Fixed Point Theory and Applications
Volume 2009 (2009), Article ID 346519, 16 pages
doi:10.1155/2009/346519
Research Article

Minimal Nielsen Root Classes and Roots of Liftings

Marcio Colombo Fenille and Oziride Manzoli Neto

Instituto de Ciências Matemáticas e de Computação, Universidade de São Paulo, Avenida Trabalhador São-Carlense, 400 Centro Caixa Postal 668, 13560-970 São Carlos, SP, Brazil

Received 24 April 2009; Accepted 26 May 2009

Academic Editor: Robert Brown

Copyright © 2009 Marcio Colombo Fenille and Oziride Manzoli Neto. 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

Given a continuous map 𝑓 ∶ 𝐾 → 𝑀 from a 2-dimensional CW complex into a closed surface, the Nielsen root number 𝑁 ( 𝑓 ) and the minimal number of roots 𝜇 ( 𝑓 ) of 𝑓 satisfy 𝑁 ( 𝑓 ) ≤ 𝜇 ( 𝑓 ) . But, there is a number 𝜇 𝐶 ( 𝑓 ) associated to each Nielsen root class of 𝑓 , and an important problem is to know when 𝜇 ( 𝑓 ) = 𝜇 𝐶 ( 𝑓 ) 𝑁 ( 𝑓 ) . In addition to investigate this problem, we determine a relationship between 𝜇 ( 𝑓 ) and 𝜇 ( 𝑓 ) , when 𝑓 is a lifting of 𝑓 through a covering space, and we find a connection between this problems, with which we answer several questions related to them when the range of the maps is the projective plane.

1. Introduction

Let 𝑓 ∶ 𝑋 → 𝑌 be a continuous map between Hausdorff, normal, connected, locally path connected, and semilocally simply connected spaces, and let 𝑎 ∈ 𝑌 be a given base point. A root of 𝑓 at 𝑎 is a point 𝑥 ∈ 𝑋 such that 𝑓 ( 𝑥 ) = 𝑎 . In root theory we are interested in finding a lower bound for the number of roots of 𝑓 at 𝑎 . We define the minimal number of roots of 𝑓 at 𝑎 to be the number 𝜇 ( 𝑓 , 𝑎 ) = m i n # 𝜑 − 1 ( 𝑎 ) s u c h t h a t 𝜑 i s h o m o t o p i c t o 𝑓 . ( 1 . 1 )

When the range 𝑌 of 𝑓 is a manifold, it is easy to prove that this number is independent of the selected point 𝑎 ∈ 𝑌 , and, from [1, Propositions  2.10 and  2.12], 𝜇 ( 𝑓 , 𝑎 ) is a finite number, providing that 𝑋 is a finite CW complex. So, in this case, there is no ambiguity in defining the minimal number of roots of 𝑓 : 𝜇 ( 𝑓 ) ∶ = 𝜇 ( 𝑓 , 𝑎 ) f o r s o m e 𝑎 ∈ 𝑌 . ( 1 . 2 )

Definition 1.1. If 𝜑 ∶ 𝑋 → 𝑌 is a map homotopic to 𝑓 and 𝑎 ∈ 𝑌 is a point such that 𝜇 ( 𝑓 ) = # 𝜑 − 1 ( 𝑎 ) , we say that the pair ( 𝜑 , 𝑎 ) provides 𝜇 ( 𝑓 ) or that ( 𝜑 , 𝑎 ) is a pair providing 𝜇 ( 𝑓 ) .

According to [2], two roots 𝑥 1 , 𝑥 2 of 𝑓 at 𝑎 are said to be Nielsen root 𝑓 equivalent if there is a path 𝛾 ∶ [ 0 , 1 ] → 𝑋 starting at 𝑥 1 and ending at 𝑥 2 such that the loop 𝑓 ∘ 𝛾 in 𝑌 at 𝑎 is fixed-end-point homotopic to the constant path at 𝑎 . This relation is easily seen to be an equivalence relation; the equivalence classes are called Nielsen root classes of 𝑓 at 𝑎 . Also a homotopy 𝐻 between two maps 𝑓 and 𝑓 provides a correspondence between the Nielsen root classes of 𝑓 at 𝑎 and the Nielsen root classes of 𝑓 at 𝑎 . We say that such two classes under this correspondence are 𝐻 -related. Following Brooks [2] we have the following definition.

Definition 1.2. A Nielsen root class ℜ of a map 𝑓 at 𝑎 is essential if given any homotopy 𝐻 ∶ 𝑓 ≃ 𝑓 starting at 𝑓 , and the class ℜ is 𝐻 -related to a root class of 𝑓 at 𝑎 . The number of essential root classes of 𝑓 at 𝑎 is the Nielsen root number of 𝑓   at   𝑎 ; it is denoted by 𝑁 ( 𝑓 , 𝑎 ) .

The number 𝑁 ( 𝑓 , 𝑎 ) is a homotopy invariant, and it is independent of the selected point 𝑎 ∈ 𝑌 , provid that 𝑌 is a manifold. In this case, there is no danger of ambiguity in denot it by 𝑁 ( 𝑓 ) .

In a similar way as in the previous definition, Gonçalves and Aniz in [3] define the minimal cardinality of Nielsen root classes.

Definition 1.3. Let ℜ be a Nielsen root class of 𝑓 ∶ 𝑋 → 𝑌 . We define 𝜇 𝐶 ( 𝑓 , ℜ ) to be the minimal cardinality among all Nielsen root classes ℜ , of a map 𝑓 , 𝐻 -related to ℜ , for 𝐻 being a homotopy starting at 𝑓 and ending at 𝑓 :

Again in [3] was proved that if 𝑌 is a manifold, then the number 𝜇 𝐶 ( 𝑓 , ℜ ) is independent of the Nielsen root class of 𝑓 ∶ 𝑋 → 𝑌 . Then, in this case, there is no danger of ambiguity in defining the minimal cardinality of Nielsen root classes of 𝑓 𝜇 𝐶 ( 𝑓 ) ∶ = 𝜇 𝐶 ( ℜ ) f o r s o m e N i e l s e n r o o t c l a s s ℜ . ( 1 . 3 )

An important problem is to know when it is possible to deform a map 𝑓 to some map 𝑓 with the property that all its Nielsen root classes have minimal cardinality. When the range 𝑌 of 𝑓 is a manifold, this question can be summarized in the following: when 𝜇 ( 𝑓 ) = 𝜇 𝐶 ( 𝑓 ) 𝑁 ( 𝑓 ) ?

Gonçalves and Aniz [3] answered this question for maps from CW complexes into closed manifolds, both of same dimension greater or equal to 3. Here, we study this problem for maps from 2 -dimensional CW complexes into closed surfaces. In this context, we present several examples of maps having liftings through some covering space and not having all Nielsen root classes with minimal cardinality.

Another problem studied in this article is the following. Let 𝑝 𝑘 ∶ 𝑌 → 𝑌 be a 𝑘 -fold covering. Suppose that 𝑓 ∶ 𝑋 → 𝑌 is a map having a lifting 𝑓 ∶ 𝑋 → 𝑌 through 𝑝 𝑘 . What is the relationship between the numbers 𝜇 ( 𝑓 ) and 𝜇 ( 𝑓 ) ? We answer completely this question for the cases in which 𝑋 is a connected, locally path connected and semilocally simply connected space, and 𝑌 and 𝑌 are manifolds either compact or triangulable. We show that 𝜇 ( 𝑓 ) ≥ 𝑘 𝜇 ( 𝑓 ) , and we find necessary and sufficient conditions to have the identity.

Related results for the Nielsen fixed point theory can be found in [4].

In Section 4, we find an interesting connection between the two problems presented. This whole section is devoted to the demonstration of this connection and other similar results.

In the last section of the paper, we answer several questions related to the two problems presented when the range of the considered maps is the projective plane.

Throughout the text, we simplify write 𝑓 is a map instead of 𝑓 is a continuous map.

2. The Minimizing of the Nielsen Root Classes

In this section, we study the following question: given a map 𝑓 ∶ 𝐾 → 𝑀 from a 2-dimensional CW complex into a closed surface, under what conditions we have 𝜇 ( 𝑓 ) = 𝜇 𝐶 ( 𝑓 ) 𝑁 ( 𝑓 ) ? In fact, we make a survey on the main results demonstrated by Aniz [5], where he studied this problem for dimensions greater or equal to 3. After this, we present several examples and a theorem to show that this problem has many pathologies in dimension two.

In [5] Aniz shows the following result.

Theorem 2.1. Let 𝑓 ∶ 𝐾 → 𝑀 be a map from an 𝑛 -dimensional CW complex into a closed 𝑛 -manifold, with 𝑛 ≥ 3 . If there is a map 𝑓 ′ ∶ 𝐾 → 𝑀 homotopic to 𝑓 such that one of its Nielsen root classes ℜ has exactly 𝜇 𝐶 ( 𝑓 ) roots, each one of them belonging to the interior of 𝑛 -cells of 𝐾 , then 𝜇 ( 𝑓 ) = 𝜇 𝐶 ( 𝑓 ) 𝑁 ( 𝑓 ) .

In this theorem, the assumption on the dimension of the complex and of the manifold is not superfluous; in fact, Xiaosong presents in [6, Section  4] a map 𝑓 ∶ 𝕋 2 # 𝕋 2 → 𝕋 2 from the bitorus into the torus with 𝜇 ( 𝑓 ) = 4 and 𝜇 𝐶 ( 𝑓 ) 𝑁 ( 𝑓 ) = 3 .

In [3, Theorem 4.2], we have the following result.

Theorem 2.2. For each 𝑛 ≥ 3 , there is an 𝑛 -dimensional CW complex 𝐾 𝑛 and a map 𝑓 𝑛 ∶ 𝐾 𝑛 → ℝ P 𝑛 with 𝑁 ( 𝑓 𝑛 ) = 2 , 𝜇 𝐶 ( 𝑓 𝑛 ) = 1 and 𝜇 ( 𝑓 𝑛 ) ≥ 3 .

This theorem shows that, for each 𝑛 ≥ 3 , there are maps 𝑓 ∶ 𝐾 𝑛 → 𝑀 𝑛 from 𝑛 -dimensional CW complexes into closed 𝑛 -manifolds with 𝜇 ( 𝑓 ) ≠ 𝜇 𝐶 ( 𝑓 ) 𝑁 ( 𝑓 ) . Here, we will show that maps with this property can be constructed also in dimension two. More precisely, we will construct three examples in this context for the cases in which the range-of the maps are, respectively, the closed surfaces ℝ P 2 (the projective plane), 𝕋 2 (the torus), and ℝ P 2 # ℝ P 2 (the Klein bottle). When the range is the sphere 𝑆 2 , it is obvious that every map 𝑓 ∶ 𝐾 → 𝑆 2 satisfies 𝜇 ( 𝑓 ) = 𝜇 𝐶 ( 𝑓 ) 𝑁 ( 𝑓 ) , since in this case there is a unique Nielsen root class.

Before constructing such examples, we present the main results that will be used.

Let 𝑓 ∶ 𝑋 → 𝑌 be a map between connected, locally path connected, and semilocally simply connected spaces. Then 𝑓 induces a homomorphism 𝑓 # ∶ 𝜋 1 ( 𝑋 ) → 𝜋 1 ( 𝑌 ) between fundamental groups. Since the image 𝑓 # 𝜋 1 ( 𝑋 ) of 𝜋 1 ( 𝑋 ) by 𝑓 # is a subgroup of 𝜋 1 ( 𝑌 ) , there is a covering space 𝑝 + ∶ 𝑌 + → 𝑌 such that 𝑝 + # 𝜋 1 ( 𝑌 + ) = 𝑓 # 𝜋 1 ( 𝑋 ) . Thus, 𝑓 has a lifting 𝑓 + ∶ 𝑋 → 𝑌 + through 𝑝 + . The map 𝑓 + is called a Hopf lift of 𝑓 , and 𝑝 + ∶ 𝑌 + → 𝑌 is called a Hopf covering for 𝑓 .

The next result corresponds to [2, Theorem  3.4].

Proposition 2.3. The sets ( 𝑓 + ) − 1 ( 𝑎 𝑖 ) , for 𝑎 𝑖 ∈ ( 𝑝 + ) − 1 ( 𝑎 ) , that are nonempty, are exactly the Nielsen root class of 𝑓 at 𝑎 and a class ( 𝑓 + ) − 1 ( 𝑎 𝑖 ) is essential if and only if ( 𝑓 + 1 ) − 1 ( 𝑎 𝑖 ) is nonempty for every map 𝑓 + 1 ∶ 𝑋 → 𝑌 + homotopic to 𝑓 + .

In [3], Gonçalves and Aniz exhibit an example which we adapt for dimension two and summarize now. Take the bouquet of 𝑚 copies of the sphere 𝑆 2 , and let 𝑓 ∶ ∨ 𝑚 𝑖 = 1 𝑆 2 → ℝ P 2 be the map which restricted to each 𝑆 2 is the natural double covering map. If 𝑚 is at least 2, then 𝑁 ( 𝑓 ) = 2 , 𝜇 𝐶 ( 𝑓 ) = 1 , and 𝜇 ( 𝑓 ) = 𝑚 + 1 .

Now, we present a little more complicated example of a map 𝑓 ∶ 𝐾 → ℝ P 2 , for which we also have 𝜇 ( 𝑓 ) ≠ 𝜇 𝐶 ( 𝑓 ) 𝑁 ( 𝑓 ) . Its construction is based in [3, Theorem   4.2].

Example 2.4. Let 𝑝 2 ∶ 𝑆 2 → ℝ P 2 be the canonical double covering. We will construct a 2-dimensional CW complex 𝐾 and a map 𝑓 ∶ 𝐾 → ℝ P 2 having a lifting 𝑓 ∶ 𝐾 → 𝑆 2 through 𝑝 2 and satisfying: (i) 𝑁 ( 𝑓 ) = 2 , (ii) 𝜇 𝐶 ( 𝑓 ) = 1 , (iii) 𝜇 ( 𝑓 ) ≥ 3 , (iv) 𝜇 ( 𝑓 ) = 1 .

We start by constructing the 2-complex 𝐾 . Let 𝑆 1 , 𝑆 2 , and 𝑆 3 be three copies of the 2-sphere regarded as the boundary of the standard 3-simplex Δ 3 :

𝑆 1 = 𝜕 ⟨ 𝑥 0 , 𝑥 1 , 𝑥 2 , 𝑥 3 ⟩ , 𝑆 2 = 𝜕 ⟨ 𝑦 0 , 𝑦 1 , 𝑦 2 , 𝑦 3 ⟩ , 𝑆 3 = 𝜕 ⟨ 𝑧 0 , 𝑧 1 , 𝑧 2 , 𝑧 3 ⟩ . ( 2 . 1 )

Let 𝐾 be the 2-dimensional (simplicial) complex obtained from the disjoint union 𝑆 1 ⊔ 𝑆 2 ⊔ 𝑆 3 by identifying [ 𝑥 0 , 𝑥 1 ] = [ 𝑦 0 , 𝑦 1 ] and [ 𝑦 0 , 𝑦 2 ] = [ 𝑧 0 , 𝑧 1 ] . Thus, each 𝑆 𝑖 , 𝑖 = 1 , 2 , 3 , is imbedded into 𝐾 so that

𝑆 1 ∩ 𝑆 2 = 𝑥 0 , 𝑥 1 = 𝑦 0 , 𝑦 1 , 𝑆 2 ∩ 𝑆 3 = 𝑦 0 , 𝑦 2 = 𝑧 0 , 𝑧 1 . ( 2 . 2 )

Then, 𝑆 1 ∩ 𝑆 2 ∩ 𝑆 3 is a single point 𝑥 0 = 𝑦 0 = 𝑧 0 . The (simplicial) 2-dimensional complex 𝐾 is illustrated in Figure 1.

346519.fig.001
Figure 1: A simplicial 2-complex.

Two simplicial complexes 𝐴 and 𝐵 are homeomorphic if there is a bijection 𝜙 between the set of the vertices of 𝐴 and of 𝐵 such that { 𝑣 1 , … , 𝑣 𝑠 } is a simplex of 𝐴 if and only if { 𝜙 ( 𝑣 1 ) , … , 𝜙 ( 𝑣 𝑠 ) } is a simplex of 𝐵 (see [7, page 128]). Using this fact, we can construct homeomorphisms ℎ 2 1 ∶ 𝑆 2 → 𝑆 1 and ℎ 3 2 ∶ 𝑆 3 → 𝑆 2 such that ℎ 2 1 | 𝑆 1 ∩ 𝑆 2 = 𝑖 𝑑 𝑒 𝑛 𝑡 𝑖 𝑡 𝑦 𝑚 𝑎 𝑝 and ℎ 3 2 | 𝑆 2 ∩ 𝑆 3 = 𝑖 𝑑 𝑒 𝑛 𝑡 𝑖 𝑡 𝑦 𝑚 𝑎 𝑝 .

Let 𝑓 1 ∶ 𝑆 1 → 𝑆 2 be any homeomorphism from 𝑆 1 onto 𝑆 2 . Define 𝑓 2 = 𝑓 1 ∘ ℎ 2 1 ∶ 𝑆 2 → 𝑆 2 and note that 𝑓 2 𝑓 ( 𝑥 ) = 1 ( 𝑥 ) for 𝑥 ∈ 𝑆 1 ∩ 𝑆 2 . Now, define 𝑓 3 = 𝑓 2 ∘ ℎ 3 2 ∶ 𝑆 3 → 𝑆 2 and note that 𝑓 3 𝑓 ( 𝑥 ) = 2 ( 𝑥 ) for 𝑥 ∈ 𝑆 2 ∩ 𝑆 3 . In particular, 𝑓 1 ( 𝑥 0 𝑓 ) = 2 ( 𝑥 0 𝑓 ) = 3 ( 𝑥 0 ) . Thus, 𝑓 1 , 𝑓 2 , and 𝑓 3 can be used to define a map 𝑓 ∶ 𝐾 → 𝑆 2 such that 𝑓 | 𝑆 𝑖 = 𝑓 𝑖 for 𝑖 = 1 , 2 , 3 .

Let 𝑓 ∶ 𝐾 → ℝ P 2 be the composition 𝑓 = 𝑝 2 ∘ 𝑓 , where 𝑝 2 ∶ 𝑆 2 → ℝ P 2 is the canonical double covering. Note that 𝑓 # 𝜋 1 ( 𝐾 ) = ( 𝑝 2 ) # 𝜋 1 ( 𝑆 2 ) . Thus, we can use Proposition 2.3 to study the Nielsen root classes of 𝑓 through the lifting 𝑓 .

Let 𝑎 = 𝑓 ( 𝑥 0 ) ∈ ℝ P 2 , and let 𝑝 2 − 1 ( 𝑎 ) = { ̃ 𝑎 , − ̃ 𝑎 } be the fiber of 𝑝 2 over 𝑎 .

Clearly, the homomorphism 𝑓 ∗ ∶ 𝐻 2 ( 𝐾 ) → 𝐻 2 ( 𝑆 2 ) is surjective, with 𝐻 2 ( 𝐾 ) ≈ ℤ 3 and 𝐻 2 ( 𝑆 2 ) ≈ ℤ . Hence, every map from 𝐾 into 𝑆 2 homotopic to 𝑓 is surjective. It follows that, for every map ̃ 𝑔 ∶ 𝐾 → 𝑆 2 homotopic to 𝑓 , we have ̃ 𝑔 − 1 ( ̃ 𝑎 ) ≠ ∅ and ̃ 𝑔 − 1 ( − ̃ 𝑎 ) ≠ ∅ . By Proposition 2.3, 𝑓 − 1 ( ̃ 𝑎 ) and 𝑓 − 1 ( − ̃ 𝑎 ) are the Nielsen root classes of 𝑓 , and both are essential classes. Therefore, 𝑁 ( 𝑓 ) = 2 .

Now, since 𝑎 = 𝑓 ( 𝑥 0 ) , either 𝑥 0 ∈ 𝑓 − 1 ( ̃ 𝑎 ) or 𝑥 0 ∈ 𝑓 − 1 ( − ̃ 𝑎 ) . Without loss of generality, suppose that 𝑥 0 ∈ 𝑓 − 1 ( ̃ 𝑎 ) . Then, by the definition of 𝑓 , we have 𝑓 − 1 ( ̃ 𝑎 ) = { 𝑥 0 } . Hence, one of the Nielsen root classes is unitary. Furthermore, since such class is essential, it follows that its minimal cardinality is equal to one. This proves that 𝜇 𝐶 ( 𝑓 ) = 1 .

In order to show that 𝜇 ( 𝑓 ) ≥ 3 , note that since each restriction 𝑓 | 𝑆 𝑖 is a homeomorphism and 𝑝 2 ∶ 𝑆 2 → ℝ P 2 is a double covering, for each map 𝑔 homotopic to 𝑓 , the equation 𝑔 ( 𝑥 ) = 𝑎 must have at least two roots in each 𝑆 𝑖 , 𝑖 = 1 , 2 , 3 . By the decomposition of 𝐾 this implies that 𝜇 ( 𝑓 ) ≥ 3 .

Moreover, it is very easy to see that 𝜇 ( 𝑓 ) = 1 , with the pair ( 𝑓 , 𝑓 ( 𝑎 0 ) ) providing 𝜇 ( 𝑓 ) .

Now, we present a similar example where the range of the map 𝑓 is the torus 𝕋 2 . Here, the complex 𝐾 of the domain of 𝑓 is a little bit more complicated.

Example 2.5. Let 𝑝 2 ∶ 𝕋 2 → 𝕋 2 be a double covering. We will construct a 2-dimensional CW complex 𝐾 and a map 𝑓 ∶ 𝐾 → 𝕋 2 having a lifting 𝑓 ∶ 𝐾 → 𝕋 2 through 𝑝 2 and satisfying the following: (i) 𝑁 ( 𝑓 ) = 2 , (ii) 𝜇 𝐶 ( 𝑓 ) = 1 , (iii) 𝜇 ( 𝑓 ) = 3 , (iv) 𝜇 ( 𝑓 ) = 1 .

We start constructing the 2-complex 𝐾 . Consider three copies 𝕋 1 , 𝕋 2 , and 𝕋 3 of the torus with minimal celular decomposition. Let 𝛼 𝑖 (resp., 𝛽 𝑖 ) be the longitudinal (resp., meridional) closed 1-cell of the torus 𝕋 𝑖 , 𝑖 = 1 , 2 , 3 . Let 𝐾 be the 2-dimensional CW complex obtained from the disjoint union 𝕋 1 ⊔ 𝕋 2 ⊔ 𝕋 3 by identifying 𝛼 1 = 𝛼 2 , 𝛼 3 = 𝛽 2 . ( 2 . 3 )

That is, 𝐾 is obtained by attaching the tori 𝕋 1 and 𝕋 2 through the longitudinal closed 1-cell and, next, by attaching the longitudinal closed 1-cell of the torus 𝕋 3 into the meridional closed 1-cell of the torus 𝕋 2 .

Each torus 𝕋 𝑖 is imbedded into 𝐾 so that

𝕋 1 ∩ 𝕋 2 = 𝛼 1 = 𝛼 2 , 𝕋 2 ∩ 𝕋 3 = 𝛼 3 = 𝛽 2 , 𝕋 1 ∩ 𝕋 3 = 𝕋 1 ∩ 𝕋 2 ∩ 𝕋 3 = 𝑒 0 , ( 2 . 4 )

where 𝑒 0 is the (unique) 0-cell of 𝐾 , corresponding to 0-cells of 𝕋 1 , 𝕋 2 , and 𝕋 3 through the identifications. The 2-dimensional CW complex 𝐾 is illustrate, in Figure 2.

346519.fig.002
Figure 2: A 2-complex obtained by attaching three tori.

Henceforth, we write 𝕋 𝑖 to denote the image of the original torus 𝕋 𝑖 into the 2-complexo 𝐾 through the identifications above.

Certainly, there are homeomorphisms ℎ 2 1 ∶ 𝕋 2 → 𝕋 1 and ℎ 3 2 ∶ 𝕋 3 → 𝕋 2 with ℎ 2 1 | 𝕋 1 ∩ 𝕋 2 = 𝑖 𝑑 𝑒 𝑛 𝑡 𝑖 𝑡 𝑦 𝑚 𝑎 𝑝   and   ℎ 3 2 | 𝕋 2 ∩ 𝕋 3 = 𝑖 𝑑 𝑒 𝑛 𝑡 𝑖 𝑡 𝑦 𝑚 𝑎 𝑝 such that ℎ 2 1 carries 𝛽 2 onto 𝛽 1 , and ℎ 3 2 carries 𝛽 3 onto 𝛼 2 . Thus, given a point 𝑥 3 ∈ 𝛽 3 we have ℎ 3 2 ( 𝑥 3 ) ∈ 𝛼 1 = 𝕋 1 ∩ 𝕋 2 . We should use this fact later.

Let 𝑓 1 ∶ 𝕋 1 → 𝕋 2 be an arbitrary homeomorphism carrying longitude into longitude and meridian into meridian. Define 𝑓 2 = 𝑓 1 ∘ ℎ 2 1 ∶ 𝕋 2 → 𝕋 2 and note that 𝑓 2 𝑓 ( 𝑥 ) = 1 ( 𝑥 ) for 𝑥 ∈ 𝕋 1 ∩ 𝕋 2 . Now, define 𝑓 3 = 𝑓 2 ∘ ℎ 3 2 ∶ 𝕋 3 → 𝕋 2 and note that 𝑓 3 𝑓 ( 𝑥 ) = 2 ( 𝑥 ) for 𝑥 ∈ 𝕋 2 ∩ 𝕋 3 . In particular, 𝑓 1 ( 𝑒 0 𝑓 ) = 2 ( 𝑒 0 𝑓 ) = 3 ( 𝑒 0 ) . Thus, 𝑓 1 , 𝑓 2 , and 𝑓 3 can be used to define a map 𝑓 ∶ 𝐾 → 𝕋 2 such that 𝑓 | 𝕋 𝑖 = 𝑓 𝑖 for 𝑖 = 1 , 2 , 3 .

Let 𝑝 2 ∶ 𝕋 2 → 𝕋 2 be an arbitrary double covering. (We can consider, e.g., the longitudinal double covering 𝑝 2 ( 𝑧 ) = ( 𝑧 2 1 , 𝑧 2 ) for each 𝑧 = ( 𝑧 1 , 𝑧 2 ) ∈ 𝑆 1 × 𝑆 1 ≅ 𝕋 2 .)

We define the map 𝑓 ∶ 𝐾 → 𝕋 2 to be the composition 𝑓 = 𝑝 2 ∘ 𝑓 .

In order to use Proposition 2.3 to study the Nielsen root classes of 𝑓 using the information about 𝑓 , we need to prove that 𝑓 # 𝜋 1 ( 𝐾 ) = ( 𝑝 2 ) # 𝜋 1 ( 𝕋 2 ) . Now, since 𝑓 # = ( 𝑝 2 ) # ∘ 𝑓 # , it is sufficient to prove that 𝑓 # is an epimorphism. This is what we will do. Consider the composition 𝑓 ∘ 𝑙 ∶ 𝕋 1 → 𝕋 2 , where 𝑙 ∶ 𝕋 1 ↪ 𝐾 is the obvious inclusion. This composition is exactly the homeomorphism 𝑓 1 , and therefore the induced homomorphism 𝑓 # ∘ 𝑙 # 𝑓 = ( 1 ) # is an isomorphism. It follows that 𝑓 # is an epimorphism. Therefore, we can use Proposition 2.3.

Let 𝑎 = 𝑓 ( 𝑒 0 ) ∈ 𝕋 2 , and let 𝑝 2 − 1 ( 𝑎 ) = { ̃ 𝑎 , ̃ 𝑎 } be the fiber of 𝑝 2 over 𝑎 . (If 𝑝 2 is the longitudinal double covering, as above, then if ̃ 𝑎 = ( ̃ 𝑎 1 , ̃ 𝑎 2 ) , we have ̃ 𝑎 = ( − ̃ 𝑎 1 , ̃ 𝑎 2 ) .)

Clearly, the homomorphism 𝑓 ∗ ∶ 𝐻 2 ( 𝐾 ) → 𝐻 2 ( 𝕋 2 ) is surjective, with 𝐻 2 ( 𝐾 ) ≈ ℤ 3 and 𝐻 2 ( 𝕋 2 ) ≈ ℤ . Hence, every map from 𝐾 into 𝕋 2 homotopic to 𝑓 is surjective. It follows that, for every map ̃ 𝑔 ∶ 𝐾 → 𝕋 2 homotopic to 𝑓 , we have ̃ 𝑔 − 1 ( ̃ 𝑎 ) ≠ ∅ and ̃ 𝑔 − 1 ( ̃ 𝑎 ) ≠ ∅ . By Proposition 2.3, 𝑓 − 1 ( ̃ 𝑎 ) and 𝑓 − 1 ( ̃ 𝑎 ) are Nielsen root classes of 𝑓 , and both are essential classes. Therefore, 𝑁 ( 𝑓 ) = 2 .

Now, since 𝑎 = 𝑓 ( 𝑒 0 ) , either 𝑒 0 ∈ 𝑓 − 1 ( ̃ 𝑎 ) or 𝑒 0 ∈ 𝑓 − 1 ( ̃ 𝑎 ) . Without loss of generality, suppose that 𝑒 0 ∈ 𝑓 − 1 ( ̃ 𝑎 ) . Then, by the definition of 𝑓 , we have 𝑓 − 1 ( ̃ 𝑎 ) = { 𝑒 0 } . Thus, one of the Nielsen root classes is unitary. Furthermore, since such class is essential, it follows that its minimal cardinality is equal to one. Therefore, 𝜇 𝐶 ( 𝑓 ) = 1 .

In order to prove that 𝜇 ( 𝑓 ) = 3 , note that since each restriction 𝑓 | 𝕋 𝑖 is a homeomorphism and 𝑝 2 ∶ 𝕋 2 → 𝕋 2 is a double covering, for each map 𝑔 homotopic to 𝑓 , the equation 𝑔 ( 𝑥 ) = 𝑎 must have at least two roots in each 𝕋 𝑖 , 𝑖 = 1 , 2 , 3 . By the decomposition of 𝐾 , this implies that 𝜇 ( 𝑓 ) ≥ 3 . Now, let 𝑥 3 be a point in 𝛽 3 , 𝑥 3 ≠ 𝑒 0 . As we have seen, ℎ 3 2 ( 𝑥 3 ) ∈ 𝛼 1 ⊂ 𝕋 1 ∩ 𝕋 2 . Write 𝑥 1 2 = ℎ 3 2 ( 𝑥 3 ) . By the definition of 𝑓 , we have 𝑓 ( 𝑥 1 2 ) = 𝑓 ( 𝑥 3 ) ≠ 𝑓 ( 𝑒 0 ) . Denote 𝑦 0 = 𝑓 ( 𝑒 0 ) and 𝑦 1 = 𝑓 ( 𝑥 1 2 ) .

Let 𝑎 ∈ 𝕋 2 be a point, and let 𝑝 2 − 1 ( 𝑎 ) = { ̃ 𝑎 , ̃ 𝑎 ′ } be the fiber of 𝑝 2 over 𝑎 . Since 𝕋 2 is a surface, there is a homeomorphism ℎ ∶ 𝕋 2 → 𝕋 2 homotopic to the identity map such that ℎ ( 𝑦 0 ) = ̃ 𝑎 and ℎ ( 𝑦 1 ) = ̃ 𝑎 . Let 𝑞 2 ∶ 𝕋 2 → 𝕋 2 be the composition 𝑞 2 = 𝑝 2 ∘ ℎ , and let 𝜑 ∶ 𝐾 → 𝕋 2 be the composition 𝜑 = 𝑞 2 ∘ 𝑓 . Then, 𝜑 is homotopic to 𝑓 and 𝜑 − 1 ( 𝑎 ) = { 𝑒 0 , 𝑥 1 2 , 𝑥 3 } . Since 𝜇 ( 𝑓 ) ≥ 3 , this implies that 𝜇 ( 𝑓 ) = 3 .

Moreover, it is very easy to see that 𝜇 ( 𝑓 ) = 1 , with the pair ( 𝑓 , 𝑓 ( 𝑒 0 ) ) providing 𝜇 ( 𝑓 ) .

Note that in this example, for every pair ( 𝜑 , 𝑎 ) providing 𝜇 ( 𝑓 ) (which is equal to 3), we have necessarily 𝜑 − 1 ( 𝑎 ) = { 𝑒 0 , 𝑥 1 , 𝑥 2 } with either 𝑥 1 ∈ 𝛼 1 and 𝑥 2 ∈ 𝛽 3 or 𝑥 1 ∈ 𝛽 1 and 𝑥 2 ∈ 𝛽 2 .

For the same complex 𝐾 of Example 2.5, we can construct a similar example with the range of 𝑓 being the Klein bottle. The arguments here are similar to the previous example, and so we omit details.

Example 2.6. Let 𝑝 2 ∶ 𝕋 2 → ℝ P 2 # ℝ P 2 be the orientable double covering. We will construct a 2-dimensional CW complex 𝐾 and a map 𝑓 ∶ 𝐾 → ℝ P 2 # ℝ P 2 having a lifting 𝑓 ∶ 𝐾 → 𝕋 2 through 𝑝 2 and satisfying the following: (i) 𝑁 ( 𝑓 ) = 2 ,(ii) 𝜇 𝐶 ( 𝑓 ) = 1 , (iii) 𝜇 ( 𝑓 ) = 3 , (iv) 𝜇 ( 𝑓 ) = 1 .

We repeat the previous example replacing the double covering 𝑝 2 ∶ 𝕋 2 → 𝕋 2 by the orientable double covering 𝑝 2 ∶ 𝕋 2 → ℝ P 2 # ℝ P 2 . Also here, we have 𝜇 ( 𝑓 ) = 1 , with the pair ( 𝑓 , 𝑓 ( 𝑒 0 ) ) providing 𝜇 ( 𝑓 ) .

Small adjustments in the construction of the latter two examples are sufficient to prove the following theorem.

Theorem 2.7. Let 𝐾 be the 2-dimensional CW complex of the previous two examples. For each positive integer 𝑛 , there are cellular maps 𝑓 𝑛 ∶ 𝐾 → 𝕋 2 and 𝑔 𝑛 ∶ 𝐾 → ℝ P 2 ℝ P 2 satisfying the following: (1) 𝑁 ( 𝑓 𝑛 ) = 𝑛 , 𝜇 𝐶 ( 𝑓 𝑛 ) = 1 and 𝜇 ( 𝑓 𝑛 ) = 2 𝑛 − 1 .(2) 𝑁 ( 𝑔 𝑛 ) = 2 𝑛 , 𝜇 𝐶 ( 𝑔 𝑛 ) = 1 and 𝜇 ( 𝑔 𝑛 ) = 4 𝑛 − 1 .

Proof. In order to prove item (1), let 𝑓 ∶ 𝐾 → 𝕋 2 be as in Example 2.5. Let 𝑝 𝑛 ∶ 𝕋 2 → 𝕋 2 be an 𝑛 -fold covering (which certainly exists; e.g., for each 𝑧 ∈ 𝕋 2 considered as a pair 𝑧 = ( 𝑧 1 , 𝑧 2 ) ∈ 𝑆 1 × 𝑆 1 , we can define 𝑝 𝑛 ( 𝑧 ) = ( 𝑧 𝑛 1 , 𝑧 2 ) ). Define 𝑓 𝑛 = 𝑝 𝑛 ∘ 𝑓 ∶ 𝐾 → 𝕋 2 . Then, the same arguments of Example 2.5 can be repeated to prove the desired result.
In order to prove item (2), let 𝑓 ∶ 𝐾 → 𝕋 2 be as in Example 2.6. Let 𝑝 𝑛 ∶ 𝕋 2 → 𝕋 2 be an 𝑛 -fold covering (e.g., as in the first item), and let 𝑝 2 ∶ 𝕋 2 → ℝ P 2 # ℝ P 2 be the orientable double covering. Define 𝑞 2 𝑛 ∶ 𝕋 2 → ℝ P 2 # ℝ P 2 to be the composition 𝑞 2 𝑛 = 𝑝 2 ∘ 𝑝 𝑛 . Then 𝑞 2 𝑛 is a 2 𝑛 -fold covering. Define 𝑓 𝑛 = 𝑞 2 𝑛 ∘ 𝑓 ∶ 𝐾 → ℝ P 2 # ℝ P 2 . Now proceed with the arguments of Example 2.6.

Observation. It is obvious that if 𝑚 and 𝑛 are different positive integers, then the maps 𝑓 𝑚 , 𝑓 𝑛 and 𝑔 𝑚 , 𝑔 𝑛 satisfying the previous theorem are such that 𝑓 𝑚 is not homotopic to 𝑓 𝑛 and 𝑔 𝑚 is not homotopic to 𝑔 𝑛 .

3. Roots of Liftings through Coverings

In the previous section, we saw several examples of maps from 2-dimensional CW complexes into closed surfaces having lifting through some covering space and not having all Nielsen root classes with minimal cardinality. In this section, we study the relationship between the minimal number of roots of a map and the minimal number of roots of one of its liftings through a covering space, when such lifting exists.

Throughout this section, 𝑀 and 𝑁 are topological 𝑛 -manifolds either compact or triangulable, and 𝑋 denotes a compact, connected, locally path connected, and semilocally simply connected spaces All these assumptions are true, for example, if 𝑋 is a finite and connected CW complex.

Lemma 3.1. Let 𝑝 𝑘 ∶ 𝑌 → 𝑌 be a 𝑘 -fold covering, and let 𝑓 ∶ 𝑋 → 𝑌 be a map having a lifting 𝑓 ∶ 𝑋 → 𝑌 through 𝑝 𝑘 . Let 𝑎 ∈ 𝑌 be a point, and let 𝑝 𝑘 − 1 ( 𝑎 ) = { 𝑎 1 , … , 𝑎 𝑘 } be the fiber of 𝑝 𝑘 over 𝑎 . Then ∑ 𝜇 ( 𝑓 , 𝑎 ) ≥ 𝑘 𝑖 = 1 𝜇 ( 𝑓 , 𝑎 𝑖 ) .

Proof. Let 𝜑 ∶ 𝑋 → 𝑌 be a map homotopic to 𝑓 such that # 𝜑 − 1 ( 𝑎 ) = 𝜇 ( 𝑓 , 𝑎 ) . Then, since 𝑝 𝑘 is a covering, we may lift 𝜑 through 𝑝 𝑘 to a map 𝜑 ∶ 𝑋 → 𝑌 homotopic to 𝑓 . It follows that 𝜑 − 1 ( 𝑎 ) = ∪ 𝑘 𝑖 = 1 𝜑 − 1 ( 𝑎 𝑖 ) , with this union being disjoint, and certainly # 𝜑 − 1 ( 𝑎 𝑖 ) ≥ 𝜇 ( 𝑓 , 𝑎 𝑖 ) for all 1 ≤ 𝑖 ≤ 𝑘 . Therefore, 𝜇 ( 𝑓 , 𝑎 ) = # 𝑘 𝑖 = 1 𝜑 − 1 𝑎 𝑖 ≥ 𝑘 𝑖 = 1 𝜇 𝑓 , 𝑎 𝑖 . ( 3 . 1 )

Theorem 3.2. Let 𝑝 𝑘 ∶ 𝑀 → 𝑁 be a 𝑘 -fold covering, and let 𝑓 ∶ 𝑋 → 𝑁 be a map having a lifting 𝑓 ∶ 𝑋 → 𝑀 through 𝑝 𝑘 . Then 𝜇 ( 𝑓 ) ≥ 𝑘 𝜇 ( 𝑓 ) . Moreover, 𝜇 ( 𝑓 ) = 0 if and only if 𝜇 ( 𝑓 ) = 0 .

Proof. Let 𝑎 ∈ 𝑁 be an arbitrary point, and let 𝑝 𝑘 − 1 ( 𝑎 ) = { 𝑎 1 , … , 𝑎 𝑘 } be the fiber of 𝑝 𝑘 over 𝑎 . Since 𝑀 and 𝑁 are manifolds, we have 𝜇 ( 𝑓 ) = 𝜇 ( 𝑓 , 𝑎 ) and 𝜇 ( 𝑓 ) = 𝜇 ( 𝑓 , 𝑎 𝑖 ) for all 1 ≤ 𝑖 ≤ 𝑘 . Hence, by the previous lemma, 𝜇 ( 𝑓 ) ≥ 𝑘 𝜇 ( 𝑓 ) . It follows that 𝜇 ( 𝑓 ) = 0 if 𝜇 ( 𝑓 ) = 0 . On the other hand, suppose that 𝜇 ( 𝑓 ) = 0 . Then 𝑁 ( 𝑓 ) = 0 and by [8, Theorem  2.3], there is a map ̃ 𝑔 ∶ 𝑋 → 𝑀 homotopic to 𝑓 such that d i m ̃ 𝑔 ( 𝑋 ) ≤ 𝑛 − 1 , (where 𝑛 is the dimension of 𝑀 and