Abstract
Given a continuous map 
from a 2-dimensional CW complex into a closed surface, the Nielsen root number 
and the minimal number of roots 
of 
satisfy 
. But, there is a number 
associated to each Nielsen root class of 
and an important problem is to know when 
. In addition to investigate this problem, we determine a relationship between 
and 
, when 
is a lifting of 
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.
be a continuous map between Hausdorff, normal, connected, locally path connected, and semilocally simply connected spaces, and let 
be a given base point. A root of 
at 
is a point 
such that 
. In root theory we are interested in finding a lower bound for the number of roots of 
at 
. We define the minimal number of roots of 
at 
to be the number 
of 
is a manifold, it is easy to prove that this number is independent of the selected point 
, and, from [
is a finite number, providing that 
is a finite CW complex. So, in this case, there is no ambiguity in defining the minimal number of roots of 
: 
is a map homotopic to 
and 
is a point such that 
, we say that the pair 
provides 
or that 
is a pair providing 
.
, 
of 
at 
are said to be Nielsen root
equivalent if there is a path 
starting at 
and ending at 
such that the loop 
in 
at 
is fixed-end-point homotopic to the constant path at 
. This relation is easily seen to be an equivalence relation; the equivalence classes are called Nielsen root classes of
at
. Also a homotopy 
between two maps 
and 
provides a correspondence between the Nielsen root classes of 
at 
and the Nielsen root classes of 
at 
. We say that such two classes under this correspondence are 
-related. Following Brooks [
of a map 
at 
is essential if given any homotopy 
starting at 
, and the class 
is 
-related to a root class of 
at 
. The number of essential root classes of 
at 
is the Nielsen root number of
at 
; it is denoted by 
.
is a homotopy invariant, and it is independent of the selected point 
, provid that 
is a manifold. In this case, there is no danger of ambiguity in denot it by 
.
be a Nielsen root class of 
. We define 
to be the minimal cardinality among all Nielsen root classes 
, of a map 
, 
-related to 
, for 
being a homotopy starting at 
and ending at 
:
is a manifold, then the number 
is independent of the Nielsen root class of 
. Then, in this case, there is no danger of ambiguity in defining the minimal cardinality of Nielsen root classes of
to some map 
with the property that all its Nielsen root classes have minimal cardinality. When the range 
of 
is a manifold, this question can be summarized in the following: when 
?
-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.
be a 
-fold covering. Suppose that 
is a map having a lifting 
through 
. What is the relationship between the numbers 
and 
? We answer completely this question for the cases in which 
is a connected, locally path connected and semilocally simply connected space, and 
and 
are manifolds either compact or triangulable. We show that 
, and we find necessary and sufficient conditions to have the identity.
is a map instead of 
is a continuous map.
from a 2-dimensional CW complex into a closed surface, under what conditions we have 
? In fact, we make a survey on the main results demonstrated by Aniz [
be a map from an 
-dimensional CW complex into a closed 
-manifold, with 
. If there is a map 
homotopic to 
such that one of its Nielsen root classes 
has exactly 
roots, each one of them belonging to the interior of 
-cells of 
, then 
.
from the bitorus into the torus with 
and 
.
, there is an 
-dimensional CW complex 
and a map 
with 
, 
and 
.
, there are maps 
from 
-dimensional CW complexes into closed 
-manifolds with 
. 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 
(the projective plane), 
(the torus), and 
(the Klein bottle). When the range is the sphere 
, it is obvious that every map 
satisfies 
, since in this case there is a unique Nielsen root class.
be a map between connected, locally path connected, and semilocally simply connected spaces. Then 
induces a homomorphism 
between fundamental groups. Since the image 
of 
by 
is a subgroup of 
, there is a covering space 
such that 
. Thus, 
has a lifting 
through 
. The map 
is called a Hopf lift of 
, and 
is called a Hopf covering for 
.
, for 
, that are nonempty, are exactly the Nielsen root class of 
at 
and a class 
is essential if and only if 
is nonempty for every map 
homotopic to 
.
copies of the sphere 
, and let 
be the map which restricted to each 
is the natural double covering map. If 
is at least 2, then 
, 
, and 
.
, for which we also have 
. Its construction is based in [
be the canonical double covering. We will construct a 2-dimensional CW complex 
and a map 
having a lifting 
through 
and satisfying:
. Let 
, 
, and 
be three copies of the 2-sphere regarded as the boundary of the standard 3-simplex 
:
be the 2-dimensional (simplicial) complex obtained from the disjoint union 
by identifying 
and 
. Thus, each 
, 
, is imbedded into 
so that 
is a single point 
. The (simplicial) 2-dimensional complex 
is illustrated in Figure 
and 
are homeomorphic if there is a bijection 
between the set of the vertices of 
and of 
such that 
is a simplex of 
if and only if 
is a simplex of 
(see [
and 
such that 
and 
.
be any homeomorphism from 
onto 
. Define 
and note that 
for 
. Now, define 
and note that 
for 
. In particular, 
. Thus, 
, 
, and 
can be used to define a map 
such that 
for 
.
be the composition 
, where 
is the canonical double covering. Note that 
. Thus, we can use Proposition 2.3 to study the Nielsen root classes of 
through the lifting 
.
, and let 
be the fiber of 
over 
.
is surjective, with 
and 
. Hence, every map from 
into 
homotopic to 
is surjective. It follows that, for every map 
homotopic to 
, we have 
and 
. By Proposition 2.3, 
and 
are the Nielsen root classes of 
, and both are essential classes. Therefore, 
.
, either 
or 
. Without loss of generality, suppose that 
. Then, by the definition of 
, we have 
. 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 
.
, note that since each restriction 
is a homeomorphism and 
is a double covering, for each map 
homotopic to 
, the equation 
must have at least two roots in each 
, 
. By the decomposition of 
this implies that 
.
, with the pair 
providing 
.
is the torus 
. Here, the complex 
of the domain of 
is a little bit more complicated.
be a double covering. We will construct a 2-dimensional CW complex 
and a map 
having a lifting 
through 
and satisfying the following:
. Consider three copies 
, 
, and 
of the torus with minimal celular decomposition. Let 
(resp., 
) be the longitudinal (resp., meridional) closed 1-cell of the torus 
, 
. Let 
be the 2-dimensional CW complex obtained from the disjoint union 
by identifying 
is obtained by attaching the tori 
and 
through the longitudinal closed 1-cell and, next, by attaching the longitudinal closed 1-cell of the torus 
into the meridional closed 1-cell of the torus 
.
is imbedded into 
so that 
is the (unique) 0-cell of 
, corresponding to 0-cells of 
, 
, and 
through the identifications. The 2-dimensional CW complex 
is illustrate, in Figure 
to denote the image of the original torus 
into the 2-complexo 
through the identifications above.
and 
with 
and 
such that 
carries 
onto 
, and 
carries 
onto 
. Thus, given a point 
we have 
. We should use this fact later.
be an arbitrary homeomorphism carrying longitude into longitude and meridian into meridian. Define 
and note that 
for 
. Now, define 
and note that 
for 
. In particular, 
. Thus, 
, 
, and 
can be used to define a map 
such that 
for 
.
be an arbitrary double covering. (We can consider, e.g., the longitudinal double covering 
for each 
.)
to be the composition 
.
using the information about 
, we need to prove that 
. Now, since 
, it is sufficient to prove that 
is an epimorphism. This is what we will do. Consider the composition 
, where 
is the obvious inclusion. This composition is exactly the homeomorphism 
, and therefore the induced homomorphism 
is an isomorphism. It follows that 
is an epimorphism. Therefore, we can use Proposition 2.3.
, and let 
be the fiber of 
over 
. (If 
is the longitudinal double covering, as above, then if 
, we have 
.)
is surjective, with 
and 
. Hence, every map from 
into 
homotopic to 
is surjective. It follows that, for every map 
homotopic to 
, we have 
and 
. By Proposition 2.3, 
and 
are Nielsen root classes of 
, and both are essential classes. Therefore, 
.
, either 
or 
. Without loss of generality, suppose that 
. Then, by the definition of 
, we have 
. 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, 
.
, note that since each restriction 
is a homeomorphism and 
is a double covering, for each map 
homotopic to 
, the equation 
must have at least two roots in each 
, 
. By the decomposition of 
, this implies that 
. Now, let 
be a point in 
, 
. As we have seen, 
. Write 
. By the definition of 
, we have 
. Denote 
and 
.
be a point, and let 
be the fiber of 
over 
. Since 
is a surface, there is a homeomorphism 
homotopic to the identity map such that 
and 
. Let 
be the composition 
, and let 
be the composition 
. Then, 
is homotopic to 
and 
. Since 
, this implies that 
.
, with the pair 
providing 
.
providing 
(which is equal to 3), we have necessarily 
with either 
and 
or 
and 
.
of Example 2.5, we can construct a similar example with the range of 
being the Klein bottle. The arguments here are similar to the previous example, and so we omit details.
be the orientable double covering. We will construct a 2-dimensional CW complex 
and a map 
having a lifting 
through 
and satisfying the following:
,
by the orientable double covering 
. Also here, we have 
, with the pair 
providing 
.
be the 2-dimensional CW complex of the previous two examples. For each positive integer 
, there are cellular maps 
and 
satisfying the following:
, 
and 
.
, 
and 
.
be as in Example 2.5. Let 
be an 
-fold covering (which certainly exists; e.g., for each 
considered as a pair 
, we can define 
). Define 
. Then, the same arguments of Example 2.5 can be repeated to prove the desired result.
be as in Example 2.6. Let 
be an 
-fold covering (e.g., as in the first item), and let 
be the orientable double covering. Define 
to be the composition 
. Then 
is a 
-fold covering. Define 
. Now proceed with the arguments of Example 2.6.
and 
are different positive integers, then the maps 
and 
satisfying the previous theorem are such that 
is not homotopic to 
and 
is not homotopic to 
.
and 
are topological 
-manifolds either compact or triangulable, and 
denotes a compact, connected, locally path connected, and semilocally simply connected spaces All these assumptions are true, for example, if 
is a finite and connected CW complex.
be a 
-fold covering, and let 
be a map having a lifting 
through 
. Let 
be a point, and let 
be the fiber of 
over 
. Then 
.
be a map homotopic to 
such that 
. Then, since 
is a covering, we may lift 
through 
to a map 
homotopic to 
. It follows that 
, with this union being disjoint, and certainly 
for all 
. Therefore, 
be a 
-fold covering, and let 
be a map having a lifting 
through 
. Then 
. Moreover, 
if and only if 
.
be an arbitrary point, and let 
be the fiber of 
over 
. Since 
and 
are manifolds, we have 
and 
for all 
. Hence, by the previous lemma, 
. It follows that 
if 
. On the other hand, suppose that 
. Then 
and by [
homotopic to 
such that 
, (where 
is the dimension of 
and 
). Let 
be the composition 
. Then 
is homotopic to 
and 
. Therefore 
.
, then the covering 
is a homeomorphism and 
.
from 2-dimensional CW complexes into closed surfaces (here 
is the projective plane, the torus, and the Klein bottle, resp.) for which we have 
from 2-dimensional CW complexes into closed surfaces having liftings 
through a double covering 
and satisfying the strict inequality 
such that, for each integer 
, there is a map 
and a map 
having liftings 
through an 
-fold covering 
and 
through a 
-fold covering 
, respectively, satisfying the relations 
and 
.
to be true. We show this after the following lemma.
be a 
-fold covering, let 
be different points of 
, and let 
be a point. Then, there is a 
-fold covering 
isomorphic and homotopic to 
such that 
.
be the fiber of 
over 
. It can occur that some 
is equal to some 
. In this case, up to reordering, we can assume that 
for 
and 
for 
, for some 
. If 
for any 
, 
, then we put 
. If 
, then there is nothing to prove. Then, we suppose that 
. For each 
, let 
be an open subset of 
homeomorphic to an open 
-ball, containing 
and 
and not containing any other point 
and 
. Let 
be a homeomorphism homotopic to the identity map, being the identity map outside 
and such that 
. Let 
be the homeomorphism 
. Then 
is homotopic to the identity map and 
for each 
. Let 
be the composition 
. Then 
is a 
-fold covering isomorphic and homotopic to 
. Moreover, 
.
be a 
-fold covering, and let 
be a map having a lifting 
through 
. Then 
if and only if, for each pair 
providing 
, each pair 
provides 
, where 
is a lifting of 
homotopic to 
and 
.
be a pair providing 
, let 
be the fiber of 
over 
, and let 
be a lifting of 
homotopic to 
. Then 
, with this union being disjoint. Hence 
. Now, 
for each 
. Therefore, 
if and only if 
for each 
, that is, each pair 
provides 
.
be a 
-fold covering, and let 
be a map having a lifting 
through 
. Then 
if and only if, given 
different points of 
, say 
, there is a map 
such that, for each 
: the pair 
provides 
.
be a pair providing 
, and let 
be a covering isomorphic and homotopic to 
, such that 
, as in Lemma 3.3.
. Let 
be a lifting of 
through 
homotopic to 
. Then, by the previous theorem, 
provides 
for each 
.
such that, for each 
, the pair 
provides 
. Let 
be the composition 
. Then 
is a lifting of 
through 
homotopic to 
and 
But, by Theorem 3.2, we have 
. Therefore 
.
be a 
-fold covering, and let 
be a map having a lifting 
through 
. Then 
if and only if, for every map 
homotopic to 
, there are at most 
points in 
whose preimage by 
has exactly 
points.
if and only if 
. Thus, a trivial argument shows that this theorem is equivalent to Theorem 3.5.
, 
and 
be the maps of Examples 2.4, 2.5, or 2.6. Then, we have proved that 
. (More precisely, in Examples 2.5 and 2.6 we have 
.) Therefore, by Theorem 3.6, if 
is a map providing 
(which is equal to 1), then there is a unique point of 
whose preimage by 
is a single point.
and its liftings 
through a covering.
be a 
-fold covering, and let 
be a map having a lifting 
through 
. Then, the following statements are equivalent:
(iv). Also, since 
for every map 
, it follows that (iii)
(i) and (iv)
(ii). On the other hand, by [
(iii) and (ii)
(iv). This completes the proof.
has a lifting through a finite fold covering. When 
has a lifting through an infinite fold covering, the problem is easily solved using the results of Gonçalves and Wong presented in [
be a map having a lifting 
through an infinite fold covering 
. Then the numbers 
, 
, 
and 
are all zero.
has infinite index in the group 
. Thus, by [
and so 
. Now, it is easy to check that also 
and so 
.
is always a compact, connected, locally path connected and semilocally simply connected space and 
and 
are topological 
-manifolds either compact or triangulable.
be a map with 
having the same properties of 
. We denote the Riedemeister number of 
by 
, which is defined to be the index of the subgroup 
in the group 
. In symbols, 
. When 
is a topological manifold (not necessarily compact), it follows from [
. Thus, if 
, then 
.
be a map with 
, let 
be a 
-fold covering and let 
be a lifting of 
through 
. Then the following statements are equivalent:
(iii)
(iv) are proved in Proposition 3.8. The implication (ii)
(i) is trivial. For a proof that (i) implies (ii) see [
be a 
-fold covering, and let 
be a map having a lifting 
. If 
, then 
.
, then all 
, 
, and 
also are zero. In this case, there is nothing to prove. Now, suppose that 
. Then, by Corollary 4.1, 
and 
and 
are both nonzero. Thus, also 
. Let 
be a Nielsen root class of 
, and let 
be a homotopy starting at 
and ending at 
. Moreover, let 
be the Nielsen root class of 
that is 
-related with 
. Let 
be a lifting of 
through 
homotopic to 
. By Proposition 2.3, 
for some point 
over a specific point 
of 
. Thus, the cardinality 
is minimal if and only if the cardinality 
is minimal; that is, 
if and only if 
.
be a 
-fold covering, and let 
be a map having a lifting 
through 
. If 
, then the following statements are equivalent:
. Thus, if one of these numbers are zero, then the three statements are automatically equivalent. Now, if 
, then 
and, by Theorem 4.2,
. This proves the desired equivalences.
to denote finite and connected 2-dimensional CW complexes, and we use 
to denote closed surfaces.
with cellular decomposition 
and the 2-sphere in the range of 
with cellular decomposition 
.
be a map with degree 
, and let 
be a point, 
. Then, there is a cellular map 
such that 
and 
.
is the north pole and so 
is the south pole.
such that 
and 
. In fact, consider the domain sphere 
fragmented in 
southern tracks by meridians 
chosen so that 
is in 
. Let 
be a map defined so that each meridian 
, for 
, is carried homeomorphically onto a same distinguished meridian 
of the range 2-sphere containing 
, and each of the 
tracks covers once the sphere 
, always in the same direction, which is chosen according to the orientation of 
, so that 
is a map of degree 
.
and 
have the same degree, they are homotopic. Moreover, 
and 
, where 
is the north pole of the domain 2-sphere, and so 
is its south pole. Therefore, we have 
. What we cannot guarantee immediately is that the homotopy between 
and 
is a homotopy relative to 
.
is a homotopy starting at 
and ending at 
, then as in [
in a small closed neighborhood 
of 
, with 
homeomorphic to a closed 2-disc and not containing 
and 
, to obtain a new homotopy 
, which is relative to 
. Let 
be the end of this new homotopy, that is, 
. Since 
and 
differ only on 
and 
and 
do not belong to 
, we have 
and 
.
be a map with zero degree and let 
be the constant map at 
. Then 
. Moreover, if 
, 
, then 
.
be a map. We say that 
is of type 
if there is a pair 
providing 
such that 
. Moreover, we say that 
is of type 
if in addition we can choose the map 
being a cellular map.
of type 
is also of the type 
.
be a map and let 
be a point such that 
provides 
and 
. We can assume that 
is in the interior of the unique 2-cell of 
. (We consider 
with a minimal cellular decomposition.) Let 
be an open neighborhood of 
in 
homeomorphic to an open 
-disc and such that the closure 
of 
in 
is contained in 
, where 
is the 1-skeleton of 
. Let 
be the attaching map of the 2-cell of 
, and let 
be a homeomorphism, where 
is the unitary closed 
-disc.
such that for each 
we have 
. Then, the maps 
and 
can be used to define a map 
such that 
and 
. Now, it is easy to see that 
is cellular and homotopic to the identity map 
.
be the composition 
and call 
. Then, 
is a cellular map homotopic to 
and 
. This concludes the proof.
and so of type 
.
be a map between closed surfaces. Suppose that 
, and let 
be a pair providing 
. Let 
. If each 
is in the interior of the 2-cell of 
, then there is nothing to prove. Otherwise, let 
be 
different points of 
belonging to its 2-cell. There is a homeomorphism 
homotopic to the identity map 
such that 
for each 
. Let 
be the composition 
. Then 
is homotopic to 
and 
. Now, we use the previous proposition to complete the proof.
be a map having a lifting 
through the double covering 
. If 
is of type 
, then 
.
is of type 
, then 
is also of type 
, by Proposition 5.4. Let 
be a cellular map, and let 
be a point different from 
such that 
and 
. Let 
be the 2-cells of 
. For each 
, we define the quotient map 
to a point 
. The image 
is naturally homeomorphic to a 2-sphere 
which inherits from 
a cellular decomposition 
, where the interior of the 2-cell 
corresponds homeomorphically to the image by 
of the interior of the 2-cell 
of the 2-complex 
.
is a cellular map, the 1-skeleton 
of 
is carried by 
into the 0-cell 
of 
. Moreover, 
is carried by 
(which is also a cellular map) into the 0-cell 
of the sphere 
, for all 
. Then we can define, for each 
, a unique cellular map 
such that 
In fact, for each 
, we define 
. Since 
is a cellular map, 
is well defined and is also a cellular map. Moreover, for each 
, we have 
.
, the set 
is in one-to-one correspondence with the set 
; in fact, we have 
. Now, by the proof of Theorem 4.1 of [
, either 
or 
is homotopic to a constant map. Then, by Lemmas 5.1 and 5.2, for each 
, there is a cellular map 
such that 
and 
. Let 
be such homotopies, 
.
, choose once and for all an index 
such that 
. Then, define 
by 
. This map is clearly well defined and cellular. Moreover, the homotopies 
, 
, can be used to define a homotopy 
starting at 
and ending at 
.
. By Theorem 3.5, we have that 
. Now, it is obvious that 
. So, by Theorem 4.3, 
.
is not of the type 
. We present an example to illustrate this fact.
be the bouquet of two 2 spheres with minimal cellular decomposition with one 0-cell 
and two 2-cells 
and 
. Let 
be a map which, restricted to each 
, 
, is homotopic to the identity map. Consider the sphere 
with its minimal cellular decomposition 
. Then, there is a cellular map 
homotopic to 
such that 
. Thus, the pair 
provides 
(
, of course). Now, it is obvious that ,for every map 
homotopic to 
, the restrictions 
, 
, are surjective. Hence, for every such map 
, the equation 
has at least one root in each 
, 
, whatever the point 
. Therefore, if 
is a root of 
belonging to the interior of one of the 2 cells of 
, then the equation 
must have a second root, which must belong to the closure of the other 2 cell of 
. But in this case, 
, and so the pair 
do not provide 
. This means that the map 
is not of type 
. Moreover, this shows that if 
is a pair providing 
, then necessarily 
. Thus, for every map 
homotopic to 
, there is at most one point in 
whose preimage by 
is a set with 
points. Now, let 
be a double covering, and let 
be the composition 
. Then 
is a lifting of 
through 
, and, by Theorem 3.6, we have 
. More precisely, 
. Moreover, 
, 
and 
.
denotes the absolute degree of the given map 
(see [
be a map inducing the trivial homomorphism on fundamental groups. Then, 
if 
and 
if 
.
is trivial, 
has a lifting 
through the (universal) double covering 
. By Proposition 5.5, 
is of type 
. Hence, by Theorem 5.6, we have 
. Now, it is well known that 
if 
and 
if 
. (see, e.g., [
. This concludes the proof.
is not the trivial homomorphism. To illustrate this, let 
be the identity map. It is obvious that this map induces the identity isomorphism on fundamental groups and 
.
is a compact, connected, locally path connected, and semilocally simply connected space.
be a map. Then 
if at least one of the following alternatives is true: < (i) 
; (ii) 
is a 2-dimensional CW complex, and 
is of type 
.
, namely, the identity covering 
and the double covering 
. Suppose that (i) is true. Then, 
, and 
is a covering corresponding to 
. Thus, either 
or 
. Now, if 
, then also 
by Proposition 3.8. If 
, then the result is obvious. Therefore, we have 
. If, on the other hand, (ii) is true and (i) is false, then we use Theorem 5.6.