Estimating Nielsen Numbers on Wedge Product Spaces

Let be a self-map of a finite polyhedron that is an aspherical wedge product space . In this paper, we estimate the Nielsen number of . In particular, we study some algebraic properties of the free products and then estimate Nielsen numbers on torus wedge surface with boundary, Klein bottle wedge surface with boundary, and torus wedge torus.


Introduction
Let X be a space and let f : X→X be a self-map. Let Fix( f ) = {x ∈ X : f (x) = x} denote the fixed point set. The Nielsen number N( f ) provides a lower bound for min #Fix(g) : g f (1.1) and is often sharp. But, in general, it is very difficult to compute N( f ) from its definition. For background on Nielsen fixed point theory, see [1][2][3]. For a given space X, algebraic properties of its fundamental group π 1 (X) are usually important to compute Nielsen numbers on it. However, if the fundamental group of X is free or a free product group, then computing Nielsen number on X is extremely difficult see [4].
In this paper, we estimate the Nielsen numbers on aspherical wedge product spaces, focusing on the following three cases: (1) torus wedge surface with boundary; (2) Klein bottle wedge surface with boundary; (3) torus wedge torus. It is well known that the fundamental group of the wedge product of spaces is the free product of the fundamental groups of the spaces. In Section 2, we present several properties of free product of groups which we use in the final section to classify all maps of the spaces above. Then we consider the Nielsen numbers on aspherical wedge product spaces in Section 3. As applications, we estimate the Nielsen numbers on the above three types of spaces in the final section.

Free products
Let A * C be the free product of two groups A and C. The groups A and C are called the free factors of A * C. A reduced sequence (or normal form) is a sequence of elements g 1 ,g 2 ,...,g n from A * C such that each g i = 1, each g i is in A or C, and successive g i , g i+1 are not in the same free factor. It is well known that each element g of A * C can be uniquely expressed as a product g = g 1 ,g 2 ,...,g n , where g 1 ,g 2 ,...,g n is a reduced sequence, which is called the reduced form (or normal form) of g.
Let g be an element of A * C with reduced form g 1 ,g 2 ,...,g n . The syllable length λ(g) of g is n and g is called cyclically reduced if g 1 and g n are from different free factors or n ≤ 1.
Proof. Suppose that u = u 1 ,u 2 ,...,u m is cyclically reduced. Since u is not in the conjugate of a free factor, we have m ≥ 2. Since u is cyclically reduced, is also cyclically reduced and λ(u k ) = km. If v is not cyclically reduced, v k is not either, and this contradicts the hypothesis u k = v k . Therefore, v is cyclically reduced and so v k is cyclically reduced. Let v = v 1 ,v 2 ,...,v n . Then, similarly to u, we have n ≥ 2 and λ(v k ) = kn.
Since u k = v k , this implies that λ(u k ) = λ(v k ) and so m = n. Furthermore, Since u k and v −k are cyclically reduced and m = n ≥ 2, we can say that u k v −k = 1 implies that each pair u i and v −1 i is in a same free factor and u i v −1 i = 1. Therefore, u i = v i for all i and hence u = v. Now, suppose that u is not cyclically reduced. Then, by Theorem 2.1, we can denote u = wsw −1 for some element w and cyclically reduced element s in A * C. Since u is not in the conjugate of a free factor, the same is true of s = w −1 uw and N. Khamsemanan and S. W. Kim 3 Since v is not in the conjugate of a free factor, this implies that w −1 vw is not either, and hence the above paragraph applied to s and w −1 vw in place of u and v shows that s = w −1 vw. Therefore, we have (1) u and v are in the same conjugate of a free factor.
(2) u and v are both powers of the same element in A * C.
We say a group is 2-torsion free if it contains no elements of order 2.
Lemma 2.6. Suppose that A and C are 2-torsion free groups. If g is in A * C and both a and gag are in A, then g is in A.
Proof. Let g = g 1 ,g 2 ,...,g r be the reduced form of g in A * C. We use induction on r. If r = 1, then g = g 1 is either in A or in C. Suppose g ∈ C. If a = 1, then gag = gg ∈ A, and if a = 1, then gag is a reduced form and has syllable length >1, and hence, gag ∈ A, contrary to hypothesis. Therefore g is in A. Suppose that the lemma is true for r = n − 1 and that r = n. We first show that g n ∈ A. If g 1 and g n are in C, then gag = g 1 ,...,g n ag 1 ,...,g n has syllable length > 1. Thus gag ∈ A, contrary to hypothesis. Suppose g 1 ∈ A and g n ∈ C.
Since the terminal g 2k cancels in gag, we have Then g 2 k+1 = 1 in A or C, which contradicts the hypothesis that A and C are 2-torsion free. Let r = 2k + 1 for some positive integer k. Similarly to the case n = 2k, we have Thus g k+1 and g k+2 are in a same free factor. This is impossible because g 1 ,g 2 ,...,g 2k+1 is a reduced sequence. Therefore, we can say that g n ∈ A and so g n a ∈ A. Then, since g 1 ,g 2 ,...,g n−1 g n a g 1 ,g 2 ,...,g n−1 = (gag)g −1 n (2.7) 4 Fixed Point Theory and Applications is in A, by the induction hypothesis, g 1 ,g 2 ,...,g n−1 is in A and hence g = (g 1 ,g 2 ,...,g n−1 )g n is in A.
The following example illustrates the fact that the requirement that A and C are 2torsion free is crucial in Lemma 2.6.
Theorem 2.8. Suppose that A and C are 2-torsion free groups. If uv = v −1 u in A * C, then at least one of the following is true.
(1) u and v are in the same conjugate of a free factor.
Proof. We show that if v = 1, then u and v are in the same conjugate of a free factor. Since By Lemma 2.4, this implies that g −1 ug is in A and hence u is in gAg −1 . Since A and C are 2-torsion free and vuv = u, using Lemma 2.6 instead of Lemma 2.4, we can similarly see that if u is in the conjugate of some free factor, then v is in the same conjugate of the free factor. Now, suppose that neither u nor v is in the conjugate of a free factor. Since uv = v −1 u, this implies that and by Lemma 2.3, we have uv = vu. Thus v = v −1 and hence v = 1, contrary to the assumption. Therefore, v = 1 implies that u and v are in the same conjugate of a free factor.
Let F be a finitely generated free group.
Proof. Let x 1 ,x 2 ,...,x k be k-free generators of F. We use induction on k. Suppose uv = v −1 u and k = 1. Then, since F = x 1 is abelian, we have uv = vu and thus v = 1. Suppose that the corollary is true for k = n − 1 and that k = n. Since F is isomorphic to the free product of two free groups A = x 1 ,x 2 ,...,x n−1 and C = x n , by Theorem 2.8 there are only three cases possible.
(1) u and v are in the same conjugate of A. Let u = waw −1 and v = wbw −1 for some w ∈ F and a,b ∈ A. Since uv = v −1 u, this implies that ab = b −1 a in A. By the induction hypothesis, we have b = 1 and hence v = ww −1 = 1.
N. Khamsemanan and S. W. Kim 5 (2) u and v are in the same conjugate of C. Let u = wx p n w −1 and v = wx q nw −1 for some w ∈ F and some integers p and q. Since uv = v −1 u, we have wx p+q n w −1 = wx p−q n w −1 . Therefore, q = 0 and hence v = 1.
(3) Neither of the above holds, but v = 1. (1) h(a) and h(b) are in the same conjugate of a free factor.
The following example illustrates the fact that if a group C has an order-2 element, then there is an endomorphism which does not satisfy both cases in Corollary 2.10.
Thus h is an endomorphism of G but h(a) and h(b) are not in the same conjugate of a free factor and clearly h(b) = 1.

Estimating Nielsen numbers on wedge product spaces
Let Y and Z be aspherical finite polyhedra and let X = Y ∨ Z be the wedge product of (Y , y 0 ) and (Z,z 0 ). We denote the intersection by x 0 . Let f : X→X be a self-map such that f (Y ) ⊆ Y . Using the homotopy extension property, we may assume that f (x 0 ) = x 0 and so f induces a homomorphism f π : π 1 (X,x 0 )→π 1 (X,x 0 ). Note that, up to homotopy, the condition f (Y ) ⊆ Y is equivalent to the condition f π (π 1 (Y ,x 0 )) ⊆ π 1 (Y ,x 0 ). We will assume that Y and Z have those properties throughout this section unless stated otherwise. Now consider a retraction q : X→Z sending Y to x 0 , that is, where id Z : Z→Z is the identity map and x 0 is the constant map at x 0 . Let f Z denote the restriction of f to Z.
Lemma 3.1. The following diagram commutes: In 1992, Woo and Kim [6] introduced the q-Nielsen number which is a lower bound for the Nielsen number. The map q : X→X q in [6] is not our special map q defined above but an arbitrary map from X to a space X q . Two fixed points x 0 and x 1 of f are in the same q-(fixed point) class if there exists a path c in X from x 0 to x 1 such that q • c q • f • c in X q . Using the ordinary fixed point index, they defined the q-Nielsen number N q ( f ). This is just the mod K-Nielsen number (see [2, Chapter III]) when K = kerq but it turns out to be more convenient when we consider the mod K-Nielsen number geometrically. Now we return to our map q : X→Z.
By the definition of the map q, Proof. Suppose that two fixed points z 1 and z 2 are in the same fixed point class of Therefore, two fixed points z 1 and z 2 are in the same q-class of f . Conversely, suppose that z 1 and z 2 are in the same q-class of f . Then there exists a path δ in X from z 1 to z 2 such that q For a fixed point x ∈ Fix( f ), let [x] (resp., [x] q ) denote the fixed point class (resp., q-class) of f containing x, and for z ∈ Fix(q • f Z ), let [z] Z denote the fixed point class of the map q • f Z containing z. Proof. The contrapositive statement of the lemma is as follows: if [z] q = [y] q for some Theorem 3.4. The q-Nielsen number has a lower bound In [7], Ferrario developed a formula for the Reidemeister trace of a pushout map. This is useful for union of spaces, quotient spaces, connected sums, and wedge product spaces. In particular, for a map f : X→X on a wedge product space X = Y ∨ Z, he proved the following theorem. We obtain it again in a different way using our previous results.
Proof. In the proof of Theorem 3.4, we show that for Since Therefore, Since the assumption of Theorem 3.5 is quite strong, it is necessary to generalize Theorem 3.5 in order to estimate the Nielsen number on wedge product spaces. First, we consider the condition p π • f π (π 1 (Z,x 0 )) = 1 instead of the condition f (Z) ⊆ Z in the assumption of Theorem 3.5. Since the condition f (Z) ⊆ Z implies that p π • f π (π 1 (Z,x 0 )) = 1 in π 1 (X,x 0 ) and the converse is not true, p π • f π (π 1 (Z,x 0 )) = 1 is a more generalized 8 Fixed Point Theory and Applications condition. But, unfortunately, under the assumptions f (Y ) ⊆ Y and p π • f π (π 1 (Z,x 0 )) = 1, the result of Theorem 3.5 in general form is no longer true, as follows.
Example 3.6. Let X = T ∨ S 1 be the wedge product of a torus and a circle at x 0 . In this example, we have Y = T and Z = S 1 . Then π 1 (X, . In order to compute the Nielsen number of f , we use the Fadell-Husseini formula for the Reidemeister trace in [8]. (See [8] or [9].) For the map f , the Reidemeister trace is . Using the technique of abelianization, we know that these terms are distinct and thus that Hence Proof. Suppose that y 1 and y 2 are in the same p-class of f , that is, there is a path γ : I→X from y 1 to y 2 such that Modifying γ slightly as necessary, we can assume that γ is the product of finite numbers of the pathes γ 1 ,γ 2 ,...,γ 2k+1 such that all γ 2i+1 with i = 0,...,k and γ 2 j with j = 1,...,k are pathes in Y and Z, respectively. Furthermore, all γ 2 j are loops at x 0 in Z and so {γ 2 j } ∈ π 1 (Z,x 0 ). Since p π • f π (π 1 (Z,x 0 )) = 1, we have p π • f π ({γ 2 j }) = 1 for all j. Since Y is aspherical, it follows that p • f • γ 2 j x 0 . Therefore, Since p = id on Y , the converse is obvious.
We now consider a map f : X→X with N. Khamsemanan and S. W. Kim 9 where w = w 1 ,w 2 ,...,w k ∈ π 1 (X,x 0 ) is cyclically reduced. We may assume that w 1 ∈ π 1 (Z,x 0 ) and w k ∈ π 1 (Y ,x 0 ) because if w 1 ∈ π 1 (Y ,x 0 ), then since X is an aspherical space, we can homotope f to a map g such that g π (·) = w −1 1 f π (·)w 1 and if w k ∈ π 1 (Z,x 0 ), then w k π 1 (Z,x 0 )w −1 k ⊆ π 1 (Z,x 0 ). For the map f , we also consider a map f : X→X which is homotopic to f such that The following theorem is a generalization of Theorem 3.5.
Theorem 3.8. If a map f satisfies the conditions of (3.14), then To extend f to all of X , we divide I into 2k + 1 equal closed intervals I 0 ,J 1 ,I 1 ,...,J k ,I k .
Then for each integer r, I 2r is mapped homeomorphically onto I from i 0 to i k , J 2r+1 is mapped onto a loop in Z at i k representing w 2r+1 , I 2r+1 is mapped homeomorphically onto I from i k to i 0 , J 2r is mapped onto a loop in Y at i 0 representing w 2r .
(3.18) By construction, there is exactly one fixed point in each interval I i and no fixed point in any interval J i . Therefore, f has k + 1 fixed points in I including i 0 and i k .
We now show that f is the same homotopy type as f . Choose a small neighborhood D of x 0 in X and let ∂ Z D denote the intersection of the boundary of D with Z. Then there is a homotopy equivalence ϕ : X→X such that ϕ(x 0 ) = i 0 and ϕ(∂ Z D) = i k . Then we see that f π • ϕ π = ϕ π • f π : π 1 X,x 0 −→ π 1 X ,i 0 , (3.19) because of the construction of f . Thus f is the homotopy type of f . Let Z = I ∨ Z ⊆ X and let p : X →Y and q : X →Z retractions sending Z to i k and Y to i 0 , respectively. Let α = N(p • f Y ) and β = N(q • f Z ). Since f (Y ) ⊆ Y , by Lemmas 3.2 and 3.3, there exist at least β − 1 essential q -classes in Z such that each of them is not On the other hand, since f (Z) ⊆ Z, by Lemmas 3.2 and 3.3, there exist at least α − 1 essential p -classes such that each of them does not contain i k and so, by Lemma 3.3, each of them does not contain any fixed points of f in Z. Since Fix( f ) ∩ I has k + 1 fixed points of f , we see that |Fix( f ) ∩ I \ {i k }| = k and thus at least (α − 1) − k essential p -classes do not intersect Z .
Consequently, there exist at least (α − 1) + (β − 1) − k essential disjoint p or q -classes of f . Therefore, (3.20) Note that f and f are of the same homotopy type and so are p • f Y (resp., q • f Z ) and p • f Y (resp., q • f Z ). Since the Nielsen number is a homotopy type invariant (see [2]), we have

Applications
Let M be a surface with boundary, which is homotopy equivalent to a bouquet of k circles, and let f : M→M be a map. The fundamental group of M is the free group on k generators. In 1999, Wagner [10] provided a method for computing the Nielsen number N( f ) for a large class of maps of M by means of an algorithm that depends only on the induced endomorphism f π of π 1 (M). In 2006, Kim [11] extended her results for another large class of maps of M. In the case of k = 2, Yi [12] extended Wagner's work using the idea of a mutant, which was introduced by Jiang in [13], and Kim [14] completed Yi's work so that the Nielsen number of all maps can be calculated.
The following spaces are examples of aspherical wedge product spaces for which we can classify the endomorphisms of the fundamental groups, and thus the self-maps. The general results in Section 3 along with existing technique allow us to estimate or calculate the Nielsen number on the following spaces: (1) torus wedge surface with boundary; (2) Klein bottle wedge surface with boundary; (3) torus wedge torus; except for some cases in (3) which satisfy the L1 condition in Theorem 4.5.  (H1) f π (a) = wa m1 b n1 w −1 and f π (b) = wa m2 b n2 w −1 for some w ∈ G; (H2) f π (a) = g s and f π (b) = g t for some element g ∈ G and integers s and t.
Proof. From Theorem 2.5, it is enough to show that if f π (a) and f π (b) are in the same conjugate of C, then both are powers of the same element in G. Suppose f π (a) = wg 1 w −1 and f π (b) = wg 2 w −1 for some w ∈ G and g 1 ,g 2 ∈ C. Since f π is an endomorphism of G, then f π (a) f π (b) = f π (b) f π (a) and therefore g 1 and g 2 are commuting elements of a free