Copyright © 2009 Seung Won Kim et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

Let X be a finite polyhedron that has the homotopy type of the wedge of the projective plane and the circle. With the aid of techniques from combinatorial group theory, we obtain formulas for the Nielsen numbers of the selfmaps of X.

1. Introduction

Although compact surfaces were the setting of Nielsen's fixed point theory in 1927 [1], until relatively recently the calculation of the Nielsen number was restricted to maps of very few surfaces. For surfaces with boundary, such calculations were possible on the annulus and Möbius band because they have the homotopy type of the circle. In 1987 [2], Kelly used the commutativity property of the Nielsen number to make calculations for a family of maps of the disc with two holes. We will discuss Kelly's technique in more detail below. The first general algorithm for calculating Nielsen numbers of maps of surfaces with boundary was published by Wagner in 1999 [3]. It applies to many maps and recent research has significantly extended the class of such maps whose Nielsen number can be calculated (see [4–7] and, especially, the survey article [8]). This approach makes use of the fact that a surface with boundary has the homotopy type of a wedge of circles. For the calculation of the Nielsen number, Wagner and her successors employ techniques of combinatorial group theory.

The key properties of surfaces with boundary that are exploited in the Wagner-type calculations are that they have the homotopy type of a wedge and that they are aspherical spaces so their selfmaps are classified up to homotopy by the induced homomorphisms of the fundamental group. The paper [9] studies the fixed point theory of maps of other aspherical spaces that have the homotopy type of a wedge, for instance the wedge of a torus and a circle. The purpose of this paper is to demonstrate that combinatorial group theory furnishes powerful tools for the calculation of Nielsen numbers, even for maps of a nonaspherical space. We investigate a setting that is not aspherical and hence fundamental group information is not sufficient to classify selfmaps up to homotopy. We obtain explicit, easily calculated formulas for the Nielsen numbers of these maps.

Denote the projective plane by and the circle by . This paper is concerned with maps of finite polyhedra that have the homotopy type of the wedge . If the polyhedron has no local cut points but is not a surface, then the Nielsen number of a map is the minimum number of fixed points among all the maps homotopic to it [10]. However, since a map of such a polyhedron has the homotopy type of a map of and the Nielsen number is a homotopy type invariant, we will assume that we are concerned only with maps of itself. We identify and with their images in and denote their intersection by . We need to consider only selfmaps of and their homotopies that preserve . The fundamental group of at is the free product of a group of order two, whose generator we denote by , and, choosing an orientation for , the infinite cyclic group generated by . To simplify notation, throughout the paper we denote the fundamental group homomorphism induced by a map by the same letter as the map because it will be clear from the context whether it represents the map or the homomorphism. Since all maps from to are homotopic to the constant map, we may assume that , the restriction of to , maps to itself.

The paper is organized as follows. We will describe in the next section a standard form for the map in which the fixed point set is minimal on and on the fixed point set consists of together with a fixed point for each appearance of or in the fundamental group element . In Section 3 we calculate the Nielsen numbers of the maps for which by proving that, in that case, equals the Nielsen number of a certain selfmap of obtained from and therefore is determined by the degree of that map. In Section 4 we obtain formulas for the Nielsen numbers of almost all maps for which . The formulas depend on integers obtained from the word in the fundamental group of . However, the nonaspherical nature of , which makes fundamental group information insufficient to determine the homotopy class of a map, requires us to find two different formulas for each word . One formula calculates in the case that is homotopic to the identity map whereas the other applies when belongs to one of the infinite number of homotopy classes that do not contain the identity map. Section 5 then considers the two exceptional cases that are not calculated in Section 4. We demonstrate there that even if the induced fundamental group homomorphisms in these cases vary only slightly from those of Section 4, their Nielsen numbers can differ by an arbitrarily large amount. Section 6 presents the proof of a technical lemma from Section 4.

This paper is the fruit of a collaboration made possible by the Research Experiences for Undergraduates program funded by the U. S. National Science Foundation through its VIGRE grant to UCLA.

2. The Standard Form of

Given a map where , we write (2.1)where and for all .

Let denote the restriction of to . By the simplicial approximation theorem, we may homotope to a map with the property that the inverse image of is a finite union of points and arcs. A further homotopy reduces the inverse image of to a finite set and we view as the union of arcs whose endpoints are mapped to . We then homotope the map restricted to each arc, relative to the endpoints, so that it is a loop in that is an embedding except at the endpoints and it represents either or . If the restriction of the map to adjacent arcs corresponds to any of or , we can homotope the map to a map constant at on both intervals and then shrink the intervals. We will continue to denote the map by . Starting with and moving along the circle clockwise until we come to a point of which we call , we denote the arc in from to by . Continuing in this manner, we obtain arcs where the endpoints of are and . As a final step, we homotope the map so that it is constant at on arcs and that form a neighborhood of in . Thus we have constructed a map, still written , that is constant on and and, otherwise, its restriction to an arc is a loop representing or according to the form of above, in the order of the orientation of .

Given a map , we may deform by a homotopy so that , its restriction to , maps to itself. We will make use of the constructions of Jiang in [11] to deform so that has a minimal fixed point set. If , then belongs to one of two possible homotopy classes and, in both cases, Jiang constructs homotopies of to a map with a single fixed point, which we may take to be . Let denote a lift of to the universal covering space, then the degree of is determined up to sign and we denote its absolute value by . If , the homotopy class of is determined by , which must be an odd natural number. If is a deformation, that is, it is homotopic to the identity map, then and Jiang constructs a map homotopic to with a single fixed point, which we again take to be . For the remaining cases, where , the Nielsen number and Jiang constructs maps homotopic to with two fixed points. We take one of those fixed points to be and denote the other fixed point by .

We also homotope so that , its restriction to , is in the form described above. The map thus obtained we call the standard form of and denote it also by . We note that, for each in there is exactly one fixed point of in , of index , and for each in there is one fixed point, of index . The fixed points and are of index , see [11]. For the rest of the paper, all maps will be assumed to be in standard form.

Our tools for calculating the Nielsen numbers come from Wagner's paper [3] which we will describe in the specific setting of selfmaps of . Let be a fixed point of in which is distinct from , then lies in an arc corresponding to an element or in ; we write or . We identify this element by writing or . The Wagner tails of the fixed point are defined by and if and by and if .

We will use the following results of Wagner.Lemma 2.1 (see [3, Lemma 1.3]). For any fixed point of on , (2.2)Lemma 2.2 (see [3, Lemma 1.5]). If and are fixed points of on , then and are in the same fixed point class if and only if there exists such that (2.3)

Wagner's Lemma 1.5 concerns the case where is a wedge of circles. However, the same proof establishes the statement of Lemma 2.2 for . When (2.3) holds, we will say that and are -Nielsen equivalent by or, when the context is clear, more briefly that and are equivalent.

3. The Case

If is an aspherical polyhedron and a map induces a homomorphism of the fundamental group that is trivial on the factor of , then is homotopic to the map where is the retraction sending to . Therefore, by the commutativity property of the Nielsen number, Since , its Nielsen number is easily calculated. This is the technique that Kelly used, with , in [2] to construct his examples. If is not aspherical, then a map that induces a homomorphism that is trivial on the factor need not be homotopic to . However, when , we will prove that it is still true that .

We note that since, in the case, all fixed points of lie in , then the fixed point sets of and of consist of the same points. Moreover, the fixed point index of each fixed point is the same whether we view it as a fixed point of or of . We will demonstrate that the fixed point classes and of are also the same, and thus the Nielsen numbers are equal.

Since is a circle with fundamental group generated by , the condition corresponding to Wagner's for and to be in the same fixed point class of in [3, Lemma 1.5] is that there exist an integer such that(3.1)That is, there exists such that(3.2)Although Wagner's paper [3] assumes reduced form for map and may not be in reduced form, in fact that condition is not used in the proof of [3, Lemma 1.5] so the existence of satisfying (3.2) is still equivalent to the statement that and are in the same fixed point class of . Corresponding to the previous terminology, in this case we will say that and are -Nielsen equivalent by .

We have(3.3)where and for all . Let be the sum of the from to . Similarly, for an element , we write(3.4)where, as before, and for all . Let be the sum of all the from to . The retraction induces such that and and thus and . For fixed points , define , then for some integer .Lemma 3.1. If , then the following are equivalent: (1) and are -Nielsen equivalent by ,(2) and are -Nielsen equivalent by ,(3).

Proof. (1)(2) If and are -Nielsen equivalent by , there exists such that(3.5)so(3.6)Every element of finite order in the fundamental group of is a conjugate of an element of finite order in or in . Therefore, implies that so we have and thus(3.7)As we noted above, (3.7) implies that and are -Nielsen equivalent by .

(2)(3) If and are -Nielsen equivalent by , then we have (3.7). Since , we see that(3.8)and conclude that .

(3)(1) Suppose that . Since , then . If , it must be that . So, if we let , then and thus(3.9)that is, and are -Nielsen equivalent by this . If , we define and, again using the hypothesis , we can write for some integer . That hypothesis also implies that(3.10)Now writing , we see that(3.11)If we let then, since , we have(3.12)Therefore,(3.13)which again means that and are -Nielsen equivalent by .

Since Lemma 3.1 has demonstrated that the fixed point classes of and of are identical and the Nielsen number of a map of the circle is determined by its degree, we haveTheorem 3.2. Let be induced by retraction. If is a map such that and , then (3.14)

4. The Case

Let be a map, where , such that . We will use Lemma 2.2 to calculate the Nielsen number of most such maps. We write(4.1)where and for all . Suppose that . Then there is a map that induces the homomorphism , that is, and . (2.3) of Lemma 2.2 is satisfied for if and only if it is satisfied for . Thus, we can assume that in and we write(4.2)where and either or is cyclically reduced, which means that is a reduced word. Then, for some integers and ,(4.3)where may be zero. If , then either or . Let when .

Now suppose that fixed points and are equivalent by(4.4)where and for all . Let denote the sum of the from to and let(4.5)be the right-hand side of the (2.3) of Lemma 2.2.

Denote the length of a word in by , where the unit element is of length zero.Lemma 4.1. Suppose and are equivalent fixed points of . If and , then or .Proof. Suppose that and . Then(4.6)

Case 1. and .

Since so that starts and ends with or , it follows that one of those elements ends and one of them starts . Since , we see that is reduced ( may be ) and therefore(4.7)This is a contradiction and thus there is no solution in this case.Case 2. and . ( and is similar.)

If there is no cancellation between and , then we can see that the solution does not exist as in Case 1. Suppose there is a cancellation between and . Suppose and write where is the part of that is cancelled by , then . By Lemma 2.1,(4.8)so , for some word , which contradicts the assumption that is cyclically reduced. Thus so we may write and we have(4.9)and thus(4.10)We have shown that cannot be negative and, if then begins with which cannot be reduced since implies that ends with either or . So suppose and cancels part of . Then must end with to cancel and, since is either or , further cancellation would cancel parts of . But is cyclically reduced and therefore we conclude that there is no further cancellation. Thus, as in Case 1, there are no solutions to this equation.Case 3. and .

If , then an argument similar to that of Case 2 applies. Thus we may assume that , which implies that or . Suppose that , then(4.11)and so . Similarly, if , then .Lemma 4.2. Suppose and are equivalent fixed points of . If and , then or .

The proof of Lemma 4.2 is similar to that of Lemma 4.1, but it requires the analysis of a greater number of cases, so we postpone it to Section 6.

Suppose are fixed points of with and , then implies because is in standard form; the same is true in the case and . In these cases, also implies . On the other hand, if and or and , then and . Thus, in our setting, the only ways that two distinct fixed points and of can be directly related in the sense of [3, page 47] are if or if . The point of Lemmas 4.1 and 4.2 is that, if two fixed points in are equivalent, then they must be directly related rather than related by intermediate fixed points. It is this property that permits the calculations of Nielsen numbers that occupy the rest of this section.

We continue to assume that is in standard form and . If is a deformation, then is the only fixed point of on . Otherwise, there is another fixed point of on denoted by and both and are of index 1, see [11]. We again write or depending on whether maps the arc containing to or to . The fixed points of on are , ordered so that lies in the arc corresponding to the first appearance of or in . Moreover, for a subword of , we write if lies in an arc corresponding to an element of . Let denote the number of fixed points such that .Lemma 4.3. Suppose is not a deformation and, if , suppose also that . If and , then and are equivalent. Otherwise, is not equivalent to any other fixed point of .Proof. Let be a fixed point of and let and denote the arcs of going from to in the clockwise and counterclockwise directions, respectively. Then and , where and are the Wagner tails of . The fixed points and are equivalent if and only if there is a path in from to such that the loops and represent the identity element of . Using a homotopy, we may assume that is of the form or where is a path in from to and is a loop in based at . Since, by [11], the fixed points and are not -Nielsen equivalent, then , the only nonidentity element of .

If , then and are equivalent by if and only if(4.12)which is equivalent to , for some which we now view as an element of . If then, similarly, and are equivalent by if and only if

There is no solution to or for which since starts with but and will start with . If , and , then there is no solution either since, again, starts with and starts with . If and , then there is no solution since starts with but starts with . If and , then there is no solution since but contains at least one or . So suppose that and . This means that with so is equivalent to by letting . However, no other fixed point is equivalent to because it would then also be equivalent to and, in this case, every starts with and no starts with so, since we assumed , we may conclude from Lemma 4.2 that no such equivalence is possible.

We now have the tools we will need to calculate the Nielsen number for almost all maps such that . (The remaining cases will be computed in Section 5.) We continue to write where .Theorem 4.4. If and is not a deformation, then (4.13)Proof. Since is cyclically reduced, if then also and thus, for where , the Wagner tail starts with and starts with so, by Lemma 4.1, no two of the fixed points are equivalent. However, and are equivalent to so, since is an essential fixed point class by Lemma 4.3, there are essential fixed point classes. If none of the fixed points on are equivalent to each other, nor is equivalent to any of them.

In standard form, each is represented by consecutive arcs in and there is a first arc and a last arc with respect to the orientation of , which correspond to the first and last appearance, respectively, of or in . We will refer to the fixed points in these arcs as the first and last fixed points in .

We say that a fixed point cancels a fixed point if and are equivalent and one is of index 1 and the other is of index .Theorem 4.5. If but and is not a deformation, then (4.14)Proof. If and then, if is not the first fixed point, it cancels one because . The only fixed point of not so cancelled is the first one. If , then all but the last fixed point of cancels a fixed point of with only the last fixed point not cancelled. One of and is cancelled by but each remaining uncancelled fixed point in and is an essential fixed point class. Thus, including , there are fixed point classes outside of . Let such that and such that . Then and are equivalent if and only if since that implies and thus to . We conclude that the number of essential fixed point classes in is if and and otherwise.Theorem 4.6. If and , and is not a deformation, then (4.15)Proof. If then, since ends with and begins with , a negative would produce cancellations in the reduced word , so we have . Since is cyclically reduced, it must be that . As in the previous proof, there are fixed points in each of and that do not cancel, is cancelled by but is an essential fixed point class. Similarly, in each of and there is one fixed point that is not cancelled. However, there exist and such that and they cancel each other, so . If then there is one uncancelled fixed point in each of and , and no fixed point in is cancelled, so . The other cases are symmetric to these.

In each of Theorems 4.4, 4.5, and 4.6, we assume that is not a deformation, so is an essential fixed point class of . If and but is a deformation, let be a map such that for all but the restriction of to is not a deformation though it induces a homomorphism mapping to itself. Then by Lemma 4.3 and can be calculated by the previous theorems. We note that, since and induce the same fundamental group homomorphism, this difference in the Nielsen numbers reflects the nonaspherical nature of . This completes the calculation of in the case that and .Theorem 4.7. Suppose and . If is not a deformation, then (4.16)If is a deformation, then (4.17)Proof. By Lemma 4.2, no two among the fixed points can be equivalent because, for each one, begins with and does not. Suppose and . If is not a deformation then, using Lemma 4.3, we see that each of is an essential fixed point class so whereas, if is a deformation, then . If and , then cancels . If is not a deformation then, by Lemma 4.3, cancels so except when . However, if is a deformation, then is an essential fixed point class so . If and then cancels whereas if is fixed by , then it is an essential fixed point class so if is not a deformation and if it is. Finally, suppose and . If is not a deformation, then cancels by Lemma 4.3 so . If is a deformation, then each of is an essential fixed point class and .

5. The Exceptional Cases

The only cases remaining occur when and for .

We will make use of the following result concerning Wagner tails.Lemma 5.1. Let and be fixed points of in . If one of or is in the kernel of then is equivalent to .Proof. Let denote the word in the hypotheses that is in the kernel of . If let , if let , if let and if let . Using Lemma 2.1, we verify that , so is equivalent to by Lemma 2.2.

If , so and , then the kernel of is the normal closure of the subgroup of generated by . Let be the quotient homomorphism, then there is a homomorphism such that . Setting and , we note that(5.1)where or . Let denote the number of appearances of in .Theorem 5.2. Suppose and (5.2)If is not a deformation, then (5.3)and, if is a deformation, then (5.4)Proof. As in the proof of Theorem 4.5, if then each fixed point of except the first one cancels a fixed point because , leaving only the first fixed point of uncancelled in this way. If , it is the last fixed point of and the last of that are the only fixed points that are not cancelled in this way. However, further cancellations take place. If is even, let and be the uncancelled fixed points of and respectively. Then is in the kernel of so the fixed points cancel by Lemma 5.1.

Suppose that and , for , are odd numbers and(5.5)is in the kernel of , and thus in the kernel of as well. Let and be fixed points in that were not cancelled in the previous step. If , then cancels and cancels whereas if , then cancels and cancels . We will demonstrate these cancellations only in the case and because the other three cases are similar. Since is in the kernel of , then and are also in the kernel, so and cancel, as do and , by Lemma 5.1.

After all the cancellations, let be adjacent fixed points in among those that remain. Writing(5.6)it must be that and are odd and . Therefore(5.7)so that and contribute two copies of to . We conclude that there are fixed points remaining in .

None of the remaining fixed points in are equivalent. Let and be two such fixed points, so . We claim that there is no solution to the equation(5.8)for any , which implies that and are not equivalent since (2.3) of Lemma 2.2 then has no solution. We first show that is not a solution to (5.8) because is not in the kernel of . Let(5.9)then cannot be in the kernel of since, otherwise, and would have been eliminated previously. If and , then whereas if and then which also cannot be in the kernel of since and are odd. If then, if is in the kernel of , there must exist with and odd, and both and are in the kernel of . But that would have eliminated these fixed points, so we have proved that is not a solution to (5.8). The argument that there is no solution to (5.8) with depends on word length considerations like those in the proofs of Lemmas 4.1 and 4.2, which we therefore omit, and we conclude that none of the remaining fixed points in are equivalent.

Suppose and let and be the first and last uncancelled fixed points in , respectively. Assume that , then either or is cancelled by . The reason is that, since , it must be that implies that . If , then so is cancelled by because so (2.3) of Lemma 2.2 is satisfied with . Similarly, if , then is cancelled by because and therefore (2.3) is satisfied by setting . On the other hand, if , then is not equivalent to any of the remaining fixed point in because, under this condition, there is no solution to (5.8) above when or . Thus, if is a deformation so there are no fixed points other than on in the standard form of , we see that if and if . If , then is the only uncancelled fixed point and .

Now suppose is not a deformation so the standard form of has a fixed point in . If then and are the only fixed point that do not cancel, so . If , then is not equivalent to any other fixed point by the following argument. Let and denote the Wagner tails of . As in the proof of Lemma 4.3, and are equivalent if and only if or for some and therefore, in the quotient group , we would have or . Since , there is no such because starts with but starts with . Since we have seen that one of or is cancelled by , we conclude that . If , then, in contrast to Lemma 4.3, does cancel a fixed point in . Let be the Wagner tail of then, since , we see that so and therefore cancels . Thus we again conclude that .Example 5.3. Let and define maps such that but and are not deformations, and . Then so, by Theorem 5.2, . On the other hand, by Theorem 4.6, . Thus, the class of maps in Theorem 5.2 are truly very exceptional in their fixed point behavior compared to those of Section 4.

In the final case, where so and , the kernel of is the normal closure of the subgroup of G generated by . Let again be the quotient group of by the normal closure of . Define by and , then there is a homomorphism such that given by and(5.10)where or . Let denote the number of appearances of in .Theorem 5.4. Suppose and . If is not a deformation, then (5.11)and, if is a deformation, then (5.12)Proof. Let be maps such that and and are maps in standard form representing homomorphisms such that and so . Let , then by the commutativity property of the Nielsen number. We note that and so or a map that induces and satisfies the hypotheses of Theorem 5.2. Since (2.3) of Lemma 2.2 is satisfied for if and only if it is satisfied for , we may assume that we can apply Theorem 5.2 to . Thus if is not a deformation, then if , and if , and if is a deformation, then if and , and otherwise, where is the number of appearances of in , where for . Since , then is a deformation if and only if is a deformation. Noting that , we have so and the conclusion of the theorem follows.Example 5.5. Let for and define maps such that but and are not deformations, and . Then, by Theorem 5.4, if is even and if is odd. On the other hand, by Theorem 4.7 and we find that the maps of Theorem 5.4 also have very different fixed point behavior compared to the maps of Section 4.

6. Proof of Lemma 4.2

Suppose and are equivalent fixed points of where and . Lemma 4.2 asserts that either or . We now present the proof of this assertion.

In the notation introduced at the beginning of Section 4, we write(6.1)(6.2)where ,(6.3)Let be the sum of the .

Case 1. There are no cancellations between and the first nor between the last and . As in Case 1 of Lemma 4.1, we will prove that there are no equivalent fixed points and for which and possess these noncancellation properties.

SubCase 1.1. . Then,(6.4)Since , all equalities must hold, and thus we have(6.5)This implies that or , and or . Since , we conclude that or , which is contrary to the hypothesis that is cyclically reduced.

SubCase 1.2. . We first consider the case of and then the case of . For , we have(6.6)

Claim 1. The inequality is obvious except for the case of and . In that case, we know that all the have the same sign and therefore either or is equal to zero. Since , if then and if then . Since , the equalities above must hold and so we have(6.7)Since , this implies that and thus all have the same sign. Suppose that and . (The case of and is similar.) Since all have the same sign, and therefore . If , that would imply either that and or and in adjacent arcs in , contrary to the assumption that is reduced. Thus which, since is in standard form, would imply , a contradiction. Now suppose that and . All the have the same sign; suppose it is negative and thus all . Then where implies that so or and so . If then either and thus or and thus , both of which contradict the assumption that . If , substituting or again leads to a contradiction, now to . If all then, similarly, all cases lead to a contradiction to the assumption that .

Suppose that . Since , we have or , which is a subword of and thus . Therefore,(6.8)This is a contradiction so, if there are no such cancellations, there cannot be a solution to (2.3) of Lemma 2.2.

SubCase 1.3. or . If , then or and the lemma is obviously true. Thus we assume that . We will consider only the case because the other is similar. Since we suppose and equivalent, we are assuming there exists such that . But then, as in Subcase 1.2, we will show that, for any choice of we have . We will do it by dividing into subwords such that their image under is reduced in and of greater word length. We first consider each subword of for which . Then contains the subword that is reduced in and(6.9)because , so .

Now consider a subword of of the form where but and . Suppose (or ) and (or ). Then contains a subword which is reduced in . In the case we have , so we consider . Since we are assuming that , it must be that for some . Since , it follows that contains a subword that is reduced in and(6.10)If, instead, and (or ), then(6.11)contains as a subword that is reduced in . The assumption that then implies that for some . The length of the image under of the subword of consisting of and is greater than that of the word itself. The same holds for the appropriate choice of subwords of when and .

Suppose instead that we consider a subword of of the form where now but and . An analysis like that just presented again leads to the conclusion that increases the word length of subwords of . Thus we have established that and consequently there are no solutions to (2.3) of Lemma 2.2.

Case 2. Suppose there is a cancellation between and the first but no cancellation between the last and . If and or and , then begins with and no such cancellation is possible. If and , an argument like that of Case 2 of Lemma 4.1 shows that a cancellation would contradict the assumption that is cyclically reduced. Thus, we can conclude that and so and so, similarly to Lemma 4.1,(6.12)There are no further cancellations and thus, as in the previous case, there is no solution to (2.3) unless so that and .

Case 3. Suppose there is no cancellation between and the first but there is cancellation between the last and . An argument similar to that of Case 2 demonstrates that but then a solution is possible only if and thus that .

Case 4. Suppose that there is a cancellation between and the first and also between and the last . Following Cases 2 and 3, we conclude that and that(6.13)There are now no cancellations so and . Since is in standard form, the condition also implies that and thus there is no solution of this type.

Acknowledgment

The first author was partially supported by Kyungsung University Research Grants in 2009.

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