A fixed point theorem for branched covering maps of the plane
Abstract.
It is known that every homeomorphism of the plane has a fixed point in a nonseparating, invariant subcontinuum. Easy examples show that a branched covering map of the plane can be periodic point free. In this paper we show that any branched covering map of the plane of degree with absolute value at most two, which has an invariant, nonseparating continuum , either has a fixed point in , or contains a minimal (by inclusion among invariant continua), fully invariant, nonseparating subcontinuum . In the latter case, has to be of degree and has exactly three fixed prime ends, one corresponding to an outchannel and the other two to inchannels.
Key words and phrases:
Fixed points; treelike continuum; branched covering map2000 Mathematics Subject Classification:
Primary 37F10; Secondary 37F50, 37B45, 37C25, 54F15Alexander Blokh] Lex Oversteegen]
1. Introduction
By we denote the plane and by the Riemann sphere. Homeomorphisms of the plane have been extensively studied. Cartwright and Littlewood [CL51] have shown that each orientation preserving homeomorphism of the plane, which has an invariant nonseparating subcontinuum , must have a fixed point in . This result was generalized to all homeomorphisms by Bell [Bel78]. The existence of fixed points for orientation preserving homeomorphisms under various conditions was considered in [Bro12, Bro84, Fat87, Fra92, Gui94], and of a point of period two for orientation reversing homeomorphisms in [Bon04].
In this paper we investigate fixed points of light open maps of the plane. By a Theorem of Stoilow [Why42], all such maps have finitely many critical points and are branched covering maps of the plane. In particular if denotes the set of critical points of , then for each , is finite and independent of . We will denote this number by . All such maps are either positively or negatively oriented (see definitions below); holomorphic maps are prototypes of positively oriented maps. If is positively oriented then the degree (of the map ), denoted by , equals and if is negatively oriented then . Easy examples, described in Section 2, show that positively and negatively oriented branched covering maps of the pane can be periodic point free.
The following is a known open problem in plane topology [Ste35]: “Does a continuous function taking a nonseparating plane continuum into itself always have a fixed point?”. Bell announced in 1984 (see also Akis [Aki99]) that the CartwrightLittlewood Theorem can be extended to holomorphic maps of the plane. This result was extended in [FMOT07] to all branched covering maps (even to all perfect compositions of open and monotone maps) which are positively oriented. Thus, if is positively oriented branched covering map of the plane and is a nonseparating continuum such that then contains a fixed point. The main remaining question concerning branched covering maps then is that for negatively oriented maps.
Given a continuum in the plane, we denote by , the topological hull of , the union of and all of the bounded components of . Also, denote by the unbounded component of . Then is a nonseparating plane continuum. In this paper we consider a branched covering map of the plane of degree and prove the following theorem.
Theorem 5.2.
Suppose that is a branched covering map of degree with absolute value at most and let be a continuum such that . Then one of the following holds.

The map has a fixed point in .

The continuum contains a fully invariant indecomposable continuum such that contains no subcontinuum with ; moreover, in this case .
It follows that in case (2) induces a covering map of the circle of prime ends of with and has exactly three fixed prime ends and for all of them their principle set is equal to . More precisely, let us consider in the uniformization plane the complement to the closed unit disk, and choose a Riemann map such that . Then one of the fixed prime ends, say, , corresponds to an outchannel (i.e., for sufficiently small crosscuts whose preimages in the uniformization plane separate from infinity, separates from infinity in ) and the other two prime ends correspond to inchannels (i.e., for sufficiently small crosscuts separating the corresponding points on the unit circle from infinity, separates from infinity in ).
Let us outline the main steps of the proof. By known results we may assume that ; we may also assume that has no fixed points in . Bell [Bel67] (see also [Sie68, Ili70]) has shown that then contains a subcontinuum with the following properties: (1) is minimal with respect to the property that , (2) is indecomposable, and (3) there exists an external ray to whose principal set is . Let be the critical point of and be the map such that and is the point with (if ). By [Bel78] we assume that . By way of contradiction we assume that is not fully invariant.
The first important step in the proof is made in Lemma 3.7 where we prove that is a first category subset of . Krasinkiewicz in [Kra74] introduced the notions of internal and external composants and described important properties of these objects. His tools are instrumental for the results of Section 3. In Section 4 we construct a modification of the map , which coincides with on and for which the external ray has an invariant tail, i.e. a part of from some point on to maps over itself, repelling points away from in the sense of the order on . In doing so we use a new sufficient condition allowing one to extend a function from the boundary of a domain over the domain. The proof of Theorem 5.2 is given in Section 5. There we study how the ray approaches and use the map on and the fact that is a first category set in in order to come up with a sequence of segments of which map one onto the other and converge to a proper subcontinuum of , a contradiction.
2. Main notions and examples
All maps considered in this paper are continuous. We begin by giving some definitions (avoiding the most standard ones). A map is monotone provided for each continuum or singleton , is a continuum or a point. A map is light provided for each point , is totally disconnected. A map is confluent if for each continuum and each component of , . It is well known [Why42] that all open maps between compacta are confluent. In the above situation components of are often called pullbacks of .
Every homeomorphism of the plane is either orientationpreserving or orientationreversing. In this section we will recall an appropriate extension of this result, which applies to open and perfect maps (see [FMOT07]).
Definition 2.1 (Degree of a map).
Let be a map from a simply connected domain into the plane. Let be a simple closed curve in , and a point. Define by
Then has a welldefined degree, denoted . Note that is the winding number of about .
Definition 2.2.
A map is from a simply connected domain is strictly positivelyoriented (strictly negativelyoriented) if for each we have (, respectively).
Definition 2.3.
A map is said to be perfect if preimages of compacta are compacta. A perfect map is oriented provided for each simple closed curve we have .
Remark 2.4.
Every strictly positively or strictly negativelyoriented map is oriented because if a point is such that for a simple closed curve , then . Also, if is a continuum then as follows from the definition of an oriented map and continuity arguments.
The following theorem was established in [FMOT07]:
Theorem 2.5.
Suppose that is a perfect map. Then the following are equivalent:

is either strictly positively or strictly negatively oriented.

is oriented.

is confluent.
Let us prove a useful lemma related to Theorem 2.5. A branched covering map of the plane is a map such that at all points, except for finitely many critical points, the map is a local homeomorphism, at each critical point the map acts as at for the appropriate , and each point which is not the image of a critical point has the same number of preimages (then equals if is positively oriented and if is negatively oriented). By a Theorem of Stoilow [Why42] an open light map of the plane is a branched covering map.
Lemma 2.6.
Suppose that is a perfect map such that for every continuum and every component of the image is not a point. Then is confluent. If in addition is light, then it is open (and hence in this case is a branched covering map).
Proof.
Let be light and show that then it is open. Suppose that is an open Jordan disk, , and . Choose a small semiopen arc in with an endpoint of at , and then choose a component of containing . By the assumptions of the lemma, is not degenerate. Choose a small disk so that . Then the component of containing is not degenerate. Now, the fact that is light implies that there are points of mapped into , a contradiction. By a Theorem of Stoilow [Why42] then is a branched covering map of the plane.
Consider the general case. We can use the socalled monotonelight decomposition. Indeed, consider the map which collapses all components of sets to points. Then it follows that where is a light map. By the above this implies that is a composition of a monotone map and an open light map of the plane. Clearly, this implies that is confluent. ∎
A translation by a vector (and a translation by a vector followed by a reflection with respect to an axis nonorthogonal to ), are obvious examples of plane homeomorphisms which are periodic point free. Clearly, any polynomial of degree strictly bigger than one, acting on the complex plane, has points of all periods. The following examples show that this is not true for all positively oriented branched covering maps of the plane.
Example 2.7.
There exists a degree two positively oriented branched covering map of the plane which is periodic point free.
We will use both polar and rectangular coordinates. Set . We will look for a map in the form with and is a continuous positive function such t hat . Before we define , let us describe the set of all points such that and have the same coordinates. Then . Hence, . So, the set consists of the axis and two radial straight lines coming out of at angles . Given , consider the point of intersection between the horizontal line of points whose coordinate is and the set .
The point is the only point on with image also on . Then the distance between the point and the point (and the origin) is . Set . Then for any point of because all points of map off by , and hence, by the construction, by . On the other hand, translates two units to the right. Hence does not contain fixed points. Moreover, since , also has no fixed points on the axis.
To see that has no periodic points^{1}^{1}1We are indebted to M. Misiurewicz for suggesting this argument., let be the set of points in whose argument is in . Then coincides with the shift of the entire set to the right by two units ( because ). Let , and let denote the imaginary part of . If , then and if , then . Let us show that a point cannot stay in . Indeed, otherwise has to converge to points of . However if were one of these points, then by continuity we would have which would imply that and hence that contradicting the assumption that stays in .
Hence, for every , there exists such that . Since , the trajectory stays in forever. If there exists such that belongs to the real line, then it converges to . To study the orbit of a point which does not belong to the real line, observe that there exists an increasing function such that if then . Therefore if does not belong to the real line then as . Hence in fact for any point and has no periodic points.
The example above can be easily modified to obtain a periodic point free branched covering map of degree .
3. Basic preliminaries
A continuum is called indecomposable if cannot be written as the union of two proper subcontinua. Also, is unshielded if . We argue by way of contradiction, therefore the following is our main assumption.
Main Assumption. The map is a branched covering map and is a continuum such that and is fixed point free.
Bell [Bel67] has shown that in this case contains a subcontinuum which is minimal with respect to the property that (then, clearly, ) and which must have the following properties (see [Sie68, Ili70] for alternative proofs):

is minimal among continua such that ;

and is fixed point free;

there exists a curve (a conformal external ray, see below) in such that (so that is unshielded and has empty interior);

is indecomposable.
We will use exclusively for a continuum with the just listed properties (A0)  (A3) which are ingredients of the standing assumption on . Our main aim is to show that then is fully invariant (i.e., ). Thus, by way of contradiction we can add the following to our standing assumption (as we progress, the standing assumption will be augmented by other ingredients as well).

The set is not fully invariant.
Below we list wellknown facts from Carathéodory theory. Good sources are the books [Mil00] and [Pom92]. Let be the open unit disk in the complex plane and . Let be a conformal map such that . An external ray is the image of the radial line segment . Clearly, an external ray is diffeomorphic to the positive real axis. If an external ray compactifies on a point, say, (i.e., then is said to land on . For convenience we extend the map onto all angles whose rays land: if the ray lands at a point , we set . Observe, that the extended map is not necessarily continuous at angles whose rays land. Still, this extension is convenient and will be used in what follows.
A crosscut (of or of ) is an open arc in whose closure is a closed arc with its endpoints in . If is a crosscut, then by the shadow of , denoted , we mean the bounded component of . Sometimes the crosscut which gives rise to a shadow is said to be the gate of the shadow. In the uniformization plane we consider as a marked continuum which allows us to talk about crosscuts of too. Moreover, given a crosscut in we can then talk about its shadow etc. It is known that if is a crosscut of , then is a crosscut of , and .
We say that a crosscut is an essential crosscut if is a single point, called the central point, and the intersection of and is transverse. A sequence of crosscuts of is a fundamental chain provided , for each , and . Two fundamental chains and are said to be equivalent if contains all but finitely many crosscuts , and contains all but finitely many crosscuts . A prime end of is an equivalence class of fundamental chains; a fundamental chain is said to belong to its prime end.
Given a fundamental chain , the set is a point ; the corresponding prime end then may be identified with the angle . Given a prime end and a corresponding fundamental chain , denote by , called the impression of , the set ; it is known that does not depend on the choice of a fundamental chain and therefore is welldefined. Also, consider the set , called the principal set of (or just of ). It is known that and that for each point there exists a fundamental chain of the prime end such that .
The last claim can be improved a little. It was shown in [BO06] that given an angle , there exists for each an essential crosscut such that as . We call such a family a defining family of crosscuts of the prime end . For convenience we order each so that if and only if the subarc of from to is contained in the subarc of from to (thus, as the points move along from infinity towards , they decrease in the sense of the order on ). Denote by the set of points in enclosed between the points . Also, set . Similarly we define semiopen and closed subsegments of when possible (e.g., if lands, it makes sense to talk of however otherwise the set is not defined). Also, similarly we define relations and . By a tail of we mean the set of all points such that (or ) for some .
It is well known that the geometry of the ray in (A2) and the continuum is quite complicated. The ray approaches so that on either side of the distance to goes to while it simultaneously accumulates upon the entire . It then follow from properties of conformal maps that round balls, disjoint from but nondisjoint from , with points of the intersection with approaching must go to in diameter. One can say that “digs a dense channel” in the plane eventually accumulating on by (A2).
As was explained in the Introduction, the main remaining question concerning fixed point problem for branched covering maps is that dealing with negatively oriented maps. Since we are interested in maps such that and thanks to a theorem of Bell [Bel78] we may make the following assumption.

From now on we assume that is of degree .
Then has a unique critical point, denoted by , and a unique critical value, denoted by . Let be the involution defined by and if , where . Clearly, (i.e., the map is an idempotent homeomorphism of the plane); sometimes we call the sibling of . Let us establish basic properties of in the following lemma (some of the properties hold in more general situations, however we do not need such generality in this paper).
Lemma 3.1.
The following facts hold.

If is a continuum then .

Suppose that is a nonseparating continuum. If then there are exactly two pullbacks of which are disjoint and map onto homeomorphically (on their neighborhoods). If then is the unique pullback of which must contain .

If is a continuum and is a pullback of then is a pullback of (and hence ). In particular, a pullback of a nonseparating continuum is nonseparating.

Suppose that is a simply connected domain such that is also simply connected. If is not a homeomorphism then must contain a critical point.
Proof.
(1) Suppose otherwise. Then there is a point such that . By Remark 2.4 . Since is open, we can then find a point such that is outside contradicting Remark 2.4.
(2) If we can take a curve from to infinity disjoint from . Then is a curve which cuts into two open halfplanes and is disjoint from . Also, each halfplane maps onto homeomorphically. Thus, in this case consists of two components each of which maps onto homeomorphically (on sufficiently small neighborhoods of the pullbacks). Suppose that . Then cannot have more than one component because is confluent (hence each pullback of maps onto ) and has a unique preimage .
(3) Let us apply (1) to . If then must be a homeomorphic pullback of which implies the desired. Let and set where by (1) is the unique pullback of . Let us show that then is the unique pullback of . Let be the boundary of , be the boundary of . It follows that . On the other hand, by (1) no point of can map to a point of , and by the construction no point from can map to a point of . Hence .
This implies that is connected. Indeed, suppose otherwise. Then there are two pullbacks of each of which maps onto . This implies that which contradicts the fact that is a continuum. Thus, . Moreover, and hence . On the other hand, by Remark 2.4, and so . Hence . If is nonseparating, then . By the above is a pullback of containing , that is . Thus, is nonseparating as desired.
(4) It immediately follows from (2) that and . However we need to show that . Take such that and connect them with an arc . Then . Since is simply connected by the assumption of the lemma, . Since , it follows that is the unique pullback of (because is confluent and both preimages of belong to ). By (2) and (3), is the unique pullback of , and . Since , the unique preimage of belongs to as desired. ∎
Suppose that . Then by Lemma 3.1 it follows that there exists a neighborhood of on which is a homeomorphism. Again, by the Bell’s results [Bel78] this implies the existence of an fixed point . (Alternatively, the proof given below in Section 5 can easily be adapted to cover this case.) Therefore we extend our standing assumption as follows.

From now on we assume that .
Since is oriented and , . By Lemma 3.1 the continuum is nonseparating and maps onto in a twotoone fashion (except for the point ). Let us list simple consequences of our standing assumption as applies to in this case.
Lemma 3.2.
The set is not fully invariant, , and .
Proof.
Let us show that is not fully invariant. By Lemma 3.1 no point from the interior of can map to (recall that is unshielded by (A2) and hence no point of belongs to . Hence if is fully invariant then so is contradicting (A4). This implies that there are points of outside , and hence cannot be contained in .
Finally, suppose that . Since there are points of outside , this implies that is disjoint from . Hence is a homeomorphism and so . However then by (A6) we have , a contradiction with and being disjoint. We conclude that as desired. ∎
For convenience let us make the conclusions of Lemma 3.2 a part of our standing assumption.

is not fully invariant, , and .
A composant of in a continuum is the union of all proper subcontinua of which contain . If is indecomposable then any two composants of are either equal or disjoint; clearly, if for some continuous map , then the image of a composant, being a connected set, either coincides with , or is contained in a composant of . It follows from the definition that if is a composant of then for each , there exists a subcontinuum such that . It is well known [Kra74] that, if is indecomposable, then each composant in a dense first category subset of . By the Baire Category Theorem there are uncountably many distinct composants in an indecomposable continuum.
Again, assume that is indecomposable. A composant of is internal if every continuum which meets and , intersects all composants of . Equivalently, a composant of is internal if and only if every continuum which meets and , intersects all composants of (indeed, if meets and then must coincide with ). A composant which is not internal is called external. We denote the union of all external composants by . Clearly, an internal composant does not contain accessible points: if is an accessible (from ) point, then we can choose an arc in the appropriate component of with an endpoint at which intersects only at , a contradiction with being internal. Thus, an external composant is a generalization of a composant containing an accessible (from ) point. The following results are due to Krasinkiewicz [Kra74].
Lemma 3.3 (Krasinkiewicz).
Let be an indecomposable continuum in the plane. Then the following claims hold.

The set is a first category subset of ; hence, the union of all internal composants is a dense set in .

Let be an internal composant of . If is any continuum which meets , the complement of and does not contain , then there exists a neighborhood of and a continuum which separates between two distinct points of .
A plane continuum is called treelike if it is onedimensional (has no interior) and nonseparating. Recall that is an invariant continuum which is minimal among continua with respect to the property that . The set has a number of properties listed in the beginning of this section, in particular it is indecomposable (and hence onedimensional) and unshielded. We now study other properties of . First we need a few technical lemmas.
Lemma 3.4.
Suppose that is a dense subset of . Then is not a first category subset of .
Proof.
First let us show that there exists a point and a small neighborhood of such that is a homeomorphism onto its image and . Indeed, it is obvious if is fully invariant. Otherwise choose a point so that (i.e., has a unique preimage in , namely , and is not critical). If now is a sufficiently small neighborhood of , then ; hence consists of points of which cannot have preimages in , but must have some preimages because . The only preimages points of are those in which implies that as desired.
Now, by the conditions of the lemma is a subset of , hence by the previous paragraph is a subset of . Therefore cannot be a first category subset of . ∎
The next lemma uses Lemma 3.4.
Lemma 3.5.
Suppose that is a union of some composants of which is a first category subset of . Then there exists a composant of such that .
Proof.
By way of contradiction suppose that for every composant the image of intersects . Take a composant whose image contains points not from . By the assumption also has points mapped into . Hence contains points of a least two distinct composants which implies that . If there is another composant such that is not contained in , then it again follows that . Since is twotoone, this implies that which is impossible because there are countably many pairwise disjoint composants of . Hence for any composant we have . However, the union of all composants except for is a subset of while the set is a first category subset of . By Lemma 3.4 this is impossible. ∎
The next lemma studies the images of composants.
Lemma 3.6.
Suppose that is an internal composant of . Then is an internal composant of .
Proof.
By way of contradiction suppose that is not an internal composant of . There are two possibilities this can fail: (1) is an external composant of or , or (2) is contained in a composant with which it does not coincide. We consider these possibilities separately.
(1) Suppose that is either an external composant or the entire . Then there exists a continuum such that and is contained in the union of some, but not all, composants of . Then all composants which intersect are external, and hence their union is a first category subset of .
Choose a point and then a point such that . Then choose the pullback of which contains . Since is an internal composant and meets , by definition intersects all composants of . Hence all composants of have points mapped into the union of some composants which is a first category subset of . By Lemma 3.5, this is impossible.
(2) Suppose that where is a composant of . Choose a subcontinuum which contains a point and a point of . Choose the pullback of which contains a point such that . Then because otherwise it would be contained in and its image would not contain . Hence meets points of and contains the point which implies that intersects all composants of (because is an internal composant). Since , then this implies that all composants of have points mapped into , and , being a composant of , is a first category subset of . Again, by Lemma 3.5 this is impossible. ∎
In what follows we will use the following lemma which studies the set . In the proof we rely upon the above developed tools.
Lemma 3.7.
The set is contained in the union of all external composants of . In particular, is a proper closed subset of with empty interior in .
Proof.
By (A7) and . Choose , then and, hence, is a proper, closed subset of . The fact that has empty interior in is much less trivial.
Let us first show that at most countably many composants of contain a subcontinuum which separates . Indeed, if is a separating continuum, we can associate to a bounded component of . Since is unshielded, the sets for distinct composants of are disjoint. Hence at most countably many composants of contain a subcontinuum which separates . Also, there is exactly one composant which contains the critical value . By Lemma 3.3(1) and since each composant is a first category subset of (see Lemma 2.1 of [Kra74]), the union of the above listed countably many composants and all external composants of is still a first category subset of . Its complement is the union of points of all composants from the collection of all internal composants of for which every subcontinuum is treelike not containing the critical value . Thus, is still a dense subset of .
By Theorem 4.2 of [CMT], is a first category subset of . Hence we can choose a point . Let be the internal composant of which contains . Choose , then is contained in an internal composant of . By Theorem 5.5 of [Rog98], such that and for each and each , there exists a subcontinuum such that . Moreover, the map is a bijection for each . Assume that . Then by Lemma 3.6 and . Hence and contain the disjoint, internal composants and , respectively.
Let us show that if is any internal composant of , then . Indeed, suppose otherwise. Then by symmetry the internal composant of intersects . Choose a point and a point ; then choose continuum which contains both and . Since is an internal composant of , this implies that intersects all composants of . On the other hand, is an internal composant of by Lemma 3.6. Thus, images of all composants of are nondisjoint from the composant contradicting Lemma 3.5.
In order to finish the proof it suffices to show that . Suppose that this is not the case. Then meets an internal composant of . It is easy to verify that Lemma 3.3 applies to this situation with playing the role of , playing the role of , and playing the role of . Thus, it follows from Lemma 3.3(2) that there exists a neighborhood of and a continuum such that separates two points of in . Since by the above and are disjoint, is contained in one component of , contradicting that is dense in . This completes the proof of the lemma. ∎
4. Creating an invariant ray
In what follows essential and defining crosscuts are called simply essential and defining. To begin with, we need a lemma which will allow us to simplify the applications of results of [FMOT07, OT08, KP94] in this section. For simplicity when talking of angles we often mean the points of with arguments equal to these angles; here is considered as the boundary of the unit disk in . Take the (Euclidean) convex hull of . Then the ray eventually enters through a crosscut so that the tail of stays inside the shadow . Observe that is a straight segment. This defines the arc of angles whose rays have tails in , and it follows that . Consider now a geodesic of connecting and and its counterpart which is a crosscut of . Clearly, eventually enters (and stays in) the shadow of .
Lemma 4.1.
There exists an essential crosscut with the following properties.

We have that and, hence, .

Let be the set of points in accessible from . Then is onetoone and . Moreover, there exist such that any defining crosscut is less than in diameter, and any essential crosscut less than in diameter maps to a crosscut which is at least distant from .
Proof.
(1) Consider a defining family of crosscuts of . Then there exists a sequence of defining crosscuts such that , points converge to , and all ’s are disjoint from (otherwise for some and all we would have implying that , a contradiction with Lemma 3.7). We may assume that .
Denote by the open component of which contains points of located between and . By way of contradiction (and refining if necessary the sequence ) we may assume that every contains points of . Choose a point and a point . Connect these points with an arc in which intersects at just one point . Then choose points so that the subarc of with the endpoints is disjoint from and contains .
It follows that is a crosscut of . Denote the endpoints of by . Also, denote by the shadow of in the sense of . Then it follows that one of the points belongs to and the other one does not. Now, consider an internal composant of . It has points close to both points and , hence it has points both inside and outside . However, by Lemma 3.7 is disjoint from any internal composant of , a contradiction. Hence indeed there exists an essential crosscut such that which implies that .
(2) We may assume that and all defining crosscuts are sufficiently small. By continuity and since is fixed point free, images of all these crosscuts are disjoint from the crosscuts themselves (each crosscut moves off itself by a bounded away from distance). Moreover, by (A2) and because of the properties of crosscuts, ’s approach all points of . Choose a crosscut among them so that is sufficiently far from and hence is sufficiently far from so that . By Lemma 3.1 then is a homeomorphic pullback of and so is a small crosscut too. Also, we can choose so that .
We claim that and is a homeomorphism. Indeed, since is open. Hence we see that . Points of cannot be mapped to outside because otherwise there will be points of not in . Considering points close to shows that some points of are in . Now the fact that implies that . Finally, if then there will have to be points of in , a contradiction to . Thus, . Hence by Lemma 3.1 is a homeomorphism.
This easily implies that is onetoone too. Indeed, suppose that for . Clearly, . Choose an arc joining to in . Then is a simple closed curve so that . Choose a small arc (in the circular order on ) so that