%Authors: Kevin M. Pilgrim and Tan Lei, % %Title:``Surgery on postcritically finite %rational maps by blowing up an arc'' % %Date: December, 1995 \documentstyle[12pt]{amsart} \textheight=574pt \textwidth=432pt \oddsidemargin=18.88pt \evensidemargin=18.88pt \topmargin=14.21pt \input epsf \epsfverbosetrue %messages will show height and width %macros for including figures easily \newcommand{\xincludeps}[3]{\begin{figure} \centerline{% \epsfxsize=#2in \epsfbox{#1} } \caption{ {\sf #3} \label{fig:#1}} \end{figure} } \newcommand{\yincludeps}[3]{\begin{figure} \centerline{% \epsfysize=#2in \epsfbox{#1} } \caption{ {\sf #3} \label{fig:#1}} \end{figure} } %miscellaneous macros %\input mathfonts%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \font\tenmsb=msbm10 scaled \magstep1 \font\sevenmsb=msbm8 \font\fivemsb=msbm6 \newfam\msbfam \textfont\msbfam=\tenmsb \scriptfont\msbfam=\sevenmsb \scriptscriptfont\msbfam=\fivemsb \def\Bbb{\fam\msbfam} \newcommand{\rs}{\mbox{$\widehat{\Bbb C}$}} \newcommand{\C}{\mbox{$\Bbb C$}} \newcommand{\N}{\mbox{$\Bbb N$}} \newcommand{\R}{\mbox{$\Bbb R$}} \newcommand{\Z}{\mbox{$\Bbb Z$}} \newcommand{\id}{\mbox{id}} %\newcommand{\to}{\mbox{$\rightarrow$}} \newtheorem{thm}{Theorem}[section] \newtheorem{defn}{Definition} \newtheorem{fact}{Fact} \newtheorem{prop}{Proposition} \newtheorem{claim}{Claim} \newtheorem{conj}{Conjecture} \newtheorem{question}{Question} \newtheorem{lemma}{Lemma} \newtheorem{cor}{Corollary} \newtheorem{example}{Example} \newcommand{\vac}{\mbox{\O}} \newcommand{\minus}{\backslash} \newcommand{\bdry}{\partial} \newcommand{\cl}{\overline} \newcommand{\gap}{\vspace{5pt}} %\newcommand{\pf}{\noindent {\bf Proof: }} \newcommand{\rmk}{\gap \noindent {\bf Remark: }} %\newcommand{\ex}{\gap \noindent {\bf Example: }} \newcommand{\intersect}{\cap} %\newcommand{\Re}{\mbox{Re}} \newcommand{\wt}{\widetilde} \newcommand{\Int}{\mbox{Int}} \newcommand{\Aut}{\mbox{Aut}} \newcommand{\lcm}{\mbox{lcm}} \newcommand{\into}{\mbox{$\hookrightarrow$}} \newcommand{\mult}{\mbox{mult}} \newcommand{\be}{\begin{enumerate}} \newcommand{\eb}{\end{enumerate}} \newcommand{\bi}{\begin{itemize}} \newcommand{\ib}{\end{itemize}} \begin{document} %authors \author{Kevin M. Pilgrim} \address{\hskip-\parindent Kevin M. Pilgrim\\ \\Mathematics Department\\Cornell University\\ White Hall\\Ithaca, NY 14853\\USA} \thanks{Research at MSRI is supported in part by NSF grant DMS-9022140} \email{pilgrim@@math.cornell.edu} \author{Tan Lei} \address{\hskip-\parindent Tan Lei\\Department of Mathematics\\ University of Warwick\\ Coventry CV4 7AL\\England} \email{tanlei@@maths.warwick.ac.uk} \title{Surgery on postcritically finite rational maps by blowing up an arc} \begin{abstract} Using Thurston's characterization of postcritically finite rational functions as branched coverings of the sphere to itself, we give a new method of constructing new conformal dynamical systems out of old ones. Let $f(z)$ be a rational map and suppose that the postcritical set $P(f)$ is finite. Let $\alpha$ be an embedded closed arc in the sphere and suppose that $f|{\alpha}$ is a homeomorphism. Define a branched covering $g$ as follows. Cut the sphere open along $\alpha$. Glue in a closed disc $D$. Map $S^{2} - \Int (D)$ via $f$ and $\Int (D)$ by a homeomorphism to the complement of $f(\alpha)$. We prove theorems which give combinatorial conditions on $f$ and $\alpha$ for $g$ to be equivalent in the sense of Thurston to a rational map. The main idea in our proofs is a general theorem which forces a possible obstruction for $g$ away from the disc $D$ on which the new dynamics is defined. \end{abstract} \maketitle %\input{introduction} \section{Introduction} \label{section:introduction} A {\it rational map} $f(z)=p(z)/q(z)$ where $p$ and $q$ are relatively prime complex polynomials determines a holomorphic map of the Riemann sphere $\rs $ to itself, and so defines a holomorphic dynamical system. The set of points $C(f)$ where the derivative of $f$ vanishes is the set of {\it critical points} of $f$; these are the points where the local degree of the map is larger than one. Counted with multiplicity, there are $2d-2$ critical points where $d$ is the degree of $f$. The {\it postcritical set} $P(f)$ of $f$ is defined by $P(f)=\cl{ \cup_{n>0} f^{n}(C(f)) }$. The map $f$ is said to be {\it postcritically finite} if $P(f)$ is finite. The Riemann sphere decomposes into the open {\it Fatou set} $F(f)$ of points whose asymptotic behavior is stable under small perturbations, and the closed {\it Julia set} $J(f)$ of points whose behavior under iteration is chaotic. A {\it branched covering} $f: S^{2} \to S^{2}$ is an orientation-preserving continuous map such that for every $x \in S^{2}$, there exist local coordinates near $x$ and $f(x)$ such that, in these coordinates, $x$ and $f(x)$ are the point $0 \in \C$ and $f$ is given by $z \mapsto z^{n}$, $n \geq 1$. The integer $n$ is called the {\it local degree} of $f$ near $n$. The sets $C(f)$ and $P(f)$ are defined analogously as for rational maps. \gap\noindent {\bf Convention:} We assume throughout this paper that all rational maps and branched coverings are postcritically finite. Two branched coverings $f$ and $g$ of the sphere to itself are said to be {\it equivalent} if there are homeomorphisms $\psi_{0}$ and $\psi_{1}$ such that $\psi_0 \circ f = g \circ \psi_1$, and such that $\psi_0$ is isotopic to $\psi_1$ through homeomorphisms $\psi_t, t \in [0,1] $ with $\psi_t | P(f) = \psi_0 | P(f)$ for all $t$. Thurston has given necessary and sufficient combinatorial conditions for a postcritically finite branched covering $g$ to be equivalent to a rational map, and has also shown that (with one understood exception) this map is determined up to conformal congugacy by $g$; see \cite{DH1}. If $f(z)$ is a rational map, suppose by cutting and pasting we obtain a branched covering $g$. When is $g$ equivalent to a rational map? This question is, by Thurston's theorem, a combinatorial one. In general, given $d$ and $N$, there are infinitely many distinct equivalence classes of branched coverings $f$ of degree $d$ with $|P(f)|=N$. Only finitely many of these can contain rational maps, so it is useful to have constructions which, when applied to rational maps, yield new rational maps. In this paper we define a certain kind of cutting and pasting process on $f(z)$ which is very robust: with fairly weak hypotheses which do not include restrictions on the degree of $f$ or its global combinatorics, the result $g$ will be equivalent to a rational map. The rough idea of our construction is the following. Let $f(z)$ be a postcritically finite rational map, let $\alpha$ be a closed, embedded topological arc in $\rs $ and suppose that $f|\alpha$ is a homeomorphism. Define a new branched covering $g$ by cutting open the sphere along $\alpha$, gluing in a disc $D$, mapping the complement of $D$ by $f$, and mapping $D$ to the complement of $f(\alpha)$ in the sphere. A local model of $g$ near $D$ is the map $z \mapsto \frac{1}{2}(z+1/z)$ on a neighborhood of the closed unit disc; this maps the unit circle onto the interval $[-1,1]$. Then $\deg(g)=\deg(f)+1$, and the local degree of $g$ is one more than the local degree of $f$ at the endpoints of $\alpha$. More generally, one can map the disc $D$ in a similar fashion $n$ times around the sphere so that $\deg(g)=\deg(f)+n$. We call this construction {\it blowing up the arc $\alpha$ $n$ times}. Note that by construction $f=g$ away from $D$. The precise conditions on $\alpha$ for this construction to be well-defined, and the statements of our main theorems, are somewhat technical and are given precisely in Sections \ref{subsection:conditions} and \ref{subsection:statements}. We prove that under these conditions, the combinatorial class of $g$ depends only on combinatorial data (Proposition \ref{prop:G well defined}). For example, we derive the following as a corollary to our first main theorem, Theorem A. To set up the statement, we recall that in a periodic Fatou component of a postcritically finite rational map $f$, there is a canonical family of arcs called {\it internal rays} which are mapped homeomorphically under $f$ to other internal rays (see Section \ref{subsection:construction of S} for the definition of internal ray). \gap \noindent {\bf Theorem: }{\it If $\alpha$ is a periodic or preperiodic internal ray of $f$, then the branched covering $g$ which is $f$ blown up $n$ times along $\alpha$ is equivalent to a rational map.} \gap We also apply the blowing up construction to obtain rational maps from M\"{o}bius transformations of finite order (see Theorem \ref{subsection:blowing up moebius transformations}). The central idea in our proofs is Theorem \ref{thm:arcs intersecting obstructions}, {\it Arcs intersecting obstructions}. This theorem asserts that under suitable hypotheses, an obstruction $\Gamma$ to the existence of a rational map equivalent to a branched covering $g$ cannot intersect any family of arcs $\Lambda$ which maps in a certain way. We use this theorem to ``push'' an obstruction away from the region $D$ in the sphere where we add new dynamics. We then conclude that the obstruction $\Gamma$ yields an obstruction for the original rational map $f$, which is impossible. Our theorems may also be thought of as {\it combination theorems} between conformal dynamical systems. Other examples of such theorems include {\it mating} (see \cite{tanlei:matings}, \cite{shishikura:tan:cubicmating}, \cite{jluo:thesis} ), where two polynomials of the same degree are glued together along their circles at infinity, and {\it tuning} (see \cite{rees:parami}, \cite{ahmadi:thesis}), where the dynamics of a polynomial is glued into the dynamics of a rational map near a superattracting cycle. In both of these examples the degree of the new map is the same as that of the old one. To date, all known combination theorems involving mating and tuning have hypotheses on the degree of the map and on the combinatorics of both the original map $f(z)$ and the new dynamics which is being glued into $f(z)$. Also, a definition of mating or tuning for branched coverings has not yet been given in such a manner as to be defined in terms of purely combinatorial data. Our techniques of proof also allow us to establish that certain blown-up quadratic polynomials may be mated (in a suitably generalized way) with an {\it arbitrary} quadratic polynomial to yield a rational map; see Section \ref{subsection:generalized matings}. Moreover, the technique generalizes easily to similar statements in higher degrees. \gap \noindent {\bf Organization of the paper.} \begin{itemize} \item \S \ref{section:theorems}: we state needed definitions, give the precise definition of the blowing up construction, state our first two main theorems (Theorems A and B) precisely, and prove that the combinatorial class of $g$ depends only on combinatorial data (Proposition \ref{prop:G well defined}). \item \S \ref{section:proofs}: we prove Theorems A and B. \item \S \ref{section:complements}: we prove other related results. \item \S \ref{section:examples}: we give examples illustrating our main theorems. \item \S \ref{appendix:arcs}: is an appendix containing a proof of Theorem \ref{thm:arcs intersecting obstructions}, {\it Arcs intersecting obstructions}. \end{itemize} \noindent {\bf Acknowledgments:} The pictures were drawn using programs written by Curt McMullen. The authors also thank MSRI, where some of this work was prepared. \gap\noindent {\bf Example: blowing up a preperiodic arc in the basilica.} Let $f(z)=z^{2}-1$. Let $\alpha = [0,1]$. The filled-in Julia set of $f(z)$ is the set of points whose orbits are bounded; this set is the black region in Figure \ref{fig:basilica.ps}. Let $g$ be the branched covering which is $f$ blown up once along $\alpha$. The Julia set of the rational map equivalent to $g$ is shown in Figure \ref{fig:semibasilica.ps}; see also \cite{kmp:jcurve} for a discussion of this example and for a description of $g$ as a branched covering. \yincludeps{basilica.ps}{1.75}{The filled-in Julia set of $f(z)=z^2-1$.} \yincludeps{semibasilica.ps}{1.75}{Julia set of the rational map corresponding to blowing up $f(z)=z^{2}-1$ along $[0,1]$. The white regions form the basin of infinity.} %\input{theorems} \section{Blowing up an arc} \label{section:theorems} \S\S 2.1 and 2.2 give definitions needed to formulate our construction and theorems precisely. \S 2.3 states conditions under which blowing up will be a well-defined operation on equivalence classes. \S 2.4 gives the precise statements of Theorems A and B. \S 2.5 gives the construction of blowing up an arc, a proof that the equivalence class of the new map $g$ depends only on combinatorial information, and a proposition ({\it Orbits off $\alpha$ persist}) which we will use in our proofs. \subsection{Marked branched covers} \begin{defn} A {\bf marked branched covering} is a pair $(f,X)$ where $f: S^2 \to S^2$ is a branched covering and $X$ is a finite set containing $P(f)$ such that $f(X) \subset X$. \end{defn} Two marked branched coverings $(f,X), (g,Y)$ are said to be {\it equivalent} if they satisfy the definition of equivalence of branched coverings with the sets $P(f)$ and $P(g)$ replaced by $X$ and $Y$ respectively. There is a functor from the category of marked branched coverings $(f,X)$ and equivalences to the category of branched coverings and equivalences. The image of a marked brached covering $(f,X)$ under this functor is $f$; the image of an equivalence $\psi_{0}:(S^{2},X) \to (S^{2},Y)$ is the map of pairs $\psi_{0}:(S^{2},P(f)) \to (S^{2},P(g))$. This functor ``forgets'' that $\psi_{0}$ must send $X-P(f)$ to $Y-P(g)$. We call this the {\it forgetful functor} sending $(f,X)$ to $f$. \gap \noindent {\bf Examples:} \begin{enumerate} \item Let $f$ be a rational map and $X$ a finite forward-invariant set containing $P(f)$. Then $(f,X)$ is a branched cover marked by the set $X$. We will call $(f,X)$ a {\it marked} rational map. Note that if $(f,X)$ is a marked rational map with $\deg(f) = 1$ and $|X| \geq 3$, then $f$ is necessarily a M\"{o}bius transformation of finite order. \item Let $p(z)=z^{2}-1$. Then $p:[-1,0] \to [-1,0]$ is an orientation-reversing homeomorphism. Postcompose $p(z)$ with a homeomorphism $h$ isotopic to the identity relative to $P(p)$ such that for $f(z)=h \circ p(z)$, $f([-1/4,0])=[-1,-3/4]$ and $f([-1,-3/4]) = [-1/4,0]$. Let $X=P(f) \cup \{-1/4, -3/4\}$. Then $(f,X)$ is a branched cover marked by the set $X$. \end{enumerate} In both examples, the image of $(f,X)$ under the forgetful functor is combinatorially equivalent to a rational map. In the first case, however, the set $X-P(f)$ of ``new'' marked points are already preperiodic under the dynamics of a rational map, whereas in the second case we have added new preperiodic points by our choice of representative of the combinatorial class of $p(z)$. \subsection{Arcs and curves in $(S^{2},X)$} Let $X$ be a finite set in the sphere. An {\it arc $\alpha$ in $(S^{2},X)$} we define to be the image of the unit interval $[0,1]$ under a map $j$ into the sphere such that the endpoints $e(\alpha):=j(\{0,1\}) \subset X$, $j|(0,1)$ is an embedding, and $j(0,1) \cap X= \vac$. Thus we allow the endpoints $e(\alpha)$ of $\alpha$ to coincide (though we will not cut along such arcs in our construction). A {\it simple closed curve in $(S^{2},X)$} is a simple closed curve in $S^{2}-X$; it is said to be {\it peripheral} if it is isotopic into every neighborhood of some point in $X$ and {\it essential} if it is not contractible in $S^2-X$. Two arcs (resp. curves) $\eta_{0}, \eta_{1}$ in $(S^2,X)$ are said to be {\it isotopic relative to $X$}, written $\eta_{0} \simeq_X \eta_{1}$, if there is a continuous, one-parameter family $\eta_{t}, t \in [0,1]$ of such arcs (respectively curves) joining $\eta_{0}$ to $\eta_{1}$. The isotopy class relative to $X$ of an arc or simple closed curve $\eta$ we will denote by $[\eta]_X$, or by $[\eta]$ when the set $X$ is understood. A set of pairwise nonisotopic arcs (essential curves) in $(S^2,X)$ will be called an {\it arc (curve) system} in $(S^2,X)$. If $\Lambda$ and $\Lambda'$ are arc or curve systems in $(S^2,X)$ we write $\Lambda \simeq_X \Lambda'$ if every element of $\Lambda$ is isotopic to a unique element of $\Lambda'$ relative to $X$, and conversely. The {\it intersection number} between a pair $\alpha, \beta$ consisting of either arcs or curves in $(S^2,X)$ is defined by \[ \alpha \cdot \beta = \min_{\alpha \simeq_X \alpha', \beta \simeq_X \beta'} \# \{(\alpha' - e(\alpha')) \cap (\beta' - e(\beta'))\}.\] \gap \noindent {\bf Invariance up to isotopy of arc systems.} Let $(f,X)$ be a marked branched covering. If $\lambda$ is an arc in $(S^2,X)$, we will call the closure $\wt{\lambda}$ of a connected component of $f^{-1}(\lambda - e(\lambda))$ a {\it lift} of $\lambda$ under $f$. Thus each arc $\lambda$ in $(S^2,X)$ has $\deg(f)$ distinct lifts $\wt{\lambda}$. If the endpoints of $\lambda$ are distinct, each lift maps homeomorphically under $f$ onto $\lambda$. If $\Lambda$ is an arc system in $(S^2,X)$, a {\it lift} $\wt{\Lambda}$ is defined to be an arc system in $(S^2,X)$ each of whose elements $\wt{\lambda}$ is a lift of some element $\lambda \in \Lambda$. An arc system $\Lambda$ in $(S^2,X)$ is said to be {\it forward invariant under $f$ up to isotopy relative to $X$} if there exists a subset $\Lambda_0 \subset \Lambda$ (possibly all of $\Lambda$) and a lift $\wt{\Lambda}_0$ of $\Lambda_0$ such that $\wt{\Lambda}_0 \simeq_X \Lambda$. If $\Lambda \simeq_X \Lambda'$ and $\Lambda$ is forward invariant up to isotopy relative to $X$, then so is $\Lambda'$. (An isotopy between an element $\lambda \in \Lambda$ and an element $\lambda' \in \Lambda'$ can be lifted to an isotopy between $\wt{\lambda} \in \wt{\Lambda}$ and an arc $\wt{\lambda}'$ which is a lift of $\wt{\lambda}'$.) The next proposition allows us to extract arc systems which are forward-invariant up to isotopy from a collection of arcs which are literally forward-invariant. \begin{prop} \label{prop:finding invariant arc systems} Suppose $\cal{L}$ is a collection of arcs $l$ in $(S^2,X)$ such that for each $l \in \cal{L}$, $f|l$ is a homeomorphism and $f(l) \in \cal{L}$. Then there is a subset $\Lambda \subset \cal{L}$ which is an arc system forward invariant under $f$ up to isotopy, and such that each $l \in \cal{L}$ is isotopic relative to $X$ to a unique element of $\Lambda$. \end{prop} \pf Let $\Lambda_0 \subset f(\cal{L})$ be an arc system chosen arbitrarily so that each arc $f(l), l \in \cal{L}$ is isotopic relative to $X$ to a unique element of $\Lambda_0$. Then for each $\lambda \in \cal{L}$, $f(\lambda) \simeq_X \lambda_0$ for a unique $\lambda_0 \in \Lambda_0$. Hence by lifting isotopies, $\lambda \simeq_X \wt{\lambda}_0$, where $\wt{\lambda}_0$ is a lift of $\lambda_0$. Hence every element of $\cal{L}$ is isotopic to a lift of an element of $\Lambda_0$. We now enlarge our set $\Lambda_0$ so that the last property holds. Let $\cal{L}$$_1$ denote the set of elements of $\cal{L}$ which are not isotopic to any element of $\Lambda_0$. Let $\Lambda_1 \subset $$\cal{L}$$_1$ be any arc system for which each element of $\cal{L}$$_1$ is isotopic to a unique element of $\Lambda_1$, and for which each element $\lambda_1 \in \Lambda_1$ is isotopic to some element of $\cal{L}$$_1$. Then $\Lambda = \Lambda_0 \cup \Lambda_1$ is an arc system with the desired properties. \qed \subsection{Conditions under which blowing up is defined} \label{subsection:conditions} Now suppose that $(f,X)$ is a branched covering of degree $\geq 1$ marked by the finite set $X$, and let $\alpha$ be an arc in $S^{2}$. Assume that the following conditions are satisfied: \gap \noindent {\bf Blowing Up Conditions:} \begin{enumerate} \item $f|\alpha$ is a homeomorphism; \item $\alpha$ is a union of arcs $\alpha_{j}, j=1,2,...,L$ in $(S^{2},f^{-1}(X))$; \item $\Int(\alpha) \cap (X \cup C(f))= \vac$. \end{enumerate} Note that the arcs $f(\alpha_j)$ are arcs in $(S^2,X)$. \gap\noindent {\bf Remarks:} Allowing $\alpha$ to consist of more than one arc in $(S^{2},f^{-1}(X))$ is indeed useful: for example, consider the example of blowing up the ``airplane'' in Section \ref{section:examples}. More generally, we may apply the blowing up construction to finite collections $\alpha^1, \alpha^2, ...$ of arcs for which $\Int(\alpha^i) \cap \alpha^j = \vac$ when $i \neq j$ and for which each arc satisfies the Blowing Up Conditions. \subsection{Statement of main results} \label{subsection:statements} We assume the following in this subsection. Let $(f,X)$ be a marked rational map with $\deg(f) \geq 2$. Let $\alpha$ be an arc in $\rs$ satisfying the Blowing Up Conditions. Let $(g,X)$ be marked branched cover which is the map $(f,X)$ blown up $n$ times along $\alpha$. \gap\noindent{\bf Theorem A} {\it If for all $j=1,2,...,L$, \be \item $f(\alpha_j)$ is isotopic relative to $X$ to an arc which is contained in a finite arc system $\Lambda_j$ which is forward invariant under $f$, and \item there is a subset $\Lambda_{j,0} \subset \Lambda_j$ with a lift $\wt{\Lambda}_{j,0}$ such that $\wt{\Lambda}_{j,0} \simeq_X \Lambda_j$ and $\wt{\Lambda}_{j,0} \cap \Int(\alpha) = \vac$, \eb then $g$ is combinatorially equivalent to a rational map.} \begin{cor} If $\alpha$ is a periodic or preperiodic internal ray of $f$, then $g$ is equivalent to a rational map. \end{cor} \begin{cor} Suppose $\alpha$ is a finite union of periodic or preperiodic internal rays of $f$. If for all $n \geq 0$, the intersection number relative to $X$ satisfies $f^n (f(\alpha)) \cdot f(\alpha) \neq 0$ only when $f^n (f(\alpha)) = f(\alpha)$, then $g$ is equivalent to a rational map. \end{cor} The definition of internal ray is given in Section \ref{subsection:construction of S}. \gap Next, assume $f(z)$ is a polynomial. According to work of Douady and Hubbard, there is a canonically defined finite topological tree $T$ contained in the filled-in Julia set of $f$, depending only on $f$ and $e(\alpha)$, such that $f(T) \subset T$ with vertices mapped to vertices. In Section \ref{subsection:proof of theorem two} we outline the construction of $T$. \gap\noindent{\bf Theorem B: (Blowing up an arc in a generalized Hubbard tree)} {\it If $\alpha \subset T$, then $g$ is combinatorially equivalent to a rational map.} \rmk The Blowing Up Conditions strongly restrict the possibilities for $\alpha$. The novel feature of this theorem, relative to the first, is that we do not require that the $\alpha_j$ are eventually periodic up to isotopy relative to $X$. \subsection{Definition of blown-up map} Let $(f,X)$ and $\alpha$ satisfy the Blowing Up Conditions of Section 2.3 . The construction is outlined in Figure \ref{fig:blowup.ps} for the case $n=4$. \xincludeps{blowup.ps}{4.5}{Construction of the blown-up map $g$ from $f$.} We first cut the sphere along the interior of $\alpha$. Let $W$ be an open disc containing $\Int(\alpha)$ bounded by a Jordan curve such that $\cl{W} \intersect f^{-1}(X) = \alpha \intersect f^{-1}(X)$ and $\bdry W \supset e(\alpha)$. Since $\Int(\alpha) \cap(X \cup C(f)) = \vac$, we may choose $W$ so that $f|\cl{W}$ is a homeomorphism. Let $h_{t}: \rs - \Int(\alpha) \hookrightarrow \rs $ be a continuous family of embeddings satisfying \begin{enumerate} \item $h_{0}=\id $; \item $h_{t}|(\rs - W) = \id $ for all $t$; \item $h_{1}(\rs - \Int(\alpha)) = \rs - (D-e(\alpha))$, where $D$ is a closed disc. \end{enumerate} The Jordan curve $\bdry D$ is a union of two closed arcs $\bdry D_{+}$ and $\bdry D_{-}$ with endpoints $e(\alpha)$ which we denote by $\{e_{-1},e_{+1}\}$. The map $h_{1}^{-1}$ extends uniquely to a continuous ``collapsing'' map $c: \rs - \Int(D) \to \rs $ sending $\rs - (D-e(\alpha))$ homeomorphically to $\rs - \Int(\alpha)$ and each arc $\bdry D_{+}$ and $\bdry D_{-}$ homeomorphically to $\alpha$. We add the following new dynamics. Let $\Delta = \{z\; |\; |z|<1\; \}$, and let $\phi_{n}: \cl{\Delta} \to \rs $ be the map given by \[ \phi_{n}(z)=A_{2}( A_{1}(z) )^{2n}, \] where $A_{1}(z)=-i \frac{z+1}{z-1}$ and $A_{2}(z)=\frac{z-1}{z+1}$. The M\"{o}bius transformation $A_{1}$ is a rigid rotation with respect to the spherical metric sending the ordered triple $(-1,0,1)$ to $(0,i,\infty)$ and sending $\cl{\Delta}$ to $Im(z) \geq 0$; the map $z \mapsto z^{2n}$ maps the half-plane $Im(z) \geq 0$ $n$ times around the sphere sending the real axis to the nonnegative real axis; the M\"{o}bius transformation $A_{2}$ is a rigid rotation sending $(0,1,\infty)$ to $(-1,0,1)$. Thus the map $\phi_{n}$ wraps the closed unit disc $\cl{\Delta}$ around the sphere $n$ times in such a way that $\phi_{n}$ is locally injective on $\cl{\Delta}-\{-1,1\}$, $\phi_{n}(-1)=-1, \phi_{n}(1)=1$, and $\phi_{n}(\bdry \Delta)=[-1,1]$. Let $\bdry \Delta_{+}, \bdry \Delta_{-}$ denote respectively the intersection of $\cl{\Delta}$ with $Im(z) \geq 0$ and $Im(z) \leq 0$. We now describe the gluing of $f$ and $\phi_{n}$. Choose a homeomorphism $$h': (\rs , [-1,1]) \to (\rs , f(\alpha))$$ which sends $-1$ to $f(e_{-1})$ and $+1$ to $f(e_{+1})$. The compositions $(\phi_{n}|_{\bdry \Delta_{\pm}})^{-1} \circ (h')^{-1} \circ f \circ c |_{\bdry D_{\pm}}: \bdry D_{\pm} \rightarrow \bdry \Delta_{\pm}$ are homeomorphisms of closed arcs and together define a homeomorphism from $\bdry D$ to $\bdry \Delta$. Since $D$ and $\cl{\Delta}$ are closed discs we can extend this to a homeomorphism $h''$ between $D$ and $\cl{\Delta}$. We now define a new branched covering $g$ by \[ g(z) = \left\{ \begin{array}{ll} f \circ c (z), & \mbox{if $ z \in \rs - \Int(D)$}, \\ h' \circ \phi_{n}\circ h''(z), & \mbox{if $ z \in \cl{D}.$} \end{array} \right. \] The two definitions agree on the boundary, by construction, hence they give a well-defined branched covering of the sphere to itself such that $\deg(g)=\deg(f)+n$. We may also apply the above construction simulaneously to finite collections of arcs $\alpha^1, \alpha^2, ...$ which satisfy the Blowing Up Conditions and which satisfy $\Int(\alpha^i) \cap \alpha^j = \vac, i \neq j$. We omit the details of this more general construction but emphasize that we may assume that for $i \neq j$, $\cl{W^i} \cap \cl{W^j} \subset e(\alpha^i) \cap e(\alpha^j) \subset X$. \begin{prop}[$g$ depends only on classes] \label{prop:G well defined} The combinatorial class of $g$ depends only on the integer $n$, the combinatorial class of $(f,X)$, and the isotopy classes of the arcs in $(S^{2},f^{-1}(X))$ comprising $\alpha$. \end{prop} \pf We prove in detail only the case of blowing up along a single arc; the case of blowing up along multiple arcs in similar, since we may choose the closed discs $\cl{W^i}$ to intersect only in $X$. If $(f_{1},X_{1})$ is equivalent to $(f_{2}, X_{2})$, then $(f_{1},f_{1}^{-1}(X_{1}))$ is equivalent to $(f_{2}, f_{2}^{-1}(X_{2}))$: an equivalence $\psi_{0}: (S^{2},X_{1}) \to (S^{2},X_{2})$ may be lifted to an equivalence $\psi_{1}:(S^{2},f_{1}^{-1}(X_{1})) \to (S^{2}, f_{2}^{-1}(X_{2}))$. Let $\phi_{0}, \phi_{1}: (S^{2},f_{1}^{-1}(X_{1})) \to (S^{2},f_{2}^{-1}(X_{2}))$ be a pair of homeomorphisms giving an equivalence between $(f_{1},f_{1}^{-1}(X_{1}))$ and $(f_{2},f_{2}^{-1}(X_{2}))$. Then $\phi_{0} \circ f_{1}=f_{2}\circ \phi_{1}$ and $\phi_{0} \simeq \phi_{1}$ relative to $f_{1}^{-1}(X_{1})$. Let $\alpha^{2}=\phi_{1}(\alpha^{1})$. Then $\alpha^{2}$ is an arc in $S^{2}$ which is comprised of arcs in $(S^{2}, f_{2}^{-1}(X_{2}))$. We must prove that if $(g_{i},X_i)$ is $(f_{i},X_{i})$ blown up $n$ times along $\alpha^{i}, i=1,2$, then $g_{1}$ and $g_{2}$ are combinatorially equivalent. For convenience, if $\beta^i, i=1,2$ are arcs which are unions of finitely many arcs $\beta^i _j$ in $(S^2,Y)$, we will write $\beta^1 \simeq_Y \beta^2$ if $\beta^1 _j \simeq_Y \beta^2 _j$ for all $j$. \gap\noindent{\bf Step 1:} We first show the following. Suppose for $i = 1,2$, $g^i$ is the map produced from the map $f$ by blowing up an arc $\alpha^i$, where $\alpha^1 \simeq_{f^{-1}X} \alpha^2$, and $W_i, D_i, h'_i, h''_i$, and $h^i _t$ are the discs and maps used in the definition of $g_i$. Then $g_1$ is equivalent to $g_2$. Since $f|\cl{W}_i$ is a homeomorphism, $f(\cl{W}_i)$ is a closed disc. There exists a homeomorphism $\psi_0: (S^2,X) \to (S^2,X)$ isotopic to the identity relative to $X$ such that $\psi_0 (f(\cl{W}_1), f(\alpha^1)) = (f(\cl{W}_2), f(\alpha^2))$. For since $\alpha^1 \simeq_{f^{-1}X}\alpha^2$, $f(\alpha^1) \simeq_X f(\alpha^2)$. By results of Epstein \cite{dbaepstein:curves}, there is an ambient isotopy of $S^2$ carrying $\bdry f(\cl{W}_1) \cup f(\alpha^1)$ to $\bdry(f(\cl{W}_2)) \cup f(\alpha^2)$ fixing $X$ pointwise. Let $\Psi: S^2 \times I \to S^2$ be such an isotopy such that $\Psi (\cdot, 0) = \id$, $\Psi(\cdot,1) = \psi_0$, and $\Psi(X,t) = \id_X$ for all $t$. Since $X \supset P(f)$, $f| S^2 - f^{-1}X \to S^2 - X$ is a covering map. Hence there is a lift $\wt{\Psi}$ of $\Psi$ which is an isotopy fixing $X$ pointwise between the identity and a map which we call $\psi^1$. Note that $\psi^1: (\cl{W}_1, \alpha^1) \to (\cl{W}_2, \alpha^2)$. Recall that $g_i | S^2 - \cl{W}_i = f| S^2 - \cl{W}_i, i=1,2$. We will define a homeomorphism $\psi_1: (S^2,X) \to (S^2,X)$ such that $g_2 \psi_1 = \psi_0 g_1$ and show $\psi_0$ is isotopic to $\psi_1$ relative to $X$. The map $\psi_1$ will be defined as four maps $\psi^{(1)}=\psi^1 | S^2 - W_1$, $ \psi^{(2)}, \psi^{(3)}, \psi^{(4)}$ which piece together. We now define $\psi^{(2)}$. Let $K_i = (g_i | D_i)^{-1}(f(\alpha^i)), i=1,2$. Then $K_i$ is a graph in $S^2$ with exactly two vertices which are the endpoints $e(\alpha^i) \subset f^{-1}(X)$ and with $n+2$ edges, each of which joins one vertex to the other. There is a well-ordering on the set of edges of $K_i$: $E < E'$ if the edge $E$ separates $\Int(E')$ from $\Int(\bdry_{-}D_i)$ in $D_i$. Write the set of edges in increasing order as $\{E^i _0, ..., E^i _{n}\}$. Since we glue in the same new dynamics $\phi_n$ in the construction of $g_1$ and $g_2$, there is a homeomorphism $\psi^{(2)}: K_1 \to K_2$ such that $g_2 \psi^{(2)} = \psi_0 g_1 $ and $\psi^{(2)}(E^1 _l) = E^2 _l, l=0,1,...,n$, i.e. $\psi^{(2)}$ preserves the well-ordering of the edges. We now define $\psi^{(3)}$. For $l=0,1,...,n-1$, the complementary component $U^i _l$ of $D^i - K_i$ bounded by $E^i _l \cup E^i _{l+1}$ is a Jordan domain mapping homeomorphically under $g_i$ onto $S^2 - f(\alpha^i)$, $i=1,2$. Hence for each $l$, there is a homeomorphism $\psi^{(3),l}: U^1 _l \to U^2 _l$ such that $g_2 \psi^{(3),l} | U^1 _l = \psi_0 g_1 | U^1 _l$. Since $\psi^{(2)}$ preserves the ordering of edges, the homeomorphisms $\psi^{(3),l}$ and $\psi^{(2)}$ agree on the common boundaries of their domains of definition. A similar argument shows that there is a homeomorphism $\psi^{(4)}: W_1 - D_1 \to W_2 - D_2$ such that $g_2 \psi^{(4)}| W_1 - D_1 = \psi_0 g_1 | W_1 - D_1$ and which agrees with $\psi^{(1)}$ on $\bdry W_1$ and with $\psi^{(2)}$ on $\bdry D_1$. Hence the maps $\psi^{(1)}, \psi^{(2)}, \psi^{(3)}, \psi^{(4)}$ define a homeomorphism of the sphere $\psi_1$ such that $\psi_0 g_1 = g_2 \psi_1$. That $\psi_0$ and $\psi_1$ are isotopic relative to $X$ follows from the fact that $\psi_1 | S^2 - W_1$ is isotopic (via $\wt{\Psi}$) to the identity relative to $X$ and the fact that $X \cap \cl{W}_1 \subset e(\alpha^1) \subset \bdry W_1$. This completes the proof of Step 1. \gap\noindent{\bf Step 2:} We now prove the general statement. Let $(f_1, f_1^{-1}X_1), (f_2, f_2^{-1}X_2) $ be marked covers and $\phi_0,\phi_1$ as above give an equivalence carrying the arc $\alpha^1$ to $\alpha^2$. By replacing $f_2$ with the conjugate $\phi_0 f_2 \phi^{-1} _0$ we may assume that $\phi_0 = \id$ and that $f_1(\alpha^1) = f_2(\alpha^2)$. By Step 1, the combinatorial class of $g_i$ is independent of the choice of discs $W, D$. Let $W_1,D_1$ be any such choice for $g_1$. Hence we may assume that $(W_2,D_2) = \phi_1 (W_1,D_1)$ and that $f_1(\cl{W}_1, \alpha^1) = f_2(\cl{W}_2,\alpha^2)$. Let $\psi_0: (S^2,X)) \to (S^2,X)$ be the identity map (after conjugation, we may assume $X=X_1 = X_2$ ). We will find a homeomorphism $\psi_1$ such that $\psi_0 g_1 = g_2 \psi_1$ and $\psi_0$ is isotopic to $\psi_1$ relative to $X$. As in the proof of the previous step, we define $\psi_1$ in four pieces $\psi_1 | S^2 - W_1 = \psi^{(1)}, \psi^{(2)}, \psi^{(3)}, \psi^{(4)}$. The details are all similar to those in the proof of the previous step. \qed \rmk A similar proof shows that when blowing up along any finite collection of arcs $\alpha^1, \alpha^2, ...$ of arcs for which $\Int(\alpha^i) \cap \alpha^j = \vac, i \neq j$, the equivalence class of the resulting map is independent of the choices and representatives. Moreover, the resulting class is the same whether we blow up one arc at a time or all of them simultaneously. The next proposition gives a condition under which arc systems in $(S^2,X)$ forward invariant under $f$ up to isotopy are also forward invariant under $g$ up to isotopy. \begin{prop}[Orbits off $\alpha$ persist] \label{prop:orbits off alpha persist} Suppose $\Lambda$ is an arc system in $(S^2,X)$ which is forward invariant under $f$ up to isotopy relative to $X$. If, under $f$, there exists a lift $\wt{\Lambda} \simeq_X \Lambda$ such that $\wt{\Lambda} \cap \Int(\alpha) = \vac$, then $\Lambda$ is forward invariant under $g$ up to isotopy relative to $X$. \end{prop} \pf Let $\Lambda_0 \subset \Lambda$ be an arc system and $\wt{\Lambda}_0$ a lift of $\Lambda_0$ under $f$ such that $\wt{\Lambda}_0 \simeq_X \Lambda$. Since $g|\rs - D = f \circ c$, if $\wt{\lambda}_0$ is a lift under $f$ of $\lambda_0 \in \Lambda_0$ such that $\wt{\lambda}_0 \cap \Int(\alpha) = \vac$, then $h_1 (\wt{\lambda}_0) = c^{-1}(\wt{\lambda}_0)$ is a lift under $g$ of $\lambda_0$. Since $h_1$ is isotopic to the identity relative to $X$, $\wt{\lambda}_0 \simeq_X h_1 (\wt{\lambda}_0)$. Hence $h_1 (\wt{\Lambda}_0)$ is a lift of $\Lambda_0$ under $g$ such that $h_1 (\wt{\Lambda}_0) \simeq_X \Lambda_0$. \qed %\input{proofs} \section{Proofs of Theorems A and B} \label{section:proofs} In \S 3.1 we state the characterization of marked rational maps as marked branched coverings as an immediate corollary to the proof of Thurston's characterization of rational maps as branched coverings. In \S 3.2 we state the main ingredient in our proof, Theorem \ref{thm:arcs intersecting obstructions}, {\it Arcs intersecting obstructions}; its proof we relegate to an Appendix. In \S 3.3 and we prove Theorem A; \S 3.4 gives the proof of Theorem B, which depends on a construction given in \S 3.5. \subsection{Characterization of marked rational maps} Suppose $(f,X)$ is a marked branched covering. To a curve system $\Gamma$ in $(S^2, X)$ we associate the {\it Thurston linear transformation} which is the linear map \[ f_{\Gamma}: \R ^{\Gamma} \rightarrow \R ^{\Gamma} \] given by \[ f_{\Gamma}(\gamma) = \sum_{\gamma' \subset f^{-1}(\gamma)} \frac{1}{\deg (f: \gamma' \to \gamma) }[\gamma']_{\Gamma}, \] where $[\gamma']_{\Gamma}$ denotes the element of $\Gamma$ isotopic to $\gamma'$, if it exists. If there are no such elements, the sum is taken to be zero. Here, we regard elements of $\Gamma$ as the basis vectors of $\R^{\Gamma}$. This transformation depends only on the isotopy classes of curves relative to $X$ and is natural with respect to iteration of $f$. Since the entries of $f_{\Gamma} $ are nonnegative, the Perron-Frobenius Theorem (\cite{gantmacher}, Chapter XIII) implies that there is a nonnegative real eigenvalue $\lambda(f_{\Gamma})$ equal to the spectral radius of $f_{\Gamma}$. A nonnegative square matrix $A_{ij}$ is said to be {\it irreducible} if for each $(i,j)$, there is an $n \geq 0$ such that $A^n _{ij} >0$. We say that $\Gamma$ is irreducible if the matrix for $f_{\Gamma}$ is irreducible. The Perron-Frobenius theory easily gives that if $\Gamma$ is any curve system with $\lambda(f_\Gamma) >0$, then then $\Gamma$ contains an irreducible curve system $\Gamma'$ for which $\lambda(f_{\Gamma'}) = \lambda(f_{\Gamma})$. A curve system $\Gamma$ is said to be a {\it multicurve} if its elements are nonperipheral and pairwise disjoint. The proof of Thurston's characterization of rational maps as branched coverings of the sphere given in \cite{DH1} extends immediately to the setting of branched covers marked by additional points. \begin{thm}[Characterization of marked rational maps] \label{thm:characterization of marked rational maps} Let $(f,X)$ be a marked branched covering. Then $(f,X)$ is combinatorially equivalent to a marked rational map $(R,Y)$ if and only if for every irreducible multicurve $\Gamma$ in $(S^2,X)$, either \begin{enumerate} \item $\lambda(f_{\Gamma})=1$ and the orbifold associated to $(f,P(f))$ has signature $(2,2,2,2)$, in which case $R$ is covered by an integral endomorphism of the torus, or \item $\lambda(f_{\Gamma}) < 1$, in which case the rational map $R$ is unique up to conjugation by elements of $\Aut ( \rs )$. \end{enumerate} \end{thm} \begin{defn}[Thurston obstruction.] If $(f,X)$ is a marked branched covering for which the orbifold associated to $(f,P(f))$ is not the $(2,2,2,2)$ orbifold, an (irreducible) multicurve $\Gamma$ for which $\lambda(f_{\Gamma}) \geq 1$ is called an (irreducible) {\it Thurston obstruction}. \end{defn} \noindent {\bf Example:} Let $\Gamma = \{\gamma_0, ...,\gamma_{p-1}\}$ be a multicurve such that the following holds: for each $0 \leq j \leq p-1$, $\gamma_{j}$ has a (not necessarily unique) lift $\delta \simeq_X \gamma_{j-1 (\bmod p)}$ and $\deg{f:\delta \to \gamma_j} = 1$. Then $\Gamma$ is called a {\it Levy cycle} and is an example of an irreducible Thurston obstruction. \gap\noindent{\bf Remarks:} \begin{enumerate} \item See \cite{DH1} for the proof and for the definition of the orbifold associated to a rational map. A rational map $R$ is said to be {\it covered by a torus endomorphism} if there is a complex torus $T$, a holomorphic covering map $\wt{R}: T \to T$, and a surjective holomorphic map $p$ such that $R \circ p = p \circ \wt{R}$. The endomorphism $\wt{R}$ is said to be {\it integral} if it has a lift to an automorphism of $\C$ of the form $z \mapsto n z$, some $n \geq 2$. The first case in the above theorem is rare and does not arise, for example, if there are periodic critical points of $f$ or if $|P(f)| \geq 5$. \item In \cite{DH1} the curve systems $\Gamma$ in the theorem are required to be {\it $f$-invariant}, i.e. every lift $\delta$ of an element $\gamma \in \Gamma$ is either inessential, peripheral, or isotopic to an element of $\Gamma$. However, by taking preimages we may always extend an irreducible multicurve $\Gamma$ to an $f$-invariant multicurve without decreasing the leading eigenvalue of the corresponding Thurston transformation. \end{enumerate} To apply Theorem \ref{thm:characterization of marked rational maps} we will need to rule out the first rare case. Let $(f,X)$ be a marked rational map $f(z)$. If $f$ is covered by an integral torus endomorphism, then $f$ is not a polynomial. Moreover, there can be no arcs in $(S^2,X)$ with distinct endpoints which are periodic under $f$ up to isotopy relative to $X$. For the Euclidean metric on $\C$ pushes down to an (orbifold) metric on $\rs$ which is uniformly expanded under $f$. Hence all inverse branches of $f$ are uniformly contracting with respect to this orbifold metric. This implies that if $\wt{\lambda}$ is any lift under $f$ of an arc $\lambda$ in $(S^2,X)$, then the diameter of $\wt{\lambda}$ is strictly less than that of $\lambda$. Since the diameter of any arc in $(S^2,X)$ with distinct endpoints is positive, this shows that there are no arcs with distinct endpoints which are periodic under $f$ up to isotopy relative to $X$. Hence to prove our two theorems, we may assume that the orbifold $O_{f}$ is not the $(2,2,2,2)$ orbifold. If $g$ is $f$ blown up along $\alpha$ then the Euler characteristic $\chi(O_{g}) < \chi(O_{f}) \leq 0$. The $(2,2,2,2)$ orbifold has Euler characteristic zero, hence $O_{g}$ cannot be the $(2,2,2,2)$ orbifold. So $g$ is combinatorially equivalent to a rational map if and only if there are no irreducible Thurston obstructions for $g$. \subsection{Arcs intersecting obstructions} \label{subsection:arcs intersecting obstructions} Let $(f,X)$ be a marked branched covering, $\Lambda$ be an arc system in $(S^{2},X)$, and $\Gamma$ a curve system in $(S^2,X)$. The {\it unweighted Thurston linear transformation} $f_{\#, \Gamma}: \R ^{\Gamma} \to \R ^{\Gamma}$ is defined by \[ f_{\# , \Gamma}(\gamma) = \sum_{\gamma' \subset f^{-1}(\gamma)} [\gamma']_{\Gamma} \; \mbox{for $\gamma \in \Gamma$,} \] where $[\gamma']_{\Gamma}$ denotes the element in $\Gamma$ homotopic to $\gamma'$ and zero otherwise. As before, this map is independent of choice of representatives and natural with respect to iteration. Note that $0 \leq (f_\Gamma)_{i,j} \leq (f_{\#,\Gamma})_{i,j}$ for all $i,j$ and that $(f_\Gamma)_{i,j} = 0$ if and only if $(f_{\#,\Gamma})_{i,j} = 0$. This implies that $f_\Gamma$ is irreducible if and only if $f_{\#,\Gamma}$ is irreducible. In analogy with the unweighted Thurston linear transformation, we define \[ f_{\#, \Lambda}: \R ^{\Lambda} \to \R^{\Lambda} \] by setting \[ f_{\#, \Lambda}(\lambda) = \sum_{\lambda' \subset f^{-1}(\lambda)} [\lambda']_{\Lambda} \; \mbox{for $\lambda \in \Lambda$}, \] where $[\lambda']_{\Lambda}$ denotes the element of $\Lambda$ homotopic to $\lambda'$ (if it exists). We say that $\Lambda$ is {\it irreducible} if $f_{\# ,\Lambda}$ is irreducible. It is straightforward to verify that the transformation $f_{\#,\Lambda}$ depends only on $[\Lambda]_X$ and not on the choice of representatives and that $f_{\#,\Lambda}$ is natural with respect to iteration of $f$, i.e. $(f^{n})_{\#,\Lambda} = (f_{\#,\Lambda})^n$. If $\alpha$ and $\beta$ are arcs in $(S^2,X)$ we define the {\it multiplicity} $\mult(f: \alpha \to \beta)$ to be the number of lifts of $\beta$ which are isotopic to $\alpha$ relative to $X$. Thus $(f_{\#,\Lambda})_{ij}=\mult(f:\lambda_i \to \lambda_j)$. The following proposition follows immediately from the decomposition theorem for nonnegative square matrices in \cite{gantmacher}, Chapter XIII. \begin{prop}[{\bf Decomposition of invariant arc systems}] \label{prop:decomposition of invariant arc systems} Let $\Lambda$ be a finite arc system forward invariant under $f$ up to isotopy relative to $X$. Then there are disjoint (as subsets of $\Lambda$) irreducible arc subsystems $\Theta_1, \Theta_2, ..., \Theta_s$ of $\Lambda$ and integers $n_1, n_2, ..., n_s$ such that for each $\lambda \in \Lambda$, there is a (not necessarily unique) $\Theta_k$ such that $\lambda$ is isotopic to a lift of an element of $\Theta_k$ under $f^{n_k}$. \end{prop} For $\Lambda$ a curve system or an arc system in $(S^2,X)$, set $\widetilde{\Lambda}$ (resp. $\widetilde{\Lambda}(f^n)$) to be the union of those components of $f^{-1}(\Lambda)$ (resp. $f^{-n}(\Lambda)$) which are isotopic to elements of $\Lambda$. If $\Lambda$ is irreducible, each component of $\Lambda$ is isotopic to some (not necessarily unique) component of $\widetilde{\Lambda}(f)$ (resp. $\widetilde{\Lambda}(f^n)$). The following theorem appears in a different form in \cite{shishikura:tan:cubicmating}. \begin{thm}[Arcs intersecting obstructions] \label{thm:arcs intersecting obstructions} Let $(f,X)$ be a marked branched covering, $\Gamma$ an irreducible Thurston obstruction in $(S^{2},X)$, and $\Lambda$ an irreducible arc system in $(S^{2},X)$. Suppose furthermore that $\#(\Gamma\cap\Lambda)=\Gamma \cdot \Lambda$. Then either \begin{enumerate} \item $\Gamma \cdot \Lambda = 0$ and $\Gamma\cdot f^{-n}(\Lambda) = 0$, $f^{-n}(\Gamma)\cdot \Lambda=0$ for all $n\ge 1$; or \item $\Gamma \cdot \Lambda \neq 0$ and \begin{enumerate} \item Each component of $\Gamma$ (resp. $\Lambda$) is isotopic to a unique component of $\widetilde{\Gamma}$ (resp. $\widetilde{\Lambda}$), the mapping $f:\widetilde{\Gamma}\to \Gamma$ (resp. $f: \widetilde{\Lambda}\to \Lambda$) is a homeomorphism, $\widetilde{\Gamma}\cap (f^{-1}(\Lambda)-\widetilde{\Lambda}))=\emptyset$, and $\widetilde{\Lambda}\cap (f^{-1}(\Gamma)-\widetilde{\Gamma})=\emptyset$. More precisely, \begin{enumerate} \item for each $\gamma \in \Gamma$, there is exactly one curve $\gamma' \subset f^{-1}(\gamma)$ such that $\gamma' \cap \widetilde{\Lambda} \neq \emptyset$. Moreover, the curve $\gamma'$ is the unique component of $f^{-1}(\gamma)$ which is isotopic to an element of $\Gamma$; \item for each $\lambda \in \Lambda$, there is exactly one connected component $\lambda'$ of $f^{-1}(\lambda)$ such that $\lambda' \cap \widetilde{\Gamma} \neq \emptyset$. Moreover, the arc $\lambda'$ is the unique component of $f^{-1}(\lambda)$ which is isotopic to an element of $\Lambda$; \end{enumerate} \item the transformations $f_{\# ,\Gamma}$ and $f_{\# ,\Lambda}$ are transitive permutations of the basis vectors. \item The above results remain true if we replace $f$ by $f^n$, for any $n \geq 1$ (though transitivity may fail). \item For any arc $\lambda$ in $(S^2,X)$ which is isotopic to an arc in $f^{-n}(\Lambda)$ for some $n>1$ but which is not isotopic to an arc in $\Lambda$, $\Gamma \cdot \lambda=0$. Similarly, for any curve $\gamma$ in $(S^2,X)$ which is isotopic to a curve in $f^{-n}(\Gamma)$ for some $n>1$ but which is not isotopic to a curve in $\Gamma$, $\Lambda \cdot \gamma = 0$. \end{enumerate} \end{enumerate} \end{thm} We prove this theorem in the Appendix. \begin{cor} If $\Lambda$ contains more than one periodic cycle, or if the multiplicity of $f$ from some element $\alpha \in \Lambda$ to $\beta \in \Lambda$ is larger than one, then there are no irreducible Thurston obstructions which intersect $\Lambda$. \end{cor} \pf If so then $f_{\#, \Lambda}$ is not a transitive permutation of the basis vectors. \qed \subsection{Proof of Theorem A} \label{subsection:proof of theorem one} By Theorem \ref{thm:characterization of marked rational maps}, {\it Characterization of marked rational maps}, and our discussion at the end of Section 3.1, $(g,X)$ is combinatorially equivalent to a marked rational map if there exist no irreducible Thurston obstructions $\Gamma$ for $g$. Let $\Gamma$ be such an obstruction and let $\widetilde{\Gamma}$ be the union of those components of $g^{-1}(\Gamma)$ isotopic to elements of $\Gamma$, relative to $X$. By the second hypothesis of the theorem and Proposition \ref{prop:orbits off alpha persist}, {\it Orbits off $\alpha$ persist}, for each $j$, $\Lambda_j$ is forward invariant under $g$ up to isotopy relative to $X$. By Proposition \ref{prop:decomposition of invariant arc systems}, {\it Decomposition of invariant arc systems}, for each $j$, there is an irreducible arc system $\Theta_j \subset \Lambda_j$ and an integer $n_j$ such that $f(\alpha_j)$ is isotopic relative to $X$ to a lift of an element of $\Theta_j$ under $g^{n_j}$. We now show that we may choose representatives of $\Gamma$ so that $\widetilde{\Gamma} \cap D = \vac$. This will complete the proof of the theorem: for since $g|\rs - (D-e(\alpha)) = f \circ c$ and $c: \rs - (D-e(\alpha)) \to \rs - Int(\alpha)$ is isotopic to the identity relative to $X$, $\Gamma$ is then an irreducible Thurston obstruction for the marked rational map $(f,X)$. By Theorem \ref{thm:characterization of marked rational maps}, {\it Characterization of marked rational maps}, and the fact that $f$ is not covered by an integral torus endomorphism, this gives a contradiction. First, we may assume $\Gamma$ minimizes the number of intersections with $f(\alpha_j)$ and $\Theta_j$ for all $j$. Write $\bdry D = \cup_{j}\alpha_{0,j}^{\pm}$, where $\alpha_{0,j}^{\pm}$ are the two unique components of $(g|D)^{-1}(f(\alpha_j))$ which are contained in $\bdry_{\pm} D$ respectively. The proof now breaks down into several cases: \gap \noindent {\bf Case 1: $\Gamma \cdot f(\alpha_j) = 0$}. Then $\Gamma \cap f(\alpha_j) = \vac$, hence $\widetilde{\Gamma} \cap g^{-1}(f(\alpha_j)) = \vac$, and so $\widetilde{\Gamma} \cap \alpha_{0,j}^{\pm} = \vac$. \gap \noindent {\bf Case 2: $\Gamma \cdot f(\alpha_j) \neq 0$}. We will again show that $\wt{\Gamma} \cap \alpha^{\pm}_j = \vac$. We first show that in this case $f(\alpha_j) \in \Theta_j$ up to isotopy. By our choice of $\Theta_j$, $f(\alpha_j)$ is isotopic to a lift of an element $\theta \in \Theta_j$ under $f^{n_j}$. First, if $\Gamma \cdot \Theta_j = 0$, then by an application of Theorem \ref{thm:arcs intersecting obstructions}, {\it Arcs intersecting obstructions}, Part 1, we would have $\Gamma \cdot f(\alpha_j) = 0$, contradicting our assumption. Hence $\Gamma \cdot \Theta_j \neq 0$. Next, if $f(\alpha_j) \not\in \Theta_j$ up to isotopy, then by Theorem \ref{thm:arcs intersecting obstructions}, {\it Arcs intersecting obstructions}, Part 2(d) ( with $\Lambda = \Theta_j$ and $\lambda=f(\alpha_j)$), $\Gamma \cdot f(\alpha_j) = 0$, contradicting our assumption. Hence $f(\alpha_j) \in \Theta_j$ up to isotopy. Neither of the arcs $\alpha_{0,j}^{+}, \alpha_{0,j}^{-}$ can be isotopic to elements of $\Theta_{j}$. For if so, then $\alpha$ must consist of a single arc in $(S^2,X)$, i.e. $L=1$ in the definition of the blown up map $g$. Then $\alpha$ itself is isotopic to an element of $\Theta_j$ since $\Theta_j$ is irreducible. It then follows that $\bdry D \subset g^{-1}(f(\alpha))$ consists of two arcs $\alpha_{0}^{+}$ and $\alpha_{0}^{-}$ in $(S^2,X)$ which are isotopic relative to $X$, by construction of the blown-up map $g$. Hence the multiplicity of $g: \alpha_{0}^{\pm} \to f(\alpha)$ is at least $n \geq 2$, where $n$ is the number of times we blow up $\alpha$. Hence the transformation $g_{\#,\Theta}$ is not a transitive permutation of the basis vectors. By Theorem \ref{thm:arcs intersecting obstructions}, {\it Arcs intersecting obstructions}, $\Gamma \cdot \Theta = 0$, hence $\Gamma \cdot f(\alpha) = 0$, contradicting our assumption. By the previous paragraph $\alpha_{0,j}^{\pm} \not\in \Theta_{j}$ up to isotopy. By Theorem \ref{thm:arcs intersecting obstructions}, Part 2a(ii), there is exactly one lift $\lambda'$ of $f(\alpha_j)$ which intersects (as a subset of $S^2$) the set $\widetilde{\Gamma}$, and this lift $\lambda'$ is isotopic to an element of $\Theta_{j}$. Hence $\alpha_{0,j}^{\pm} \cap \widetilde{\Gamma} = \vac$. \gap In both cases we have shown that for all $j$, $\widetilde{\Gamma} \cap \alpha_{0,j}^{\pm} = \vac$. Thus $\widetilde{\Gamma} \cap \bdry D = \vac$, which implies that $\widetilde{\Gamma} \cap D = \vac$: there can be no components of $\Gamma$ contained entirely in $D$ since $\Int(D) \cap X = \vac$. \qed \gap\noindent{\bf Proof of Corollary 1} Suppose $\alpha$ is a periodic or preperiodic internal ray of $f$. Points in the interior of an internal ray lie in Fatou components and have infinite forward orbits under $f$. Hence no points in the interior of an internal ray can be preperiodic under $f$. We may therefore assume that $\alpha$ consists of a single arc in $(S^2,f^{-1}X)$, and hence that $f(\alpha)$ is a periodic or preperiodic internal ray with endpoints in $X$. Two internal rays either coincide, or intersect only in their endpoints. Hence if $l_i = f^i (f(\alpha)), i > 0$, then $l_i \cap \Int(f(\alpha)) \neq \vac $ if and only if $\alpha=l_i$ for some $l_i$, i.e. $f(\alpha)$ is a periodic internal ray. Let $\cal{L}$ $ = \{l_i\}_{i>0}$. Then $\cal{L}$ is a finite collection of arcs in $(S^2,X)$ which, as a subset of the sphere, is forward invariant under $f$. By Proposition \ref{prop:finding invariant arc systems} there exists a subset $\Lambda \subset \cal{L}$ which is an arc system forward invariant under $f$ such that each element of $\cal{L}$ is isotopic relative to $X$ to a unique element of $\cal{L}$. By the definition of forward invariance there is a subset $\Lambda_0 \subset \Lambda$ and a lift $\wt{\Lambda}_0 \simeq_X \Lambda$. If $f(\alpha) \not\in \Lambda$, then $\Int(f(\alpha)) \cap \Lambda = \vac$, hence $\Int(\alpha) \cap f^{-1}(\Lambda) = \vac$. Since $\wt{\Lambda}_0 \subset f^{-1}(\Lambda)$, $\wt{\Lambda}_0 \cap \Int(\alpha) = \vac$. The Corollary then follows by Theorem A. If $f(\alpha) \in \Lambda$, let $\lambda'$ be any arc which is isotopic to $f(\alpha)$ relative to $X$ and such that $\Int(f(\alpha)) \cap \lambda' = \vac$. Let $\Lambda'$ be the arc system $(\Lambda - \{f(\alpha)\}) \cup \{\lambda'\}$. Then $\Lambda \simeq_X \Lambda'$ and so $\Lambda'$ is forward invariant under $f$ up to isotopy relative to $X$. By construction $\Lambda' \cap \Int(f(\alpha)) = \vac$, hence $f^{-1}(\Lambda') \cap \Int(\alpha) = \vac$. If $\Lambda' _0 $ is the unique subset of $\Lambda'$ for which $\Lambda' _0 \simeq_X \Lambda_0$, and if $\wt{\Lambda}' _0$ is the lift of $\Lambda' _0$ obtained by lifting isotopies between elements of $\Lambda' _0$ and $\Lambda_0$, then $\wt{\Lambda}' _0 \simeq_X \wt{\Lambda}_0 \simeq_X \Lambda$. Since $\wt{\Lambda}'_0 \subset f^{-1}(\Lambda')$, $\wt{\Lambda}' _0 \cap \Int(\alpha) = \vac$. The Corollary then follows by Theorem A. \qed \noindent{\bf Proof of Corollary 2} The argument is similar to the one given above. The hypothesis guarantees the existence of the arc $\lambda'$ used in the last paragraph. \qed \subsection{Proof of Theorem B} \label{subsection:proof of theorem two} Let $f(z)$ be a postcritically finite monic polynomial and $X$ a finite set such that $f(X) \subset X$ and $X \supset P(f)$ . Let $K(f)$ and $F(f)$ denote the filled-in Julia set and Fatou set of $f$, respectively. In \cite{poirier:pcfpii}, \S I.1 (originally in \cite{DH2}), it is shown that $X-\{\infty\}$ determines uniquely a finite topological tree $T$ with vertices $V(T)$ satisfying the following properties. \begin{enumerate} \item $T \subset K(f)$ and $T \cap F(f)$ only in internal rays; \item $f(T) \subset T$ and $f(V(T)) \subset V(T)$; \item $V(T) \supset X-\{\infty\}$. \end{enumerate} The definition of internal and external rays is given in the following section. Our first step in the proof is to construct a ``spider'' $S$ for $T$. The construction of $S$ is given in the following section. $S$ is a finite topological graph with vertices $V(S)$ and edges $E(S)$ satisfying the following properties: \begin{enumerate} \item $V(S) = X \supset P(f)$; \item Every edge of $S$ is a topological arc joining a finite vertex to infinity; \item $f(S) \subset S$; \item $f$ maps edges of $S$ homeomorphically to edges of $S$; \item $S \cap T = X-\{\infty\}$; \end{enumerate} Note that the edges of $S$ are arcs in $(S^{2},X)$. Given the existence of the spider $S$ with the properties above, we now prove Theorem B. The edges $E(S)$ form a finite collection of arcs $l$ in $(S^2,X)$ such that $f(l) \in E(S)$ for each $l \in E(S)$. By Proposition \ref{prop:finding invariant arc systems}, there is an arc system $\Lambda \subset E(S)$ which is forward invariant under $f$ up to isotopy relative to $X$, i.e. there is a subset $\Lambda_0 \subset \Lambda$ and a lift $\wt{\Lambda}_0 \simeq_X \Lambda$. Note that since $\wt{\Lambda}_0 \simeq_X \Lambda \subset X$, the endpoints of $\wt{\Lambda}$ are points in $X$. We now show that $\Lambda$ is forward invariant under $g$ up to isotopy relative to $X$. By Proposition \ref{prop:orbits off alpha persist}, {\it Orbits off $\alpha$ persist}, it is enough to show that $\wt{\Lambda}_0 \cap \Int(\alpha) = \vac$. Since $\alpha \subset T$ and $f(T) \subset T$, $f(\alpha)\subset T$. Since $T \cap S \subset X$, $f(\alpha) \cap S \subset X$. Hence $\alpha \cap f^{-1}(S) \subset f^{-1}(X)$. Now, $\wt{\Lambda}_0 \subset f^{-1}(S)$, and as remarked above, the endpoints of $\wt{\Lambda}_0$ are points in $X$, hence $\alpha \cap \wt{\Lambda}_0 \subset X$. But $\Int(\alpha) \cap X = \vac$, hence $\Int(\alpha) \cap \wt{\Lambda}_0 = \vac$. Hence $\Lambda$ is forward invariant under $g$ up to isotopy relative to $X$. By Proposition \ref{prop:decomposition of invariant arc systems}, there are disjoint (as subsets of $\Lambda$) irreducible arc systems $\Theta_k \subset \Lambda$ and integers $n_k$ such that each element of $\Lambda$ is a lift of an element of $\Theta_k$ under $n_k$ for some $k$. Let $\Gamma$ be an irreducible Thurston obstruction for $g$ and let $\widetilde{\Gamma}$ be the union of components of $g^{-1}(\Gamma)$ which are isotopic relative to $X$ to elements of $\Gamma$. We may assume that $\Gamma$ minimizes the number of intersections with each $f(\alpha_j)$ and with each $\lambda \in \Lambda$. As in the previous proof it suffices to show that $\widetilde{\Gamma} \cap \bdry D = \vac$, where $D$ is the disc used in the construction of $g$. Let $\cal{K}$ $ = (g|D)^{-1}(\Lambda)$; this is a finite, connected graph in $D$ with vertices equal to $g^{-1}(X) \cap D \subset \cal{K}$. We first reduce to the case when $\widetilde{\Gamma} \cap \cal{K} = \vac$. As in the previous proof, there are two cases. Let $\lambda \in \Lambda$. \gap \noindent {\bf Case 1:} $\Gamma \cdot \lambda = 0$. Then $\Gamma \cap \lambda = \vac$ by our choice of representatives of $\Gamma$, so $\widetilde{\Gamma} \cap (g|D)^{-1}(\lambda)= \vac$. \gap \noindent {\bf Case 2:} $\Gamma \cdot \lambda \neq 0$. Let $\wt{\lambda}$ be any lift of $\lambda$ under $g$ which is contained in $D$ (and hence in $\cal{K}$). By Proposition \ref{prop:decomposition of invariant arc systems}, $\lambda$ is isotopic to a lift of an element of some irreducible arc system $\Theta_j \subset \Lambda$ under $f^{n_k}$, some $n_k$. Since $\lambda \in \Lambda \subset E(S)$, $\lambda$ is an edge of the spider $S$ and hence joins a point of $X-\{\infty\}$ to the point $\infty$. By construction $D \cap \infty = \vac$. Since $\wt{\lambda} \subset D$ is a lift of $\lambda$ under $g$, one of the endpoints of $\wt{\lambda}$ is contained in $D$ and maps onto $\infty$. Hence this endpoint is strictly preperiodic under $g$. Hence $\wt{\lambda}$ cannot be contained in the irreducible arc system $\Theta_k$. By Theorem \ref{thm:arcs intersecting obstructions}, {\it Arcs intersecting obstructions}, $\wt{\Gamma} \cap \wt{\lambda} = \vac$. Together, the two cases show that $\wt{\Gamma} \cap \cal{K}=\vac$. \gap So we may assume $\widetilde{\Gamma} \cap \cal K = \vac$. We conclude the proof by showing that if $\wt{\Gamma} \cap \bdry D \neq \vac$, then $\Gamma$ does not have minimal intersection number with $f(\alpha)$, contrary to our assumption. Suppose $\wt{\gamma} \subset \wt{\Gamma}$ intersects $\bdry D$. Since $\wt{\Gamma} \cap \cal{K} = \vac$, $\wt{\gamma} \cap \Int(D)$ is contained in a component $V$ of $D-\cal{K}$. Since $V(S) = X \supset P(f)$, and since each edge $E(S)$ is isotopic relative to $X$ to an element of some $\Lambda_k$, each component of $S^2 - \Lambda$ is an open disc containing no critical values of $g$. Hence each component of $g^{-1}( S^2 - \Lambda)$ maps homeomorphically under $g$ to its image. Since $V$ is contained in such a component, $g|V$ is a homeomorphism. Let $W \subset V$ be any component of $V - \wt{\gamma}$ which ``outermost'', i.e. not separated from $\bdry D$ by any component of $\wt{\gamma} \cap D$. Then $W$ is a disc in $S^2-g^{-1}X$ bounded by exactly one subarc of $\wt{\gamma}$ and one subarc of $\bdry_+ D$ or $\bdry_- D$. Since $g|V$ is a homeomorphism and $W \subset V$, $g(W)$ is a disc in $S^2-X$ bounded by one subarc of $f(\alpha)$ and one subarc of $\gamma \subset \Gamma$, contradicting the assumption that $\Gamma \cap f(\alpha)$ is minimized. \qed \subsection{Construction of $S$} \label{subsection:construction of S} We first recall some facts about the dynamics of postcritically finite polynomials from \cite{MIL1}. Let $f(z)$ be a postcritically finite polynomial. The {\it filled-in Julia set} $K(f)=\{z | f^{n}(z) \not \to \infty\}$ is a connected, compact subset of $\C$ whose complement is connected. The Julia set $J(f) = \bdry K(f)$ is locally connected. For each Fatou component $\Omega$, there is exactly one point $x \in \Omega$ such that $f^{n}(x) \in P(f)$ for some $n$. Let $\Omega_{x}$ denote the Fatou component containing such a point $x$. A classical theorem of B\"{o}ttcher implies the following: there are holomorphic isomorphisms $\phi_{x}: (\Delta,0) \to (\Omega_{x},x)$ such that for all $w \in \Delta$, $\phi_{f(x)}(w^{d_{x}}) = f(\phi_{x}(w))$, where $d_{x}$ is the local degree of $f$ near $x$. Since $J(f)$ is locally connected, each Fatou component has locally connected boundary, and a theorem of Carath\'eodory implies that the maps $\phi_{x}$ extend continuously to $S^{1}=\bdry \Delta$. Let $R_{t}$ denote the radial line $\{r \exp(2\pi i t) | 0 \leq r \leq 1\}$. The image $R_{x,t}=\phi_{x}(R_{t})$ we call the {\it ray of angle $t$ in $\Omega_{x}$}; if $x=\infty$ $R_{x, t}$ is called an {\it external ray}; otherwise we call it an {\it internal ray}. Each periodic point $x \in J(f)$ is the landing point of at least one and at most finitely many external rays. The landing point of an external ray is periodic if and only if $x$ is periodic, though the periods may differ. Internal and external rays map homeomorphically onto their images. \gap\noindent {\bf Construction of $S$.} Our spider will be a union \[ S = \bigcup_{x \in X-\{\infty\}} E_{x}, \] where $E_{x}$ denotes the set of edges of the graph $S$ incident to $x$. \begin{itemize} \item If $x \in J(f) \cap (X-\{\infty\})$, each element $e \in E_{x}$ will be an external ray landing at $x$, and $E_{x}$ will be the union of all external rays landing at $x$. \item If $x \in F(f) \cap (X-\{\infty\})$, each element $e \in E_{x}$ will be the union of exactly one internal ray in $\Omega_{x}$ and one external ray landing at a common point $q \in \bdry \Omega_{x}$. The set $E_{x}$ will be a finite union of such pairs. \end{itemize} Since external and internal rays map homeomorphically onto their images, and the image of an internal (external) ray is again an internal (external) ray, edges will map homeomorphically to edges. Each $x \in J(f) \cap (X-\{\infty\})$ is eventually periodic, so the set $E_{x}$ of external rays landing at $x$ is finite (see \cite{MIL1}, \S 18), and $f(E_{x}) = E_{f(x)}$. Since $T \subset K(f)$, $E_{x} \cap T \subset X-\{\infty\}$. We now define $E_{x}$ for $x \in F(f) \cap (X-\{\infty\})$. First, choose one element $x$ from each periodic cycle in $(X-\{\infty\}) \cap F(f)$. Let $p$ be the period of $x$. Then $T \cap \Omega_{x}$ consists of a finite collection of eventually periodic internal rays. Thus there exists a periodic internal ray $R_{x}$, of some period $kp>0$, such that $R_{x} \cap T = \{x\}$. Let $q$ be the landing point of the internal ray $R_{x}$; then $q$ is periodic of period $k$. There exists a periodic external ray $R_{\infty}$ landing at $q$. Since $q$ is not a critical point, $f^{k}$ is a local homeomorphism near $q$ sending $R_{x}$ to itself, hence the ray $R_{\infty}$ also has period $k$. Let $e_{x}=R_{x} \cup R_{\infty}$; this will be one edge in the collection $E_{x}$. Define \[ E_{x}=\bigcup_{n=0}^{k-1} f^{np}e(x). \] $E_{x}$ is finite since $q$ is periodic. For $y \in f^{n}(x), 1 \leq n < p$, define \[ E_{y} = f^{n}(E_{x}). \] Thus $E_{y}$ is finite. Since $R_{x}$ is periodic, $R_{x} \cap T = \{x\}$, $f(X-\{\infty\}) \subset X-\{\infty\}$, and $f(T) \subset T$, we have that $E_{y} \cap T \subset X-\{\infty\}$ for all $y$. We now inductively define $E_{x}$ for strictly preperiodic $x \in X-\{\infty\} \cap F(f)$. Suppose the collection $E_{y}$ has been defined, where $f(x)=y$. Choose any element $e_{y} \in E_{y}$. The edge $e_{y}$ is a union of an internal ray $R_{y}$ and external ray $R_{\infty}$; let $q$ be their common landing point. Choose any preimage $R_{x}$ of $R_{y}$ joining $x$ to a point $q'$ in $\bdry \Omega_{x}$. Since a small disc near $q$ pulled back to $q'$ is again a disc, there is a preimage $R'_{\infty}$ of $R_{\infty}$ joining $\infty$ to $q'$. Let $e_{x} = R_{x} \cup R'_{\infty}$ and $E_{x}=\{e_{x}\}$. Then $f: e_{x} \to e_{y}$ and $E_{x} \cap T = \{x\}$. This completes the construction of $S$. %\input{complements} \section{Complements} \label{section:complements} \subsection{Blowing up multiple arcs} Theorems A and B generalize in the obvious way to blowing up along finite collections $\alpha^1, \alpha^2, ...$ of arcs for which $\Int(\alpha^i) \cap \alpha^j = \vac, i \neq j$. \subsection{Blowing up M\"{o}bius transformations} \label{subsection:blowing up moebius transformations} Let $G$ be a finite connected graph in $S^2$ with vertex set $X$ and edge set $E$. Suppose in addition that no edge joins a vertex to itself and that no two vertices are joined by more than one edge. Assign to each edge $e \in E$ an integer $m(e) \geq 0$. Let $(M,X)$ be a marked M\"{o}bius transformation of finite order and suppose $M(G)=G$. \gap\noindent{\bf Theorem: (Blowing up M\"obius transformations)} \label{thm:blowing up moebius transformations} {\it Let $(g,X)$ be the marked branched covering $(M,X)$ blown up $m(e)$ times along the edge $e \in E$. Let $G'$ denote the subgraph of $G$ which is the union of edges $e'$ (and their vertices) for which $m(M^{i}e')>0$ for some $i$. Then $(g,X)$ is equivalent to a marked rational map if and only if \begin{enumerate} \item $G'$ is connected, and \item if $G'$ is nonempty, no component of $S^{2}-G'$ contains more than one point of $X$. \end{enumerate} } \pf If $G'$ is empty there is nothing to prove. Otherwise, there is at least one edge which is blown up. The vertices of this blown up edge become periodic critical points for $g$, hence the orbifold of the (unmarked) branched covering $g$ is not the $(2,2,2,2)$ orbifold. Hence by Theorem \ref{thm:characterization of marked rational maps}, {\it Characterization of marked rational maps}, $(g,X)$ is equivalent to a marked rational map if and only if $(g,X)$ has no Thurston obstructions. Note that $G'$ is invariant under $M$. Let $X'$ denote the vertices of $G'$. Then for each connected component $G''$ of $G'$, $G''$ has at least two vertices. We now prove the necessity of (1) and (2). First, assume that $(g,X)$ is equivalent to a marked rational map. Then $(g^{\circ n}, X)$ is also equivalent to a marked rational map for every $n \geq 1$. If (2) fails, then there is a component $U$ of $S^2-G'$ such that $|U \cap (X - X')| \geq 2$. Let $\gamma$ be any simple closed curve in $U$ separating $G'$ from $U \cap (X - X')$. Then $\gamma$ is nonperipheral since $|X'| \geq 2$ and $|U \cap(X-X')| \geq 2$ by assumption. Choose $n \geq 1$ so that $M^{\circ n} (\gamma) = \gamma$. We may further assume that $\gamma$ is disjoint from the discs $W$ within which we alter the map $(M,X)$ to obtain the map $(g,X)$. It then follows that $g^{\circ n}(\gamma) = \gamma$ and that $(g^{\circ n})|\gamma$ is a homeomorphism. Then $\gamma$ is a Levy cycle for $(g^{\circ n},X)$, so $(g^{\circ n},X)$ is not equivalent to a rational map. If (1) fails, then there is a simple closed curve $\gamma$ in $S^2 - (G' \cup X)$ separating $G'$ into two subsets, each of which has at least two elements of $X$. Hence $\gamma$ is nonperipheral. The same argument as in the previous paragraph then shows that $(g^{\circ n},X)$ is not equivalent to a rational map. We now prove the sufficiency of (1) and (2). We may assume $|X| \geq 4$, since otherwise there are no nonperipheral simple closed curves in $(S^2,X)$ and hence no Thurston obstructions. Next, we may assume that for every $e \in E$ and for every $i > 0$, $M^i e \simeq_X e $ if and only if $M^i e = e$, by our requirement that there is at most one edge between any two vertices and the assumption $M(G)=G$. For each orbit of an edge in $E'$ under $M$, choose one edge $e' _j$. Let $\epsilon_j \simeq_X e' _j$ be an arc in $(S^2,X)$ chosen so that $\epsilon_j \cap G = e(\epsilon_j)$. Then in particular, $\Int(\epsilon_j) \cap e' = \vac$ for each $e' \in E'$. Let $\Lambda_j$ denote the arc system which is the orbit of $\epsilon_j$ under $M$. Then $\Lambda_j$ is forward-invariant under $M$ and the interior of every element of $\Lambda_j$ is disjoint from the edges along which we blow up. Hence by Proposition \ref{prop:orbits off alpha persist}, {\it Orbits off $\alpha$ persist}, each $\Lambda_j$ is forward-invariant up to isotopy under $(g,X)$. The transformations $g_{\#, \Lambda_j}$ are irreducible but are not transitive permutations of the basis vectors. The transformation $M_{\#, \Lambda_j}$ is a transitive permutation of the basis vectors since $\Lambda_j$ is the orbit of a single arc under $M$ and is therefore irreducible. Hence as matrices $g_{\#, \Lambda_j} \geq M_{\#, \Lambda_j} \geq 0$, so $g_{\#, \Lambda_j}$ is irreducible since $M_{\#, \Lambda_j}$ is irreducible. Moreover, if $\lambda \in \Lambda_j$ and $\lambda \simeq_X e'$ where $m(e') \geq 1$, then $\mult(g: \lambda \to g(\lambda)) = m(e') + 1 \geq 2$, and so the entry of $g_{\#, \Lambda_j}$ corresponding to the ordered pair $(\lambda, g(\lambda))$ is at least two. Let $\Gamma$ be an irreducible Thurston obstruction for $(g,X)$. We may assume that the elements of $\Gamma$ are chosen so that $\Gamma \cdot \Lambda_j = | \Gamma \cap \Lambda_j|$ for each $j$. By Theorem \ref{thm:arcs intersecting obstructions}, {\it Arcs intersecting obstructions}, $\Gamma \cap \Lambda_j = \vac$ for each $j$. Since $\cup_j \Lambda_j = G'$, $\Gamma \subset S^2 - G'$. Conditions (1) and (2) imply that every component of $S^2-(G' \cup X)$ is either simply-connected or is a once-punctured disc. Hence there are no nonperipheral simple closed curves in $S^2-(G' \cup X)$, and so there are no Thurston obstructions for $(g,X)$. \qed \gap\noindent{\bf Example:} Let $f_{p/q}(z)=z^{2}+c_{p/q}$, $q \geq 2$, where $p 0$. Let $X=\{0,c,f(c)\}$. Then $f$ is ``starlike'', so there is a tree $T$ (shown in Figure \ref{fig:rabbit_tree.ps}) with vertices consisting of the set $X$ and the alpha fixed point $p$ of $f$ and which is mapped homeomorphically to itself under $f$ with rotation number $1/3$. $T$ is a union of the three internal rays for $f$ which are fixed under $f^{\circ 3}$. \yincludeps{rabbit_tree.ps}{1}{The tree for the Douady Rabbit.} We let $\alpha$ be the union of the two internal rays joining $0$ and $c$ to $p$. By Corollary 2, blowing up once along $\alpha$ yields a branched covering equivalent to a rational map. The Julia set of $f$ is shown in Figure \ref{fig:rabbit.ps}, and that of $f$ blown up along $\alpha$ is shown in Figure \ref{fig:rabbit_blow.ps}. \yincludeps{rabbit.ps}{1.75}{Douady's Rabbit.} \yincludeps{rabbit_blow.ps}{1.75}{The Douady Rabbit blown up along $\alpha$.} \smallbreak \noindent{\bf Blowing up the ``airplane''.} Let $f(z)=z^{2}+c$, where $c \approx -1.75488$ is chosen so that the critical point at zero is periodic of period three with orbit lying in the real line. Then $c < 0 < f(c)$. Let $X=\{c,0,f(c)\}$. The tree $T_{X}$ is the interval $[c,f(c)]$ with vertices $X$. The polynomial $f$ sends the invervals $[c,0]$ and $[0,f(c)]$ homeomorphically to the interval $[c,f(c)]$. We let $\alpha=[c,0] \subset T_{X}$. Then the Blowing Up conditions are satisfied, and by Theorem B, $(f,X)$ blown up once along $\alpha$ is equivalent to a rational map. The Julia set of $f$ is shown in Figure \ref{fig:airplane.ps}. The Julia set of the result of blowing up $f$ along $\alpha$ is shown in Figure \ref{fig:airplane_blow.ps}; the map is given by $g(z)=a \frac{(z-1)^{3}}{1-3z}$, where $a \approx -2.37123$, normalized so that $1 \mapsto 0$ by local degree three, $0 \mapsto -a$ by local degree two, and $-a \mapsto 1$ by local degree one. \xincludeps{airplane.ps}{4.5}{The airplane.} \xincludeps{airplane_blow.ps}{4.5}{The airplane blown up along $\alpha$.} \gap\gap \noindent {\bf A degree four map with $S_3$ symmetry.} Let $G$ be a topological graph in $S^2$ which is a triangle with vertices $X$ and edges $E$. Let $M$ be the identity map. By Theorem \ref{thm:blowing up moebius transformations}, {\it Blowing up M\"obius transformations}, the marked covering $(g,X)$ blown up once along each edge in $E$ is equivalent to a degree four rational map $g$ with three fixed critical points each of mulitiplicity two. The full mapping class group of orientation-preserving homeomorphisms $h$ of $S^{2}-X$ up to isotopy is isomorphic to $S_3$ and commutes with $g$ up to equivalence since the blowing up construction in this case is performed symmetrically. That is, $h\circ g$ is equivalent to $g \circ h$. It follows that the conformal automorphism group of $g$ is $S_3$. A formula for this map is given in \cite{ctm:iterative} and is used there for giving a generally convergent iterative algorithm for solving the cubic. \gap\noindent{\bf Remark:} More generally one can apply Theorem \ref{thm:blowing up moebius transformations}, {\it Blowing up M\"obius transformations} to other symmetric graphs and the identity map to produce symmetric postcritically finite rational maps; cf. \cite{ctm:pd:quintic} for a detailed discussion of rational maps with icosahedral symmetry. %\input{appendix1} \section{Appendix: Proof of Theorem {\it Arcs intersecting obstructions} } \label{appendix:arcs} % %Tan Lei's macros % \def\g{\gamma} \def\G{\Gamma} \def\la{\lambda} \def\La{\Lambda} \def\pf{postcritically finite } \def\pfbc{postcritically finite branched covering } \def\pfbcs{postcritically finite branched coverings } \def\cc{connected component } \def\ccs{connected components } \def\gf{geometrically finite } \def\pc{postcritical set } \def\lc{locally connected } \def\beginp{ {\noindent\em Proof.} } \def\endp{\hspace*{\fill}$\rule{.55em}{.55em}$ \smallskip} %\leqno We will use the following easy fact: \begin{lemma} \label{lemma:crucial} Let $f$ be a branched covering from $S^2$ to $S^2$. Let $B$ be a subset of $S^2$ such that $f|_B:B\to f(B)$ is a bijection (resp. a $k$-to-$1$ mapping). Let $A$ be any subset of $S^2$. Then $\# f^{-1}(A)\cap B=\# A\cap f(B)$ (resp. $\# f^{-1}(A)\cap B=k\cdot \# A\cap f(B)$). \end{lemma} For $\La$ a curve system or an arc system in $(S^2,X))$, set $\widetilde{\Lambda}$ (resp. $\widetilde{\Lambda}(f^n)$) to be the union of those components of $f^{-1}(\Lambda)$ (resp. $f^{-n}(\La)$) which are isotopic to elements of $\Lambda$. If $\La$ is irreducible, each component of $\La$ is isotopic to some (possibly non-unique) component of $\widetilde{\La}(f)$ (resp. $\widetilde{\La}(f^n)$). By the definition of irreducible arc system, there exists at least one subset $\widetilde{\La}'$ of $\widetilde{\La}$ such that $\widetilde{\La}'$ is isotopic to $\La$ and $f:\widetilde{\La}'-X \to \La-X$ is a homeomorphism. Then \begin{equation} \label{eqn:1} f_{\#,\G}(\g_j) \cdot \La \le \sum_{\g'\subset f^{-1}(\g_j)} \#(\g'\cap \widetilde{\La}') =\#( f^{-1}(\g_j)\cap \widetilde{\La}') = \#(\g_j\cap \La) = \g_j\cdot \La\ \end{equation} where the first inequality follows since $f:\widetilde{\La}'-X \to \La-X$ is a homeomorphism and the second to last equality follows by the lemma. Assume $f_{\#,\G}(\g_j)=\sum_i b_{ij}\g_i$. Then by (1), $ (\sum_j\sum_i b_{ij}\g_i) \cdot \La \le \G\cdot \La\ .$ But \begin{equation} \label{eqn:2} (\sum_j\sum_i b_{ij}\g_i) \cdot \La =(\sum_i(\sum_j b_{ij})\g_i) \cdot \La \geq \sum_i \g_i \cdot \La = \G\cdot \La\ , \end{equation} where the last inequality is obtained as follows. By irreducibility, $f_\G$ has no zero row no zero column, so the same is true for $f_{\#,\G}$. But each entry of $f_{\#,\G}$ is a non negative integer, so $b_{ij}\ge 1 $ if $b_{ij}\ne 0$. We have then either A) $\G\cdot \La=0$ or B) $\G\cdot \La \neq 0$ and all inequalities above become equalities. In case that $\G\cdot \La=0$, we claim that $\G\cdot f^{-1}(\La)=0 $. Let $\g_i\in \G$. Then $\g_i\sim \g'$ with $\g'$ a curve in $f^{-1}(\g_j)$ for some $\g_j\in \G$. Then $\g_i \cdot f^{-1}(\La) \le \#(\g'\cap f^{-1}(\La))=k\#( \g_j\cap \La) = 0$. The case for $n > 1$ is proved similarly. This prove part one of the theorem. In case B), equalities in Equation \ref{eqn:2} imply that for each $i$, there is a unique $j$ such that $b_{ij}\ne 0$ and that $b_{ij}=1$. Thus for each $\g_i$, there is a unique $\g_j$ such that some component of $f^{-1}(\g_j)$ is isotopic to $\g_i$. Also, there is a unique such component $\g'$ of $f^{-1}(\g_j)$ isotopic to $\g_i$. Since $f_{\#,\G}$ is irreducible, this implies that $f_{\#,\G}$ is a transitive permutation of the basis vectors. As a consequence, $\g'$ is also the unique curve in $f^{-1}(\g_j)$ isotopic to an element of $\G$. Equalites in Equation \ref{eqn:1} tells us that $f_{\#,\G}(\g_j)\cdot \La =\#(f^{-1}(\g_j)\cap \widetilde{\La}') =\g_j\cdot \La$. Combining with the facts that $f_{\#,\G}(\g_j) =[\g']_\G=\g_i$, $\g'\cdot \La\le \#(\g'\cap \widetilde{\La}')$ and $\g'\cap \widetilde{\La}'\subset f^{-1}(\g_j)\cap \widetilde{\La}'$, we get $\g_i\cdot \La=\g_j\cdot \La$ and $ \g'\cap \widetilde{\La}' = f^{-1}(\g_j)\cap \widetilde{\La}'$. As a consequence, $\g'$ is also the unique curve in $f^{-1}(\g_j)$ such that $\g'\cap \widetilde{\La}'\ne \emptyset$. Moreover $f^{-1}(\G)\cap \widetilde{\La}'=\widetilde{\G}\cap \widetilde{\La}'$. The matrix $f_\G$ is a permutation with entry $a_{ij}=1/k_j$, where $k_j$ is the degree of $f:\g'\to \g_j$. Since $\la(f_\G)\geq 1$, we have $k_j=1$ for all $j$ and $f_\G=f_{\#,\G}$. Thus $f:\widetilde{\G}\to \G$ is a homeomorphism. Note that in the above argument we only used the following properties of the triple $(\G,\La,\widetilde{\La}')$: the matrix $f_{\G}$ has neither zero row nor zero column, with leading eigenvalue at least $1$, the set $\widetilde{\La}'$ is isotopic to $\La$ and $f:\widetilde{\La}'-X\to \La-X$ is a homeomorphism. By what we just proved, the triple $(\La,\G,\widetilde{\G})$ verifies also these properties. So we can redo the same argument for this new triple. We conclude then for each $\la\in \La$, there is exactly one curve $\la'$ in $f^{-1}(\la)$ with $\la'\cap \widetilde{\G}\ne \emptyset$. Moreover, $\la'$ is the unique curve in $f^{-1}(\la)$ which is isotopic to an element of $\La$. The matrix $f_{\#,\La}$ is a transitive permutation. As a consequence, $\widetilde{\La}'=\widetilde{\La}$. Replace $\widetilde{\La}'$ by $\widetilde{\La}$ in the beginning of the proof, we conclude finally that $\widetilde{\G}\cap f^{-1}(\La)=\widetilde{\G}\cap \widetilde{\La} = f^{-1}(\G)\cap \widetilde{\La}$ and $\#\widetilde{\G}\cap \widetilde{\La}=\#\G\cap \La=\G\cdot \La$. As a consequence, $\widetilde{\G}\cap (f^{-1}(\La)- \widetilde{\La})=\emptyset$ and $(f^{-1}(\G)-\widetilde{\G})\cap \widetilde{\La}=\emptyset$. Now take any arc $\la$ isotopic to some arc in $f^{-1}(\La)$ but to none of the arcs in $\La$, it is isotopic to an arc $\la'\in f^{-1}(\La)-\widetilde{\La}$. So $\G\cdot \la\le \#(\widetilde{\G}\cap \la')=0$. This proves the part $2(a),2(b)$ and $(2(d),n=1)$ of the theorem. To get $2(c)$ and $(2(d),n>1)$, we just need to note that if $\G,\La$ are irreducible for $f$, they are also irreducible for $f^n$, for all $n\ge 1$. This gives the theorem. \qed The following theorem can be proved in exactly the same manner as Theorem \ref{thm:arcs intersecting obstructions}. \begin{thm}[Arcs intersecting arcs] Let $\Lambda_{1}, \Lambda_{2}$ be two irreducible systems of arcs. Then either 1. $\Lambda_{1} \cdot \Lambda_{2} = 0$ and $ f^{-n}(\Lambda_{1}) \cdot f^{-n}(\Lambda_{2}) = 0$ for all $n\ge 1$, or 2. $\Lambda_{1} \cdot \Lambda_{2} \neq 0$ and $( f^{-n}(\Lambda_{1})-\widetilde{\Lambda}_{1}(f^n) ) \cdot \La_2=0= \La_1\cdot ( f^{-n}(\Lambda_{2}) -\widetilde{ \Lambda}_{2}(f^n))$ for all $n\ge 1$. In this case we also have (a) The transformations $f_{\sharp ,\Lambda_{i}}$ are transitive permutations of the basis vectors; (b) for each $\lambda \in \Lambda_{i}$, there is a unique component of $f^{-1}(\lambda)$ intersecting $\Lambda_{i'}$, $i \neq i' \in \{1,2\}$; (c) (a) and (b) remain true if we replace $f$ by $f^n, n \geq 1$. \end{thm} \begin{cor} If $\Lambda_{1}$ contains more than one periodic cycle, or if some element of $\Lambda$ maps with multiplicity greater than one, then $\Lambda_{1} \cdot \Lambda_{2} = 0$ for any irreducible arc system $\Lambda_{2}$. \end{cor} %\bibliography{/u/pilgrim/text/tex/macros/refs} %\bibliographystyle{/u/pilgrim/text/tex/macros/math} \begin{thebibliography}{McM} \bibitem[Ahm]{ahmadi:thesis} Davood Ahmadi. \newblock {\em Dynamics of Certain Rational Maps of Degree Two}. \newblock PhD thesis, University of Liverpool, 1991. \bibitem[BD]{branner:douady:surgery} Bodil Branner and Adrien Douady. \newblock {Surgery on complex polynomials}. \newblock In {\em Holomorphic Dynamics: Proceedings, Mexico}. 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