Proof. By Proposition 2.5, one has $\| H-\mathrm {Id}_{\mathbb {R}^2} \|_{C^0}<C_H$ . Then, for any $k\in \mathbb {N}$ ,

$$ \begin{align*} &d(A^k(H(x+n)-n),A^k\circ H(x))\\&\quad =d(H(F^k(x+n))-A^k(n),H\circ F^k(x))\\&\quad =d(H(F^k(x)+A^k(n))-A^k(n),H\circ F^k(x))\\&\quad \leqslant d(H(F^k(x)+A^k(n))-A^k(n),F^k(x))+d(F^k(x),H\circ F^k(x))\\&\quad <2C_H, \end{align*} $$

where d is the natural Euclidean distance on $\mathbb {R}^2$ . This uniform upper bound indicates that there is no ‘unstable component’ for the difference between $H(x+n)-n$ and $H(x)$ , that is, they are in the same stable leaf for any $n\in \mathbb {Z}^2$ .

Moreover, for any $n^*\in A^k\mathbb {Z}^2$ , one has

$$ \begin{align*}F^{-k}(x+n^*)-F^{-k}(x)=A^{-k}(n^*)\in\mathbb{Z}^2,\end{align*} $$

so $H (F^{-k}(x)+A^{-k}(n^*))$ and $H\circ F^{-k}(x)+A^{-k}(n^*)$ are in the same stable leaf. Note that

$$ \begin{align*} H(x+n^*)&=A^k\circ H\circ F^{-k}(x+n^*)=A^k\circ H (F^{-k}(x)+A^{-k}(n^*)),\\ H(x)+n^*&=A^k\circ ( H\circ F^{-k}(x)+A^{-k}(n^*)). \end{align*} $$

Hence, one can get

$$ \begin{align*} | H(x+n^*)-H(x)-n^* |<2C_H\cdot \|A|_{L^s}\|^k. \\[-37pt] \end{align*} $$

Proof. By [Reference Hall and Hammerlindl27], this proposition holds for the $cu$ -DA case. Specifically, the dynamical coherence follows from [Reference Hall and Hammerlindl27, Theorem A], the global product structure is given by [Reference Hall and Hammerlindl27, Proposition 2.12], and the quasi-isometry of the foliations $\mathcal {F}^u_F$ and $\mathcal {F}^c_F$ are respectively given by [Reference Hall and Hammerlindl27, Proposition 2.10(2)] and [Reference Hall and Hammerlindl27, Lemma 3.3].

Now, we consider the $sc$ -DA $f:\mathbb {T}^2\to \mathbb {T}^2$ . The proofs for the dynamical coherence, the global product structure, and the quasi-isometry of the foliation $\mathcal {F}^s_F$ are similar to the proofs in [Reference Hall and Hammerlindl27]. Here, we briefly explain how it works for the $sc$ -DA case. Consider the inverse map $F^{-1}:\mathbb {R}^2\to \mathbb {R}^2$ of the lift of $sc$ -DA f. Note that $F^{-1}$ is a partially hyperbolic diffeomorphism with invariant unstable bundle and is semi-conjugate to the hyperbolic diffeomorphism $A^{-1}:\mathbb {R}^2\to \mathbb {R}^2$ . Since the argument in [Reference Hall and Hammerlindl27, Theorem A, Propositions 2.12, and 2.10(2)] are completely in the universal cover $\mathbb {R}^2$ , these properties hold for $F^{-1}$ and $\mathcal {F}^{\,c/u}_{F^{-1}}$ . Hence, we get that these properties also hold for F and $\mathcal {F}^{\,s/c}_{F}$ . Then, it remains to prove the quasi-isometry of $\mathcal {F}^c_F$ .

The quasi-isometry of $\mathcal {F}^c_F$ in the $sc$ -DA case is not as direct as above. Note that the proof of [Reference Hall and Hammerlindl27, Lemma 3.3], that is, the quasi-isometry of $\mathcal {F}^c_F$ in the $cu$ -DA case, needs $\mathcal {F}^c_F$ commuting with the $\mathbb {Z}^2$ -actions. This does not a priori hold in the $sc$ -DA case. However, combining the argument in [Reference Hammerlindl29, Reference Hall and Hammerlindl27], $\mathcal {F}^c_F$ is also quasi-isometric in the $sc$ -DA case. Note that though the paper [Reference Hammerlindl29] deals with the $3$ -nilmanifold case, the argument can be adapted for the $2$ -torus. We refer to the proof of [Reference Hall and Hammerlindl27, Proposition 2.10], where the authors of both [Reference Hall and Hammerlindl27, Reference Hammerlindl29] explain how the argument in [Reference Hammerlindl29] works for the $2$ -torus as well.

Now, we give a brief proof of the quasi-isometry of $\mathcal {F}^c_F$ in the $sc$ -DA case. Denote by $\pi ^{s/u}:\mathbb {R}^2\to \widetilde {\mathcal {L}}^{s/u}(0)$ the projections along foliations $ \widetilde {\mathcal {L}}^{u/s}$ . The following properties are immediate from [Reference Hall and Hammerlindl27, Reference Hammerlindl29] to the $cu$ -partially hyperbolic diffeomorphism $F^{-1}$ .

1. (1) ([Reference Hall and Hammerlindl27, Lemma 2.3], [Reference Hammerlindl29, Lemma 4.10]) There exists $R>0$ such that for any center segment $J^c$ , the interval $\pi ^s(J^c)$ has length at most R. (This can be done by using the argument in [Reference Hall and Hammerlindl27, Lemma 2.3] or [Reference Hammerlindl29, Lemma 4.10] for the $cu$ -partially hyperbolic diffeomorphism $F^{-1}$ .)

2. (2) [Reference Hall and Hammerlindl27, Corollary 2.5] There exists $R>0$ such that for any stable segment $J^s$ , the interval $\pi ^u(J^s)$ has length at most R. (This can be done by using the argument in [Reference Hall and Hammerlindl27, Corollary 2.5] for the $cu$ -partially hyperbolic diffeomorphism $F^{-1}$ .)

3. (3) ([Reference Hall and Hammerlindl27, Lemma 2.9], [Reference Hammerlindl29, Lemma 4.7]) For any $C>0$ , there exists $l>0$ such that any center segment $J^c$ of length greater than l is such that $\pi ^u(J^c)$ has length greater than C. (This can be proved by the same proof of [Reference Hall and Hammerlindl27, Proposition 2.8 and Lemma 2.9], where one just needs to replace [Reference Hall and Hammerlindl27, Corollary 2.5] by the property given by the above item.)

Then, by modifying the proof of [Reference Hammerlindl29, Lemma 4.13], we can get the following claim.

Proof. Let $J^c$ be a center curve with length great enough and $\alpha :[0,1]\to \mathbb {R}^2$ be its parameterization. Let R be as in items (1) and (2) above. Taking $C=M+2R$ and l in item (3) above, we get that there are $s,t\in [0,1]$ satisfying

(2.7) $$ \begin{align} \pi^u\alpha(t)-\pi^u\alpha(s)>C=M+2R. \end{align} $$

Without loss of generality, we assume $s<t$ (see Figure 1). Here, we identify $\widetilde {\mathcal {L}}^u(0)$ with  $\mathbb {R}$ .

Figure 1

The projections of endpoints of $J^c$ on $\widetilde {\mathcal {L}}^u$ have distance bigger than M. (Color online.)

Figure 1 Long description

A line diagram on a white background featuring horizontal and vertical axes. A central horizontal band is bounded by two parallel black lines labeled tilde L super s on the left, with a vertical double-headed arrow inside labeled M. A vertical black line on the right is labeled tilde L super u (0).

In the upper region:

- A wavy blue curve labeled script F sub F super s (alpha(t)) spans the width.

- A red curve labeled script F sub F super c (alpha(t)) starts at a red dot alpha(1) and intersects the blue curve at a blue dot alpha(t).

- A horizontal dashed line extends from alpha(t) to the vertical axis at a black dot labeled pi super u alpha(t).

- A vertical double-headed arrow between this dashed line and the central band is labeled R.

In the lower region:

- A second wavy blue curve labeled script F sub F super s (alpha(s)) spans the width.

- The red curve continues downward, passing through a red dot alpha(0) and intersecting the lower blue curve at a blue dot alpha(s).

- A horizontal dashed line extends from alpha(s) to the vertical axis at a black dot labeled pi super u alpha(s).

- A vertical double-headed arrow between this dashed line and the central band is also labeled R.

By item (2), for any $x\in \mathbb {R}^2$ , the complement $\mathbb {R}^2 \setminus \mathcal {F}^s_F(x)$ has connected components $B_-(x)$ and $B_+(x)$ satisfying

$$ \begin{align*} H_R^-(x):=\{y\in\mathbb{R}^2\mid \pi^u(y)<\pi^u(x)-R \} \subset B_-(x),\\ H_R^+(x):=\{y\in\mathbb{R}^2\mid \pi^u(y)>\pi^u(x)+R \} \subset B_+(x). \end{align*} $$

Combining item (2) with the global product structure, we get that the curve $\alpha ((t,1])$ is located in the component $B_+(\alpha (t))$ of $\mathbb {R}^2 \setminus \mathcal {F}^s_F(\alpha (t))$ , namely, $\alpha ((t,1])\subset B_+(\alpha (t))$ . Hence, we have $\alpha ((t,1])\subset \mathbb {R}^2\setminus H^{-}_R(\alpha (t))$ , which implies $\pi ^u\alpha (1)-\pi ^u\alpha (t)>-R$ . Similarly, we have $\alpha ([0,s))\subset \mathbb {R}^2\setminus H^{+}_R(\alpha (s))$ , which implies $\pi ^u\alpha (s)-\pi ^u\alpha (0)>-R$ . Then, by (2.7), we get

$$ \begin{align*} \pi^u\alpha(1)-\pi^u\alpha(0) &= (\pi^u\alpha(t)-\pi^u\alpha(s))+ (\pi^u\alpha(1)-\pi^u\alpha(t))+ (\pi^u\alpha(s)-\pi^u\alpha(0))\\ &> (M+2R)+(-R)+(-R)=M. \end{align*} $$

This ends the proof of Claim 2.8.

Now, we show that Claim 2.8 is sufficient to get the quasi-isometry of $\mathcal {F}^c_F$ . Indeed, let constants M and l be as in Claim 2.8, and $J^c$ be a center curve with endpoints $x,y$ and with length $L=kl+r$ , where $k\in \mathbb {N}$ and $0\leqslant r<l$ . Then, by Claim 2.8, ${|\pi ^u(x)-\pi ^u(y)|\geqslant kM}$ . Thus, there is a constant $\beta>0$ determined by the angle between $\widetilde {\mathcal {L}}^s$ and $\widetilde {\mathcal {L}}^u$ such that $d(x,y)\geqslant k\beta M$ . Then, for $L\geqslant 2l$ , one has $L-r\geqslant {L}/{2}$ and

$$ \begin{align*}d(x,y)\geqslant k\beta M =\frac{M}{l}\beta kl = \frac{M}{l}\beta (L-r) \geqslant\frac{M}{2l}\beta L.\end{align*} $$

Hence, $d_{\mathcal {F}^c_F}(x,y)=L\leqslant ({2l}/{M\beta })d(x,y)$ for $L\geqslant 2l$ . In particular, for any $L>0$ , one has

$$ \begin{align*} d_{\mathcal{F}^c_F}(x,y)=L\leqslant\frac{2l}{M\beta}d(x,y)+2l.\end{align*} $$

This concludes the quasi-isometry of $\mathcal {F}^c_F$ , and ends the proof of Proposition 2.7.

Proof. It essentially follows the proof dealing with the $cu$ -DA case in [Reference Hall and Hammerlindl27], while the details in [Reference Hammerlindl and Shi31] for DA diffeomorphisms are strongly related as well. Just as in the last proposition, we briefly show the proof for the lifted $sc$ -DA map.

By the fact that the semi-conjugacy H is close to the identity and uniformly continuous, one can get H restricted to the hyperbolic leaf

$$ \begin{align*} H:\mathcal{F}_F^{s}(x) \to \widetilde{\mathcal{L}}^{s}(H(x)), \end{align*} $$

is a homeomorphism. Then, according to the global product structure, it follows that H preserves the center leaf, and maps two distinct center leaves to the different lines in $\widetilde {\mathcal {L}}^{u}$ , namely, H is a bijection on the space of center leaves. Thus, H is surjective when restricted to center leaves.

Let us equip ‘ $\prec $ ’ the order corresponding to the relative orientation on each leaf of $\mathcal {F}^c_F$ . Without loss of generality, we assume that F preserves this orientation. Now, we prove that the semi-conjugacy H is monotonic along center leaves. If not, taking $x\prec y\prec z$ along a center leaf $J^c$ with $H(x)=H(z)\neq H(y)$ , the forward iterates of A will expand the distances between $H(y)$ and $H(x)$ , $H(z)$ . By the fact that H is a finite distance from the identity, one has

$$ \begin{align*} d_{\mathcal{F}_F^c}(F^n(x),F^n(y))\to \infty \quad\text{and}\quad d_{\mathcal{F}_F^c}(F^n(y),F^n(z))\to \infty. \end{align*} $$

In addition, the order $F^n(x)\prec F^n(y)\prec F^n(z)$ implies

$$ \begin{align*} d_{\mathcal{F}_F^c}(F^n(x),F^n(z))\to \infty.\end{align*} $$

Using facts that center leaves are quasi-isometrically embedded and H is uniformly continuous,

$$ \begin{align*} |A^nH(x)-A^nH(z)|\to \infty,\end{align*} $$

which contradicts $H(x)=H(z)$ .

Based on the monotonicity of H along center leaves, for any $x\in \mathbb {R}^2$ , the non-empty set $H^{-1}(x)$ , which is a compact subset of some center leaf $J^c$ , will be connected; hence, it consists of either a single point or a compact segment inside the center leaf $J^c$ . Denote the two endpoints of $H^{-1}(x)$ by $x_1$ and $x_2$ . Since $\mathcal {F}^c_F$ is quasi-isometric (Proposition 2.7), one has

$$ \begin{align*} d_{\mathcal{F}^c_F}(x_1,x_2)\leqslant C_1|x_1-x_2|+C_2. \end{align*} $$

It follows from $\|H-\mathrm {Id}_{\mathbb {R}^2}\|_{C^0}<C_H$ that

$$ \begin{align*} d_{\mathcal{F}^c_F}(x_1,x_2)\leqslant 2C_HC_1+C_2. \end{align*} $$

Then, we obtain $C\triangleq 2C_HC_1+C_2$ , the uniform upper bound of the length of $H^{-1}(x)$ .

Proof. It suffices to prove the existence of the center bundle $E^c_f$ . If so, the unique integrability of $E^c_f$ and the property of $h^{-1}$ follow immediately from Propositions 2.7 and 2.9.

When f is $cu$ -DA, the existence of the $Df$ -invariant center bundle $E^c_f$ is guaranteed by iterating backward the center cone-field along positive orbits.

Now, let f be $sc$ -DA. Since f is semi-conjugate to A, the lifted semi-conjugacy $H:\mathbb {R}^2\to \mathbb {R}^2$ satisfies $H(x+n)=H(x)+n$ for any $x\in \mathbb {R}^2$ and $n\in \mathbb {Z}^2$ . Hence, one has

$$ \begin{align*} H(\mathcal{F}^c_F(x+n)) &= \widetilde{\mathcal{L}}^{u}(H(x+n))=\widetilde{\mathcal{L}}^{u}(H(x)+n)\\ &=\widetilde{\mathcal{L}}^{u}(H(x))+n=H(\mathcal{F}^c_F(x))+n =H(\mathcal{F}^c_F(x)+n). \end{align*} $$

Since H is homeomorphic along $\mathcal {F}^s_F$ , it follows from the global product structure that

$$ \begin{align*}\mathcal{F}^c_F(x+n) =\mathcal{F}^c_F(x)+n\end{align*} $$

for any $x\in \mathbb {R}^2$ and $n\in \mathbb {Z}^2$ . Note that the center direction of f is continuous with respect to f-orbits, and the natural projection of the set of F-orbits is dense in $\mathbb {T}^2_f$ . Hence, one can get the existence of the $Df$ -invariant center bundle $E^c_f$ .

Proof. For the first item, let $x\in \Gamma $ . Then,

$$ \begin{align*} H^{-1}\circ H(F^{\pm 1} (x)) = H^{-1}\circ A^{\pm 1}\circ H(x)= F^{\pm 1} \circ H^{-1}\circ H(x) = \{F^{\pm 1} (x) \} .\end{align*} $$

Hence, $F^{\pm 1}(x)\in \Gamma $ . It follows from the continuity that $\bar {\Gamma }$ is $F^{\pm 1}$ -invariant.

For the second item, let $x\in \Gamma $ and $n\in \mathbb {Z}^2$ . We claim that $H^{-1}\circ H(x+n)=\{x+n\}$ . Indeed, let $y\in H^{-1}\circ H(x+n)$ , then $H(y)=H(x+n)=H(x)+n$ , that is, ${H(y-n)=H(x)}$ . Since $x\in \Gamma $ , one has $y=x+n$ . Hence, $\Gamma +\mathbb {Z}^2=\Gamma $ and $\bar {\Gamma }+\mathbb {Z}^2=\bar {\Gamma }$ .

Now, the set $\Lambda :=\pi (\bar {\Gamma })$ satisfies $\pi ^{-1}(\Lambda )=\bar {\Gamma }$ and is $f^{\pm 1}$ -invariant by the first item. Indeed, let $x\in \Lambda $ . Then, as a set, $f^{\pm 1}(x)=\pi \circ F^{\pm 1} \circ \pi ^{-1}(x)$ . Since $ \pi ^{-1}(x)\subset \bar {\Gamma }$ and $\bar {\Gamma }$ is $F^{\pm 1}$ -invariant, one has $F^{\pm 1} \circ \pi ^{-1}(x)\subset \bar {\Gamma }$ and $f^{\pm 1}(x)\subset \Lambda $ .

Note that $\Lambda =\pi (\bar {\Gamma })=\overline {\pi (\Gamma )}$ since $\bar {\Gamma }$ is $\mathbb {Z}^2$ -periodic. It is clear that

(2.9) $$ \begin{align} \pi(\Gamma) =\{x\in\mathbb{T}^2\mid h^{-1}\circ h(x) =\{x\}\}, \end{align} $$

since H commutes with the deck transformations. However, by Proposition 2.9,

(2.10) $$ \begin{align} \bar{\Gamma}= \bigcup_{x\in\mathbb{R}^2} \{ y\in \mathbb{R}^2\mid y\ \text{is an endpoint of} \ H^{-1}\circ H(x)\}, \end{align} $$

since for any $x\in \mathbb {R}^2\setminus \Gamma $ , every endpoint of $H^{-1}\circ H(x)$ can be approached by points in $\Gamma $ . Combining (2.9) and (2.10), one just has (2.8).

For the third item, by Proposition 2.9, $\bar {\Gamma }$ is $\mathcal {F}^{s/u}_F$ -saturated. Let us consider $f|_{\Lambda }$ for the cases of $sc$ -DA and $cu$ -DA.

• • In the $sc$ -DA case, it follows from Corollary 2.10 that f is special on the whole $\mathbb {T}^2$ . In addition, the foliation $\mathcal {F}^s_f$ exists and is the projection of $\mathcal {F}^s_F$ . Hence, $\Lambda $ is $\mathcal {F}^s_f$ -saturated.

• • In the $cu$ -DA case, let $x\in \Gamma $ and $n\in \mathbb {Z}^2$ . Then,

  $$ \begin{align*} H( \mathcal{F}^u_F(x+n)-n ) &= H(\mathcal{F}^u_F(x+n))-n \\ &= \widetilde{\mathcal{L}}^u (H(x)+n)-n = \widetilde{\mathcal{L}}^u (H(x))=H(\mathcal{F}^u_F(x)). \end{align*} $$

  Since $x\in \Gamma $ , one has $\mathcal {F}^u_F(x+n)-n = \mathcal {F}^u_F(x)$ for any $n\in \mathbb {Z}^2$ . By the continuity of $\mathcal {F}^u_F$ , $\mathcal {F}^u_F(z+n) = \mathcal {F}^u_F(z)+n$ for any $z\in \bar \Gamma $ and $n\in \mathbb {Z}^2$ . It follows that $E^u_F(z+n)= DT_n(E^u(z))$ for any $z\in \bar \Gamma $ and $n\in \mathbb {Z}^2$ . Recall that $T_n:\mathbb {R}^2\to \mathbb {R}^2$ is the deck transformation. Hence,

  $$ \begin{align*}T_{\Lambda}=E^c_f\oplus D\pi(E^u_F)\end{align*} $$
  is an invariant hyperbolic splitting, $f|_{\Lambda }$ is special, and $\Lambda $ is $\mathcal {F}^u_f$ -saturated.

For the fourth item, we just prove for $\sigma =+$ and $\mathcal {F}^u_f$ , since other cases are similar. Note that we do not know $f|_{\mathbb {T}^2\setminus \Lambda }$ is special yet, so it will be convenient to deal with this case in $\mathbb {R}^2$ . We equip the same orientation to $\mathcal {F}^c_F$ . Assume that there exist $x=x^+\in \bar \Gamma $ and $y\in \mathcal {F}^u_F(x)$ with $y\neq y^+$ . Then, one has $y\prec y^+$ . Let $z = \mathcal {F}^u_F(y^+)\cap \mathcal {F}^c_F(x^+)$ , one has $x^+\prec z$ by the holonomy map along unstable foliation. However,

$$ \begin{align*}H(\mathcal{F}^u_F(z))=H(\mathcal{F}^u_F(y^+))=H(\mathcal{F}^u_F(y))=H(\mathcal{F}^u_F(x^+)),\end{align*} $$

which implies $H(x^+)=H(z)$ . It contradicts the choice of $x^+$ .

For the fifth item, let U be an open set with respect to the topology of $\Lambda $ . Then, $h(U)$ is an open set in $\mathbb {T}^2$ (see a precise proof in the next independent Lemma 2.12). It follows that there exists $p'\in \mathrm {Per}(A)\cap h(U)$ . Hence, the extreme point(s) of $H^{-1}(p')$ will be periodic in U.

Proof. The existence of center leaf $\mathcal {F}^c_f(x)$ for every $x\in \mathbb {T}^2$ is guaranteed by Corollary 2.10.

For the first item, let $x=x^+$ and $y\in \mathcal {F}^c_f(x)$ such that $x^+\prec y$ . By contradiction, we assume that for every $z\in \mathcal {F}^c_f(x)$ with $x^+\prec z\prec y$ , the set $h^{-1}\circ h(z)$ is a closed interval. From this assumption, two such adjacent intervals intersect with each other; hence, the union of these intervals is an interval that we denote by I. In particular, for every point $z\in I\subset \mathcal {F}^c_f(x)$ , one has $h(z)=h(z_0)$ , where $z_0\in \mathcal {F}^c_f(x)$ is a given point such that ${x^+\prec z_0\prec y}$ . Then, one can take $z_n\in I$ such that $z_n\to x^+$ with $x^+\prec z_n$ and $h(z_n)=h(z_0)$ . Since h is continuous,

$$ \begin{align*}h(x^+)=\lim_{n\to +\infty}h(z_n)=\lim_{n\to +\infty}h(z_0)=h(z_0).\end{align*} $$

It contradicts the fact that $x^+\prec z$ .

For the second item, we just prove the case of $\sigma =+$ and the other case is similar. Let $x\in \mathbb {T}^2$ and $y_n\in \mathcal {F}^c_f(x)$ with $y_n\to x^+$ . Taking a subsequence, one has $x^+\prec y_n$ or $y_n\prec x^+$ for all $n\in \mathbb {N}$ .

If $x^+\prec y_n$ for all $n\in \mathbb {N}$ , by the first item, there exists $z_n\in \mathcal {F}^c_f(x)$ such that ${x^+\prec z_n\prec y_n}$ and $h^{-1}\circ h(z_n)=\{z_n\}$ . Here, one can assume that

$$ \begin{align*}x^+\prec y_{n+1}^- \preceq y_{n+1}^+\prec z_n\prec y_n^-.\end{align*} $$

Note that $y_n \to x^+$ , one has $z_n\to x^+$ . Hence, $y_{n+1}^+\to x^+$ and $y_{n+1}^-\to x^+$ as $n\to +\infty $ .

Let $y_n\prec x^+$ for all $n\in \mathbb {N}$ . If $h^{-1}\circ h(x)$ is an interval, then for large n, we have ${y_n\in h^{-1}\circ h(x)}$ ; hence, $y_n^+=x^+$ . If $h^{-1}\circ h(x)=\{x\}$ , by the first item again, there exists ${z_n\in \mathcal {F}^c_f(x)}$ such that

$$ \begin{align*}y_n\prec z_n\prec x^- \quad\text{and}\quad h^{-1}\circ h(z_n)=\{z_n\}.\end{align*} $$

Hence, one can assume that

$$ \begin{align*}y_n\preceq y_n^+\prec z_n\prec y_{n+1}^- \preceq y_{n+1}^+\prec x^-=x^+.\end{align*} $$

It follows from $y_n\to x=x^+=x^-$ that $y^-_{n+1}\to x^+$ and $y^+_{n+1}\to x^+$ as $n\to +\infty $ .

Finally, we prove the third item. Denote by $\mathcal {L}^c$ the foliation corresponding to $H(\mathcal {F}^c_F)$ on $\mathbb {R}^2$ . Note that H is homeomorphic along each leaf of the hyperbolic foliation $\mathcal {F}^{s/u}_F$ , and the center foliation $\mathcal {F}^c_F$ and hyperbolic one $\mathcal {F}^{s/u}_F$ admit the product structure. Hence, the third item follows from that for every $x\in \Lambda $ and ${\varepsilon }>0$ , the ${\varepsilon }$ -ball $B_{\varepsilon }^c(x)$ on center leaf $\mathcal {F}^c_f(x)$ satisfies that $h(B_{\varepsilon }^c(x)\cap \Lambda )$ has non-empty interior on $\mathcal {L}^c(h(x))$ .

Indeed, let ${\varepsilon }>0$ and $x=x^+$ (or symmetrically $x=x^-$ ). By the second item, there exists $y_n\in \mathcal {F}^c_f(x) \cap \Lambda $ such that $y_n=y_n^+\to x$ as $n\to \infty $ . In particular, there exists ${y_0=y_0^+\in B_{\varepsilon }^c(x)\cap \Lambda }$ and $x\prec y_0$ . Since h is continuous and monotonic along the center leaf, $h( [x,y_0] \cap \Lambda )=h( [x,y_0])$ is an interval contained in $h(B_{\varepsilon }^c(x)\cap \Lambda )$ .