Proof. Choose small $\nu $ and consider the image of the left-hand side and the right-hand side of $\bar \Pi _\nu $ under $H_a \circ G$ . Join endpoints of these curves by segments. If ${\nu =\nu (\Omega ^{{\mathbb {R}}})>0}$ is sufficiently small, a is sufficiently close to zero and $G\in \Omega $ is sufficiently close to $\Omega ^{{\mathbb {R}}}$ , the bounds on $G', G"$ imply that these curves bound a curvilinear rectangle K with non-self-intersecting boundary and $K\subset H_a \circ G(\bar \Pi _\varepsilon )$ .

The map $H_a\circ G ( G^{-1}\circ H^{-1}_a(z)+1)$ takes the left-hand side of K to its right-hand side. Consider a smooth map $\Xi $ that depends analytically on $a, G$ , takes $\bar \Pi _\nu $ to K and conjugates $z\to z+1$ to $H_a\circ G ( G^{-1}\circ H^{-1}_a(z)+1)$ . This map $\Xi $ is a linear homotopy between the restrictions of the map $ H_a\circ G$ to the left and to the right boundary of $\bar \Pi _\nu $ : that is,

$$ \begin{align*} \Xi (x+iy)= H_a G (-0.5+iy)\cdot (0.5-x) + H_a G (0.5+iy) \cdot (0.5+x). \end{align*} $$

Note that if G preserves the real line and a is real, then $\Xi $ preserves the real line.

We verify that the fraction $| \Xi _{\bar z} / \Xi _z|$ is bounded away from one. We have

$$ \begin{align*} \begin{aligned} \Xi_x &= H_a G (0.5+ iy) - H_a G(-0.5+iy), \\ -i\Xi_y &= (H_a G)'(-0.5+iy) \cdot (0.5-x) + (H_a G)'(0.5+iy)\cdot (0.5+x) \end{aligned} \end{align*} $$

and

$$ \begin{align*} \frac {\Xi_{\bar z}} {\Xi_z} = \frac{\Xi_x+i\Xi_y}{\Xi_x-i\Xi_y}. \end{align*} $$

Owing to the bound on $G'$ , the values $H_0G(0.5)$ , $-H_0G(-0.5)$ , $(H_0G)'(\pm 0.5)$ are positive, uniformly bounded and bounded away from zero for $G\in \Omega ^{{\mathbb {R}}}$ . Owing to our bound on $G"$ , we can choose small $\Omega \supset \Omega ^{{\mathbb {R}}}$ , $\nu $ and $\delta $ so that both $\Xi _x$ and $-i\Xi _y$ are close to ${\mathbb {R}}^+$ , uniformly bounded and bounded away from zero for all $|a|<\delta , (x,y)\in \bar \Pi _\nu $ . Thus, the fraction $| \Xi _{\bar z} / \Xi _z|$ is uniformly bounded away from one.

Define a Beltrami differential $\mu $ on the plane by extending $ \Xi _{\bar z} / \Xi _z$ to the horizontal strip of width $\nu $ periodically and by setting $\mu =0$ outside the strip. Let $\psi $ be a solution of the Beltrami equation $\psi _{\bar z}/\psi _z =\mu $ with the Beltrami differential $\mu $ , normalized by $\psi (0)=0, \psi (1)=1, \psi (\infty )=\infty $ ; it exists owing to the Ahlfors–Bers–Boyarski theorem [Reference Ahlfors and BersAB60]. Set $\pi _{a, G} = \psi \circ \Xi ^{-1}$ . By construction, $\pi _{a,G}$ is an analytic univalent map in K that extends continuously to its boundary and conjugates

$$ \begin{align*}H_a\circ G ( G^{-1}\circ H^{-1}_a(z)+1)\end{align*} $$

to the unit translation; thus, $\pi _{a,G}\circ H_{a} \circ G$ commutes with the unit translation on its domain. If $G\in \Omega ^{{\mathbb {R}}}$ , then $\Xi $ is ${\mathbb {R}}$ -symmetric, and thus the solution of the Beltrami equation is ${\mathbb {R}}$ -symmetric. We conclude that, in this case, $\pi _{a,G}$ preserves the real axis.

We prove that the map $\pi _{a, G}$ depends analytically on $a,G$ . Indeed, $\Xi $ (and thus $\mu $ ) depends on $a,G$ analytically by construction, and $\psi $ depends on $a,G$ analytically due to the Ahlfors–Bers–Boyarski theorem. It is easy to see that an analytic compositional difference of two (non-analytic) maps depends analytically on a parameter if both maps depend analytically on a parameter. This implies the statement.

If $\delta $ , $\Omega \supset \Omega ^{{\mathbb {R}}}$ are sufficiently small, then there exists $\tau _1>0$ such that $H_a G (\bar \Pi _{\tau _1}) \subset K$ for $|a|<\delta ,G\in \Omega $ . Thus, $\pi _{a,G} H_a G$ is well defined in $\bar \Pi _{\tau _1}$ .

A posteriori, we know that the solution of the Beltrami equation

$$ \begin{align*}\psi = \pi_{a,G}\circ\Xi\end{align*} $$

is smooth, with non-degenerate differential in $\bar \Pi _{\nu }$ . Since it depends analytically on G in $\Omega $ , the norm of its differential is bounded from below on $\bar \Pi _{\nu }$ for $G\in \Omega ^{{\mathbb {R}}}$ . The set $\Xi ^{-1} (H_0G(\bar \Pi _{\tau _1}))\subset \bar \Pi _{\nu }$ contains a strip around $[-0.5, 0.5]$ of width uniformly bounded from below, and thus the projection of $\pi _{0,G}(H_0G(\bar \Pi _{\tau _1})) = \psi (\Xi ^{-1} (H_0G(\bar \Pi _{\tau _1}))$ to ${\mathbb {C}}/{\mathbb {Z}}$ contains a strip around ${\mathbb {R}}/{\mathbb {Z}}$ of width uniformly bounded from below. Hence, we can fix ${\tau _2>0}$ so that, for small $\delta $ and $\Omega \supset \Omega ^{{\mathbb {R}}}$ , for $|a|<\delta , G\in \Omega $ , the projection of $\pi _{a, G}H_aG(\bar \Pi _{\tau _1})$ to ${\mathbb {C}}/{\mathbb {Z}}$ contains $\Pi _{\tau _2}$ .

If $H_a G$ commutes with the unit translation, then $H_a\circ G ( G^{-1}\circ H^{-1}_a(z)+1)$ coincides with the unit translation on the left-hand side of the rectangle K. Then $\Xi =\psi =\mathrm {id}$ and thus $\pi _{a,G}=\mathrm {id}$ .

Proof. Set $F (z)= z+f(0)$ , $G(z) = \sqrt [3]{f(z)-f(0)}$ . For G, we choose the branch of the cubic root that preserves the real line for critical circle maps $f\in \mathcal D_{\varepsilon }^{cr, {\mathbb {R}}}$ , which defines the branch unambiguously in a neighborhood of $\mathcal D_{\varepsilon }^{cr, {\mathbb {R}}}$ in $\mathcal D_{\varepsilon }^{cr}$ . Set $j( f) = (F, H_0, G)$ . Then G is well defined in a strip $\Pi _{\varepsilon }$ , and since it contains no preimages of $f(0)$ under f except zero, the map $H_0\circ G=f(z)-f(0)$ commutes with the unit translation, and thus $\pi _{0, G}=\mathrm {id}$ . We get $F\circ \pi _{0, G} \circ H_0 \circ G = f$ .

We have $G(0)=0$ ; to guarantee $G'(0)=1$ , we have to replace G by $c\cdot G$ for an appropriate c. This replaces $\pi _{0, G}$ by $z\mapsto z/c^3 $ and does not affect the equality $F\circ \pi _{0, G} \circ H_0 \circ G = f$ .

Proof. Slightly abusing the notation, we keep the notation $f_\mu $ for a lift $f_\mu \colon \bar \Pi _\varepsilon \to {\mathbb {C}}$ .

We need to satisfy

$$ \begin{align*} h_\mu F_\mu \pi_{a(\mu),G_\mu} H_{a(\mu)} G_\mu h_\mu^{-1}(z) = f_\mu(z)\quad\text{for } G_\mu(0)=0, G^{\prime}_\mu(0)=1. \end{align*} $$

First, we satisfy these equalities without the requirement $G_\mu '(0)=1$ , and then we make the final adjustments.

Since $f_\mu $ for $\mu _1=0$ is a cubic critical map, for small $\mu _1\neq 0$ , $f_{\mu }$ has a pair of quadratic critical points $c_1, c_2$ close to zero, given by $(f_\mu )'(c_{1,2})=0$ . When $\mu _1$ makes one turn around zero, these critical points make half of a turn each and exchange; this follows from the fact that $(f_\mu )'(0)$ has a simple root at $\mu _1=0$ . Namely, it is clear that the pair of critical points depends analytically on $\mu $ when $\mu _1\neq 0$ , and thus we only need to count the number of turns; since

$$ \begin{align*} (c_1)^{\prime}_{\mu_1} = -(f^{\prime\prime}_{z\mu_1}/f^{\prime\prime}_z)|_{c_1}\!\sim \text{ const}/ c_1, \end{align*} $$

we have $c_1\sim \sqrt {\mu _1}$ and this number is 1/2.

Let images of $c_1, c_2$ under f be $d_1, d_2$ . Then $d_1$ , $d_2$ make $3/2$ of the turn each around $f(0)$ when $\mu _1$ makes one turn around zero, and they also exchange since

$$ \begin{align*}f_\mu(c_1)-f_\mu(0) \sim f^{\prime}_z(c_1) c_1 \sim \mu_1^{3/2}.\end{align*} $$

Also, they are complex conjugate for real $\mu $ , $\mu _1<0$ .

Both maps $F_\mu , h_\mu $ will be translations. We set

$$ \begin{align*}F_\mu h_\mu (z)=z+(d_1+d_2)/2.\end{align*} $$

Note that this composition depends analytically on $\mu $ (the translations $F_\mu , h_\mu $ themselves will be chosen later). Then the critical values of the map $(F_\mu h_\mu )^{-1}f_\mu $ are symmetric with respect to $0$ and purely imaginary for real $\mu , \mu _1<0$ . Also, they make $3/2$ of the turn each around $0$ when $\mu _1$ makes one turn around zero. The difference between these points is still $d_1-d_2$ .

Set

$$ \begin{align*} A(\mu)= \bigg(\frac{3\sqrt{3}(d_1-d_2)}{4}\bigg)^{2/3}. \end{align*} $$

We choose the branch of the cubic root that is real for $\mu _1\in {\mathbb {R}}_{-}$ (that is, when $d_1- d_2$ is purely imaginary). Then the critical values of $H_A$ , namely, $\pm {2A}/{3}\sqrt {A/3}$ , coincide with the critical values $\pm {(d_1-d_2)}/{2}$ of $(F_\mu h_\mu )^{-1}f_\mu $ . Since $d_1-d_2$ makes $3/2$ turns around the origin as $\mu _1$ makes one turn around the origin, A depends analytically on $\mu $ and has a simple zero at $\mu _1=0$ .

Consider the three-valued map $H_A^{-1}(F_\mu h_\mu )^{-1}f_\mu $ . It is defined in $\bar \Pi _{\varepsilon }$ . For each $\mu $ , it does not branch at critical points of $f_\mu $ since critical values of $H_A$ and $(F_\mu h_\mu )^{-1}f_\mu $ coincide. It might branch at other preimages of critical values of $f_\mu $ , but our assumptions imply that the strip $\Pi _{\varepsilon }$ does not contain such points. Also, it is bounded on $\bar \Pi _\varepsilon $ . Thus, this formula for each $\mu $ defines three distinct holomorphic functions in $\bar \Pi _\varepsilon $ that depend analytically on $\mu $ . We choose the one that is real for real $\mu $ , $\mu _1<0$ , and we let it be equal to $G_\mu h_\mu $ . Finally, we choose the translation $h_\mu $ so that $G_\mu (0)=0$ . Since $F_\mu h_\mu $ was chosen above, this uniquely defines $F_\mu $ .

Note that, for $\mu _1=0$ , we get

$$ \begin{align*} A=0,\quad F_\mu h_\mu (z)=z+f_{\mu}(0),\quad G_{\mu}(z) = H_0^{-1}(f_\mu(z)-f_\mu(0))\quad \text{and}\quad h_\mu=\mathrm{id} \end{align*} $$

so, for critical maps in the family $f_\mu $ , the lift coincides with $j(f_\mu )$ from Lemma 4.3.

We now have that $h_\mu F_\mu H_{A(\mu )} G_\mu h_\mu ^{-1}(z) = f_\mu (z)$ , all maps depend analytically on $\mu $ and $G_{\mu }(0)=0$ . Since $H_{A(\mu )}\circ G_\mu $ is a circle map, $\pi _{A(\mu ),G_\mu }=\mathrm {id}$ (see the last statement of Lemma 4.2).

It remains to adjust our triple to achieve $G_\mu '(0)=1$ . Note that $G_\mu '(0)\neq 0$ for $\mu _1=0$ since f has a cubic critical point and $H_0^{-1}(z)=\sqrt [3]{z}$ , so $G_\mu '(0)=c_\mu $ is a non-zero analytic function for small $\mu $ . Since

$$ \begin{align*}H_A \circ cz = c^3 H_{A/c^2}z,\end{align*} $$

we can make this adjustment while preserving analytic dependence of $A, F, G$ on $\mu $ : we replace $G_\mu $ by ${1}/{c_\mu }G_\mu $ and A by $A/c_\mu ^2$ , $F_\mu $ does not change and $\pi _{a,G} $ becomes $z\mapsto z\cdot c_\mu ^3$ . This completes the proof.

Proof. Indeed,

$$ \begin{align*} p ( {\mathcal{R}}_N (F, H_a, G)) = \tilde F \circ \pi_{\tilde a, \tilde G} H_{\tilde a} \tilde G = {\mathcal{R}}_{\text{cyl},N} f ={\mathcal{R}}_{\text{cyl},N} (p (F, H_a, G)).\\[-38pt] \end{align*} $$

Proof. Note that, for any $f\in \mathcal V$ , the renormalized map ${\mathcal {R}}_{\text {cyl},N} f$ is defined in a certain strip $\Pi _{c}$ . Choose $\varepsilon <c/2$ ; then the map ${\mathcal {R}}_{\text {cyl},N} f$ is defined in $\Pi _{2\varepsilon }$ . We use this $\varepsilon $ in the definition of D. The value of $\varepsilon '$ will be chosen later.

Since the map $\pi _{a,G}$ is analytic on $a, G$ , the projection p is analytic. Since the uniformizing coordinate $\Psi =\Psi _{f,N}$ is defined in $\mathcal V$ and analytic on f, the operator ${\mathcal {R}}_N$ is defined and analytic on the intersection of the domain of $\pi _{a,G}$ with $p^{-1}(\mathcal V)$ .

We prove the last assertion of the lemma. Let $(\tilde F, H_{\tilde a}, \tilde G) = {\mathcal {R}}_N(F, H_a, G)$ . Lemma 2.3 implies that, for sufficiently large N, the map $\tilde G= \Psi '(0) G \Psi ^{-1}$ is defined in a rectangle $\{-1<\mathop {\mathrm {Re}} z <1, -2\varepsilon <\mathop {\mathrm {Im}} z<2\varepsilon \}$ that covers $\bar \Pi _\varepsilon $ . Note that

$$ \begin{align*} \tilde G'(z) = \Psi'(0) G'|_{\Psi^{-1}(z)} \frac{1}{\Psi'(\Psi^{-1}(z))}\end{align*} $$

and

(4.4) $$ \begin{align} \tilde G"(z) = \Psi'(0) G"|_{\Psi^{-1}(z)} \frac{1}{\Psi'(\Psi^{-1}(z))^2} - \Psi'(0) G'|_{\Psi^{-1}(z)} \frac{\Psi"|_{\Psi^{-1}(z)} }{\Psi'(\Psi^{-1}(z))^3}. \end{align} $$

Let

$$ \begin{align*}\mathrm{dis}\, \Psi = \sup_{\mathcal V}\sup_{ a, b\in \Psi^{-1}(\bar \Pi_{\varepsilon})}\bigg(\frac{\Psi'(a)}{\Psi'(b)}, \frac{\Psi"(a) }{\Psi'(a)^2}\bigg) \end{align*} $$

be the maximal distortion of $\Psi $ ; owing to Lemma 2.3, this is finite and independent of N. We consider any C with $C>2 (\mathrm {dis}\, \Psi )^2$ , and we use it in the definition of $\Omega ^{{\mathbb {R}}}$ .

Choose N so that $\Psi '(0)> 2 C \cdot \mathrm {dis}\, \Psi $ ; this is possible owing to Lemma 2.3. Owing to the normalization $G(0)=0$ and $G'(0)=1$ , we conclude that if $G\in \Omega ^{{\mathbb {R}}}$ , then, in $\bar \Pi _{\varepsilon }$ ,

$$ \begin{align*}|\tilde G'|> (\mathrm{dis}\, \Psi)^{-1} \cdot |G'|_{\Psi^{-1}(z)}| > (\mathrm{dis}\, \Psi)^{-1} \bigg(1-\frac{C \cdot \mathrm{dis}\, \Psi}{|\Psi'(0)|}\bigg) > 0.5 (\mathrm{dis}\, \Psi)^{-1} > \frac{1}{C}\end{align*} $$

and

$$ \begin{align*}|\tilde G"|<C\frac{(\mathrm{dis}\, \Psi)^2}{ |\Psi'(0)|}+ (\mathrm{dis}\, \Psi)^2 \bigg(1 + \frac{C \cdot \mathrm{dis}\, \Psi}{|\Psi'(0)|}\bigg) < 0.5 (\mathrm{dis}\, \Psi)^2 + 1.5 (\mathrm{dis}\, \Psi)^2<C.\end{align*} $$

Thus, $G\in \Omega ^{{\mathbb {R}}}$ implies that $\tilde G\in \Omega ^{{\mathbb {R}}}$ , and we have proved the last statement of the lemma.

Now we prove compactness statements. We use Lemma 4.2 to define $\pi _{a,G}$ and to choose $\delta , \Omega \supset \Omega ^{{\mathbb {R}}}$ . Set $\varepsilon '=\tau _2/2$ . Then, for all $G\in \Omega $ , we have $\pi _{0, G}H_0 G(\bar \Pi _{\varepsilon })\supset \Pi _{2\varepsilon '} $ due to Lemma 4.2. We use this $\varepsilon '$ in the definition of the space of triples $D=D_{\varepsilon , \varepsilon '}$ .

Since ${\mathcal {R}}_{\text {cyl},N} f = \tilde F\circ \pi _{\tilde a,\tilde G}\circ H_{\tilde a}\circ \tilde G$ is defined in $\Pi _{2\varepsilon }$ , we conclude that $\tilde F$ analytically extends to $\pi _{\tilde a,\tilde G}\circ H_{\tilde a}\circ \tilde G(\bar \Pi _{\varepsilon }) \supset \Pi _{2\varepsilon '}$ . We have shown above that $\tilde G$ is analytic in the rectangle $\{-1<\mathop {\mathrm {Re}} z <1, -2\varepsilon <\mathop {\mathrm {Im}} z<2\varepsilon \}$ that covers $\bar \Pi _\varepsilon $ . Since we restrict $\tilde F, \tilde G$ to smaller domains $\bar \Pi _\varepsilon , \Pi _{\varepsilon '}$ in the definition of ${\mathcal {R}}_N$ , the operator ${\mathcal {R}}_N$ has compact derivative. We have also proved that the set ${\mathcal {R}}_N(U_{\Omega , \delta }\cap p^{-1}(\mathcal V))$ is pre-compact.

Lemma 4.2 implies that, for sufficiently small $\Omega \supset \Omega ^{{\mathbb {R}}}$ and $\delta $ , on the image of ${\hat f\in U_{\Omega , \delta }\cap p^{-1}(\mathcal V)}$ under ${\mathcal {R}}_N$ , the map $\pi _{\tilde a, \tilde G} $ is defined and $p({\mathcal {R}}_N \hat f)=\pi _{\tilde a,\tilde G}\circ H_{\tilde a}\circ \tilde G$ is well defined in a certain strip $\Pi _{\tau _1}$ around ${\mathbb {R}}/{\mathbb {Z}}$ . This map extends to the strip $\Pi _\varepsilon $ since it equals ${\mathcal {R}}_{\text {cyl},N} f$ . Thus, $p\colon {\mathcal {R}}_N(U_{\Omega , \delta }\cap p^{-1}(\mathcal V)) \to \mathcal D_{\varepsilon }$ is defined. This completes the proof.

Proof. The estimate (4.4) applied to $G_1, G_2$ and Lemma 2.3 implies that, by choosing sufficiently large N, we can guarantee that

(4.5) $$ \begin{align}\mathrm{dist}\,_{C^0(\bar \Pi_\varepsilon)} (\tilde G_{1}", \tilde G_{2} ") \le \mu_0\mathrm{dist}\,_{C^2(\bar \Pi_\varepsilon)} (G_1, G_2)\end{align} $$

for some $\mu _0<1$ . Indeed, the first summands in (4.4) are uniformly small, and, in the second summands, $G_1'-G_2'$ is computed on a uniformly small neighborhood of zero and multiplied by a bounded function. Since $G_1'(0)=G_1'(0)=1$ , we can estimate $G_1'-G_2'$ on a small neighborhood of zero in terms of $G_1"-G_2"$ and get (4.5).

Since we only consider triples normalized by $G(0)=0$ and $G'(0)=1$ , this implies that G-components of the triples with the same projection approach each other exponentially quickly.

Since G-components of ${\mathcal {R}}_N^m(F_{1,2}, H_0, G_{1,2})$ approach each other exponentially quickly, the same holds for the corresponding maps $\pi _{0,G_{1,2}}$ due to the fact that the dependence of $\pi _{0,G}$ on G is analytic and thus Lipshitz on $\Omega ^{{\mathbb {R}}}$ . Since the triples have the same projection, their F-components approach each other exponentially quickly. This implies the statement.

Proof. For each critical circle map $f\in \mathcal I$ , construct its lift $j(f) = (F,H_a, G)$ to D using Lemma 4.3: namely, set

$$ \begin{align*} G(z) = \sqrt[3]{f(z)-f(0)}/c\quad \text{and}\quad F(z) = z+f(0). \end{align*} $$

Then $p((F, H_0,G))=f$ .

All resulting maps G for all $f\in \mathcal I$ are ${\mathbb {R}}$ -preserving. Choose $\varepsilon $ small, so that all these maps extend to $\bar \Pi _\varepsilon $ univalently with uniform estimates on $\max _{\bar \Pi _\varepsilon } |G"|$ , $\min G'|_{[-0.5, 0.5]}$ ; consider C so that the set $\Omega ^{{\mathbb {R}}}$ from Lemma 4.7 contains all G-components of all lifts $j(f)$ of critical maps $f\in \mathcal I$ . Decrease $\varepsilon $ , if needed, to satisfy the assumptions of Lemma 4.7, and use this lemma to choose $\varepsilon '$ . These values of $\varepsilon , \varepsilon '$ will be used in the definition of the space D. Finally, fix $\delta $ , $\Omega \supset \Omega ^{{\mathbb {R}}}$ and N so that assertions of Lemmas 4.7 and 4.8 hold true.

The space of triples D is now fixed, and the projection p is defined. We now write ${\mathcal {R}}={\mathcal {R}}_N$ , ${\mathcal {R}}_{\text {cyl}} = {\mathcal {R}}_{\text {cyl},N}$ for brevity.

For each $f\in \mathcal I$ , consider critical circle maps $f_{-k}\in \mathcal I$ such that

$$ \begin{align*}{\mathcal{R}}_{\text{cyl}}^k f_{-k}=f.\end{align*} $$

Note that this uniquely defines $f_{-k}$ .

For each $f_{-k}$ , consider truncated fibers of the projection, $p^{-1}(f_{-k})\cap U_{\Omega ,0}$ . Owing to the choice of $\Omega ^{{\mathbb {R}}}$ , these sets are non-empty for all $g\in \mathcal I$ : every such set contains a lift $j( f_{-k})$ of $f_{-k}$ .

Since ${\mathcal {R}}$ contracts on truncated fibers (Lemma 4.8), the diameters of the sets ${\mathcal {R}}^k(p^{-1} (f_{-k})\cap U_{\Omega , 0}) \subset p^{-1}(f)$ tend to zero. All these sets are non-empty since they contain ${\mathcal {R}}^k (j( f_{-k}))$ . Since each set is pre-compact and $F,G$ -components of the triples in these sets are analytic in strips larger than $\Pi _\varepsilon $ , $\Pi _{\varepsilon '}$ , respectively (Lemma 4.7), there exists a unique accumulation point $i( f)$ of these sets, and $F,G$ -components of the triple $i(f)$ are defined in strips larger than $\Pi _\varepsilon $ , $\Pi _{\varepsilon '}$ , respectively. Thus, the projection p is defined on $i(f)$ and, by continuity,

$$ \begin{align*}p(i(f))=f.\end{align*} $$

The construction implies that

$$ \begin{align*}i({\mathcal{R}}_{\text{cyl}} f) = {\mathcal{R}} i( f).\end{align*} $$

Thus, the set $\Lambda = i(\mathcal I)$ is an invariant set for ${\mathcal {R}}$ , and the dynamics of ${\mathcal {R}}|_{\Lambda }$ is conjugate to the dynamics of ${\mathcal {R}}_{\text {cyl}}|_{\mathcal I}$ via the continuous map p.

Since $i(f)$ is the limit point of $ {\mathbb {R}}$ -preserving triples ${\mathcal {R}}^k(j( f_{-k}))$ , it is ${\mathbb {R}}$ -preserving.

All maps ${\mathcal {R}}^k(p^{-1}(f_{-k})\cap U_{\Omega ^{{\mathbb {R}}},0})$ belong to $U_{\Omega ^{{\mathbb {R}}},0}$ due to the last statement of Lemma 4.7. In particular, $\Lambda \subset U_{\Omega ^{\mathbb {R}}, 0}$ . Thus, Lemma 4.7 implies that ${\mathcal {R}}$ is an analytic operator with a compact differential on a neighborhood of $\Lambda $ .

It remains to prove that i is continuous, that is, the projection p is a homeomorphism between $ \Lambda $ and $\mathcal I$ .

Let $f\in \mathcal I$ , $\delta>0$ . We show that the lifts of sufficiently close points of $\mathcal I$ are $\delta $ -close to $i(f)$ . Let $0<\unicode{x3bb} <1, c$ be the same as in Lemma 4.8 and choose k so that $c\unicode{x3bb} ^k\cdot 2C<\delta /3$ . For a map $f_{-k}$ , take its neighborhood U in $\mathcal I$ , and let $g\in U$ . Construct the lifts $j( g), j( f_{-k})$ using Lemma 4.3; clearly, the lifts are close in the space of triples if U is small enough. Since ${\mathcal {R}}$ is analytic and thus continuous, we also have that ${\mathcal {R}}^k (j( g))$ and ${\mathcal {R}}^k(j( f_{-k}))$ are $\delta /3$ -close if U is small enough.

Note that the diameters of the sets

$$ \begin{align*}{\mathcal{R}}^{k} (U_{\Omega,0}\cap p^{-1}(g))\quad \text{and}\quad {\mathcal{R}}^{k} (U_{\Omega,0}\cap p^{-1}(f_{-k}))\end{align*} $$

are smaller than $c\unicode{x3bb} ^k \cdot \operatorname {diam} (U_{\Omega ,0})$ , due to Lemma 4.8, and that the latter is smaller than $\delta /3$ due to the choice of k. Also, the closure of the first set contains both ${\mathcal {R}}^k(j( g))$ and $i({\mathcal {R}}_{\text {cyl}}^kg)$ , and the closure of the second set contains both ${\mathcal {R}}^k(j( f_{-k}))$ and $i(f)$ . Since ${\mathcal {R}}^k (j( g))$ and ${\mathcal {R}}^k(j( f_{-k}))$ are $\delta /3$ -close, we conclude that $i(f)$ and $i({\mathcal {R}}_{\text {cyl}}^kg)$ are $\delta $ -close for $g\in U$ . This completes the proof.

Proof. Roughly speaking, since ${\mathcal {R}}$ contracts on the fibers of the projection p and ${\mathcal {R}}|_{D_0}$ projects, via p, to the hyperbolic operator ${\mathcal {R}}_{\text {cyl}}$ on $\mathcal D^{cr}_{\varepsilon }$ , the operator ${\mathcal {R}}$ must be hyperbolic on $D_0$ with one unstable direction that projects to the unstable direction of ${\mathcal {R}}_{\text {cyl}}$ and has the same expansion rate. Since ${\mathcal {R}}^k$ takes a to $(\Psi ^{\prime }_{m,k}(0))^2 a$ , it expands in the direction transversal to $D_0=\{a=0\}$ , and thus it must have one more unstable direction in D with maximal expansion rate $\unicode{x3bb} _2$ , as above.

Below, we provide detailed proofs of these statements, by constructing the stable foliation and unstable invariant cone fields for ${\mathcal {R}}$ .

Proof. Note that codimension-one spaces $\{L_{\hat f} v=0\}$ are preimages under p of the spaces $\{L_f v=0\}$ in $\mathcal D_{\varepsilon }$ . Recall that, for circle diffeomorphisms, the condition $L_f v=0$ defines tangent spaces to $\{\mathrm {rot}\, f=\alpha \}$ , and thus ${\mathcal {R}}$ takes $\{L_{\hat f} v=0\}$ to $\{L_{{\mathcal {R}}\hat f} v=0\}$ if $p(\hat f)$ is a circle diffeomorphism. The limit transition, possible owing to Theorem 3.5, implies that the same holds if $p(\hat f)$ is a cubic critical circle map, in particular, for all $\hat f \in \Lambda $ . Also, Theorem 3.5 implies that $\{L_{\hat f}v=0\}$ intersects $D_0$ on the stable subspace of ${\mathcal {R}}$ , and thus is uniformly transversal to $\eta _{\hat f}$ .

The previous lemma implies that ${\mathcal {R}}$ has a two-dimensional unstable distribution $l_{\hat f}$ . The intersection of $l_{\hat f}$ with $\{L_{\hat f} v=0\}$ is a one-dimensional invariant distribution $\xi _{\hat f}$ that is uniformly transversal to $\eta _{\hat f}$ . Since ${\mathcal {R}}$ multiplies the projection of any vector to $d/da$ by $\Psi '(0)$ , the maximal expansion rate along $\xi _{\hat f}$ is the same as in the previous lemma. This completes the proof.