Fixed points for branched covering maps of the plane

TL;DR

This paper establishes new sufficient conditions for degree 2 branched covering maps of the plane to have fixed points, extending classical fixed point results for plane homeomorphisms.

Contribution

It introduces specific criteria involving invariant sets and their topological properties that guarantee the existence of fixed points in branched covering maps.

Findings

Provides conditions involving totally invariant sets and their separation properties.

Shows that certain invariant continua containing critical points ensure fixed points.

Extends Brouwer's fixed point theory to branched covering maps with critical points.

Abstract

A well-known result from Brouwer states that any orientation preserving homeomorphism of the plane with no fixed points has an empty non-wandering set. In particular, an invariant compact set implies the existence of a fixed point. In this paper we give sufficient conditions for degree 2 branched covering maps of the plane to have a fixed point, namely: A totally invariant compact subset such that it does not separate the critical point from its image An invariant compact subset with a connected neighbourhood $U$, such that $\mathrm{Fill}(U \cup f(U))$ does not contain the critical point nor its image. An invariant continuum such that the critical point and its image belong to the same connected component of its complement.

Equations

$$t_0=\sup \{t\in [0,1]:{\tilde{V}}_t\cap f^{-1}({\tilde{V}}_t)=\varnothing \}.$$

$$h_1\circ f{|}_{U_0}(c+(\rho ,\theta ))=(f(c)+(h_{11}(\theta )\rho ,2\theta )):{\sqrt{2}}^{-1}\le h_{11}(\theta )\le \sqrt{2}$$

$$h_{22}(x,y)=\{\begin{array}{ll}\frac{y}{4}& \text{si }\frac{81}{10}\le x\le \frac{99}{10}\text{; }|y|\le \frac{1}{2},\\ \frac{7y-3}{4}& \text{si }\frac{81}{10}\le x\le \frac{99}{10}\text{; }|y|>\frac{1}{2},\\ (10x-99)y+(100-10x)\frac{y}{4}& \text{si }x>\frac{99}{10}\text{; }|y|\le \frac{1}{2},\\ (10x-99)y+(100-10x)\frac{7y-3}{4}& \text{si }x>\frac{99}{10}\text{; }|y|>\frac{1}{2}.\\ (81-10x)y+(10x-80)\frac{y}{4}& \text{si }x<\frac{81}{10}\text{; }|y|\le \frac{1}{2},\\ (81-10x)y+(10x-80)\frac{7y-3}{4}& \text{si }x>\frac{99}{10}\text{; }|y|>\frac{1}{2}.\end{array}$$

$$h_{31}(x,y)=\{\begin{array}{ll}|y|x+(1-|y|)\frac{x}{9}& \text{si }x\le 9\\ |y|x+(1-|y|)(9x-80)& \text{si }x>9\end{array}$$

$$W_1:=\{(x,y)\in f(U_0):x\le \frac{81}{10}\};W_2:=\{(x,y)\in f(U_0):x\ge \frac{99}{10}\}$$

Full text

. A totally invariant compact subset such that it does not separate the critical point from its image

. An invariant compact subset with a connected neighbourhood $U$, such that $\mathrm{Fill}(U\cup f(U))$ does not contain the critical point nor its image.

. An invariant continuum such that the critical point and its image belong to the same connected component of its complement.

1. Introduction

The existence of periodic and fixed points for continuous maps of the plane has been extensively studied. A key theorem for the development of this area was given by Brouwer in 1912, [Brou12]:

Theorem 1.1**.**

Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be an orientation preserving homeomorphism such that $\mathrm{Fix}(f)=\varnothing$. Then every point is wandering.

This result triggered a great amount of research, and addresses the simplest case of an open question for plane dynamics: Does a continuous function of the plane taking a non-separating continuum into itself, necessarily have a fixed point? [Ste35]

Cartwright and Littlewood proved in 1951 that if $f$ is an orientation preseving homeomorphism of the plane with an invariant non-separating continuum, then it has a fixed point in it (see [CL51]). A brief elegant proof of the same result was given two decades later by Brown in [Bro77].

Bell generalized this result to all homeomorphisms in 1978 (see [Bel78]), and announced in 1984 that the theorem could be extended to all holomorphic maps (see also [Aki99]) -note that these are a particular type of branched covering maps of the plane, see section 2 for definitions-. Kuperberg extended the previous theorem for orientation reversing homeomorphisms in 1991, taking out the hypothesis of the continuum being non-separating- [Kup91].

More recently, the Cartwright-Littlewood theorem was further extended to all orientation preserving branched covering maps of the plane by Fokkink, Mayer, Oversteegen and Tymchatyn in 2007 [FMOT07].

We want to find sufficient conditions for branched covering maps of the plane to have fixed (or periodic) points. In particular, we are interested in the next question: Does a branched covering map of the plane with an invariant compact subset, have a fixed point?

In this paper we study the dynamics of degree 2 branched covering maps of the plane. Simple examples can be made for these maps to be periodic point free. See, for example, [BO09].

A similar problem for branched covering maps on the sphere was addressed by Iglesias, Portela, Rovella and Xavier in 2016 ([IPRX16]), proving the next result:

Theorem 1.2**.**

Let $f$ be a branched covering map of the sphere, with $\lvert d\rvert>1$. Suppose there exists a simply connected open set $U$, whose closure is disjoint from the set of critical values, and such that $\overline{f^{-1}(U)}\subset U$. Then $f$ has the rate.

Note that in this case we can restrict the dynamics of $f$ to a branched covering map of the plane -with the same degree-, and get a totally invariant non-separating compact set $K=\displaystyle\bigcap_{n\in\mathbb{N}}f^{-n}(U)$. Note that this set has infinitely many connected components.

By their definition, the rate means that $\displaystyle\limsup_{n\rightarrow\infty}\frac{1}{n}log(\#\mathrm{Fix}(f^{n}))\geq log(d).$

Equivalently, it means there is a subsequence of iterates $n_{k}\rightarrow\infty$ such that $\#\mathrm{Fix}(f^{n_{k}})$ grows exponentially as $d^{n_{k}}$, which is the biggest growth to be expected in a non-degenerate case. They actually prove a stronger result, namely, that $f^{n}$ has at least $d^{n}$ fixed points, (in particular $\mathrm{Fix}(f)\neq\varnothing$).

The same mathematicians in 2016 gave sufficient conditions for covering maps of the annulus to have the rate (see [IPRX16.2]), and as a consequence got the next results, which will be key tools in our proof.

Theorem 1.3**.**

Let $f$ be a covering map of the annulus, with degree $d>1$, such that there exists a compact totally invariant subset. Then $f^{n}$ has at least $d^{n}-1$ fixed points.

Theorem 1.4**.**

Let $f$ be a covering map of the annulus, with degree $d>1$, with an invariant essential continuum. Then $f^{n}$ has at least $d^{n}-1$ fixed points in $\mathrm{Fill}(K)$.

2. Notation and preliminaries

Throughout this paper, a surface $S$ will be a two dimensional orientable topological manifold. We will say that $S$ is respectively a plane, annulus or sphere if it is homeomorphic to $\mathbb{C},\mathbb{C}\backslash\{0\}$ or $\mathbb{S}^{2}$. A set $U\subset S$ will be a disc if it is homeomorphic to $\mathbb{D}=\{z\in\mathbb{C}:{z}<1\}$. To lighten notation, we will define $m_{k}:\mathbb{C}\rightarrow\mathbb{C}$ as the map such that $m_{k}(z)=z^{k}$.

All maps considered in this paper are continuous.

Definition 2.1**.**

Given two points $p,p^{\prime}\in S$, a path or curve from $p$ to $p^{\prime}$ will be a function $\gamma:[0,1]\rightarrow S$ such that $\gamma(0)=p$, $\gamma(1)=p^{\prime}$. We will say the path is simple when $\gamma$ is injective, and closed when $p=p^{\prime}$.

We will say $\gamma$ is respectively a segment, line, circle if it is homeomorphic to $[0,1],\mathbb{R},\mathbb{S}^{1}$.

Definition 2.2**.**

Given a surface $S$, an oriented topological foliation $\mathcal{F}$ is a partition of $S$ in one-dimensional manifolds such that for each $p\in S$, there exists a neighbourhood $U_{p}$ and a homeomorphism $h:U_{p}\rightarrow(-1,1)\times(-1,1)$ which preserves orientation and sends $\mathcal{F}$ into the foliation by vertical lines, oriented from bottom to top.

Definition 2.3**.**

Let $U$ be a subset of $\mathbb{C}$, $c\in U$, and take $f:U\rightarrow\mathbb{C}$. We say that $f$ is geometrically conjugate to $m_{k}$ if there exist foliations $\mathcal{F}$ in $U\backslash\{c\}$ and $\mathcal{F}^{\prime}$ in $f(U)\backslash\{f(c)\}$, and two homeomorphisms $\phi:U\rightarrow\mathbb{D}$, $\phi^{\prime}:f(U)\rightarrow\mathbb{D}$, such that both $\mathcal{F}$ and $\mathcal{F}^{\prime}$ are mapped into the radial foliation in $\mathbb{D}$, and such the next diagram commutes:

{U}$${f(U)}$${\mathbb{D}}$${\mathbb{D}}$$\scriptstyle{f}$$\scriptstyle{\phi}$$\scriptstyle{\phi^{\prime}}$$\scriptstyle{m_{2}}

Definition 2.4**.**

A branched covering map $f:S\rightarrow S$ (or simply branched covering), is a map that is a local homeomorphism at each point $p\in S$, except for finitely many critical points, each of them having a neighbourhood such that $f$ is geometrically conjugate to $m_{k}$, with $k\in\mathbb{Z}^{+}$.

In that context, the degree of a critical point will be $k$. We will define $\mathrm{Crit}(f)$ as the set of critical points of $f$. Each point which is not the image of a critical point has the same amount of preimages $d$, from which we deduce the degree of $f$ is $d$ if $f$ is orientation preserving, and $-d$ if $f$ is orientation reversing. If $\mathrm{Crit}(f)=\varnothing$, we will say $f$ is a covering map.

3. Goals and sketch of the proofs

The goal of this paper is to prove the following three results:

Theorem 3.1**.**

Let $f$ be a degree 2 branched covering map of the plane, and $c\in\mathrm{Crit}(f)$. Suppose there exists a compact set $K$ with $f^{-1}(K)=K$, and such that $c$ and $f(c)$ belong to the same connected component of $\mathbb{C}\backslash K$. Then $\mathrm{Fix}(f)\neq\varnothing$.

Theorem 3.2**.**

Let $f$ be a degree 2 branched covering map of the plane, with a compact invariant set $K$ (not necessarily totally invariant). Suppose there exists $U$ a connected neighbourhood of $K$, such that $\mathrm{Fill}((U)\cup f(U))$ does not contain neither $c$ nor $f(c)$. Then $\mathrm{Fix}(f)\neq\varnothing$.

Theorem 3.3**.**

Let $f$ be a degree 2 branched covering map of the plane, with an invariant continuum $K$, such that it does not separate $c$ from $f(c)$. Then $f$ has a fixed point in $\mathrm{Fill}(K)$.

In order to prove these, we will dedicate the next Section to build our main tool: a perturbation $h$ of the map $f$ with good properties, given below.

Proposition 3.4**.**

Let $f$ be a degree 2 branched covering map of the plane, and $c\in\mathrm{Crit}(f)$. Suppose there exists a compact set $K$ with $f^{-1}(K)=K$, and such that $c$ and $f(c)$ belong to the same connected component of $\mathbb{C}\backslash K$. Suppose in addition that $\mathrm{Fix}(f)=\varnothing$. Then, there exists an orientation preserving homeomorphism $h:\mathbb{C}\rightarrow\mathbb{C}$ with $h|_{K}=\mathrm{Id}$, and $\mathrm{Fix}(h\circ f)=\{c\}$.

Let us outline the roadmap for the proof of this Proposition.

Given that $c$ is a degree $2$ critical point, we know that there exists a neighbourhood of $c$, such that $f$ is locally geometrically conjugate to $m_{2}:\mathbb{C}\rightarrow\mathbb{C}$ -see section 2 for definitions-. In Lemma 4.1 we build a neighbourhood $U$ of $c$ with that property, and such that $\overline{U}\cap\overline{f(U)}\subset\partial U$. We use Lemma 4.2 to control the shape of $U$, letting us asume that $U=\mathbb{D}$, $c=0$.

In Lemma 4.3 we then construct $V$ the domain of the perturbation $h$ in , which will be a neighbourhood of a path from $c$ to $f(c)$, which is contained in $U\cap f(U)$ except for one point (remember that $U\cap f(U)=\varnothing$). We then change coordinates one more time, so that the path becomes a horizontal line -with $c$ on the left of $f(c)$, and the whole neighbourhood $V$ becomes a rectangle, in such a way that we can control the dynamics of $f|_{V}$.

The last part of the proof is to define the perturbation in the rectangle -defined as the identity map in its boundary-, which heuristically is the composition of a vertical contraction towards the path, composed with a translation to the left in each horizontal line.

4. The good perturbation

Lemma 4.1**.**

There exists a disc $U$ containing $c$, such that

. $U\cap K=\varnothing$.

. $f|_{U}$ is geometrically conjugate to $m_{2}$.

. $\overline{U}\cap\overline{f(U)}\subset\partial U$.

Proof. Given that $d(c,K)>0$, we may take a disc $U_{0}$ satisfying the first and second property. Let us take a path $\gamma:[0,1]\mapsto\mathbb{R}^{2}$ from $f(c)$ to $c$, such that $d(\gamma,K)>0$. Then there exists $V^{\varepsilon}$ a neighbourhood of $\gamma$, homeomorphic to $[0,1]\times(-\varepsilon,\varepsilon)$ by the homomorphism $h^{\varepsilon}$. We define $V^{\varepsilon}_{t}:=h^{\varepsilon^{-1}}([0,t]\times(-\varepsilon,\varepsilon))$, $\tilde{V}_{t}:=V^{\varepsilon}_{t}\cup f(U_{0})$. Let us take

[TABLE]

Then we may define $U:=f^{-1}(\tilde{V}_{t_{0}})$, and note that the desired intersection is nonempty and is included in the boundary, from which $U$ satifies the third property. ∎

Lemma 4.2**.**

Modulo change of coordinates, we may assume that $c=0$, $U=\mathbb{D}$, and that the radial foliation in $\mathbb{D}\backslash\{0\}$ is mapped 2-1 by $f$ into a foliation by lines in $f(U)\backslash f(0)$.

Proof. Given that $\partial U$ is a simple closed curve by construction, the Jordan-Schoenflies theorem states a homeomorphism $g:\mathbb{C}\mapsto\mathbb{C}$ such that $g(U)=\mathbb{D}$, $g(\overline{U}^{c})=\overline{\mathbb{D}}^{c}$ (we may assume $g(c)=0$). If we define $f^{\prime}:=g\circ f\circ g^{-1}$, we note it is dynamically conjugate to $f$, from which $f^{\prime}|_{\mathbb{D}}:\mathbb{D}\rightarrow f^{\prime}(\mathbb{D})$ is geometrically conjugate to $m_{2}$, and we obtain a foliation by lines $\mathcal{F}$ in $\mathbb{D}\backslash\{0\}$ as in the end of section 2. Finally, we may conjugate $\mathcal{F}$ to the radial foliation: given that each leaf $\phi\in\mathcal{F}$ gets out of $\mathbb{D}$, simply take the point of the leaf which is in the boundary of $\mathbb{D}$, and define the image of the leaf as the ray which goes through that point. ∎

We will now define the domain of the perturbation, that is, $V\subset\mathbb{R}^{2}$ such that $h|_{\mathbb{R}^{2}\backslash V}=\mathrm{Id}$. We start by taking $U_{0}\subset U$ a neighbourhood of $c$ satisfying the first two properties of Lemma 4.1

Lemma 4.3**.**

There exists a disc $V$ such that

. $V\cap K=\varnothing$,

. $V\cap U$ and $V\cap f(U)$ are discs,

. $U_{0}\cup f(U_{0})\subset V$,

. $f(V\cap U)\cap V=f(U_{0})$,

. $f(V\backslash(U\cup f(U)))\cap V=\varnothing$.

Proof. Let us take $z\in\partial U\cap\partial f(U)$ (given by the lemma 4.1). We then take $\hat{z}\in\partial U$ such that $f(\hat{z})=z$. Let $\gamma$ and $\hat{\gamma}$ be respectively the rays from $z$ to [math], and from [math] to $\hat{z}$, parametrized by arc’s length. We define $\gamma^{\prime}:=f(\hat{\gamma}).\gamma$.

For ease, let us call $V^{\varepsilon}$ to the neighbourhoods of $\gamma^{\prime}$ (as we did with $\gamma$ in Lemma 4.1), and define $\gamma^{\prime}_{1}:=\gamma^{\prime}|_{[0,\frac{1}{2}]}$; $\gamma^{\prime}_{2}:=\gamma^{\prime}|_{[\frac{1}{2},1]}$. Note that there exists $\varepsilon_{1}$ such that $V^{\varepsilon_{1}}\cap K=\varnothing$, because $d(\gamma^{\prime},K)>0$.

Given $\mathrm{Fix}(f)=\varnothing$, we know that $f(z)\neq z$. We then get that $z$ is the only point of $\gamma^{\prime}$ which is outside the discs $U$ and $f(U)$, so there exists $\varepsilon_{2}$ such that $(V^{\varepsilon_{2}}\backslash(U\cup f(U)))\cap V^{\varepsilon_{2}}=\varnothing$.

Moreover, we have $f(\gamma^{\prime}_{2})\cap\gamma^{\prime}_{1}=f(c)$, from which there also exists $\varepsilon_{3}$ such that $(f(V^{\varepsilon_{3}}\cap U)\cap V)\subset f(U_{0})$. Finally, taking $\varepsilon_{0}:=\mathrm{min}\{\varepsilon_{1},\varepsilon_{2},\varepsilon_{3}\}$, we conclude that $V=U_{0}\cup V^{\varepsilon_{0}}\cup f(U_{0})$ satisfies the desired properties. ∎

Remark 4.4**.**

$\mathrm{Fix}(h\circ f)\subset\mathrm{Fix}(f)\cup(V\cap f^{-1}(V))$.

Remark 4.5**.**

With a new change of coordinates, we may assume:

. $V$ is the rectangle $(0,10)\times(-1,1)$.

. $U_{0}=(0,2)\times(-1,1)$, $f(U_{0})=(8,10)\times(-1,1)$,

. $c=(1,0)$, $f(c)=(9,0)$.

. $\gamma^{\prime}$ is the segment between $c$ and $f(c)$.

. $f|_{U_{0}}(c+(\rho,\theta))=(f(c)+(f_{1}(\rho,\theta),2\theta))$.

. If $(x,y)\in U\cap V$, $0<x^{\prime}<x$, then $(x^{\prime},y)\in U\cap V$.

. If $(x,y)\in f(U)\cap V$, $x<x^{\prime}<10$, then $(x^{\prime},y)\in f(U)\cap V$.

Note that the last two properties in the Remark state that there is a well-defined notion of left and right inside the rectangle, namely, $U\cap V$ is on the left of $f(U)\cap V$.

We now proceed to build the perturbation $h$.

Proof of Proposition 3.4: We begin by taking $h_{1}$ a perturbation on the rays of $f(U_{0})$, such that $(h_{1}\circ f)|_{U_{0}}$ is, restricted to each ray, the affine map which sends $r_{c,\theta}\cap V$ into $r_{f(c),2\theta}\cap V$. Note that this function may be extended as the identity to the boundary of $f(U_{0})$, from which we can extend it to the whole plane, and get $h_{1}|{\mathbb{R}^{2}\backslash f(U_{0})}=\mathrm{Id}$. Observe that the expansion (or contraction) of $h_{1}\circ f$ in each ray is uniformly bounded, precisely:

[TABLE]

.

We will now define $h_{2}$, which is also the identity outside $f(U_{0})$, so we will define it on the square $(8,10)\times(-1,1)$. Heuristically, we want a strong contraction towards the horizontal $y=0$, but we need to adjust it so it becomes the identity in the boundary, so we will define a piecewise affine map. We impose simmetry with respect to the line $x=9$, and make it preserve verticals, that is, $h_{2}(x,y)=(x,h_{22}(x,y))$, where

[TABLE]

Note that we may extend it as the identity to the boundary of $f(U_{0})$.

Finally, we define $h_{3}$ supported in $V$, preserving horizontals and sending $f(c)$ into $c$ , in a similar fashion to how we defined the map $h_{2}$. Let $h_{3}(x,y)=(h_{31}(x,y),y)$, with

[TABLE]

Note that it may also be extended as the identity yo the boundary of $V$.

Let us define $h:=h_{3}\circ h_{2}\circ h_{1}$. It only remains to prove:

Lemma 4.6**.**

$\mathrm{Fix}(h\circ f)=\{c\}$.

Proof. Let $w\in\mathrm{Fix}(h\circ f)$. We start by proving $w\in f(U_{0})$.

Given that $h$ is supported in $V$, we have that $w\in(V\cap f^{-1}(V))$. By Lemma 4.3, $f(V\backslash(U\cup f(U)))\cap V=\varnothing$, from which $w\in U$ or $w\in f(U)$.

If $w\in f(U)$, then $f(w)\notin f(U)$, thus we deduce $h\circ f(w)=h_{3}\circ f(z)$. Moreover, by the last property of Remark 4.5, and recalling that $h_{3}$ preserves horizontals, we conclude that $w\notin\mathrm{Fix}(h_{3}\circ f)$. On the other hand, if $w\in U$, we use that $f(V\cap U)\cap V=f(U_{0})$ (Lemma 4.3) to conclude $w\in U_{0}$.

Let $w=(x,y)$. Note that necessarily $w\in h\circ f(U_{0})$. Since $h_{1}$ and $h_{2}$ are bijections of $f(U_{0})$, we obtain $h\circ f(U_{0})=h_{3}\circ f(U_{0})$, and therefore $w\in(U_{0}\cap(h_{3}\circ f)(U_{0}))$, so $\lvert{y}\rvert\leq\frac{\sqrt{2}}{8}$. We then get $w\in U_{1}:=[0,2]\times[\frac{-\sqrt{2}}{8},\frac{\sqrt{2}}{8}]$. Let us define

[TABLE]

. Observe that:

. $h_{1}\circ f(U_{1})\subset[8,10]\times[\frac{-1}{2},\frac{1}{2}]$.

. $f(U_{1})\cap W_{1}=\varnothing$.

. $h_{3}(W_{2})\cap U=\varnothing$.

Given that outside the vertical stripes $W_{1}$, $W_{2}$, the map $h_{2}$ divides the height by $4$ in $f(U_{1})$, we obtain $h\circ f(U_{1})\subset U_{2}:=[0,2]\times[\frac{-1}{8},\frac{1}{8}]$, so $w\in U_{2}$.

Proceeding inductively, we define $U_{n}:=[0,2]\times[\frac{-\sqrt{2}^{n}}{2^{n+2}},\frac{\sqrt{2}^{n}}{2^{n+2}}]$, and conclude $w\in\bigcap_{n\geq 0}U_{n}$, from which $y=0$. Furthermore, we have that $x\geq 1$ (since $h_{1}\circ f(\rho,\theta)=(\rho,2\theta)$), so the candidates $w$ are in the interval $I$ of ends $c$ and $(2,0)$. There is only left to observe that $f|_{I}(x,0)=(9x-8,0)$ has $c=(1,0)$ as its unique fixed point, which concludes the proof. ∎

Remark 4.7**.**

The only moment in which we use the fact that $\mathrm{Fix}(f)=\varnothing$ is in Lemma 4.3, when we ask $z$ not to be fixed. We could exchange it for the (weaker) hypothesis: “there exists a path from $f(c)$ to $c$ with no fixed point of $f$, contained in $\mathbb{C}\backslash K$”. The proof works exactly the same way, and in this context we can conclude $\mathrm{Fix}(h\circ f)\backslash\mathrm{Fix}(f)=\{c\}$. As we are searching for fixed points of $f$, both hypothesis are equally as good.

5. Proof of the main results

This Section is devoted to the proofs of the three theorems stated in Section 3.

Proof of Theorem 3.1. The idea is to take a perturbation $h$ as in the Proposition 3.4 (without modificating the dynamics in $K$ and such that the critical point becomes fixed, but we do not generate any other fixed point in this process). Given that $h\circ f$ has degree 2, we have that $(h\circ f)^{-1}(c)=c$, from which it follows that we can puncture the plane in the critical point, and restrict the dynamics to a covering map $g$ of the open annulus $\mathbb{C}\backslash\{c\}$.

The set $K$ is then totally invariant for this covering map $g$, so we are on the hypothesis of Theorem 1.3, which lets us infer that $\mathrm{Fix}(g)\neq\varnothing$. We then conclude that $\{c\}$ is a proper subset of $\mathrm{Fix}(h\circ f)$, which finishes the proof. ∎

In order to prove Theorem 3.2, we will use the following result, which can also be found in [LeC07].

Lemma 5.1**.**

Let $f$ be an orientation preserving homeomorphism of the plane, with a compact invariant set $K$. Let $U$ be a connected neighbourhood of $K$. Then $f$ has a fixed point in $V=\mathrm{Fill}(U\cup f(U))$.

Proof. We know that $f$ has at least one fixed point, by Theorem 1.1. Suppose by contradiction that none of them are in $V=\mathrm{Fill}(U\cup f(U))$. Take $S=\mathbb{C}\backslash\mathrm{Fix}(f)$. Given that $\mathrm{Fix}(f)$, is a (closed) invariant set, we may restrict $f$ to $S$. We may also assume that $S$ is connected (otherwise, $f$ preserves the connected component which contains $K$).

We then take $\tilde{S}$ the universal covering of $S$, which is a plane. Given that $V$ is simply connected, every connected component of $\tilde{\pi}^{-1}(V)$ is homeomorphic to $V$. For the same reason, for any lift $\tilde{f}:\tilde{S}\rightarrow\tilde{S}$ of $f$, and $\tilde{V}$ and any connected component of $\tilde{\pi}^{-1}(V)$, we have that $\tilde{f}(\tilde{V})$ intersects exactly one connected component of $\tilde{\pi}^{-1}(V)$.

Therefore, if $\tilde{K}$ is the lift of $K$ contained in $\tilde{V}$, we may take a lift $\tilde{f}_{0}$ of $f$ such that $\tilde{f}_{0}(\tilde{V})$ only intersects $\tilde{V}$, thus having $\tilde{f}_{0}(\tilde{K})=\tilde{K}$. Again by Theorem 1.1, we get that $\tilde{f}_{0}$ has a fixed point, which is a contradiction. ∎

Proof of Theorem 3.2. We build the same perturbation $h$ as in Proposition 3.4, except in this case we take into account not altering the dynamics in $U$. It is enough to take $U^{\prime}$ a connected nieghbourhood of $K$ such that it does not contain $c$ nor $f(c)$, and note that there exists $U$ as in the hypothesis, with $(U\cup f(U))\subset U^{\prime}$. We then build the perturbation as in Proposition 3.4, such that its domain does not intersect $U^{\prime}$.

We have that $h\circ f$ has no new fixed points other than $c$. Let us restrict the dynamics of $h\circ f$ to the annulus $S=\mathbb{C}\backslash\{c\}$, and take the universal covering $\tilde{S}$ (which is a plane). As in Lemma 5.1, any lift $\tilde{V}$ of $V$ is homeomorphic to $V$ (in particular it is bounded), and we can take a lift $\tilde{f}_{0}$ of $h\circ f$ such that for a lift $\tilde{K}$ of $K$, we get $\tilde{f}_{0}(\tilde{K})=\tilde{K}$.

Recalling that $\tilde{f}_{0}$ is an orientation preserving homemorphism of $\tilde{S}$, we get that it has a fixed point $\tilde{x}$ in $\tilde{V}$, from which we deduce that $x=\tilde{\pi}(\tilde{x})$ is fixed by $h\circ f$ and belongs to $V=\tilde{\pi}(\tilde{V})$, so it is fixed by $f$. ∎

Remark 5.2**.**

If $K$ is a non-separating continuum not containing $c$ nor $f(c)$, the hypothesis of Theorem 3.2 are immediately verified. Given $K$ is compact and $f$ is continuous, we may control the size of $U\cup\mathrm{Fill}(U)$ and conclude that the fixed point must be found in $K$, thus giving an elementary proof -for degree 2- of a result mentioned in Section 1 and given in [FMOT07], namely: a positively oriented branched covering map of the plane with an invariant non-separating continuum $K$, has a fixed point in $K$.

We now give the proof of the last of our three results.

Proof of Theorem 3.3. We divide the proof in two cases:

1. The point c belongs to the non-bounded component of $\mathbb{C}\backslash K$.

We build the perturbation $h$ in the same fashion as in Proposition 3.4. We puncture the plane and go to the annulus $S=\mathbb{C}\backslash c$. As in the last proof, we may take a lift $\tilde{f}_{0}$ of $h\circ f$ which leaves invariant $\tilde{K}$ a lift of $K$. As in the previous Proof, this implies the existence of a fixed point of $h\circ f$ different from $c$, and so, fixed by $f$. By the same argument used in Remark 5.2, we conclude that the fixed point is in $\mathrm{Fill}(K)$.

2. The point c belongs to a bounded component $C$ of $\mathbb{C}\backslash K$.

We build the same perturbation, again puncture the plane and take the annulus $A=\mathbb{C}\backslash c$. Observe that $K$ is essential in $A$, then by Theorem 1.4, we get that $h\circ f|_{A}$ has a fixed point in $\mathrm{Fill}(K)$, which is also fixed by $f$. Note that in this case we get a little stronger result: the fixed point is actually in $\mathrm{K}\backslash C$ (it may be in $K$ or in one of the bounded components of the complement, in the annulus A).

Remark 5.3**.**

As in Theorem 3.2, we may apply this result when $K$ is an invariant non-separating continuum. In Theorem 3.3 we actually do not need $K$ to be non-separating, but we do need to impose that $c$ and $f(c)$ belong to the same component of $\mathbb{C}\backslash K$.

6. Room for improvement

The result proved in this paper leads naturally to other questions, namely:

• •

Can we build the perturbation so that it does no alter the set of periodic points of $f$? In that case, we will have proved that under the hypothesis of Theorem 3.1, $f$ has the rate.

• •

The same happens if we take Theorem 1.3 and improve the result to find $f$ has the rate in $K$. Keep in mind that the perturbation we built does not modify the dynamics in that set.

• •

Can we adjust the techniques to the case where the degree of $f$ is bigger, and get similar results? (This is quite audacious a priori, given that the amount of critical points may grow, resulting in cases with more complicated dynamics).