Proof. We first prove that in every sufficiently fine refinement, the periodic points in the periodic orbit are in distinct rectangles.

Notice that taking backward refinements makes the height of the rectangles uniformly (exponentially) small. Also, the largeness of the stable manifold of periodic points ensures that taking forward refinement makes the width of the rectangles to which the periodic points belong uniformly small. Thus, by taking sufficiently large (both in forward and backward) refinements, we can assume that the rectangles which contain a point of an orbit of a periodic point are uniformly small. In particular, none of them can coincide.

Now let us see how to construct the paths of transition rectangles. First, by the argument of the first step, we can assume that for each rectangle of a cycle of periodic rectangles, they contain homo/heteroclinic points only in the local stable/unstable manifolds by taking sufficiently fine refinements. Then, notice that while taking refinements increases the number of rectangles, such newly created rectangles are contained in the initial periodic rectangles and they contain at most one homo/heteroclinic point of a given homo/heteroclinic orbit. Thus, we only need to prove that by taking refinements, we can separate the points of homo/heteroclinic points outside the periodic rectangles. Since the diameters of periodic rectangles tend to zero by taking refinements and the transition rectangles in the refinements are images of periodic rectangles, we can assume that the diameters of rectangles containing the point of homo/heteroclinic points outside the periodic rectangles also tend to zero as we take finer refinements. Thus, they are separated into different rectangles for every sufficiently fine refinements.

This completes the proof.

Proof. We prove it for $(0, 1)$ -refinement. The general case is similar. Let $C^{\prime }_1$ , $C^{\prime }_2$ be rectangles in $\mathbf {R}\cap f(\mathbf {R})$ and assume $f(C^{\prime }_1) \cap C^{\prime }_2 \neq \emptyset $ . We will show $f^{-1}(f(C^{\prime }_1) \cap C^{\prime }_2) = C^{\prime }_1 \cap f^{-1}(C^{\prime }_2)$ is connected. Let us take a rectangle $C_1$ of $\mathbf {R}$ which contains $C^{\prime }_1$ . Then, $C^{\prime }_1$ is a vertical subrectangle of $C_1$ and $f^{-1}(C^{\prime }_2)$ is a horizontal subrectangle of $C_1$ (see [Reference Bonatti and DíazBS₂ , §2.3] for the details). Thus, they intersect and there is a unique connected component.

Proof. Let $\sigma $ be the local unstable manifold of $f^{d}(q_1)$ . It contains x and y. Let $C_i$ be the rectangle which contains $f^i(x)$ and $f^i(y)$ for $i=0,\ldots ,T$ . By using the invariance of the cone field and the generating property of $\mathbf {R}$ , we see that the part of $\sigma $ whose image under f is in $C_1$ is a connected curve. Inductively, the part of $\sigma $ whose image under $f^i$ is in $C_i$ for every $0\leq i \leq k$ is a connected curve for every $0\leq k \leq T$ . Let us denote the curve by $\sigma _k$ , and consider $\sigma _T$ . By definition, we know $x, y \in \sigma _T$ and $f^T(\sigma _T)$ has unique intersection with the local stable manifold of $f^a(q_2)$ due to the invariance of the cone field. Thus, we know $f^T(x) = f^T(y)$ and consequently $x=y$ .

Proof of Proposition 2.12

The proof is similar to [Reference Bonatti and DíazBD₃ , Proposition 3.6].

First, we apply Franks’ lemma along the orbit of q. More precisely, we take a diffeomorphism $f_1$ that satisfies the following.

• • For each $j \in \mathbb {Z}$ , $f^j_1(q) = f^j(q)$ . In particular, q is a periodic point for $f_1$ with the same orbit.

• • $\mathrm {supp}(f_1, f) \subset W$ .

• • For each j, there exists a coordinate neighborhood $V_j$ of $f_1^j(q) = f^j(q)$ such that the dynamics of $f_1$ on $V_j$ is given by the linear map $df(f^j(q))$ .

We fix such $f_1 \in {\cal U}$ . This is the only part where we perform the perturbation along ${\cal O}(q)$ . Note that this does not change the derivatives along $\mathcal {O}(q)$ .

By choosing $f_1$ sufficiently close to f, we may assume that $\mathbf {R}$ is still a filtrating Markov partition and q has a large stable manifold for $f_1$ as well. Now let us take the refinements. As we take finer refinements, due to the fact that q has a large stable manifold, the rectangle to which $f_1^i(q)$ belongs shrinks to $\{f_1^i(q)\}$ . Thus, we may assume that each corresponding periodic rectangle is contained in the linearized coordinates. Notice that in this coordinate system, $f_1^i(q)$ is mapped to the origin. We assume that the $xy$ -plane coincides with the $E^{cs}$ direction and the z-axis coincides with the $E^u$ direction.

We show that by taking the refinement and slightly perturbing $f_1$ , we have that the periodic rectangles are product rectangles in the linearized coordinates. First, let us see how to make the lid boundary flat. For each i, the intersection between $W_{\mathrm {loc}}^u(f_1^i(q))$ and $\partial _l(K_{f^i(q)})$ consists of two points. We denote them by $y_{i, +}$ and $y_{i, -}$ . Now we perform a $C^1$ -perturbation whose support is contained in a small neighborhood of $f_1(y_{i, \pm })$ so that for $f_2$ (the perturbed diffeomorphism), the preimage of the image of the lid boundary near $y_{i \pm }$ is parallel to the $xy$ -plane. Note that the size of the perturbation can be chosen arbitrarily small by taking refinement (due to the partial hyperbolicity near the periodic point), and by taking sufficiently fine refinement in advance, we can guarantee that the support of the perturbation is contained in W.

Since the periodic rectangles converges to $W_{\mathrm {loc}}^u(f_2^i(q))$ by taking the forward refinement and the property that the lid boundary is flat near $y_{i, \pm }$ is not affected by taking the forward refinement (note that taking the forward refinement does not change the repelling set but just replace the attracting set), we have that the periodic rectangles for $\{f_2^i(q)\}$ has a flat lid boundary by taking a sufficiently fine refinement.

Next, let us see how to make the side boundary flat. The argument is essentially the same. Let us consider the rectangle $K_{f_2^i(q)}$ in the linearized coordinates. Then in the linearized coordinates, $W_{\mathrm {loc}}^s(f_2^i(q))$ is a flat plane which coincides with the $xy$ -plane locally. For $K_{f_2^i(q)}$ , we fix a $C^1$ -circle $B_i := \partial _s K_{f_2^i(q)}\cap W_{\mathrm {loc}}^s(f_2^i(q))$ . Now near $f^{-1}_{2}(B_i)$ , we perform a $C^1$ -small perturbation so that for $f_3$ (the perturbed map), the image of the preimage of the side boundary near $B_i$ is flat. Now we take backward refinement: notice that taking backward refinement does not destroy the flatness of the lid boundary. However, by the uniform contraction property, we know that as the number of the refinement tends to infinity, we have that the rectangle containing $K_{f_3^i(q)}$ converges to the local stable manifold of $f_3^i(q)$ . Thus, at some moment, all the side boundaries of $\{K_{f_3^i(q)} \}$ turn to be flat. In particular, the rectangle containing $f_3^i(q)$ is now flat.

Thus, letting $g = f_3$ , we obtain the conclusion.

Proof. The proof is very similar to the proof of Proposition 2.12, but it requires some extra care.

Step 1. A perturbation along the transition map. First, we give an auxiliary perturbation. Let us consider the $(m, n)$ -refinement of $\mathbf {R}$ . Then Q is substituted by $f^{-m}(Q)$ and its transition time is $T+m+n$ . Then consider a perturbation whose support is contained in a small neighborhood of $f^{-m}(Q)$ and $f^{M+n-1}(Q)$ such that $D(f^{m+M+n})(Q)$ preserves the center stable direction and the strong unstable direction of the linearized coordinates. Note that, by letting m and n be large, the $C^1$ -size of the perturbation tends to zero, thanks to the hyperbolicity near the periodic orbits. Thus, by giving this perturbation, we may assume that the transition map $Df^T(Q)$ preserves the center stable direction and the strong unstable direction of the linearized coordinates from the very beginning. Note that this property is preserved by taking refinements.

Step 2. Linearizing local dynamics. Now let us consider the orbit $\{f^k(Q)\}_{k=0,\ldots ,T-1}$ . By applying Franks’ lemma along $\{f^k(Q)\}$ , we obtain a diffeomorphism $f_1$ close to f for which we have linearized coordinates around $\{f_1^k(Q)\}$ for every k. By taking the support sufficiently small, we may assume that the support is contained in W and the perturbation does not disturb the linearization property of the periodic rectangles. We also assume that in the linearized coordinates, $f_1^k(Q)$ is mapped to the origin, the $xy$ -plane coincides with the center-stable direction, and the z-axis coincides with the unstable direction. Note that since $Df^T(Q)$ preserves the strong unstable and the center-stable direction in the linearized coordinates (see Step 1), these new coordinates are compatible with those in the periodic rectangles.

Step 3. Obtaining the product rectangles. Then we take refinements. Due to the largeness of the stable manifolds of $q_j$ , the diameter of transition rectangles goes to zero as we take refinements. Thus, by taking sufficiently fine refinements, we have that the corresponding transition rectangles outside the periodic rectangles (those before the refinement) are contained in the domain of linearized coordinates. In particular, we may assume that they are in W.

One thing which is different from the proof of Proposition 2.12 is that taking refinements increases the number of transition rectangles. Notice that these new rectangles are contained in the periodic rectangles (those before the refinement). Thus for these newly produced rectangles, we can furnish linearized coordinates just by restricting the linearized coordinates for the cycle of periodic rectangles and thus the increase of rectangles does not cause any problem for constructing linearized coordinates.

Now let us see how to make the lid and the side boundaries of the rectangles flat. The argument is almost the same as the proof of Proposition 2.12, so we discuss only for the lid boundary. First, for each $L_{f^k_1(Q)}$ , we take the intersection of $W_{\mathrm {loc}}^u(f_1^k(Q))$ and the lid boundary. There are two such points. Then we slightly perturb $f_1$ so that the lid boundary near the intersection points is flat plane.

Then we take a forward refinement. Notice that taking $(0, n)$ -refinement increases the number of transition rectangles by n. However, the rectangles which are newly created are included in the linearized region of the periodic rectangles. Thus, we know that their lid boundaries are flat. For the rest of the rectangles, since taking the forward refinement shrinks the cylinders in the center stable direction, we know that up to some sufficiently large forward refinement, we have that all of the transition rectangles have flat lid boundaries.

By the same argument, we can make the side boundary of the transition rectangles flat as well.

Proof of Proposition 2.17

For a circuit of rectangles $\mathbf {R}(S)$ , by applying Proposition 2.18 to each path one by one, we linearize all the homo/heteroclinic orbits. For that, we need to confirm that we can apply Proposition 2.18 without destroying the linearized coordinates which are already obtained. Let us explain how to avoid the interference.

First, in the very beginning, we apply the perturbation along each homo/heteroclinic orbit so that the derivative of each transition map preserves the center-stable and the unstable direction of the linearized coordinates for periodic rectangles.

Then, let us consider a path of transition rectangles $\{L_{f^j(Q)}\}_{1\leq j \leq T-1}$ for a homo/heteroclinic point Q. If $\{L_{f^j(Q)}\}$ does not contain any transition rectangles which are already linearized, then by applying Proposition 2.18, we can linearize the homo/heteroclinic orbit $\{ f^j(Q) \}$ .

If not, we consider the refinements. Since we assume that every periodic point of the circuit has a large stable manifold, as we take refinements, every rectangle shrinks to a point. Thus, we may assume that the (finitely many) rectangles $L^{\prime }_{Q}, \ldots , L^{\prime }_{f^T(Q)}$ (these Q and T are the same as the initial ones, we do not take the corresponding points and transition times) are distinct and none of them contain any other homo/heteroclinic points of the circuit.

Then, we follow the procedure of Proposition 2.18. By applying Franks’ lemma along $Q, \ldots , f^T(Q)$ , we linearize the local dynamics along Q. Note that taking refinements increases the number of transition rectangles, but by assumption we know that for these newly created rectangles, we can endow linearized coordinates just by taking restrictions. Thus, at this moment, the dynamics along the orbit of Q is linearized. Then we need to make the boundaries of the transition rectangles flat, but this can be done in the same way as in the proof of Proposition 2.18.

Repeating this argument, we can obtain the desired coordinates for every rectangle of the circuit.

Proof. Let us see the perturbations which make rectangles in a cycle of periodic rectangles and a path of transition rectangles adapted. Then applying these perturbations one by one, we obtain the conclusion.

First, note that due to the largeness of the stable manifolds of periodic points in S, every sufficiently fine refinement $\mathbf {R}'$ satisfies $\mathbf {R}'(S) \subset W$ . Thus, we may assume that this holds from the very beginning.

Let us see how to make the cycles of periodic rectangles adapted. Let $\{K_{f^i(q)}\}$ be a cycle of periodic rectangles for q. We consider $f(\mathbf {R}) \cap K_{f^i(q)}$ . Recall that as we take backward refinements, the height of the rectangle to which $f^i(q)$ belongs decreases and tends to zero, while $W^{s}_{\mathrm {loc}}(q)$ does not change. By assumption, the image rectangles $f(\mathbf {R}) \cap K_{f^i(q)}$ containing a point of S are product rectangles. For the other image rectangles, they automatically have flat lid boundaries but possibly with non-flat side boundary. Note that if we take enough forward refinement, each rectangle in $f(\mathbf {R}') \cap K^{\prime }_{f^i(q)}$ is almost vertical thanks to the partial hyperbolicity near ${\cal O}(q)$ .

Then by giving a $C^1$ -small perturbation, we can construct a diffeomorphism $f_1$ close to f such that each connected component of $f_1(\mathbf {R}') \cap K^{\prime }_{f_1^i(q)}$ has flat boundaries near $W_{\mathrm {loc}}^s(f_1^i(q))$ . Since these perturbations are only for the boundary of rectangles which are not periodic rectangles and transition rectangles, we may assume that the support is contained in W and disjoint from S. Also, we may assume that these perturbations do not disturb the linearization property on the periodic rectangles and transition rectangles.

Now, by taking backward refinement, we have the product property for $f_1(\mathbf {R}^{\prime \prime }) \cap K^{\prime \prime }_{f_1^i(q)}$ . Thus, we can obtain the adaptedness for a cycle of periodic rectangles. By repeating this perturbation, we may assume that every cycle of periodic rectangles is adapted.

The proof for the transition rectangles can be done similarly. The only thing to which we need to pay extra attention is that by taking refinements, the number of transition rectangles increases. Meanwhile, once we have the adaptedness for periodic rectangles, this does not bring any problem for the following reasons.

• • Taking the $(0, n)$ -refinement increases the number of transition rectangles by n. It adds n rectangles to the tail of the path. However, the first n transition rectangles are images of periodic rectangles which are adapted. Thus the number of rectangles which are not adapted are the same.

• • Taking $(m, 0)$ -refinement increases the number of transition rectangles by m. It adds m rectangles in the head. However, the first m transition rectangles are contained in periodic rectangles which are adapted. Thus, the number of rectangles which are not adapted are the same.

In short, some of newly added rectangles are automatically adapted. Hence, to obtain the adaptedness for transitions, we only need to repeat the argument for finitely many rectangles.

Proof. The construction of g can be done by standard arguments involving the partition of unity and the closeness of G to F in the $C^1$ -distance. We just give a sketch of the proof.

Assume that on $D_1 \times J$ , the map $\phi _2 \circ f \circ \phi _1 ^{-1}$ is given by

$$ \begin{align*} (x, y, z) \mapsto (F(x,y), \unicode{x3bb} z) \end{align*} $$

in some neighborhood of $D_1 \times J$ . Then, we choose a $C^1$ -function $\rho $ defined in the interval $J'$ which contains J in the interior and satisfies the following:

• • $0 \leq \rho (z) \leq 1$ ;

• • $\rho (z) \equiv 1$ on J;

• • $\rho (z)\equiv 0$ near the endpoints of $J'$ .

Then, given G, consider the following map:

$$ \begin{align*} (x, y, z) \mapsto ((1-\rho(z))F(x,y) + \rho (z)G(x,y), \unicode{x3bb} z). \end{align*} $$

Considering the fact that $F \equiv G$ near the boundary of $D_1$ , this map is equal to $\phi _2 \circ F~\circ ~\phi _1 ^{-1}$ near the boundary of $D_1 \times J'$ , thus extends to outside $D_1 \times J'$ so that it coincides with the unperturbed map. If $\varepsilon _0$ and $\varepsilon _1$ are sufficiently small, then one can check that this defines a diffeomorphism on each slice by the $xy$ -plane. The surjectivity of the map is the consequence of a standard algebraic topological argument. The injectivity follows if we choose F sufficiently close to G.

Now we measure the $C^1$ -size of this perturbation. By a direct calculation, the difference of the derivatives of f and g in the local coordinates is given by

$$ \begin{align*} \begin{pmatrix} \rho (z)D_{x, y}(G(x, y) -F(x,y)) & 0 \\ \rho' (z)(G(x, y) -F(x,y)) & 0 \end{pmatrix}, \end{align*} $$

where $D_{x, y}$ denotes the Jacobi matrix with respect to x and y. This calculation shows that the $C^1$ -distance is given by two terms $|\rho (z)D_{x,y}(G(x, y) -F(x,y))|$ and $|\rho ' (z)(G(x, y) -F(x,y))|$ . The supremum norm of the first one is proportional to the $C^1$ -distance between F and G, and the second one is to $|\rho ' (z)|$ times the $C^0$ -distance between F and G. Thus, the $C^1$ -distance of the perturbation itself is given by the form $\varepsilon _1 + K \varepsilon _0$ , where K is determined by $\rho '$ , which depends only on the shape of $J'$ .

Proof. We prove the case $n =1$ , that is, $F^{-1}(R_1 \cap F({\cal D}) ) \subset \mathrm {Int}_t( R_1)$ . The general case follows by induction. Notice that

$$ \begin{align*} R_{n+1} &= F^{-1}(F^{-n}(R \cap F^{n+1}({\cal D}) ))\\ &= F^{-1}(F^{-n}(R\cap F^{n}({\cal D}) \cap (F^{n+1}({\cal D})))=F^{-1}(R_n\cap F({\cal D})). \end{align*} $$

Consider a point $x \in F^{-1}(R_1 \cap F({\cal D}) )$ . By definition, there is a point $y \in {\cal D}$ such that $x =F^{-2}(y)$ . Then consider the point $F^{-1}(y)$ , which belongs to $R_1 \cap F({\cal D})$ . We know $F^{-1}(y) \in F({\cal D})$ . By the fact that R is a relatively repelling region and $F^{-1}(y) \in R_1$ , we deduce that $F^{-1}(y) \in \mathrm {Int}_t(R)$ . Then, by taking the inverse image, we have $x =F^{-2}(y) \in \mathrm {Int}_t(F^{-1}(R \cap F({\cal D}))) = \mathrm {Int}_t(R_1)$ .

Proof. We prove the case $n=1$ . Then the general case follows by induction. Consider the set $\tilde {R}:=R\cup \overline {({\cal D}\setminus F({\cal D}))}$ and take a point $x \in \tilde {R}$ . If $x \not \in F({\cal D})$ , then there is nothing we need to prove. Suppose $x \in F({\cal D})$ . Then, we have either $x \in F^2({\cal D})$ or not. For the first case, by the definition of a relatively repelling region, we know that $F^{-1}(x) \in \mathrm {Int}_t(R) \subset \mathrm {Int}_t(\tilde {R})$ . For the latter case, $F^{-1}(x) \in {\cal D} \setminus F({\cal D}) \subset \mathrm {Int}_t(\tilde {R})$ . This shows the relatively repelling property of $\tilde {R}$ .

Proof. For each connected component $\gamma _i$ of $\Gamma $ , we take a smooth vector field satisfying the following:

• • it is perpendicular to $\gamma _i$ and has the form ${dx}/{dt} = [\log (1+ \kappa )]x$ except near the endpoints;

• • the support of the vector field is in a small neighborhood of $\gamma _i$ ;

• • the vector field is null near the endpoints.

Choosing the vector field adequately, the time- $1$ map of this vector field satisfies the conclusion.

Proof. The proof consists of two steps.

Step 1. Construction of a relatively repelling region. We apply Proposition 3.18 to $\Gamma $ letting $\kappa = 2\eta $ . We obtain a family of diffeomorphisms $\{\xi _i\}$ satisfying the conclusion. Then, by direct calculation, one can check that for $F_1:= \{\xi _{\mathrm {im}(f_j)} \circ f_j \}$ , $\Gamma $ is a family of normally repelling curves with normal strength at most $3\eta $ , if $\eta $ is sufficiently close to $0$ . Notice that the condition that $\xi _i$ is not necessarily expanding near the endpoints does not affect the conclusion, since the points near the endpoints go out from the discs by the backward iteration. By Remark 3.14, this gives us a repelling region R near $\Gamma $ and it can be chosen arbitrarily close to $\Gamma $ .

Step 2. Construction of an attracting region in R. Then, for this $F_1$ , we apply Proposition 3.18 letting $\kappa = -3\eta $ . It gives another set of diffeomorphisms $\{\theta _i\}$ such that for the Markov IFS $F_2 := \{ \theta _{\mathrm {im}(f_j)}\circ \xi _{\mathrm {im}(f_j)} \circ f_j \mid f_j \in F \}$ , $\Gamma $ is a family of univalent, normally attracting, contracting invariant curves. We choose $\{\theta _i\}$ in such a way that its support is contained in the relatively repelling region R which we constructed in Step 1. Since $\Gamma $ is contracting in the tangential direction, for each $\gamma _i$ , we can find a disc $A_i$ in R containing $\gamma _i$ such that $\bigcup A_i$ is a relative attracting region with respect to R and $F_2$ is uniformly contracting. Thus by letting $\tau _i := \theta _i \circ \xi _i$ , we complete the proof.