}}_{c_{-j_{i}}}=\varnothing.$

∎

For $Q\in\mathcal{C}$, define its critical depth
$\operatorname{cd}(Q)\in\mathbb{N}\cup\\{\varnothing,0\\}$ as follows.

  1. i)

If $Q=\mathcal{T}_{c_{-m}}$ for some $m\geq 0$, then
$\operatorname{cd}(Q):=m$.

  2. ii)

If $Q=\mathcal{T}_{c_{n}}$ for some $n\geq 1$, then
$\operatorname{cd}(Q):=\operatorname{cd}(c_{n})$.

  3. iii)

Otherwise, $\operatorname{cd}(Q):=\varnothing$.

Let $X_{Q}$ be the set of all points $p\in\Lambda\cap Q$ such that
$\operatorname{cd}(p)=\operatorname{cd}(Q)$.

######  Proposition 10.5.

Let $Q\in\mathcal{C}$. Then for all $p\in X_{Q}$, the curve $\hat{W}^{v}(p)$
is vertically proper in $Q$. If $c_{-n}\in X_{Q}$ for some $n\geq 0$, then for
all $q\in\mathcal{T}_{c_{-n}}\subset Q$, the curve $W^{v}_{c_{-n}}(q)$ is
vertically proper in $Q$.

###### Proof.

If $Q$ is a regular box, then the result is clearly true. If
$Q=\mathcal{T}_{c_{1}}$, then the result follows immediately from Theorem 9.8,
Proposition 6.12 and Lemma 10.1. If $Q=\mathcal{T}_{c_{-m}}$ or some $m\geq 0$
or $Q=\mathcal{T}_{c_{k}}$ for some $k>1$, then the result follows immediately
from the previous case and Lemma 10.2. ∎

###  10.4. Total ordering with blowups

Let $(X,\leq_{X})$ be a totally ordered set; $U$ a subset of $X$; and
$\\{(Y_{u},\leq_{Y_{u}})\\}_{u\in U}$ a collection of totally ordered sets.
Define $\mathcal{Y}:U\to\\{Y_{u}\\}_{u\in U}$ by $\mathcal{Y}(u):=Y_{u}.$ The
blow up of $X$ by $\mathcal{Y}$ is defined as the set

$X[\mathcal{Y}]:=(X\setminus U)\sqcup\bigsqcup_{u\in U}Y_{u}$

with the total order $\leq_{X[\mathcal{Y}]}$ given by the following
properties. Let $x,y\in X[\mathcal{Y}]$. Then $x\leq_{X[\mathcal{Y}]}y$ if one
of the following holds:

  1. i)

$x,y\in X\setminus U$ and $x\leq_{X}y$;

  2. ii)

$x\in X\setminus U$ and $y\in Y_{u}$ with $x\leq_{X}u$;

  3. iii)

$x\in Y_{u}$ and $y\in X\setminus U$ with $u\leq_{X}y$;

  4. iv)

$x\in Y_{u}$ and $y\in Y_{v}$ with $u\leq_{X}v$; or

  5. v)

$x,y\in Y_{u}$ and $x\leq_{Y_{u}}y$.

For $n\geq 0$, consider sets $Y^{n}$ and $U^{n}$, and a set-valued function
$\mathcal{Y}^{n}$ obtained inductively as follows. First, let $U^{0}$ be a
subset of $Y^{0}:=X$. Suppose that $Y^{n-1}$ and $U^{n-1}$ are given. Let
$\\{(Y^{n}_{u},\leq_{Y^{n}_{u}})\\}_{u\in U^{n-1}}$ be a collection of totally
ordered sets, and define

$Y^{n}:=\bigsqcup_{u\in U^{n-1}}Y^{n}_{u}.$

Next, let $\mathcal{Y}^{n}:U^{n-1}\to\\{Y^{n}_{u}\\}_{u\in U^{n-1}}$ be given
by $\mathcal{Y}^{n}(u):=Y^{n}_{u}.$ Lastly, let $U^{n}$ be a subset of
$Y^{n}$.

Let $X^{\wedge 0}:=X$, and for $n\geq 0$, inductively define an $n$th blow-up
of $X$ by

$X^{\wedge n}:=X^{\wedge(n-1)}[\mathcal{Y}^{n}]$

with the total order $\leq_{X^{\wedge n}}$ defined as above. Next, we define
an infinite blow-up of $X$ by

$X^{\wedge\infty}:=\\{x\in X^{\wedge n}\setminus U^{n}\;\text{for some}\;n\geq
0\\}$

with the total order $\leq_{X^{\wedge\infty}}$ given by the following
property. Let $x,y\in X^{\wedge\infty}$. Then $x\leq_{X^{\wedge\infty}}y$ if
$x\leq_{X^{\wedge}n}y$, where $n$ is any integer such that $x,y\in
X^{n}\setminus U^{n}$.

A sequence $u_{\infty}:=\\{u_{n}\\}_{n=0}^{\infty}$ such that $u_{0}\in
U^{0}\cap X$ and $u_{n}\in U^{n}\cap Y^{n}_{u_{n-1}}$ for $n>0$ is called a
limit point of $X^{\wedge\infty}$. Define the completion $X^{\wedge\omega}$ of
$X^{\wedge\infty}$ as the disjoint union of $X^{\wedge\infty}$ and the set of
all its limits points, together with the total order $\leq_{X^{\wedge\omega}}$
given by the following property. Let $x,y\in X^{\wedge\omega}$. Then
$x\leq_{X^{\wedge\omega}}y$ if

  1. i)

$x,y\in X^{\wedge\infty}$ and $x\leq_{X^{\wedge\infty}}y$;

  2. ii)

$x\in X^{\wedge n}\setminus U^{n}$ and $y=\\{u_{i}\\}_{i=0}^{\infty}$ with
$x\leq_{X^{\wedge n}}u_{n}$;

  3. iii)

$x=\\{u_{i}\\}_{i=0}^{\infty}$ and $y\in X^{\wedge n}\setminus U^{n}$ with
$u_{n}\leq_{X^{\wedge n}}y$; or

  4. iv)

$x=\\{u_{i}\\}_{i=0}^{\infty}$ and $y=\\{v_{i}\\}_{i=0}^{\infty}$, and
$u_{n}\leq_{X^{\wedge n}}v_{n}$, where $n$ is the smallest integer such that
$u_{n}\neq v_{n}$.

###  10.5. Total order on $\Lambda_{Q}$

Let $Q\in\mathcal{C}$, and let $\mathcal{G}$ be a collection of disjoint
Jordan curves that are vertically proper in $Q$. Then we define a total order
on $\mathcal{G}$ as follows. If $Q=\mathcal{T}_{c_{n}}$ for some $n\geq 1$,
let $\hat{Q}:=\mathcal{T}_{c_{n}}^{f}$. Otherwise, let $\hat{Q}:=Q$. Note that
for any $\Gamma_{0}\in\mathcal{G}$, the complement
$\hat{Q}\setminus\Gamma_{0}$ is the disjoint union of exactly two connected
components, say $\hat{Q}^{+}_{\Gamma_{0}}$ and $\hat{Q}^{-}_{\Gamma_{0}}$.
Thus, there is a unique (strict) total order $<_{\mathcal{G}}$ on
$\mathcal{G}$ such that for any $\Gamma\in\mathcal{G}$ distinct from
$\Gamma_{0}$, we have $\Gamma<_{\mathcal{G}}\Gamma_{0}$ if
$\Gamma\cap\hat{Q}\subset\hat{Q}^{-}_{\Gamma_{0}}$ and
$\Gamma_{0}<_{\mathcal{G}}\Gamma$ if
$\Gamma\cap\hat{Q}\subset\hat{Q}^{+}_{\Gamma_{0}}$.

For $Q\in\mathcal{C}$, let $Y_{Q}$ be the set consisting of all elements $x$
that satisfy one of the following properties.

  1. i)

$x=p\in X_{Q}$ such that $p\neq c_{-n}$ for all $n\geq 0$. In this case,
define $\Gamma^{v}_{x}:=W^{v}_{p}(p)$.

  2. ii)

$x=\mathcal{T}_{c_{-n}}$ for some $n\geq 0$ such that $c_{-n}\in X_{Q}$. In
this case, define $\Gamma^{v}_{x}:=W^{v}_{c_{-n}}(c_{-n})$.

Let $U_{Q}$ be the set of all elements in $Y_{Q}$ of type $ii)$. By
Propositions 10.3 and 10.5, the collection
$\Gamma^{v}_{Q}:=\\{\Gamma^{v}_{x}\\}_{x\in Y_{Q}}$ consists of pairwise
disjoint curves that are vertically proper in $Q$. As discussed above, this
implies that $\Gamma^{v}_{Q}$ has a total order $\leq_{\Gamma^{v}_{Q}}$ (non-
strict, since we may have $\Gamma^{v}_{x}=\Gamma^{v}_{y}$ with $x\neq y$),
which induces a total order $\leq_{Y_{Q}}$ on $Y_{Q}$.

Let $Y^{0}:=Y_{Q}$ and $U^{0}:=U_{Q}$. Proceeding inductively, suppose that
$Y^{n-1}$ and $U^{n-1}$ are given. For $x\in U^{n-1}$, define
$Y^{n}_{x}:=Y_{x}$; $U^{n}_{x}:=U_{x}$; and

$Y^{n}:=\bigsqcup_{x\in U^{n-1}}Y^{n}_{x}\hskip 14.22636pt\text{and}\hskip
14.22636ptU^{n}:=\bigsqcup_{x\in U^{n-1}}U^{n}_{x}.$

By the construction given in Subsection 10.4, we obtain the completion
$Y_{Q}^{\omega}$ of an infinite blow-up of $Y_{Q}$ with a total ordering
$<_{Y_{Q}^{\wedge\omega}}$.

######  Theorem 10.6.

For $Q\in\mathcal{C}$, let $\Lambda_{Q}:=\Lambda\cap Q$. Then

$\Lambda_{Q}=Y_{Q}^{\wedge\omega},$

and the set of limit points in $Y_{Q}^{\wedge\omega}$ is given by
$\Lambda_{Q}\cap\Delta^{c}_{c_{0}}$. Thus, $\Lambda_{Q}$ has a total order
$\leq_{Q}$ such that for $x,y\in\Lambda_{Q}$, we have $x\leq_{Q}y$ and
$y\leq_{Q}x$ if and only if $y\in W^{s}(x)$.

###### Proof.

The result follows immediately from Proposition 6.8. ∎

Let $Q\in\mathcal{C}$. A point $x\in X\subset\Lambda_{Q}$ is called an
extremal point of $X$ if either

  * •

$x\leq_{Q}y$ for all $y\in X$; or

  * •

$x\geq_{Q}y$ for all $y\in X$.

The following three propositions record observations that follow immediately
from the construction of the total order on $\Lambda_{Q}$.

######  Proposition 10.7.

Let $Q,Q^{\prime}\in\mathcal{C}$, and let
$X:=\Lambda_{Q}\cap\Lambda_{Q^{\prime}}$. Then $\leq_{Q}$ and
$\leq_{{Q^{\prime}}}$ induces the same total order on $X$ (up to orientation).
That is, either

  * •

$x\leq_{{Q^{\prime}}}y$ for all $x,y\in X$ with $x\leq_{Q}y$; or

  * •

$x\geq_{{Q^{\prime}}}y$ for all $x,y\in X$ with $x\leq_{Q}y$.

######  Proposition 10.8.

Let $Q,Q^{\prime}\in\mathcal{C}$ with $Q\neq\mathcal{T}_{c_{0}}$, and let
$X\subset\Lambda_{Q}$ be a set satisfying $F(X)\subset\Lambda_{Q^{\prime}}$.
Then $F$ preserves the total order on $X$ (up to orientation). That is, either

  * •

$F(x)\leq_{{Q^{\prime}}}F(y)$ for all $x,y\in X$ such that $x\leq_{Q}y$; or

  * •

$F(x)\geq_{{Q^{\prime}}}F(y)$ for all $x,y\in X$ such that $x\leq_{Q}y$.

######  Proposition 10.9.

Let $X\subset\Lambda_{\mathcal{T}_{c_{0}}}$. Then the following holds.

  1. i)

If $c_{0}\leq_{{\mathcal{T}_{c_{0}}}}x_{0}$ for all $x\in X$, then for all
$y,z\in X$ such that $y\leq_{{\mathcal{T}_{c_{0}}}}z$, we have
$F(y)\leq_{{\mathcal{T}_{c_{1}}}}F(z)$.

  2. ii)

If $x\leq_{{\mathcal{T}_{c_{0}}}}c_{0}$ for all $x\in X$, then for all $y,z\in
X$ such that $y\leq_{{\mathcal{T}_{c_{0}}}}z$, we have
$F(z)\leq_{{\mathcal{T}_{c_{1}}}}F(y)$.

Let $Q\in\mathcal{C}$. We define a larger covering element $\hat{Q}\supset Q$
as follows. If $Q=\mathcal{T}_{c_{m}}$ for some $m\in\mathbb{Z}$, then let
$\hat{Q}:=Q$. Otherwise, $Q=\bar{\mathcal{B}}_{p}$ for some
$p\in\Lambda^{P}_{L,\delta}$. Recall that $\bar{\mathcal{B}}_{p}$ is a
quadrilateral bound by four nearly linear arcs for which the opposite sides
are nearly parallel. Moreover, the horizontal width and the vertical height of
$\bar{\mathcal{B}}_{p}$ are approximately $1/L_{1}$ and $1/L_{2}$
respectively, where

${\bar{L}}<L_{1}\hskip 14.22636pt\text{and}\hskip
14.22636pt\overline{L_{1}}<L_{2}.$

Let $\hat{Q}$ be a strictly larger encased regular box at $p$ whose horizontal
width and the vertical height are approximately $1/L_{1}^{\prime}$ and
$1/L_{2}^{\prime}$ respectively, where

${\bar{L}}<L_{1}^{\prime}<\underline{L_{1}}\hskip 14.22636pt\text{and}\hskip
14.22636pt\overline{L_{1}^{\prime}}<L_{2}^{\prime}<\underline{L_{2}}.$

######  Proposition 10.10.

For $0\leq i\leq n$, let
$Q_{i}\in\mathcal{C}\setminus\\{\mathcal{T}_{c_{0}}\\}$. Suppose there exist
$x_{0},y_{0},z_{0}\in\Lambda_{Q_{0}}$ such that

$x_{0}\leq_{{Q_{0}}}y_{0}\leq_{{Q_{0}}}z_{0},$

and $x_{i},z_{i}\in\Lambda_{Q_{i}}$ for all $1\leq i\leq n$. Then
$y_{n}\in\Lambda_{\hat{Q}_{n}}$ and

$x_{n}\leq_{{\hat{Q}_{n}}}y_{n}\leq_{{\hat{Q}_{n}}}z_{n}.$

###### Proof.

The result clearly holds if $Q_{i}=\mathcal{T}_{c_{m}}$ for some
$m\in\mathbb{Z}$. Suppose the $Q_{i}$’s are encased regular boxes. Let
$\Gamma(x_{n})$ and $\Gamma(z_{n})$ be two proper sufficiently vertical arcs
in $\hat{Q}_{n}$ containing $x_{n}$ and $z_{n}$ respectively. By Proposition
3.19, the preimages $\Gamma(x_{0})$ and $\Gamma(z_{0})$ of $\Gamma(x_{n})$ and
$\Gamma(z_{n})$ under $F^{-n}$ are proper and sufficiently vertical in
$Q_{0}$. Let $Q_{0}(x_{0},z_{0})$ be the component of
$Q_{0}\setminus(\Gamma(x_{0})\cup\Gamma(z_{0}))$ containing $y_{0}$. Applying
Proposition 3.19 again, we see that the image of $Q_{0}(x_{0},z_{0})$ under
$F^{i}$ is still nearly a parallelogram whose height increased by no more than
a factor of ${\bar{L}}$. It follows that
$F^{i}(Q_{0}(x_{0},z_{0}))\subset\hat{Q}_{i}$. The claim about the order of
$x_{n},y_{n}$ and $z_{n}$ follows immediately from Proposition 10.8. ∎

##  11\. Connected Components of the Limit Set

Let $F$ be the mildly dissipative, infinitely renormalizable and regularity
unicritical diffeomorphism considered in Section 10.

The forward/backward/Pesin $\delta$-regularity of $p\in\Lambda$ is the
smallest value $\mathcal{L}^{+/-/P}_{\delta}(p)\in[1,\infty]$ such that $p$ is
forward/backward/Pesin $(\mathcal{L}^{+/-/P}_{\delta}(p),\delta)$-regular. We
say that $p$ is forward/backward/Pesin $\delta$-regular if
$\mathcal{L}^{+/-/P}_{\delta}(p)<\infty$ (i.e.
$p\in\Lambda^{+/-/P}_{\delta}$).

Let $Z$ be the connected component of $\Lambda$. Observe that either
$Z\cap\\{c_{m}\\}_{m\in\mathbb{Z}}=\varnothing,$ or there exists a unique
integer $m\in\mathbb{Z}$ such that $c_{m}\in Z$. In the latter case, we say
that $Z$ is a critical component if $m\leq 0$, or a valuable component if
$m\geq 1$.

We say that $z\in Z$ is an extremal point of $Z$ if for any unicritical
covering element $Q\in\mathcal{C}$ containing $z$, the point $z$ is an
extremal point for $(Z_{Q},\leq_{Q})$, where $Z_{Q}$ is a connected component
of $Z\cap Q$ containing $z$. Denote

$\mathring{Z}:=Z\setminus\\{\text{extremal points of }Z\\}.$

######  Lemma 11.1.

Let $Z$ be a connected component of $\Lambda$. Then
$\mathcal{L}^{+/-/P}_{\delta}(p)$ varies lower semi-continuously on $p\in Z$.
Moreover, $\mathcal{L}^{+/-}_{\delta}(p)<\infty$ for $p\in\mathring{Z}$.
Lastly, if $\mathcal{L}^{P}_{\delta}(p)=\infty$ for some $p\in\mathring{Z}$,
then $p=c_{m}$ for some $m\in\mathbb{Z}$.

###### Proof.

For $p\in Z$, let $t_{\pm}(p)\in[0,\mathbf{t}]$ with $\mathbf{t}:=L_{1}^{-1}$
be the largest value such that

$p\notin\bigcup_{i=1}^{\infty}\mathcal{T}_{\pm i}(t_{\pm}(p)).$

Clearly, $t_{\pm}(p)$ depends continuously on $p$. The first claim now follows
from Theorem 9.8.

Let $n\geq 0$ such that $c_{-n}\notin Z$ and
$Z\cap\mathcal{T}_{-n}\neq\varnothing.$ We claim that $\mathcal{T}_{c_{-n}}$
must contain an extremal point of $Z$. Otherwise, the two extremal points of
$Z\cap\Lambda_{\mathcal{T}_{c_{-n}}}$ coincide with the two extremal points of
$\mathcal{T}_{c_{-n}}$. However, this is not possible unless $c_{-n}\in Z$.
Consequently, if there exists $p\in Z\cap\Delta^{c}_{c_{0}},$ then $p$ must be
an extremal point in $Z$. By a similar argument, we conclude that if there
exists $p\in Z\cap\Delta^{v}_{c_{1}},$ then $p$ must be an extremal point in
$Z$. The second and third claims now follow from Theorem 9.8. ∎

######  Theorem 11.2 (Non-trivial components are center curves).

Every non-trivial component $Z_{0}$ of $\Lambda$ is a $C^{r}$-curve. Moreover,
$Z_{0}^{\prime}(t)\in E^{c}_{Z_{0}(t)}$ for all $t\in(0,|Z_{0}|)$.

The proof of Theorem 11.2 is based on the following three lemmas.

######  Lemma 11.3.

There exists a uniform constant $\kappa>0$ such that the following holds. Let
$Z_{0}$ be a non-valuable connected component of $\Lambda$, and let $p_{0}\in
Z_{0}$ be a Pesin $\delta$-regular point. If there exists $y_{0}\in(Z_{0}\cap
W^{ss}(p_{0}))\setminus\\{p_{0}\\},$ then there exists a Pliss moment $-N<0$
in the backward orbit of $p_{0}$ such that

$\operatorname{dist}(F^{-N}(y_{0}),F^{-N}(p_{0}))>\kappa.$

###### Proof.

Since $\operatorname{cd}(p_{0}),\operatorname{vd}(p_{0})<\infty$, by replacing
$p_{0}\in Z_{0}$ with $p_{-i}\in Z_{-i}$ for some $i\geq 0$, we may assume
that $\operatorname{cd}(p_{0})=\operatorname{vd}(p_{0})=\varnothing.$

If $p_{-i}\in\Lambda^{P}_{L,\delta}$ for $0\leq i\leq I,$ then we have

$\operatorname{dist}(y_{-I},p_{-I})>{\bar{L}}^{-1}\lambda^{-(1-{\bar{\delta}})I}\operatorname{dist}(y_{0},p_{0}).$

Let $-m<0$ be a moment such that $\operatorname{cd}(p_{-m})=0,$ and let $-j<0$
be the tunnel escape moment for $p_{-m}$. By Proposition 6.13, Proposition
6.13 and Theorem 9.8, $p_{-m+1}$ and $p_{-m-j}$ are forward
$(1,\delta)$-regular. If $y_{-m+1}\in W^{ss}_{\operatorname{loc}}(p_{-m+1})$,
then

$\operatorname{dist}(y_{-m+1},p_{-m+1})>k\lambda^{-(1-{\bar{\delta}})m}\operatorname{dist}(y_{0},p_{0})$

for some uniform constant $k>0$. Observe that in this case, $y_{-m+1}$ is
contained in the same connected component of
$W^{ss}_{\operatorname{loc}}(p_{-m+1})\cap\mathcal{T}_{c_{1}}$ as $p_{-m+1}$.
Otherwise, $Z_{-m+1}$ must contain $c_{1}$, which contradicts the assumption
that $Z_{0}$ is not valuable. It follows that $y_{-m-i}\in
W^{ss}_{\operatorname{loc}}(p_{-m-i})$ for $0\leq i\leq j,$ and

$\operatorname{dist}(y_{-m-j},p_{-m-j})>k\lambda^{-(1-{\bar{\delta}})(m+j)}\operatorname{dist}(y_{0},p_{0}).$

Proceeding inductively, we obtain the desired result. ∎

######  Lemma 11.4.

Let $Z_{0}$ be a connected component of $\Lambda$, and let $p_{0}\in Z_{0}$ be
a Pesin $\delta$-regular point. Then there does not exist a sequence

$\\{y^{i}_{0}\\}_{i=1}^{\infty}\subset(W^{ss}(p_{0})\cap
Z_{0})\setminus\\{p_{0}\\}$

such that $\operatorname{dist}(y^{i}_{0},p_{0})\searrow 0$ as $i\to\infty$.

###### Proof.

Suppose towards a contradiction that such a sequence exists. By Lemma 11.3,
there exists a subsequence $\\{y^{i_{k}}_{0}\\}_{k=1}^{\infty}$ and a
decreasing sequence $\\{-N_{k}\\}_{k=1}^{\infty}$ of Pliss moments in the
backward orbit of $p_{0}$ such that

$\operatorname{dist}(y^{i}_{-N_{k}},p_{-N_{k}})>\kappa.$

Consider the unicritical cover $\mathcal{C}$ of $\Lambda$ defined in
Subsection 10.3. We may assume that the vertical height of each element
$Q\in\mathcal{C}$ are bounded by $\kappa/2$. This means that if
$p\in\Lambda_{Q}$ and $q\in W^{ss}_{\operatorname{loc}}(p)$, and
$\operatorname{dist}(p,q)>\kappa/2,$ then $q$ cannot be contained in
$\Lambda_{Q}$.

By compactness of $\Lambda$, we may choose a finite subcover
$\mathcal{C}_{0}\subset\mathcal{C}$. Observe that if a connected component $Z$
of $\Lambda$ intersects multiple elements from $\mathcal{C}_{0}$, then $Z$
must contain least one of the two extremal points for each of these elements.
Hence, we see that there are at most finitely many components that can
intersect multiple covering elements. However, by the previous observations,
the component $Z_{-N_{k}}$ for all $k\geq 1$ cannot be contained in a single
covering element. This is a contradiction. ∎

######  Lemma 11.5.

Let $Z_{0}$ be a non-valuable connected component of $\Lambda$. Then there
exists a constant $\kappa_{Z_{0}}>0$ such that if $p_{0}\in Z_{0}$ is Pesin
$\delta$-regular, and $y_{0}\in(W^{ss}(p_{0})\cap Z_{0})\setminus\\{p_{0}\\},$
then $\operatorname{dist}(y_{0},p_{0})>\kappa_{Z_{0}}.$

###### Proof.

Consider the finite cover $\mathcal{C}_{0}$ of $\Lambda$ given in the proof of
Lemma 11.4. Let $N=N(Z_{0})\in\mathbb{N}$ be the smallest integer such that
for all $n\geq N$, the component $Z_{-n}$ does not intersect multiple covering
elements in $\mathcal{C}_{0}$. Then it follows from Lemma 11.3 that if
$p_{0}\in Z_{0}$ is Pesin $\delta$-regular, then

$W^{ss}_{\operatorname{loc}}(p_{-n})\cap Z_{-n}=\\{p_{-n}\\}\hskip
14.22636pt\text{for}\hskip 14.22636ptn\geq N.$

Suppose there exists

$y_{0}\in(W^{ss}_{\operatorname{loc}}(p_{0})\cap Z_{0})\setminus\\{p_{0}\\}.$

Then by Lemma 11.3, there exists $0\leq m<N$ such that
$\operatorname{dist}(y_{-m},p_{-m})>\kappa.$ Consequently,

$\operatorname{dist}(y_{0},p_{0})>K\lambda^{(1+\eta)N}\kappa,$

for some uniform constant $K\geq 1$. ∎

###### Proof of Theorem 11.2.

It suffices to prove the result for a non-valuable connected component $Z$.

Let $p_{0}\in Z_{0}$ be Pesin $(K,\delta)$-regular, and let
$\\{\Phi_{p_{m}}:\mathcal{U}_{p_{m}}\to U_{p_{m}}\\}_{m\in\mathbb{Z}}$ be a
linearization of the full orbit of $p_{0}$. By Proposition 11.1 and Lemma
11.5, if $W_{0}\subset\mathcal{U}^{0}_{p_{0}}$ is a sufficiently small
neighborhood of $p_{0}$, and $\Gamma_{0}$ is the the connected component of
$Z_{0}\cap W_{0}$ containing $p_{0}$, then
$\gamma_{0}:=\Phi_{p_{0}}(\Gamma_{0})$ is the horizontal graph of a
$C^{0}$-map $g_{0}:\mathbb{I}(l)\to\mathbb{R}$ for some sufficiently small
$l>0$.

We claim that $g_{0}$ is differentiable at $0$, and that
$g_{0}^{\prime}(0)=0$. Let $\\{-n_{i}\\}_{i=1}^{\infty}$ be the decreasing
sequence of Pliss moments in the backward orbit of $p_{0}$. As in the proof of
Lemma 11.5, there exists $N=N(Z_{0})\geq 0$ such that for all
$q_{0}\in\Gamma_{0}$, we have

$W^{ss}_{\operatorname{loc}}(q_{-n})\cap Z_{-n}=\\{q_{-n}\\}\hskip
14.22636pt\text{for}\hskip 14.22636ptn\geq N.$

For $n_{i}>N$, let

$\\{\Phi_{p_{-n_{i}+m}}^{i}:\mathcal{U}_{p_{-n_{i}+m}}^{i}\to
U_{p_{-n_{i}+m}}^{i}\\}_{m\in\mathbb{Z}}$

be a linearization of the full orbit of $p_{-n_{i}}$, and let
$\Gamma_{-n_{i}}^{i}$ be the connected component of
$\Gamma_{-n_{i}}\cap\mathcal{U}_{p_{-n_{i}}}^{{\bar{\delta}}n_{i}}$ containing
$p_{-n_{i}}$. Observe that

$\gamma^{i}_{0}:=\Phi_{p_{0}}\circ F^{n_{i}}(\Gamma_{-n_{i}}^{i})$

is a neighborhood of $0$ in $\gamma_{0}$. Thus, by Theorem 3.6 and Proposition
3.9, we see that

$|g_{0}(x)|<{\bar{K}}\lambda^{(1-{\bar{\delta}})n_{i}}\hskip
14.22636pt\text{for}\hskip
14.22636pt|x|<{\bar{K}}^{-1}\lambda^{{\bar{\delta}}n_{i}}.$

By Proposition 4.4 and Proposition 4.3, we conclude that

$\gamma_{0}\subset\\{(x,y)\in
U_{p_{0}}\;|\;|x|<l\;\text{and}\;|y|<|x|^{1/{\bar{\delta}}}\\}.$

The result follows. ∎

##  12\. Recurrence of the Critical Orbit

Let $F$ be the mildly dissipative, infinitely renormalizable and regularity
unicritical diffeomorphism considered in Section 10.

For any $C^{r}$-curve $\gamma\subset\Omega$, let
$\phi_{\gamma}:(0,|\gamma|)\to\gamma$ be its parameterization by arclength.
Recall that $\|\gamma\|_{C^{r}}:=\|\phi_{\gamma}\|_{C^{r}}.$ Let
$F_{\gamma}:=\phi_{f(\gamma)}^{-1}\circ F|_{\gamma}\circ\phi_{\gamma}.$ The
distortion of $F$ along $\gamma$ is defined as

$\operatorname{Dis}(F,\gamma):=\sup_{t,s\in(0,|\gamma|)}\frac{|F_{\gamma}^{\prime}(t)|}{|F_{\gamma}^{\prime}(s)|}.$

The following statement is a straightforward generalization of 1D Denjoy
Lemma.

######  Proposition 12.1.

Let $\gamma_{0}\subset\Omega$ be a $C^{2}$-curve. Suppose for all $0\leq i<n$,
the curve $\gamma_{i}:=F^{i}(\gamma_{0}).$ is well-defined and
$\|\gamma_{i}\|_{C^{2}}=O(1)$. Then

$\log(\mathcal{D}(F^{n}|_{\gamma}))=O\left(\sum_{i=1}^{n}|F^{i}(\gamma)|\right).$

######  Theorem 12.2.

The critical value $c_{1}$ is recurrent.

###### Proof.

The proof is by contradiction. Suppose there exists $R>0$ such that

$\mathbb{D}_{c_{1}}(R)\cap\\{c_{m}\\}_{m\in\mathbb{Z}}=\\{c_{1}\\}.$ (12.1)

######  Lemma 12.3.

Suppose (12.1) holds. Then the connected component $Z_{c_{1}}$ of $\Lambda$
containing the critical value $c_{1}$ is a non-trivial arc.

###### Proof.

Suppose not. Then $\mathcal{D}^{n}_{1}\subset\mathbb{D}_{c_{1}}(R)$ for $n$
sufficiently large. However, $c_{1+lR_{n}}\in\mathcal{D}^{n}_{1}$ for all
$l\in\mathbb{Z}$. ∎

By Theorem 11.2, $Z_{c_{1}}$ is a $C^{1}$ Jordan arc. Consider the
parameterization of $Z_{c_{1}}$ by its arclength. Denote
$\mathring{Z}_{c_{1}}:=Z_{c_{1}}(0,|Z_{c_{1}}|).$ Choose ${t}\leq R$ in
Theorem 9.8, and let $L:=L(t)\geq 1$ be the regularity factor. Note that we
have $\Lambda\cap\mathcal{T}_{c_{m}}\subset\mathring{Z}_{c_{m}}.$
Consequently, the points $z_{m}^{0}:=Z_{c_{m}}(0)$ and
$z_{m}^{1}:=Z_{c_{m}}(|Z_{c_{m}}|)$ are Pesin $(L,\delta)$-regular. Moreover,
if $m\neq l$, then $\mathcal{T}_{c_{m}}\cap\mathcal{T}_{c_{l}}=\varnothing.$

######  Lemma 12.4.

Suppose (12.1) holds. If $\mathcal{T}_{c_{-n}}\subset
Q:=\bar{\mathcal{B}}_{p}$ for some $n\geq 0$ and $p\in\Lambda^{P}_{L,\delta}$,
then either

  * •

$z_{-n}^{0}<_{Q}p<_{Q}z_{-n}^{1}$ for all $p\in\mathcal{T}_{c_{-n}}$; or

  * •

$z_{-n}^{1}<_{Q}p<_{Q}z_{-n}^{0}$ for all $p\in\mathcal{T}_{c_{-n}}$.

For $p\in\Lambda^{P}_{L,\delta}$, recall the definition of
$\bar{\mathcal{B}}_{p}(a)$ with $a>0$ (see Section 7). Rather than setting
$\bar{\mathcal{B}}_{p}:=\bar{\mathcal{B}}_{p}(1/2)$, we modify the
construction of an encased regular box in the following way. By Lemma 12.4,
$p$ is accumulated by other Pesin $(L,\delta)$-regular points. If

$q<_{{\bar{\mathcal{B}}_{p}(1/2)}}p\hskip 14.22636pt\text{or}\hskip
14.22636ptp<_{{\bar{\mathcal{B}}_{p}(1/2)}}q\hskip 14.22636pt\text{for
all}\hskip
14.22636ptq\in\Lambda^{P}_{L,\delta}\cap\bar{\mathcal{B}}_{p}(1/2),$

then define $\bar{\mathcal{B}}_{p}$ as the quadrilateral bounded by
$\bar{\partial}\bar{\mathcal{B}}_{p}$, $W^{ss}_{\operatorname{loc}}(p)$ and
$W^{ss}_{\operatorname{loc}}(q)$. Otherwise, choose
$q_{-},q_{+}\in\Lambda^{P}_{L,\delta}\cap\bar{\mathcal{B}}_{p}(1/2)$ such that

$q_{-}<_{{\bar{\mathcal{B}}_{p}(1/2)}}p<_{{\bar{\mathcal{B}}_{p}(1/2)}}q_{+},$

and define $\bar{\mathcal{B}}_{p}$ as the quadrilateral bounded by
$\bar{\partial}\bar{\mathcal{B}}_{p}$, $W^{ss}_{\operatorname{loc}}(q_{-})$
and $W^{ss}_{\operatorname{loc}}(q_{+})$. Then

$\bar{\mathcal{B}}_{p}\subset\bar{\mathcal{B}}_{p}(1/2)\subset\bar{\mathcal{B}}_{p}(1).$

Choose a finite unicritical cover of $\Lambda$

$\mathcal{C}_{0}=\\{\bar{\mathcal{B}}_{p}\\}_{p\in\mathcal{P}}\cup\\{\mathcal{T}_{c_{m}}\\}_{|m|\leq
M}$

where $\mathcal{P}\subset\Lambda^{P}_{L,\delta}$ is a finite set, and
$0<M<\infty$.

######  Lemma 12.5.

Suppose (12.1) holds. There exists $I\geq 1$ such that the following holds.
For all $i\geq I$, there exists $p_{i}\in\mathcal{P}$ and $N_{i}\geq 1$ such
that for all $n\geq N_{i}$, we have
$Z_{c_{i}}\subset\mathcal{D}^{n}_{i}\subset\bar{\mathcal{B}}_{p_{i}}.$
Moreover, $\mathcal{D}^{n}_{i}$ does not contain an extremal point of
$\bar{\mathcal{B}}_{p_{i}}$.

###### Proof.

Clearly, $Z_{c_{i}}$ cannot intersect $\mathcal{T}_{c_{m}}$ for $|m|\leq M$ if
$i>M$. If $Z_{c_{i}}$ intersects multiple covering elements in
$\mathcal{C}_{0}$, then it must contain at least one of the extremal points in
each element. Since this can happen only finitely many times, we obtain the
desired integer $I\geq 1$. Since $\mathcal{D}^{n}_{i}$ converges in Hausdorff
topology to $Z_{c_{i}}$, the result follows. ∎

For $n\geq N_{I}$, let $I_{n}\geq I$ be the largest integer such that for
$I\leq i\leq I_{n}$, we have

$\Lambda^{n}_{i}:=\mathcal{D}^{n}_{i}\cap\Lambda\subset\bar{\mathcal{B}}_{p_{i}}(1).$

Denote the extremal points in $\Lambda^{n}_{i}$ with respect to
$<_{{\bar{\mathcal{B}}_{p_{i}}(1)}}$ by $a^{n}_{i}$ and $b^{n}_{i}$. It
follows from Lemma 12.4 that $a^{n}_{i}$ and $b^{n}_{i}$ are Pesin
$(L,\delta)$-regular. Let $\hat{\mathcal{D}}^{n}_{i}$ be the quadrilateral
bounded by $\bar{\partial}\bar{\mathcal{B}}_{p_{i}}$,
$W^{ss}_{\operatorname{loc}}(a^{n}_{i})$ and
$W^{ss}_{\operatorname{loc}}(b^{n}_{i})$.

######  Lemma 12.6.

Suppose (12.1) holds. Let $\Gamma_{I}$ be a sufficiently horizontal $C^{r}$
Jordan arc in $\bar{\mathcal{B}}_{p_{I}}$ with uniformly bounded curvature.
For $n\geq N_{I}$ and $I\leq i\leq I_{n}$, define

$\Gamma_{I}^{n}:=\Gamma_{I}\cap\hat{\mathcal{D}}^{n}_{I}\hskip
14.22636pt\text{and}\hskip 14.22636pt\Gamma_{i}^{n}:=F^{i-I}(\Gamma_{I}^{n}).$

Then $\Gamma_{i}^{n}$ is sufficiently horizontal in
$\bar{\mathcal{B}}_{p_{I}}(1)$, and has uniformly bounded curvature. Moreover,
$\displaystyle\sum_{i=I}^{I_{n}}|\Gamma_{i}^{n}|=O(1).$

###### Proof.

The first claim is an immediate consequence of Proposition 3.19.

Observe that
$\\{W^{ss}_{\operatorname{loc}}(a^{n}_{i}),W^{ss}_{\operatorname{loc}}(b^{n}_{i})\\}_{i=I}^{I_{n}}$
partitions $\\{\bar{\mathcal{B}}_{p}(1)\\}_{p\in\mathcal{P}}$ by nearly linear
vertical arcs. Since $\hat{\mathcal{D}}^{n}_{i}$ occupies each cell in this
partition at most once, and $\\{\bar{\mathcal{B}}_{p}(1)\\}_{p\in\mathcal{P}}$
clearly has uniformly bounded total horizontal length, the second claim
follows. ∎

For $n\geq N_{I}$ and $I\leq i\leq I_{n}$, let $\hat{Z}_{i}$ be the
quadrilateral bounded by $\bar{\partial}\bar{\mathcal{B}}_{p_{i}}(1)$,
$W^{ss}_{\operatorname{loc}}(z_{i}^{0})$ and
$W^{ss}_{\operatorname{loc}}(z_{i}^{1})$. Write

$\Gamma_{i}^{n}:=A_{i}^{n}\cup\hat{\Gamma}_{i}\cup B_{i}^{n},$

where $\hat{\Gamma}_{i}:=F^{i-I}(\Gamma_{I}\cap\hat{Z}_{I}),$ and $A_{i}^{n}$
and $B_{i}^{n}$ are the two components of
$\Gamma_{i}^{n}\setminus\hat{\Gamma}_{i}$.

Since $\hat{\mathcal{D}}^{n}_{I}$ converges to $\hat{Z}_{I}$ as $n\to\infty$,
we see that $|A_{I}^{n}|,|B_{I}^{n}|\to 0$ as $n\to\infty.$ On the other hand,
either $|A_{I_{n}}^{n}|$ or $|B_{I_{n}}^{n}|$ must be uniformly bounded below,
while $|\hat{\Gamma}_{I_{n}}|\to 0$ as $n\to\infty$. Applying Lemma 12.6 and
Proposition 12.1, we obtain a contradiction. Hence, (12.1) cannot hold, and
Theorem 12.2 follows. ∎

##  13\. Total Order on Renormalization Pieces

Let $F$ be the mildly dissipative, infinitely renormalizable and regularity
unicritical diffeomorphism considered in Section 10.

Recall that the renormalization limit set of $F$ is given by

$\Lambda:=\bigcap_{n=0}^{\infty}\bigcup_{i=0}^{R_{n}-1}F^{i}(D^{n}_{1}),$

where

$r_{n}:=R_{n+1}/R_{n}\geq 2,$

and $D^{n}_{1}$ is an $R_{n}$-periodic Jordan domain containing the critical
value $c_{1}$. Denote

$D^{n}_{i}:=F^{i-1}(D^{n}_{1})\hskip 14.22636pt\text{and}\hskip
14.22636pt\Lambda^{n}_{i}:=\Lambda\cap D^{n}_{i}\hskip
14.22636pt\text{for}\hskip 14.22636pt1\leq i\leq R_{n}.$

We refer to $\Lambda^{n}_{i}$ as a renormalization piece of depth $n$. By
Theorem 12.2, there exists $N_{1}\geq 0$ such that

$\mathcal{D}^{n}_{1}\subset\mathcal{U}_{c_{1}}\hskip
14.22636pt\text{and}\hskip
14.22636pt\Lambda^{n}_{1}\subset\mathcal{T}_{c_{1}}\hskip
14.22636pt\text{for}\hskip 14.22636ptn\geq N_{1}.$

######  Proposition 13.1.

Let $n\geq N_{1}$. If $\Lambda^{n}_{i}\subset\Lambda^{N_{1}}_{0}$ and
$\Lambda^{n}_{i}\neq\Lambda^{n}_{0}$, then either

  * •

$x<_{{\mathcal{T}_{c_{0}}}}c_{0}$ for all $x\in\Lambda^{n}_{i}$; or

  * •

$c_{0}<_{{\mathcal{T}_{c_{0}}}}x$ for all $x\in\Lambda^{n}_{i}$.

###### Proof.

For $m\geq n\geq N_{1}$, let $J(m,n)\subset\\{0,\ldots,R_{m}-1\\}$ be the set
of all indexes such that $\mathcal{D}^{m}_{j}\subset\mathcal{D}^{n}_{0}$ for
$j\in J(m,n).$ By Theorem 9.7, for $n\geq N_{1}$, there exists $M_{n}>n$ such
that for any $m\geq M_{n}$ and $j\in J(m,N_{1})\setminus J(m,n)$, we have
$\mathcal{D}^{m}_{j}\subset\mathcal{T}_{c_{0}}\setminus\mathcal{D}^{n}_{0}.$
Since $\Lambda^{n}_{i}$ is a union of $\mathcal{D}^{m}_{j}$’s, the result
follows. ∎

Consider a finite unicritical cover of $\Lambda$:

$\hat{\mathcal{C}}=\\{\bar{\mathcal{B}}_{p}\\}_{p\in\mathcal{P}}\cup\\{\mathcal{T}_{m}\\}_{|m|<\hat{M}},$

where $\mathcal{P}\subset\Lambda^{P}_{L,\delta}$ is a finite set, and
$0<\hat{M}<\infty$.

######  Proposition 13.2.

There exist at most a finite set $\mathcal{Z}$ of non-trivial connected
components of $\Lambda$ that cannot be contained in a single covering element
in $\mathcal{C}_{0}$. Moreover, let $Z\in\mathcal{Z}$. Then there exist
$Q_{1},\ldots,Q_{l}\in\mathcal{C}_{0}$ and

$0=:t_{0}<t_{1}<\ldots<t_{l}:=|Z|$

such that

$Z([t_{i-1},t_{i}])\subset Q_{i}\hskip 14.22636pt\text{for}\hskip
14.22636pt1\leq i\leq l.$

In this case, $Q_{i}$’s are all regular boxes, except possibly exactly one of
$Q_{0}$ and $Q_{k}$, which may be a critical tunnel or a valuable crescent.
Moreover,

$\operatorname{cd}(Z(t))=\operatorname{vd}(Z(t))=\varnothing\hskip
14.22636pt\text{for}\hskip 14.22636ptt\in[t_{1},t_{l-1}].$

######  Proposition 13.3.

There exists $\hat{N}\geq N_{1}$ such that for all $n\geq\hat{N}$ and $0\leq
i<R_{n}$, one of the following cases hold.

  1. i)

There exists $Q\in\mathcal{C}_{0}$ such that $\Lambda^{n}_{i}\subset Q$.

  2. ii)

The piece $\Lambda^{n}_{i}$ contains exactly one component $Z$ in
$\mathcal{Z}$. In this case, let $Q_{1},\ldots,Q_{l}\in\mathcal{C}_{0}$ be the
covering elements given in Proposition 13.2. Then

$\Lambda^{n}_{i}\setminus Z\subset Q_{1}\cup Q_{l}.$

###### Proof.

Let $\bar{\mathcal{B}}\in\mathcal{C}_{0}$ be a regular box. If the two
extremal components of $\bar{\mathcal{B}}$ are not contained $\mathcal{Z}$,
then $\Lambda_{\bar{\mathcal{B}}}$ is a compact subset of $\bar{\mathcal{B}}$.
Hence, there exists $K_{1}\geq 0$ such that any renormalization domain
$\mathcal{D}^{n}_{i}$ of depth $n\geq K_{1}$ that intersects
$\Lambda_{\bar{\mathcal{B}}}$ must be contained in $\bar{\mathcal{B}}$.

Let $\mathcal{T}_{c_{m}}\in\mathcal{C}_{0}$ be a critical tunnel or crescent.
By Theorem 12.2, there exists $K_{2}\geq 0$ such that
$c_{m}\in\Lambda^{K_{2}}_{m}\subset\Lambda_{\mathcal{T}_{c_{m}}}.$ Denote

$\tilde{m}:=m\;(\operatorname{mod}\;R_{K_{2}}).$

If the two extremal components of $\mathcal{T}_{c_{m}}$ are not contained
$\mathcal{Z}$, then
$\Lambda_{\mathcal{T}_{c_{m}}}\setminus\Lambda^{K_{2}}_{\tilde{m}}$ is a
compact subset of
$\mathcal{T}_{c_{m}}\setminus\mathcal{D}^{K_{2}}_{\tilde{m}}$. Hence, there
exists $K_{3}\geq K_{2}$ such that any renormalization domain
$\mathcal{D}^{n}_{i}$ of depth $n\geq K_{3}$ that intersects
$\Lambda_{\mathcal{T}_{c_{m}}}\setminus\Lambda^{K_{2}}_{\tilde{m}}$ must be
contained in $\mathcal{T}_{c_{m}}$.

The result follows by the finiteness of $\mathcal{C}_{0}$. ∎

######  Theorem 13.4.

For $n\geq\hat{N}$ and $1\leq i\leq R_{n}$, there exists a total order
$<_{\Lambda^{n}_{i}}$ on $\Lambda^{n}_{i}$ such that following holds. For
$Q\in\mathcal{C}_{0}$, either

  * •

$x<_{\Lambda^{n}_{i}}y$ for all $x,y\in\Lambda^{n}_{i}\cap Q$ with $x<_{Q}y$;
or

  * •

$y<_{\Lambda^{n}_{i}}x$ for all $x,y\in\Lambda^{n}_{i}\cap Q$ with $x<_{Q}y$.

Moreover, if $1\leq i<R_{n}$, then $F(x)<_{\Lambda^{n}_{i+1}}F(y)$ for all
$x,y\in\Lambda^{n}_{i}$ with $x<_{\Lambda^{n}_{i}}y$.

###### Proof.

Suppose $\Lambda^{n}_{i}$ contains a component $Z\in\mathcal{Z}$
(parameterized by its arclength $|Z|$). Let
$Q_{1},\ldots,Q_{l}\in\mathcal{C}_{0}$ be the covering elements given in
Proposition 13.2. Define a total order on $\Lambda^{n}_{i}$ by the following
property. For $j\in\\{1,l\\}$, let $p_{j},q_{j}\in Q_{j}\setminus Z$. Then

  1. i)

$p_{j}<_{\Lambda^{n}_{i}}q_{j}$ if $p_{j}<_{{Q_{j}}}q_{j}$;

  2. ii)

$p_{0},q_{0}<_{\Lambda^{n}_{i}}Z(0)$;

  3. iii)

$Z(t)<_{\Lambda^{n}_{i}}Z(s)$ if $0\leq t<s\leq|Z|$; and

  4. iv)

$Z(|Z|)<_{\Lambda^{n}_{i}}p_{l},q_{l}$.

The result now follows by Proposition 10.7, 10.8 and 10.9. ∎

Let $\hat{N}\leq m<n$ and $i\geq 0$. Observe that

$\Lambda^{m}_{i}=\bigsqcup_{k=0}^{R_{n}/R_{m}-1}\Lambda^{n}_{i+kR_{m}}.$

For $a,b\in\Lambda^{m}_{i}$ with $a\leq_{\Lambda^{m}_{i}}b$, denote

$[a,b]_{\Lambda^{m}_{i}}:=\\{p\in\Lambda^{m}_{i}\;|\;a\leq_{\Lambda^{m}_{i}}p\leq_{\Lambda^{m}_{i}}b\\}.$

This notation generalizes to $(\cdot,\cdot)_{\Lambda^{m}_{i}}$ in the obvious
way.

A combinatorial component $\mathcal{K}$ of $\Lambda^{n}_{i+kR_{m}}$ in
$\Lambda^{m}_{i}$ is a maximal subset of $\Lambda^{n}_{i+kR_{m}}$ of the form
$\mathcal{K}=[a,b]_{\Lambda^{m}_{i}}$ for some $a,b\in\Lambda^{m}_{i}$. We say
that $\Lambda^{n}_{i+kR_{m}}$ is combinatorially connected in
$\Lambda^{m}_{i}$ if $\Lambda^{n}_{i+kR_{m}}$ has a unique combinatorial
component $\mathcal{K}$ in $\Lambda^{m}_{i}$ such that
$\Lambda^{n}_{i+kR_{m}}=\mathcal{K}.$

######  Theorem 13.5.

Let $\hat{N}\leq m<n$; $i\geq 0$ and $0\leq k<R_{n}/R_{m}$. Then
$\Lambda^{n}_{i+kR_{m}}$ is combinatorially connected in $\Lambda^{m}_{i}$.

For the proof of Theorem 13.5, we need the following lemma.

######  Lemma 13.6.

The renormalization piece $\Lambda^{n}_{i+kR_{m}}$ has at most finitely many
combinatorial components in $\Lambda^{m}_{i}$ which permute cyclically under
$F^{R_{n}}$.

###### Proof.

Suppose $\Lambda^{n}_{i+kR_{m}}$ has infinitely many combinatorial components.
Then one can find two sequences

$\\{a_{l}\\}_{l=1}^{\infty}\subset\Lambda^{n}_{i+kR_{m}}\hskip
14.22636pt\text{and}\hskip
14.22636pt\\{b_{l}\\}_{l=1}^{\infty}\subset\Lambda^{n}_{i+jR_{m}}\neq\Lambda^{n}_{i+kR_{m}}$

such that

$a_{i}<_{\Lambda^{m}_{i}}b_{i}<_{\Lambda^{m}_{i}}a_{i+1}\hskip
14.22636pt\text{for}\hskip 14.22636pti\geq 1.$

By compactness, the unique limit $a_{\infty}=b_{\infty}$ of these sequences
must belong to both $\Lambda^{n}_{i+kR_{m}}$ and $\Lambda^{n}_{i+jR_{m}}$,
which is impossible.

Since $\\{c_{-l}\\}_{l=0}^{\infty}$ accumulates to every connected component
of $\Lambda$, we see that every combinatorial component of
$\Lambda^{n}_{i+kR_{m}}$ must contain $c_{-l}$ for some $l\geq 0$.
Consequently, every combinatorial component of $\Lambda^{n}_{i+kR_{m}}$ must
eventually map to the unique one containing $c_{-R_{n}+i+kR_{m}}$. The result
follows. ∎

###### Proof of Theorem 13.5.

It suffices to prove the result for $\Lambda^{n+1}_{1}$ in $\Lambda^{n}_{1}$.
Suppose there exist combinatorial components
$\mathcal{B}_{1}\subset\Lambda^{n+1}_{1}$ and
$\mathcal{X}_{1}\subset\Lambda^{n+1}_{j}\neq\Lambda^{n+1}_{1};$ and points
$a^{0}_{1}\in\Lambda^{n+1}_{1}$ and $y^{0}_{1}\in\Lambda^{n+1}_{j}$ such that

$a^{0}_{1}<_{\Lambda^{n}_{1}}\mathcal{X}_{1}<_{\Lambda^{n}_{1}}\mathcal{B}_{1}<_{\Lambda^{n}_{1}}y^{0}_{1}.$

Proceeding by induction, suppose for $i\geq 0$, there exist points
$a^{i}_{1}\in\Lambda^{n+1}_{1}$ and $y^{i}_{1}\in\Lambda^{n+1}_{j}$ such that

$a^{i}_{1}<_{\Lambda^{n}_{1}}\mathcal{X}_{1+iR_{n}}<_{\Lambda^{n}_{1}}\mathcal{B}_{1+iR_{n}}<_{\Lambda^{n}_{1}}y^{i}_{1}.$

By Theorem 13.4, Proposition 10.9 and Proposition 13.1, we see that after
applying $F^{R_{n}}$, the following holds.

  1. i)

If $i=-1\;(\operatorname{mod}\;r_{n})$, then either

$c_{1}=:a^{i+1}_{1}<_{\Lambda^{n}_{1}}\mathcal{X}_{1+(i+1)R_{n}}<_{\Lambda^{n}_{1}}\mathcal{B}_{1+(i+1)R_{n}}<_{\Lambda^{n}_{1}}y^{i+1}_{1}:=y^{i}_{1+R_{n}}$

or

$a^{i}_{1+R_{n}}=:a^{i+1}_{1}<_{\Lambda^{n}_{1}}\mathcal{X}_{1+(i+1)R_{n}}<_{\Lambda^{n}_{1}}\mathcal{B}_{1+(i+1)R_{n}}<_{\Lambda^{n}_{1}}y^{i+1}_{1}:=y^{i}_{1+R_{n}}<_{\Lambda^{n}_{1}}c_{1}.$

  2. ii)

If $i=-k-1\;(\operatorname{mod}\;r_{n})$, then either

$c_{1}<_{\mathcal{T}_{c_{0}}}a^{i}_{1+R_{n}}=:a^{i+1}_{1}<_{\Lambda^{n}_{1}}\mathcal{X}_{1+R_{n+1}}<_{\Lambda^{n}_{1}}\mathcal{B}_{1+R_{n+1}}<_{\Lambda^{n}_{1}}y^{i+1}_{1}:=y^{i}_{1+R_{n}}$

or

$a^{i}_{1+R_{n}}=:a^{i+1}_{1}<_{\Lambda^{n}_{1}}\mathcal{X}_{1+R_{n+1}}<_{\Lambda^{n}_{1}}\mathcal{B}_{1+R_{n+1}}<_{\Lambda^{n}_{1}}y^{i+1}_{1}:=c_{1}.$

  3. iii)

Otherwise, either

$c_{1}<_{\mathcal{T}_{c_{0}}}a^{i}_{1+R_{n}}=:a^{i+1}_{1}<_{\Lambda^{n}_{1}}\mathcal{X}_{1+R_{n+1}}<_{\Lambda^{n}_{1}}\mathcal{B}_{1+R_{n+1}}<_{\Lambda^{n}_{1}}y^{i+1}_{1}:=y^{i}_{1+R_{n}}$

or

$a^{i}_{1+R_{n}}=:a^{i+1}_{1}<_{\Lambda^{n}_{1}}\mathcal{X}_{1+(i+1)R_{n}}<_{\Lambda^{n}_{1}}\mathcal{B}_{1+(i+1)R_{n}}<_{\Lambda^{n}_{1}}y^{i+1}_{1}:=y^{i}_{1+R_{n}}<_{\Lambda^{n}_{1}}c_{1}.$

Thus, we see that $\mathcal{B}_{1+iR_{n}}$ does not contain $c_{1}$ for all
$i\geq 0$. This contradicts Lemma 13.6. ∎

######  Proposition 13.7.

Let $n\geq\hat{N}$. Then the two extremal points of $\Lambda^{n}_{1}$ with
respect to $<_{\Lambda^{n}_{1}}$ are $c_{1}$ and $c_{1+R_{n}}$. That is, for
all $p\in\Lambda^{n}_{1}\setminus\\{c_{1},c_{1+R_{n}}\\}$, we have

$c_{1}<_{\Lambda^{n}_{1}}p<_{\Lambda^{n}_{1}}c_{1+R_{n}}.$

###### Proof.

Suppose $c_{1+R_{n}}<_{\Lambda^{n}_{1}}c_{1-R_{n}}.$ Then

$F^{R_{n}}([c_{1},c_{1-R_{n}}]_{\Lambda^{n}_{1}})=[c_{1},c_{1+R_{n}}]_{\Lambda^{n}_{1}}\subset[c_{1},c_{1-R_{n}}]_{\Lambda^{n}_{1}}.$

Consequently,

$\\{c_{1+iR_{n}}\\}_{i=1}^{\infty}\subset[c_{1+2R_{n}},c_{1+R_{n}}]_{\Lambda^{n}_{1}}.$

Since

$c_{1}\notin[c_{1+2R_{n}},c_{1+R_{n}}]_{\Lambda^{n}_{1}},$

this contradicts Theorem 12.2.

Let $z_{1}\neq c_{1}$ be an extremal point of $\Lambda^{n}_{1}$. Suppose
$c_{1+R_{n}}<_{\Lambda^{n}_{1}}z_{1}.$ Then

$F^{R_{n}}([c_{1},c_{1-R_{n}}]_{\Lambda^{n}_{1}})=[c_{1},c_{1+R_{n}}]_{\Lambda^{n}_{1}}\neq[c_{1},z_{1}]_{\Lambda^{n}_{1}}.$

This implies that

$F^{R_{n}}([c_{1-R_{n}},z_{1}]_{\Lambda^{n}_{1}})=[c_{1},z_{1}]_{\Lambda^{n}_{1}}.$

Since $z_{1}$ cannot be a fixed point, this is a contradiction. ∎

##  14\. Regular Hénon-like Returns

Let $F$ be the mildly dissipative, infinitely renormalizable and regularity
unicritical diffeomorphism considered in Section 10.

Let $\mathcal{C}$ be the unicritical cover of $\Lambda$ defined in Subsection
10.3. By compactness of $\Lambda$, we may choose a finite subcover

$\mathcal{C}_{0}=\\{\bar{\mathcal{B}}_{p}\\}_{p\in\mathcal{P}}\cup\\{\mathcal{T}_{c_{-m}}\\}_{m=0}^{\hat{M}_{-}}\cup\\{\mathcal{T}_{c_{m}}\\}_{m=1}^{\hat{M}_{+}}\subset\mathcal{C},$

where $\mathcal{P}\subset\Lambda^{P}_{L,\delta}$ is a finite set of Pesin
$(L,\delta)$-regular points, and $\hat{M}_{\pm}\in\mathbb{N}$. By increasing
$\hat{M}_{\pm}$ if necessary, we may assume that $\hat{M}_{\pm}$ is a Pliss
moment in the forward/backward orbit of $c_{1}$/$c_{0}$, and that we have
$\mathcal{T}_{c_{\hat{M}_{\pm}}}\subset\bar{\mathcal{B}}_{p^{\pm}}$ for some
Pesin regular point $p^{\pm}\in\mathcal{P}$. For $n\in\mathbb{N}$, recall that
the $n$th renormalization domain containing $c_{1}$ is denoted
$\mathcal{D}^{n}_{1}$. The valuable and the critical pieces of depth $n$ are
defined as

$\Lambda^{n}_{1}=\Lambda\cap\mathcal{D}^{n}_{1}\ni c_{1}\hskip
14.22636pt\text{and}\hskip
14.22636pt\Lambda^{n}_{0}:=\Lambda^{n}_{R_{n}}=\Lambda\cap\mathcal{D}^{n}_{R_{n}}\ni
c_{0}$

respectively.

Let

$\\{\Phi_{c_{m}}:(\mathcal{U}_{c_{m}},c_{m})\to(U_{c_{m}},0)\\}_{m\in\mathbb{Z}}$

be a uniformization of the orbit of $c_{0}$ given in Theorem 5.8. For
$p_{0}\in\mathcal{U}_{c_{0}}$, let $E^{v}_{p_{0}}\in\mathbb{P}^{2}_{p_{0}}$
and $E^{h}_{p_{1}}\in\mathbb{P}^{2}_{p_{1}}$ be the directions given by

$D\Phi_{c_{0}}(E^{v}_{p_{0}})=E^{gv}_{\Phi_{c_{0}}(p_{0})}\hskip
14.22636pt\text{and}\hskip
14.22636ptD\Phi_{c_{1}}(E^{h}_{p_{1}})=E^{gh}_{\Phi_{c_{1}}(p_{1})}=DF(E^{v}_{p_{0}}).$

Let $\hat{N}\in\mathbb{N}$ be the integer given in Proposition 13.3. In the
rest of this section, we assume that $n\geq\hat{N}$. The Main Theorem stated
in Section 1 is an immediate consequence of the following result.

######  Theorem 14.1 (Valuable chart).

There exist $\tau_{n}>0$ and a $C^{r}$-diffeomorphism
$\Psi^{n}:(\Phi_{c_{1}}(\mathcal{V}^{n}_{1}),0)\to(V^{n}_{1},0)$, where

$I^{n}:=(-\tau_{n},\tau_{n})\hskip 14.22636pt\text{and}\hskip
14.22636ptV^{n}_{1}:=I^{n}\times(-l_{c_{1}},l_{c_{1}}),$

such that the following properties hold.

  1. i)

We have
$F^{R_{n}}(\mathcal{V}^{n}_{1})\Subset\mathcal{V}^{n}_{1}\subset\mathcal{U}_{c_{1}},$
and consequently $F^{R_{n}-1}(\mathcal{V}^{n}_{1})\subset\mathcal{U}_{c_{0}}$
and $\Lambda^{n}_{1}\subset\mathcal{V}^{n}_{1}.$

  2. ii)

We have

$\|\Psi^{n}-\operatorname{Id}\|_{C^{r}}=O(\lambda^{(1-{\bar{\delta}})R_{n}})\hskip
14.22636pt\text{and}\hskip
14.22636pt\Psi^{n}|_{I^{n}\times\\{0\\}}\equiv\operatorname{Id}.$

  3. iii)

For $p_{1}\in\mathcal{V}^{n}_{1}$, let

$E^{v,n}_{p_{1}}:=DF^{-R_{n}+1}(E^{v}_{p_{R_{n}}})\in\mathbb{P}^{2}_{p_{1}}.$

Then

$D\bar{\Psi}^{n}(E^{h}_{p_{1}})=E^{gh}_{\bar{\Psi}^{n}(p_{1})}\hskip
14.22636pt\text{and}\hskip
14.22636ptD\bar{\Psi}^{n}(E^{v,n}_{p_{1}})=E^{gv}_{\bar{\Psi}^{n}(p_{1})},$

where $\bar{\Psi}^{n}:=\Psi^{n}\circ\Phi_{c_{1}}$.

  4. iv)

Let $p_{1}\in\mathcal{V}^{n}_{1}$. Then $p_{1}$ is $(R_{n}-1)$-times forward
$(1,\delta)_{v}$-regular along $E^{v,n}_{p_{1}}$, and $p_{R_{n}}$ is
$(R_{n}-1)$-times backward $(1,\delta)_{v}$-regular along $E^{v}_{p_{R_{n}}}$.

  5. v)

Define

$H_{n}:=\Phi_{c_{0}}\circ F^{R_{n}-1}\circ(\bar{\Psi}^{n})^{-1}.$

Then we have

$H_{n}(x,y)=(h_{n}(x),e_{n}(x,y))\hskip 14.22636pt\text{for}\hskip
14.22636pt(x,y)\in V^{n}_{1},$

where $h_{n}:I^{n}\to h_{n}(I^{n})$ is a $C^{r}$-diffeomorphism, and
$e_{n}:V^{n}_{1}\to\mathbb{R}$ is a $C^{r}$-map such that

$e_{n}(\cdot,0)\equiv 0\hskip 14.22636pt\text{and}\hskip
14.22636pt\|e_{n}\|_{C^{r}}=O(\lambda^{(1-{\bar{\delta}})R_{n}}).$

######  Lemma 14.2.

For $p\in\Lambda^{n}_{1}$, write $\Phi_{c_{1}}(p)=(x_{p},y_{p}).$ Denote

$\chi_{n+1}:=\inf_{p\in\Lambda^{n}_{1}\setminus\Lambda^{n+1}_{1}}|x_{p}|.$

Then we have

$\sup_{p\in\Lambda^{n+1}_{1}}|x_{p}|<\chi_{n+1}+O(\chi_{n+1}^{1/{\bar{\varepsilon}}}).$

###### Proof.

Let $q_{1}\in\Lambda^{n}_{1}\setminus\Lambda^{n+1}_{1}$ be a point satisfying
$|x_{q_{1}}|=\chi_{n+1}$. If $\operatorname{cd}(q_{1})=\varnothing$, then by
Theorem 9.8, $q_{1}$ is forward $(1,\delta)$-regular. Let $N$ be an integer
satisfying

$N\asymp{\bar{\varepsilon}}^{-1}\log_{\lambda}|x_{q_{1}}|.$

Then by Lemma 3.8, $q_{i}\in\mathcal{U}_{c_{i}}$ for $1\leq i\leq N$. Note
that $\lambda^{(1-{\bar{\delta}})N}<|x_{q_{1}}|^{1/{\bar{\varepsilon}}}$. By
Proposition 3.11, we see that $W^{ss}_{\operatorname{loc}}(q_{1})$ is a
$O(|x_{q_{1}}|^{1/{\bar{\varepsilon}}})$-vertical curve in
$\mathcal{U}_{c_{1}}$. By Theorem 13.5, $\Lambda^{n+1}_{1}$ is contained in
the vertical strip bounded between $W^{ss}_{\operatorname{loc}}(q_{0})$ and
$W^{ss}_{\operatorname{loc}}(c_{1})$. Thus, the result holds in this case.

If $\operatorname{cd}(q_{1})\neq\varnothing$, let
$q_{1}\in\mathcal{T}_{c_{-m}}$ for some $m>0$ with
$\operatorname{cd}(c_{-m})=\varnothing$. By Lemma 9.5, we see that

$\operatorname{dist}(q_{1},c_{-m})<|x_{q_{1}}|^{1/{\bar{\varepsilon}}}.$

Moreover, arguing as above, we see that the vertical manifold
$\hat{W}^{v}(c_{-m})$ at $c_{-m}$ is a
$O(|x_{q_{1}}|^{1/{\bar{\varepsilon}}})$-vertical curve in
$\mathcal{U}_{c_{1}}$. The set $(c_{1},c_{-m})_{\mathcal{T}_{c_{1}}}$ is
contained in the vertical strip bounded between $\hat{W}^{v}(c_{-m})$ and
$W^{ss}_{\operatorname{loc}}(c_{1})$. Moreover,
$\Lambda^{n+1}_{1}\subset(c_{1},c_{-m})_{\mathcal{T}_{c_{1}}}\cup\mathcal{T}_{c_{-m}}.$
The result follows. ∎

######  Proposition 14.3.

We have $\mathcal{T}_{c_{-R_{n}+1}}\cap\Lambda^{n+1}_{1}=\varnothing$ and
$\mathcal{T}_{c_{R_{n}}}\cap\Lambda^{n+1}_{0}=\varnothing.$

###### Proof.

Suppose there exists
$p_{1}\in\mathcal{T}_{c_{-R_{n}+1}}\cap\Lambda^{n+1}_{1}$. Let

$q_{0}:=p_{R_{n}}\in
F^{R_{n}-1}(\mathcal{T}_{c_{-R_{n}+1}})\cap\Lambda^{n+1}_{R_{n}}\subset\mathcal{T}_{c_{0}}.$

Observe that

$\operatorname{dist}(q_{0},c_{0})<\lambda^{-\eta
R_{n}}\operatorname{diam}(\mathcal{T}_{c_{-R_{n}+1}})<K\lambda^{(1-{\bar{\delta}})\delta
R_{n}}$

for some uniform constant $K\geq 1$. Since
$\operatorname{dist}(q_{1},c_{1})\asymp\operatorname{dist}(q_{0},c_{0}),$ we
conclude by Lemma 14.2 that $\chi_{n+1}<K\lambda^{(1-{\bar{\delta}})\delta
R_{n}}.$

Write $(x,y):=\Phi_{c_{1}}(c_{-R_{n}+1}).$ We have

$\operatorname{dist}(c_{-R_{n}+1},c_{1})\asymp|x|<\operatorname{diam}(\mathcal{T}_{c_{-R_{n}+1}})+\chi_{n+1}<K\lambda^{(1-{\bar{\delta}})\delta
R_{n}}\ll\operatorname{diam}(\mathcal{U}_{c_{-R_{n}+1}}).$

This contradicts Proposition 5.9.

The second claim can be proved using the same proof mutatis mutandis. ∎

Let $p\in\Omega$. For $m\geq 0$, define

$\operatorname{cd}^{m}(p):=\left\\{\begin{array}[]{cl}\operatorname{cd}(p)&:\text{
if }0\leq\operatorname{cd}(p)<m\\\ \varnothing&:\text{
otherwise}\end{array}\right..$

Similarly, for $k\geq 1$, define

$\operatorname{vd}^{k}(p):=\left\\{\begin{array}[]{cl}\operatorname{vd}(p)&:\text{
if }1\leq\operatorname{vd}(p)<k\\\ \varnothing&:\text{
otherwise}\end{array}\right..$

######  Lemma 14.4.

For $p_{0}\in\Lambda^{n}_{0}$, let $z_{0}:=\Phi_{c_{0}}(p_{0})$ and
$E^{v}_{p_{0}}:=D_{z_{0}}\Phi_{c_{0}}^{-1}(E^{gv}_{z_{0}}).$ Then for $0\leq
i<R_{n}$ such that $\operatorname{cd}^{i}(p_{-i})=\varnothing$, the point
$p_{-i}$ is $i$-times forward $(O(L),\delta)_{v}$-regular along
$E^{v,n}_{p_{-i}}=DF^{-i}(E^{v}_{p_{0}}).$ If, additionally, we have
$p_{-i}\in\mathcal{T}_{c_{1}}$, then $p_{-i}$ is $i$-times forward
$(1,\delta)_{v}$-regular along $E^{v,n}_{p_{-i}}$.

###### Proof.

First, suppose that $\operatorname{cd}(p_{0})=0.$ Then for all $0\leq
i<R_{n}$, we have $\operatorname{cd}^{i}(p_{-i})=\operatorname{cd}(p_{-i}).$

Let $0<m<R_{n}$ be an integer such that $p_{-m}\in\mathcal{T}_{c_{0}}$ and
$\operatorname{cd}^{m}(p_{-m})=0.$ Let $-j<0$ and $l>1$ be the tunnel and
crescent escape moments for $p_{-m}$ and $p_{-m+1}$ respectively. Then by
Proposition 6.9, we have

$\operatorname{cd}^{m-l}(p_{-m+l})=\operatorname{cd}^{m+j}(p_{-m-j})=\varnothing.$

Suppose the statement holds for $p_{-m+l}$, so that $p_{-m+l}$ is
$(m-l)$-times forward $(O(L),\delta)_{v}$-regular along $E^{v,n}_{p_{-m+l}}$.
Then by Lemma 9.12, Proposition 6.12 and Proposition 6.13, we see that
$p_{-m-j}$ is $(m+j)$-times forward $(O(1),\delta)_{v}$-regular along
$E^{v,n}_{p_{-m-j}}$. Moreover, by Theorem 9.8 and Proposition 6.13,
$p_{-m-j}$ is forward $(O(1),\delta)_{v}$-regular along
$\hat{E}^{v}_{p_{-m-j}}$. Thus, by Proposition 3.11, we see that

$\measuredangle(E^{v,n}_{p_{-m-j}},\hat{E}^{v}_{p_{-m-j}})<O(\lambda^{(1-{\bar{\delta}})(m+j)})\ll
1.$

Let $m_{1}<R_{n}$ be the smallest integer such that $m_{1}>m+j$ and
$p_{-m_{1}}\in\mathcal{T}_{c_{0}}$. This means that for $m+j<i<m_{1}$, we have
$\operatorname{cd}^{i}(p_{-i})=\varnothing$. Consequently, by Theorem 9.8,
$p_{-i}$ is $i$-times forward $(L,\delta)_{v}$-regular along
$\hat{E}^{v}_{p_{-i}}$. Thus, by Proposition 3.9, $p_{-i}$ is $i$-times
forward $(O(L),\delta)_{v}$-regular along $E^{v,n}_{p_{-i}}$. Proceeding by
induction, the claim follows.

If $\operatorname{cd}(p_{0})\neq 0,$ then let $0<n_{1}<\ldots<n_{K}\leq\infty$
be the increasing sequence of all integers such that
$p_{0}\in\mathcal{T}_{c_{-n_{k}}}$ for $k\geq 1$. Furthermore, let
$-(l_{k}+n_{k})<0$ be the tunnel escape moment of $c_{-n_{k+1}+n_{k}}$. For
$0\leq i<R_{n}$, let $q_{-i}:=c_{-n_{k_{i}}-i}$, where $k_{i}$ is the smallest
integer such that $l_{k_{i}}>i$. This means that

$p_{-i}\in\mathcal{T}_{c_{-n_{k_{i}}-i}}\hskip 14.22636pt\text{and}\hskip
14.22636pt\operatorname{cd}^{i}(c_{-n_{k_{i}}-i})=\operatorname{cd}(c_{-n_{k_{i}}-i}).$

If $K<\infty$, let $q_{-i}:=p_{-i}$ for $i\geq l_{K}$.

Observe that for $1\leq k\leq K$, we have

$\operatorname{cd}(c_{-n_{k}-l_{k}})=\operatorname{cd}(c_{-n_{k+1}-l_{k}})=\varnothing.$

Hence, $c_{-n_{k}-l_{k}}$ is forward $(O(1),{\bar{\varepsilon}})_{v}$-regular
along $\hat{E}^{v}_{c_{-n_{k}-l_{k}}}$, and by Proposition 6.13,
$c_{-n_{k+1}-l_{k}}$ is forward $(O(1),\delta)_{v}$-regular along
$\hat{E}^{v}_{c_{-n_{k+1}-l_{k}}}$. Moreover, if $c_{-n_{k}-l_{k-1}}$ is
$l_{k-1}$-times forward $(O(1),\delta)_{v}$-regular along
$E^{v,n}_{c_{-n_{k}-l_{k-1}}}$, then by Proposition 3.11 and 3.9, it follows
that $c_{-n_{k+1}-l_{k}}$ is $l_{k}$-times forward $(O(1),\delta)_{v}$-regular
along $E^{v,n}_{c_{-n_{k+1}-l_{k}}}$.

The result now follows by replacing $p_{-i}$ in the argument for the previous
case (where $\operatorname{cd}(p_{0})=\varnothing$) with $q_{-i}$. ∎

######  Lemma 14.5.

For $p_{1}\in\Lambda^{n}_{1}$, let $z_{1}:=\Phi_{c_{1}}(p_{1})$ and
$E^{h}_{p_{1}}:=D\Phi_{c_{1}}^{-1}(E^{gh}_{z_{1}}).$ For $1\leq i\leq R_{n}$
such that $\operatorname{vd}^{i}(p_{i})=\varnothing$, the point $p_{i}$ is
$i$-times backward $(O(L),\delta)_{h}$-regular along

$E^{h,n}_{p_{i}}:=DF^{i-1}(E^{h}_{p_{1}}).$

If, additionally, we have $p_{i}\in\mathcal{T}_{c_{0}}$, then $p_{i}$ is
$i$-times backward $(1,\delta)_{h}$-regular along $E^{h,n}_{p_{i}}$.

Let $p\in\Lambda^{P}_{L,\delta}$. Recall that an encased regular box
$\bar{\mathcal{B}}_{p}$ at $p$ is nearly a parallelogram, whose horizontal
width and vertical height are commensurate with $1/L_{1}$ and $1/L_{2}$
respectively, where ${\bar{L}}<L_{1}<\underline{L_{2}}.$ Let $\mathcal{B}_{p}$
be a (non-encased) regular box at at $p$ whose horizontal width and vertical
height are commensurate with $1/L_{1}^{\prime}$ and $1/L_{2}^{\prime}$
respectively, where ${\bar{L}}<L_{1}^{\prime}<\underline{L_{1}}$ and
$\overline{L_{1}^{\prime}}<L_{2}^{\prime}<\underline{L_{2}}.$ If
$-j<-\hat{M}_{-}$ is a Pliss moment in the backward orbit of $c_{0}$, let
$\mathcal{B}_{c_{-j}}$ be a regular box at $c_{-j}$ whose horizontal width and
vertical height are uniformly bounded below.

Let $Q\in\mathcal{C}_{0}$. We define a larger covering element $\hat{Q}\supset
Q$ as follows. If $Q=\bar{\mathcal{B}}_{p}$ for some $p\in\mathcal{P}$, let
$\hat{Q}:=\mathcal{B}_{p}$. Otherwise, we have $Q=\mathcal{T}_{c_{m}}$ for
some $-\hat{M}_{-}\leq m\leq\hat{M}_{+}$. In this case, let
$\hat{Q}:={\hat{\mathcal{T}}}_{c_{m}}$, where ${\hat{\mathcal{T}}}_{c_{m}}$ is
a thickened tunnel/crescent at $c_{m}$ defined in (10.3)/(10.4).

For $p_{0}\in\Lambda^{n}_{0}$, let $(x_{0},y_{0}):=\Phi_{c_{0}}(p_{0})$ and

$W^{v}(p_{0}):=\Phi_{c_{0}}^{-1}(\mathbb{L}^{v}_{x_{0}}\cap U_{c_{0}}).$

For $0\leq i<R_{n}$, define

$W^{v,n}(p_{-i}):=F^{-i}(W^{v}(p_{0})).$

######  Lemma 14.6.

Let $p_{0}\in\mathcal{T}_{c_{0}}$. Let $-j<0$ and $i>1$ be the tunnel and
crescent escape time for $p_{0}$ and $p_{1}$ respectively. Let

$\\{\hat{\Phi}_{c_{-j+m}}:\hat{\mathcal{U}}_{c_{-j+m}}\to\hat{U}_{c_{-j+m}}\\}_{m=-\infty}^{j}$

be a linearization of the $(-\infty,j)$-orbit of $c_{-j}$ with vertical
direction $\hat{E}^{v}_{c_{-j}}$. Similarly, let

$\\{\check{\Phi}_{c_{i+m}}:\check{\mathcal{U}}_{c_{i+m}}\to\check{U}_{c_{i+m}}\\}_{m=-i+1}^{\infty}$

be a linearization of the $(-i+1,\infty)$-orbit of $c_{i}$ with vertical
direction $\hat{E}^{v}_{c_{i}}$. Denote the connected components of
$\hat{\mathcal{U}}_{c_{-j}}^{{\bar{\varepsilon}}j}\setminus{\hat{\mathcal{T}}}_{c_{-j}}$
that are disjoint from $\hat{W}^{v}(c_{-j})$ by $\hat{\mathcal{U}}^{-}$ and
$\hat{\mathcal{U}}^{+}$. Then the following statements hold.

  1. i)

Let $\Gamma_{0}^{v}$ be $C^{1}$-curve so that $\Gamma_{i}^{v}$ is sufficiently
vertical in $\check{\mathcal{U}}_{c_{i}}$ and the distances between $p_{i}$
and the endpoints of $\Gamma_{i}^{v}$ are at least $1/\overline{L_{2}}$. Then
$\Gamma_{-j}^{v}$ is sufficiently vertical and proper in
$\mathcal{B}_{c_{-j}}$.

  2. ii)

We have

$F^{j}(\hat{\mathcal{U}}^{-}\cup{\hat{\mathcal{T}}}_{c_{-j}}\cup\hat{\mathcal{U}}^{+})\subset{\hat{\mathcal{T}}}_{c_{0}}.$

  3. iii)

Let $\Gamma^{h}_{-j}\ni p_{-j}$ be a sufficiently horizontal curve in
$\hat{\mathcal{U}}_{c_{-j}}$. Then

$\Gamma_{i}^{\pm}:=F^{j+i}(\Gamma_{-j}^{h}\cap\hat{\mathcal{U}}^{\pm})$

is a sufficiently horizontal curve in $\check{\mathcal{U}}_{c_{i}}$.

###### Proof.

Write $(x_{1},y_{1}):=\Phi_{c_{1}}(p_{1}).$ Note that

$i>(1+{\bar{\delta}})\delta^{-1}\log_{\lambda}|x_{1}|.$

By Theorem 3.6 and Proposition 3.9, we see that $\Gamma_{1}^{v}$ is
$|x_{1}|^{(1+{\bar{\delta}})/\delta}$-vertical in $\mathcal{U}_{c_{1}}$.
Furthermore,

$\operatorname{diam}(\Gamma_{1}^{v})>|x_{1}|^{-(1+{\bar{\delta}})/\delta}/\overline{L_{2}}>1\hskip
14.22636pt\text{and}\hskip
14.22636pt\operatorname{dist}(\Gamma_{1}^{v})\asymp|x_{1}|.$

This implies that $\Gamma_{1}^{v}$ is sufficiently stable-aligned in
$\mathcal{U}_{c_{1}}$, and vertically proper in ${\hat{\mathcal{T}}}_{c_{1}}$.

Let $q_{0}\in\Gamma_{0}^{v}\cap\bar{\partial}{\hat{\mathcal{T}}}_{c_{0}}$, and
write $(x_{0},y_{0}):=\Phi_{c_{0}}(p_{0})$ and
$(u_{0},v_{0}):=\Phi_{c_{0}}(q_{0}).$ Note that

$j>(1+{\bar{\delta}})\delta^{-1}\log_{\lambda}|x_{0}|.$

By Theorem 3.6 and Proposition 3.9, we have

$|v_{-j}|>\lambda^{-(1-{\bar{\varepsilon}})j}|v_{0}|>\lambda^{-(1-{\bar{\varepsilon}})j}|x_{0}|^{1/\hat{\delta}}>1.$

Additionally, Proposition 6.13 implies that $\Gamma_{-j}$ is sufficiently
vertical in $\hat{\mathcal{U}}_{c_{-j}}$. The first claim follows.

The second claim follows from Theorem 3.6 and Proposition 3.9.

The third claim follows from Proposition 6.12 and Proposition 6.14. ∎

######  Proposition 14.7.

Let $p_{0}\in\Lambda^{n}_{0}$. For $\hat{M}_{-}\leq i<R_{n}$, if $p_{-i}\in
Q\in\mathcal{C}_{0}$ and $\operatorname{cd}^{i}(p_{-i})=\varnothing,$ then
$W^{v,n}(p_{-i})$ is sufficiently vertical and vertically proper in $\hat{Q}$.
If, additionally, $Q=\mathcal{T}_{c_{1}}$, then $W^{v,n}(p_{-i})$ is
sufficiently stable-aligned in $\mathcal{U}_{c_{1}}$.

###### Proof.

Denote $\Gamma_{-i}^{v}:=W^{v,n}(p_{-i}).$ Let
$0=:m_{0}<m_{1}<\ldots<m_{K}<R_{n}$ be an increasing sequence of integers such
that $p_{-m_{k}}\in\mathcal{T}_{c_{0}}$ and
$\operatorname{cd}^{m_{k}}(p_{-m_{k}})=0.$ Let $(m_{k}-1-l_{k})>1$ and
$-(j_{k}-m_{k})<0$ be the crescent and tunnel escape time for $p_{-m_{k}+1}$
and $p_{-m)l}$ respectively. By Proposition 6.9, we see that

$\operatorname{cd}^{l}(p_{-l_{k}})=\operatorname{cd}^{j}(p_{-j_{k}})=\varnothing\hskip
14.22636pt\text{and}\hskip 14.22636ptj_{k-1}<l_{k}<m_{k}<j_{k}<l_{k-1}.$

If

$\operatorname{diam}(\Gamma_{-l_{k}}^{v})>1/\overline{L_{2}},$

then by Lemma 14.6 i) and Proposition 3.18 i), $\Gamma_{-j_{k}}^{v}$ is
sufficiently vertical and proper in any regular box $\hat{Q}$ for which we
have $p_{-j_{k}}\in Q$.

The result now follows from Proposition 3.19 ii). ∎

Let $p_{1}^{-},p_{1}^{+}\in\Lambda^{n}_{1}$. By Proposition 14.7,
$W^{v,n}(p_{1}^{\pm})$ is vertical proper in $\mathcal{T}_{c_{1}}$ and
sufficiently stable-aligned in $\mathcal{U}_{c_{1}}$. Let
$\mathcal{V}^{n}_{1}(p_{1}^{-},p_{1}^{+})$ be the connected component of
$\mathcal{U}_{c_{1}}\setminus(W^{v,n}(p_{1}^{-})\cup W^{v,n}(p_{1}^{+}))$ that
contain $p_{1}^{-}$ and $p_{1}^{+}$ in its boundary. For $1\leq i\leq R_{n}$,
denote

$\mathcal{V}^{n}_{i}(p_{i}^{-},p_{i}^{+}):=F^{i-1}(\mathcal{V}^{n}_{1}(p_{1}^{-},p_{1}^{+})).$

######  Proposition 14.8.

There exist a uniform constant $\hat{K}\in\mathbb{N}$, and a set
$\\{q^{0}_{1},\ldots,q^{\hat{K}}_{1}\\}\subset\Lambda^{n}_{1}$ such that

$c_{1}=:q^{0}_{1}<_{\mathcal{T}_{c_{1}}}\ldots<_{\mathcal{T}_{c_{1}}}q^{\hat{K}}_{1}:=c_{1+R_{n}},$

and for all $0\leq k<\hat{K}$ and $\hat{M}_{+}\leq i\leq R_{n}$, we have
$\mathcal{V}^{n}_{i}(q^{k}_{i},q^{k+1}_{i})\subset\hat{Q}$ for some
$Q\in\mathcal{C}_{0}$.

###### Proof.

Let $Z$ be a non-trivial component in the finite set $\mathcal{Z}$ given in
Proposition 13.2, so that

$Z([t_{i-1},t_{i}])\subset Q_{i}\in\mathcal{C}_{0}\hskip
14.22636pt\text{for}\hskip 14.22636pt1\leq i\leq l.$

Observe that if $\Gamma_{1}\ni Z(t_{1})$ is a sufficiently vertical proper
Jordan arc in $Q_{1}$, and $\tilde{Q}_{1}$ is the component of
$Q_{1}\setminus\Gamma_{1}$ containing $Z([t_{0},t_{1}))$, then

$\Lambda\cap\tilde{Q}_{1}=\\{p\in\Lambda_{Q_{1}}\;|\;p\leq_{Q_{1}}Z(t_{1})\\}.$

Moreover, the analogous property holds at $Z(t_{l-1})$ in $\Lambda_{Q_{l}}$.

Let

$\\{z^{1},\ldots,z^{\hat{K}-1}\\}\subset\bigcup_{Z\in\mathcal{Z}}Z$