††thanks: The first author was supported by CAPES - Brazil, the second author
was partially supported by FAPESP, Projeto Temático: Topologia Algébrica,
Geométrica e Diferencial, 2016/24707-4

# Nielsen-Borsuk-Ulam number for maps between tori

Givanildo Donizeti de Melo<EMAIL_ADDRESS>Daniel Vendrúscolo
Universidade Federal de São Carlos
São Carlos - SP
Brazil<EMAIL_ADDRESS>
(Date: January 1, 2004)

###### Abstract.

We compute the Nielsen-Borsuk-Ulam number for any selfmap of $n-$torus,
$\mathbb{T}^{n}$, as well as any free involution $\tau$ in $\mathbb{T}^{n}$,
with $n\leqslant 3$. Finally, we conclude that the tori, $\mathbb{T}^{1}$,
$\mathbb{T}^{2}$ and $\mathbb{T}^{3}$, are Wecken spaces in Nielsen-Borsuk-
Ulam theory. Such a number is a lower bound for the minimal number of pair of
points such that $f(x)=f(\tau(x))$ in a given homotopy class of maps.

###### Key words and phrases:

Borsuk-Ulam Theorem, Nielsen-Borsuk-Ulam number, Involutions, Torus

###### 1991 Mathematics Subject Classification:

Primary 55M20

##  1\. Introduction

In the literature one can find many different generalizations of the classical
Borsuk-Ulam Theorem for maps from the sphere $S^{n}$ in the Euclidean space
$\mathbb{R}^{n}$. For a good historical review, along with several other
aspects, we refer to [10].

One possible generalization can be the following: given two topological spaces
$X$ and $Y$ and a free involution $\tau$ on $X$ we can ask if the triple
$(X,\tau;Y)$ has the Borsuk-Ulam Property, i. e., if for any continuous map
$f:X\to Y$ there exists a point $x\in X$ such that $f(x)=f(\tau(x))$.

In this context, the classical Theorem just states that $(S^{n},antipodal\\\
map,\mathbb{R}^{n})$ has the Borsuk-Ulam Property. In [4] this approach was
used to study Borsuk-Ulam Property for surfaces with maps on $\mathbb{R}^{2}$,
and it indicated that the answer may depend on the involution, i. e., the same
surface can have this property in respect to an involution $\tau_{1}$, but not
for another involution $\tau_{2}$.

Accordingly, [5] discusses the triples $(X,\tau,S)$ with Borsuk-Ulam Property
where $X$ is a $CW$-complex, $\tau$ is a cellular involution and $S$ is a
connected closed surface, using braids group of surfaces.

More recently (in [6]) the Borsuk-Ulam Property was stated not for a triple
$(X,\tau,Y)$ but for each homotopy class of selfmaps of surfaces with Euler
characteristic zero. It must be noted that for maps on $\mathbb{R}^{n}$ there
is only one such class, however, this is not generally the case.

From this perspective, while investigating for which homotopy class of maps
$f:X\to Y$ it is true that for any $f^{\prime}$ in such class, there exists a
point $x\in X$ such that $f^{\prime}(x)=f^{\prime}(\tau(x))$, the studies [2,
3] have taken on a different approach. Using ideas from Nielsen fixed point
theory, Nilsen Borsuk-Ulam classes and a Nilsen Borsuk-Ulam number were
defined, for a homotopy class of maps between triangulated, orientable, closed
manifolds. Such invariant is a lower bound, in the homotopy class, for the
number of pairs of points satisfying $f(x)=f(\tau(x))$. In the present study
we compute the Nielsen-Borsuk-Ulam number for selfmaps of tori until dimension
$3$.

We will define the $n-$torus, denoted by $\mathbb{T}^{n}$, as the quotient
space ${\mathbb{R}^{n}}/{\mathbb{Z}^{n}}$ and we remember that
$\pi_{1}(\mathbb{T}^{n})=\mathbb{Z}^{n}$, and that $n-$torus is a $K(\pi,1)$
space. So the set of homotopy classes of selfmaps of the $n-$torus,
$[\mathbb{T}^{n},\mathbb{T}^{n}]$, is in bijection with the set of
homomorphisms, $\hom(\mathbb{Z}^{n},\mathbb{Z}^{n})$, and this set can be
regarded as the $n\times n$-integer matrices. Occasionally, there will be a
slight notation abuse between an equivalence class and some of their
representatives. We will always see tori as abelian groups, therefore, we will
use additive notation for the group operation.

Interestingly, the results presented here show that, for $n\leqslant 3$, the
triples $(\mathbb{T}^{n},\tau,\mathbb{T}^{n})$ do not have the Borsuk-Ulam
Property, for any involution $\tau$, and also, that tori are Wecken spaces in
Nielsen-Borsuk-Ulam theory. In each case, we present maps that realize
Nielsen-Borsuk-Ulam number.

In the proofs of Theorems 3.1, 4.3 and 5.1 a similar reasoning was adopted,
given a map $f$ that represents a homotopy class, we show that there is a
small perturbation of $f$ in this same class, usually called $f^{\prime}$,
such that $f^{\prime}$ realizes the Nielsen-Borsuk-Ulam number.

This study consists of 5 sections, this introduction included. Section $2$
gives a brief overview of Nielsen-Borsuk-Ulam theory (following [2, 3]),
Sections $3$, $4$ and $5$ deal with the computation of the Nielsen-Borsuk-Ulam
number for selfmaps of tori in dimension $1$, $2$ and $3$, respectively, and
Section $6$ presents some considerations for tori of greater dimension.

##  2\. Nielsen-Borsuk-Ulam theory

For practical reasons, we will reproduce in this section some definitions and
propositions from [2] and [3].

Denoting by $Coin(f,f\circ\tau)$ the coincidence set of the pair
$(f,f\circ\tau)$, [2, Theorem 2.1] shows, in the context of simplicial
complexes, that we can suppose $Coin(f,f\circ\tau)$ finite. Moreover [2,
Theorem 3.5] shows that if two homotopic maps, $f$ and $g$, are such that
$Coin(f,f\circ\tau)$ and $Coin(g,g\circ\tau)$ are both finite, then there
exists a homotopy between them with such set finite in each level.

######  Definition 2.1.

[3, 2.1] Let $(X,\tau;Y)$ be a triple where $X$ and $Y$ are finite
$n$-dimensional complexes, $\tau$ is a free simplicial involution on $X$ for
any map $f:X\to Y$ with
$Coin(f,f\circ\tau)=\\{x_{1},\tau(x_{1}),\ldots,x_{m},\tau(x_{m})\\}$, we
define the Borsuk-Ulam coincidence set for the pair $(f,\tau)$, as the set of
pairs:

$BUCoin(f;\tau)=\\{(x_{1},\tau(x_{1}));\ldots;(x_{m},\tau(x_{m}))\\}$

and we say that two pairs $(x_{i},\tau(x_{i})),(x_{j},\tau(x_{j}))$ are in the
same Borsuk-Ulam coincidence class if there exists a path $\gamma$ from a
point in $\\{x_{i},\tau(x_{i})\\}$ to a point in $\\{x_{j},\tau(x_{j})\\}$
such that $f\circ\gamma$ is homotopic to $f\circ\tau\circ\gamma$ with fixed
endpoints.

######  Definition 2.2.

[3, 2.2] A Borsuk-Ulam coincidence class $C$ is called single if for one (or
any) pair $(x,\tau(x))\in C$ there exists a path $\gamma$ from $x$ to
$\tau(x)$ such that $f\circ\gamma$ is homotopic to $f\circ\tau\circ\gamma$
with fixed endpoints.

######  Proposition 2.3.

[3, 2.4] A Borsuk-Ulam coincidence class $C$ is single if, and only if, it is
composed of just one usual coincidence class of the pair $(f,f\circ\tau)$.
Moreover, if $C$ is a finite Borsuk-Ulam coincidence class of the pair
$(f,\tau)$ which is not single (called double), then we can change the labels
of the elements of $C$ in a way that:

  * •

$C=\\{(x_{1},\tau(x_{1})),\dots,(x_{k},\tau(x_{k}))\\}$;

  * •

$C=C_{1}\cup C_{2}$ where $C_{1}$ and $C_{2}$ are usual coincidence classes of
the pair $(f,f\circ\tau)$;

  * •

$C_{1}=\\{x_{1},\dots,x_{k}\\}$ and
$C_{2}=\\{\tau(x_{1}),\dots,\tau(x_{k})\\}$.

At the begining of page 615 of [3] there is an observation about the relation
between the local indices $ind(f,f\circ\tau;c)$ and
$ind(f,f\circ\tau;\tau(c))$ in orientable manifolds. However, in this
observation the effect of dimension was neglected. In fact, we have:

$ind(f,f\circ\tau;c)=\left\\{\begin{array}[]{rcl}(-1)^{n}ind(f,f\circ\tau;\tau(c))&\mbox{if}&\tau\mbox{
preserves orientation,}\\\
(-1)^{n-1}ind(f,f\circ\tau;\tau(c))&\mbox{if}&\tau\mbox{ reverses
orientation.}\end{array}\right.$

where $ind(f,f\circ\tau;c)$ is the usual local index for coincidence and $n$
is the dimension of the manifold.

This change does not affect the results in [3], however, it will be necessary
to reorganize the definition of pseudo-index used there.

######  Definition 2.4.

[3, Compare with Definition 2.5] Let $X$ and $Y$ be closed orientable
triangulable $n$-manifolds, $\tau$ a free involution on $X$ and $f:X\to Y$ a
continuous map such that $BUCoin(f,\tau)$ is finite. If
$C=\\{(x_{1},\tau(x_{1})),\dots,\\\ (x_{k},\tau(x_{k}))\\}$ is a Borsuk-Ulam
coincidence class of the pair $(f,\tau)$, we define the pseudo-index of $C$,
denoted $|ind|(f,\tau;C)$, by:

$\left\\{\begin{array}[]{cl}\displaystyle\sum_{x_{i}\in
C}ind(f,f\circ\tau;x_{i})\ mod\ 2&\mbox{ (if }C\mbox{ is single) and }\\\
&\mbox{(}\tau\mbox{ reverses orientation and }n\mbox{ is even;}\\\ &\mbox{ or
}\\\ &\tau\mbox{ preserves orientation and }n\mbox{ is odd.)}\\\ &\\\
\dfrac{ind(f,f\circ\tau;C)}{2}&\mbox{ (if }C\mbox{ is single) and }\\\
&\mbox{(}\tau\mbox{ preserves orientation and }n\mbox{ is even;}\\\ &\mbox{ or
}\\\ &\tau\mbox{ reverses orientation and }n\mbox{ is odd.)}\\\ &\\\
|ind(f,f\circ\tau;C_{1})|&\mbox{(if }C\mbox{ is double, }C=C_{1}\cup
C_{2}\mbox{) and }\\\ &\mbox{(}\tau\mbox{ reverses orientation and }n\mbox{ is
even;}\\\ &\mbox{ or }\\\ &\tau\mbox{ preserves orientation and }n\mbox{ is
odd.)}\\\ &\\\ ind(f,f\circ\tau;C_{1})&\mbox{(if }C\mbox{ is double,
}C=C_{1}\cup C_{2}\mbox{) and }\\\ &\mbox{(}\tau\mbox{ preserves orientation
and }n\mbox{ is even;}\\\ &\mbox{ or }\\\ &\tau\mbox{ reverses orientation and
}n\mbox{ is odd.)}\\\ \end{array}\right.$

where $C_{1}$ and $C_{2}$ are disjoint usual coincidence classes of the pair
$(f,f\circ\tau)$.

We call a Borsuk-Ulam coincidence class $C$ essential if $|ind|(f,\tau;C)\neq
0$ and we define $NBU(f,\tau)$, the Nielsen-Borsuk-Ulam number of the pair
$(f,\tau)$, as the number of essential Borsuk-Ulam coincidences classes. The
definitions above are exactly what we need in order to prove:

######  Proposition 2.5.

[3, 2.7] If $f^{\prime}$ is a map homotopic to $f$ then $f^{\prime}$ has at
least $NBU(f,\tau)$ pairs of Borsuk-Ulam coincidence points.

##  3\. Nielsen-Borsuk-Ulam number in $\mathbb{T}^{1}$

Our objective here is to use the classification of free involutions and
compute, using good representatives, the Nielsen-Borsuk-Ulam number for each
homotopy class of maps.

We know that the $1$-torus, $\mathbb{T}^{1}$, is homeomorphic to $S^{1}$, with
the only free involution in $S^{1}$ being the antipodal:

$\tau(x)=x+\frac{1}{2}.$

Let $f:S^{1}\to S^{1}$ be a map, up to homotopy we can consider $f(x)=ax$ with
$a\in\mathbb{Z}$. If $a$ is an odd number, we have
$\operatorname{Coin}(f,f\circ\tau)=\emptyset$, implying that the Nielsen-
Borsuk-Ulam number is zero. Now, if $a$ is an even number then
$\operatorname{Coin}(f,f\circ\tau)=S^{1}$.

We will use a small perturbation of $f$, in order to do so, we define
$\epsilon:\mathbb{T}^{1}\to\mathbb{T}^{1}$ given by
$\epsilon(t)=\frac{1}{3}\operatorname{sen}(2\pi t)$ and
$f^{\prime}(x)=ax+\epsilon(x)$. Then
$\operatorname{Coin}(f^{\prime},f^{\prime}\circ\tau)=\\{0,\frac{1}{2}\\}$ and
$\operatorname{BUCoin}(f^{\prime},\tau)=\\{(0,\frac{1}{2})\\}$, both
coincidence points in the same coincidence class, so we have a unique single
Borsuk-Ulam coincidence class with pseudo-index
$|\operatorname{ind}|(f^{\prime},f^{\prime}\circ\tau;C)=1$. Therefore,
$\operatorname{NBU}(f,\tau)=1$. By doing this, we have just proved the
following result.

######  Theorem 3.1.

Let $f:S^{1}\to S^{1}$ be a map, with $f_{\\#}=\begin{pmatrix}a\end{pmatrix}$
and $a\in\mathbb{Z}$. Then

$NBU(f,\tau)=\left\\{\begin{array}[]{cl}1&\text{if a is even};\\\ 0&\text{if a
is odd},\end{array}\right.$

where $f_{\\#}$ is the homomorphism induced from $f$ in the fundamental group.
Moreover, for each map $f:S^{1}\to S^{1}$ there exists $f^{\prime}$ homotopic
to $f$ that realizes the $\operatorname{NBU}(f,\tau)$, i.e., $S^{1}$ is a
Wecken type space in Nielsen-Borsuk-Ulam theory.

##  4\. Nielsen-Borsuk-Ulam number in $\mathbb{T}^{2}$

We must remember that two free involutions $\tau_{1}$, $\tau_{2}$ in $M$
(closed, connected and orientable $n-$manifolds) are equivalent if there
exists a homeomorphism $h:M\to M$ such that $h\circ\tau_{2}=\tau_{1}\circ h$,
i.e., the diagram

$\textstyle{M\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tau_{2}}$$\scriptstyle{h}$$\textstyle{M\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{h}$$\textstyle{M\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tau_{1}}$$\textstyle{M}$

is commutative.

In [5, Proposition 30, Proposition 32] we can find a classification for free
involutions on $\mathbb{T}^{2}$, i.e, the authors proved that there are two
free involutions in the torus $\mathbb{T}^{2}$, up to equivalence. Consider
the following free involutions:

$\displaystyle\tau_{1}:\mathbb{T}^{2}$
$\displaystyle\to\mathbb{T}^{2}\qquad\qquad\qquad\qquad\tau_{2}:\mathbb{T}^{2}\to\mathbb{T}^{2}$
$\displaystyle(x,y)$
$\displaystyle\mapsto\left(x+\frac{1}{2},y\right),\qquad\qquad(x,y)\mapsto\left(-x,y+\frac{1}{2}\right).$

The free involutions $\tau_{1}$ and $\tau_{2}$ are not equivalent, because the
space of orbits
$\mathbb{T}_{\tau_{1}}^{2}:=\frac{\mathbb{T}^{2}}{x\sim\tau_{1}(x)}$ is
homeomorphic to $\mathbb{T}^{2}$ and the space of orbits
$\mathbb{T}_{\tau_{2}}^{2}:=\frac{\mathbb{T}^{2}}{x\sim\tau_{2}(x)}$ is
homeomorphic to $\mathbb{K}^{2}$ (the Klein bottle). The free involution
$\tau_{1}$ preserves orientation, while the free involution $\tau_{2}$
reverses orientation.

The following theorems by [9, Theorem 2.2.3, Theorem 2.3.10] (see [6] for
similar results), indicate which homotopy class of
$[\mathbb{T}^{2},\mathbb{T}^{2}]$ has the Borsuk-Ulam property with respect to
$\tau_{1}$ and $\tau_{2}$ involutions.

######  Theorem 4.1.

Let $\tau_{1}$ be the free involution above. So every class of homotopy
$\beta\in[\mathbb{T}^{2},\mathbb{T}^{2}]$ does not have the Borsuk-Ulam
property with respect to $\tau_{1}$.

######  Theorem 4.2.

Let $\tau_{2}$ be the aforementioned involution and
$\beta\in[\mathbb{T}^{2},\mathbb{T}^{2}]$ a homotopy class. Then $\beta$ has
the Borsuk-Ulam property with respect to $\tau_{2}$ if, and only if,
homomorphism
$\beta_{\\#}:\mathbb{Z}\oplus\mathbb{Z}\to\mathbb{Z}\oplus\mathbb{Z}$ is given
by

$\beta_{\\#}=\begin{pmatrix}m_{1}&m_{2}\\\ n_{1}&n_{2}\end{pmatrix}$

where $(m_{1},n_{1})\neq(0,0)$, $m_{2}$ and $n_{2}$ are even.

Those results will help us with the following:

######  Theorem 4.3.

Let $f:\mathbb{T}^{2}\to\mathbb{T}^{2}$ be a map. Then,

$\operatorname{NBU}(f,\tau_{1})=0$

and

$\operatorname{NBU}(f,\tau_{2})=\left\\{\begin{array}[]{cl}2&\text{if
}f_{\\#}=\begin{pmatrix}p&2k\\\ q&2l\end{pmatrix};\\\
0&\text{otherwise},\end{array}\right.$

with $p,q,k,l\in\mathbb{Z}$, $(p,q)\neq(0,0)$ and $f_{\\#}$ being the
homomorphism induced from $f$ in the fundamental group. Moreover, for each map
$f:\mathbb{T}^{2}\to\mathbb{T}^{2}$ there exists $f^{\prime}$ homotopic to $f$
which realizes the $\operatorname{NBU}(f,\tau_{i})$, i.e., the torus
$\mathbb{T}^{2}$ is a Wecken type space in Nielsen-Borsuk-Ulam theory.

###### Proof.

The Theorem 4.1 shows that $NBU(f,\tau_{1})=0$ for any
$f:\mathbb{T}^{2}\to\mathbb{T}^{2}$ and Theorem 4.2 describes homotopy classes
such that $\operatorname{NBU}(f,\tau_{2})=0$, easily realizing the Nielsen-
Borsuk-Ulam numbers in these cases.

Now, we just need to compute $\operatorname{NBU}(f,\tau_{2})$ for maps such
that:

$f_{\\#}=\begin{pmatrix}p&2k\\\ q&2l\end{pmatrix},$

with $(p,q)\neq(0,0)$.

  1. (1)

Suppose that $p=0$ or $q=0$. Without loss of generality, assume $q=0$. Up to
homotopy we can assume that $f$ can be written as $f(x,y)=(px+2ky,2ly)$ with
$p,k,l\in\mathbb{Z}$ and $p\neq 0$.

The coincidence set of pair $(f,f\circ\tau_{2})$ is

$\operatorname{Coin}(f,f\circ\tau_{2})=\left\\{(x,y)\in\mathbb{T}^{2}\mid
x=\dfrac{i}{2p}\text{ with }i=0,1,\ldots,2p-1\right\\},$

because

$f(x,y)=(f\circ\tau_{2})(x,y)\Leftrightarrow\left\\{\begin{array}[]{l}px+2ky=-px+2ky+k\\\
2ly=2ly+l\end{array}\right.\Leftrightarrow\left\\{\begin{array}[]{l}2px=0\\\
0=0\end{array}.\right.$

Using coverings, it is easy to see that each connected component of the set
$\operatorname{Coin}(f,f\circ\tau_{2})$ is exactly one usual coincidence class
of the pair $(f,f\circ\tau_{2})$.

With a small perturbation
$f^{\prime}(x,y)=(px+2ky+\epsilon(y),2ly+\epsilon(x))$, in which
$\epsilon:\mathbb{T}^{1}\to\mathbb{T}^{1}$ is given by
$\epsilon(t)=\frac{1}{n_{0}}\operatorname{sen}(2\pi t)$, $n_{0}\in\mathbb{N}$
and $n_{0}\geqslant 3$, we have $f^{\prime}$ homotopic to $f$ and

$f^{\prime}(x,y)=(f^{\prime}\circ\tau_{2})(x,y)\Leftrightarrow\left\\{\begin{array}[]{l}2px+\epsilon(y)=\epsilon(y+\frac{1}{2})\\\
\epsilon(x)=\epsilon(-x)\end{array}\right.\Leftrightarrow\left\\{\begin{array}[]{l}y=0,\frac{1}{2}\\\
x=0,\frac{1}{2}\end{array}.\right.$

Therefore,

$\operatorname{Coin}(f^{\prime},f^{\prime}\circ\tau_{2})=\left\\{\left(0,0\right),\left(0,\frac{1}{2}\right),\left(\frac{1}{2},0\right),\left(\frac{1}{2},\frac{1}{2}\right)\right\\}$

and the set of Borsuk-Ulam coincidences of the pair $(f^{\prime},\tau_{2})$ is

$\operatorname{BUCoin}(f^{\prime},\tau_{2})=\left\\{\left(\left(0,0\right),\left(0,\frac{1}{2}\right)\right),\left(\left(\frac{1}{2},0\right),\left(\frac{1}{2},\frac{1}{2}\right)\right)\right\\}.$

So there are two Nielsen coincidence classes
$C_{1}=\left\\{\left(0,0\right),\left(0,\frac{1}{2}\right)\right\\}$ and
$C_{2}=\left\\{\left(\frac{1}{2},0\right),\left(\frac{1}{2},\frac{1}{2}\right)\right\\}$,
and two single Borsuk-Ulam coincidences classes
$D_{1}=\left\\{\left(\left(0,0\right),\left(0,\frac{1}{2}\right)\right)\right\\}$
and
$D_{2}=\left\\{\left(\left(\frac{1}{2},0\right),\left(\frac{1}{2},\frac{1}{2}\right)\right)\right\\}.$

To compute the local coincidence index we use a version that generalizes the
fixed point for smooth maps presented in [8, Definition (2.2.2)]. Thus, the
local coincidence indices are shown in figure 1.

$-1$$1$$1$$-1$ Figure 1. The Borsuk-Ulam coincidence classes for $f^{\prime}$
with the local coincidence index of each point.

The pseudo-index of the Borsuk-Ulam coincidence classes $D_{1}$ and $D_{2}$
are:

$|\operatorname{ind}|(f,\tau_{2};D_{1})=\operatorname{ind}(f,f\circ\tau_{2};(0,0))\mod
2=1\quad\text{and}$
$|\operatorname{ind}|(f,\tau_{2};D_{2})=\operatorname{ind}\left(f,f\circ\tau_{2};\left(\frac{1}{2},0\right)\right)\mod
2=1$

So $D_{1}$ and $D_{2}$ are essential Borsuk-Ulam coincidence classes.
Therefore,

$\operatorname{NBU}(f,\tau_{2})=2.$

  2. (2)

Suppose that $p\neq 0$ and $q\neq 0$. Then

$\displaystyle f(x,y)=(f\circ\tau_{2})(x,y)$
$\displaystyle\Leftrightarrow\left\\{\begin{array}[]{l}2px=0\\\
2qx=0\end{array}\right.$
$\displaystyle\Leftrightarrow\left\\{\begin{array}[]{lll}x=\frac{i}{2p}&\text{with}&i=0,1,\ldots,2p-1\\\
x=\frac{j}{2q}&\text{with}&j=0,1,\ldots,2q-1,\end{array}\right.$

i.e., the coincidence set of the pair $(f,f\circ\tau_{2})$ is the set of
points $(x,y)\in\mathbb{T}^{2}$ such that $x=\frac{i}{2p}=\frac{j}{2q}$. For
example, the set $\left(0,y\right)$ and $\left(\frac{1}{2},y\right)$ are
contained in the $\operatorname{Coin}(f,f\circ\tau_{2})$, so
$\operatorname{Coin}(f,f\circ\tau_{2})\neq\emptyset$.

Let $a=\gcd\\{p,q\\}$, then $p=a\alpha$ and $q=a\beta$ for some
$\alpha,\beta\in\mathbb{Z}$. We claim that

$\operatorname{Coin}(f,f\circ\tau_{2})=\left\\{(x,y)\in\mathbb{T}^{2}\mid
x=\frac{k}{2a}\quad\text{with}\quad k=0,1,\ldots,2a-1\right\\}.$

Indeed, given $\left(\frac{k}{2a},y\right)\in\mathbb{T}^{2}$, we have

$\frac{k}{2a}=\frac{k\beta}{2a\beta}=\frac{k\beta}{2q}\quad\text{and}\quad\frac{k}{2a}=\frac{k\alpha}{2a\alpha}=\frac{k\alpha}{2p}\quad\Rightarrow\quad\frac{k}{2a}=\frac{k\alpha}{2p}=\frac{k\beta}{2q},$

soon $\left(\frac{k}{2a},y\right)\in\operatorname{Coin}(f,f\circ\tau_{2})$.
Then,

$\left\\{(x,y)\in\mathbb{T}^{2}\mid x=\frac{k}{2a}\quad\text{with}\quad
k=0,1,\ldots,2a-1\right\\}\subset\operatorname{Coin}(f,f\circ\tau_{2}).$

Now, if $(x,y)\in\operatorname{Coin}(f,f\circ\tau_{2})$ then

$x=\frac{i}{2p}=\frac{j}{2q}\Leftrightarrow 2qi=2pj\Leftrightarrow 2a\beta
i=2a\alpha j\Leftrightarrow\beta i=\alpha j.$

Let us analyze the values of $i$ and $j$ which satisfy the above equality. For
$i=0$ consider $j=0$, for $i=\alpha$ consider $j=\beta$ and successively, for
$i=k\alpha$ consider $j=k\beta$, with $k=0,1,\ldots,2a-1$. Thereby,

$\operatorname{Coin}(f,f\circ\tau_{2})\subset\left\\{(x,y)\in\mathbb{T}^{2}\mid
x=\frac{k}{2a}\quad\text{with}\quad k=0,1,\ldots,2a-1\right\\},$

proving the claim.

If $(x,y)\in\operatorname{Coin}(f,f\circ\tau_{2})$, then

$f(x,y)=(f\circ\tau_{2})(x,y)\Leftrightarrow\left\\{\begin{array}[]{l}x=\frac{k}{2a}\\\
y=y\end{array}\right.$

Define $f^{\prime}:\mathbb{T}^{2}\to\mathbb{T}^{2}$ by
$f^{\prime}(x,y)=(px+2ky+\epsilon(y),2ly+\epsilon(x))$. Thereby, $f^{\prime}$
is homotopic to $f$ and

$\displaystyle f^{\prime}(x,y)=(f^{\prime}\circ\tau_{2})(x,y)$
$\displaystyle\Leftrightarrow\left\\{\begin{array}[]{l}x+\epsilon(y)=\frac{k}{2a}+\epsilon(y+\frac{1}{2})\\\
\epsilon(x)=\epsilon(-x)\end{array}\right.$ (3)
$\displaystyle\Leftrightarrow\left\\{\begin{array}[]{l}x+\epsilon(y)=\frac{k}{2a}+\epsilon(y+\frac{1}{2})\\\
x=0,\frac{1}{2}\end{array}.\right.$ (6)

Replacing $x=0$ in the first equation of system $\eqref{st2}$, we have
$\epsilon(y)=\frac{k}{2a}+\epsilon\left(y+\frac{1}{2}\right)$. If $k=0$, the
solutions in this equation are $y=0$ and $y=\frac{1}{2}$. Otherwise, for
$n_{0}$ big enough in the definition of the map $\epsilon$, we have
$\epsilon(y)\neq\frac{k}{2a}+\epsilon\left(y+\frac{1}{2}\right)$. Replacing
$x=\frac{1}{2}$ in the first equation of system $\eqref{st2}$, we have
$\frac{1}{2}+\epsilon(y)=\frac{k}{2a}+\epsilon\left(y+\frac{1}{2}\right)$. If
$k=a$, then $y=0$ and $y=\frac{1}{2}$ are the only solutions to this equation.
Otherwise, we have
$\frac{1}{2}+\epsilon(y)\neq\frac{k}{2a}+\epsilon\left(y+\frac{1}{2}\right)$
for $n_{0}$ big enough. Thus,

$\operatorname{Coin}(f^{\prime},f^{\prime}\circ\tau_{2})=\left\\{\left(0,0\right),\left(0,\frac{1}{2}\right),\left(\frac{1}{2},0\right),\left(\frac{1}{2},\frac{1}{2}\right)\right\\}$

and

$\operatorname{BUCoin}(f^{\prime},\tau_{2})=\left\\{\left(\left(0,0\right),\left(0,\frac{1}{2}\right)\right),\left(\left(\frac{1}{2},0\right),\left(\frac{1}{2},\frac{1}{2}\right)\right)\right\\}.$

Calculating the Nielsen-Borsuk-Ulam number in the same way as in the previous
case, we conclude that

$\operatorname{NBU}(f,\tau_{2})=2.$

Moreover, in both the cases $f^{\prime}$ realizes the
$\operatorname{NBU}(f,\tau_{2})=2$. ∎

##  5\. Nielsen-Borsuk-Ulam number in $\mathbb{T}^{3}$

In [7] and [1] we can find classifications for free involutions on
$\mathbb{T}^{3}$. Up to equivalence there are four such involutions. These are
described in [7] by:

$\displaystyle h_{1}(x,y,z)$ $\displaystyle=\left(x,y,z+\frac{1}{2}\right)$
$\displaystyle h_{2}(x,y,z)$ $\displaystyle=\left(-x,-y,z+\frac{1}{2}\right)$
$\displaystyle h_{3}(x,y,z)$ $\displaystyle=\left(x,-y,z+\frac{1}{2}\right)$
$\displaystyle h_{4}(x,y,z)$
$\displaystyle=\left(x+y,-y,z+\frac{1}{2}\right).$

######  Theorem 5.1.

Let $f:\mathbb{T}^{3}\to\mathbb{T}^{3}$ be a map such that
$f_{\\#}:\pi_{1}(\mathbb{T}^{3})\to\pi_{1}(\mathbb{T}^{3})$ is represented by
the matrix $f_{\\#}=\begin{pmatrix}a&b&c\\\ r&s&t\\\ u&v&w\end{pmatrix}.$
Then,

$\operatorname{NBU}(f,h_{1})=0,\qquad\operatorname{NBU}(f,h_{3})=0,\quad\operatorname{NBU}(f,h_{4})=0\quad\text{and}$
$\operatorname{NBU}(f,h_{2})=\left\\{\begin{array}[]{cl}4&\text{if
}c,t,w\text{ are even},(a,r,u)\neq(0,0,0),\\\ &(b,s,v)\neq(0,0,0)\text{ and
}(p,q)\neq(0,0)\\\ &\text{or}\\\ &\text{if }c,t,w\text{ are
even},(a,r,u)\neq(0,0,0),\\\ &(b,s,v)\neq(0,0,0),(p,q)=(0,0)\text{ and }u=0\\\
0&\text{otherwise},\end{array}\right.$

with $p=\det\begin{pmatrix}r&s\\\ u&v\end{pmatrix}$ and
$q=\det\begin{pmatrix}a&b\\\ u&v\end{pmatrix}$. Moreover, the torus
$\mathbb{T}^{3}$ is a Wecken type space in the Nielsen-Borsuk-Ulam theory.

######  Remark 5.2.

The fact that the torus $\mathbb{T}^{3}$ is a Wecken space in the Nielsen-
Borsuk-Ulam theory has already been demonstrated, see [3, Theorem 3.5].

###### Proof.

We can suppose (up to homotopy) that $f:\mathbb{T}^{3}\to\mathbb{T}^{3}$ is
given by

$f(x,y,z)=(ax+by+cz,rx+sy+tz,ux+vy+wz).$

The proof of the theorem will be presented in the following subsections:

###  5.1. $h_{1}(x,y,z)=\left(x,y,z+\frac{1}{2}\right)$,
$h_{3}(x,y,z)=\left(x,-y,z+\frac{1}{2}\right)$ and
$h_{4}(x,y,z)=\left(x+y,-y,z+\frac{1}{2}\right)$

The free involutions $h_{1}$, $h_{3}$ and $h_{4}$ can be written as

$h_{2i+ij+1}(x,y,z)=\left(x+ijy,(-1)^{i}y,z+\frac{1}{2}\right)\text{ where
}i,j\in\\{0,1\\}.$

The equation $f(x,y,z)=(f\circ h_{2i+ij+1})(x,y,z)$ corresponds to the
following system in the group $\mathbb{T}^{3}$:

$\left\\{\begin{array}[]{l}ax+by+cz=ax+aijy+b(-1)^{i}y+cz+\frac{c}{2}\\\
rx+sy+tz=rx+rijy+s(-1)^{i}y+tz+\frac{t}{2}\\\
ux+vy+wz=ux+uijy+v(-1)^{i}y+wz+\frac{w}{2}\end{array}\right.\Leftrightarrow$
$\left\\{\begin{array}[]{l}(b-aij-b(-1)^{i})y=\frac{c}{2}\\\
(s-rij-s(-1)^{i})y=\frac{t}{2}\\\
(v-uij-v(-1)^{i})y=\frac{w}{2}\end{array}\right.\Leftrightarrow\left\\{\begin{array}[]{l}A_{ij}y=\frac{c}{2}\\\
B_{ij}y=\frac{t}{2}\\\ C_{ij}y=\frac{w}{2}\end{array},\right.$

where $A_{ij}=b-aij-b(-1)^{i}$, $B_{ij}=s-rij-s(-1)^{i}$ and
$C_{ij}=v-uij-v(-1)^{i}$. This way, the $\operatorname{NBU}(f,h_{2i+ij+1})$
calculation will be divided into a number of cases:

  1. (1)

For $A_{ij}=B_{ij}=C_{ij}=0$ consider
$f^{\prime}:\mathbb{T}^{3}\to\mathbb{T}^{3}$ defined by

$f^{\prime}(x,y,z)=\left(ax+by+cz,rx+sy+tz+\epsilon(z),ux+vy+wz+\delta(z)\right),$

where $\epsilon,\delta:\mathbb{T}^{1}\to\mathbb{T}^{1}$ are given by
$\epsilon(z)=\frac{1}{n_{0}}\operatorname{sen}(2\pi z)$ and
$\delta(z)=\frac{1}{n_{0}}\operatorname{sen}(\pi z),$ with
$n_{0}\in\mathbb{N}$ conveniently chosen. Note that the map $f^{\prime}$ is
homotopic to $f$ and the coincidence set
$\operatorname{Coin}(f^{\prime},f^{\prime}\circ h_{2i+ij+1})$ is formed by the
points $(x,y,z)\in\mathbb{T}^{3}$ such that

$f^{\prime}(x,y,z)=(f^{\prime}\circ
h_{2i+ij+1})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}0=\frac{c}{2}\\\
\epsilon(z)=\frac{t}{2}+\epsilon(z+\frac{1}{2})\\\
\delta(z)=\frac{w}{2}+\delta(z+\frac{1}{2})\end{array}.\right.$ (7)

If the integer $c$ is odd then the system (7) has no solution, so we have
$\operatorname{Coin}(f^{\prime},f^{\prime}\circ h_{2i+ij+1})=\emptyset$. For
$c$ even, we will solve system (7) in two cases:

    1. (a)

$t$ odd or $w$ odd: Assume, without loss of generality, that $t$ is odd. Then,
the second equation of the system (7) is

$\epsilon(z)=\frac{1}{2}+\epsilon\left(z+\frac{1}{2}\right)\qquad\Leftrightarrow\qquad\frac{2}{n_{0}}\operatorname{sen}(2\pi
z)=\frac{1}{2}.$

Choosing $n_{0}\geqslant 5$, we have $\frac{2}{n_{0}}\operatorname{sen}(2\pi
z)<\frac{1}{2}$. Thereby,
$\epsilon(z)\neq\frac{1}{2}+\epsilon\left(z+\frac{1}{2}\right)$ for all $z$,
concluding that $\operatorname{Coin}(f^{\prime},f^{\prime}\circ
h_{2i+ij+1})=\emptyset$ in this case.

    2. (b)

$t$ even and $w$ even: Then,

$f^{\prime}(x,y,z)=(f^{\prime}\circ
h_{2i+ij+1})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}0=0\\\
\epsilon(z)=\epsilon(z+\frac{1}{2})\\\
\delta(z)=\delta(z+\frac{1}{2})\end{array}\right.\Leftrightarrow\left\\{\begin{array}[]{l}0=0\\\
z=0,\frac{1}{2}\\\ z=\frac{1}{4},\frac{3}{4}\end{array}.\right.$

concluding that $\operatorname{Coin}(f^{\prime},f^{\prime}\circ
h_{2i+ij+1})=\emptyset$ in this case.

  2. (2)

Two of the integers $A_{ij}$, $B_{ij}$ and $C_{ij}$ are null. Assume, without
loss of generality, that $A_{ij}=0$, $B_{ij}=0$ and $C_{ij}\neq 0$. Then,

$f(x,y,z)=(f\circ
h_{2i+ij+1})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}0=\frac{c}{2}\\\
0=\frac{t}{2}\\\ C_{ij}y=\frac{w}{2}\end{array}.\right.$ (8)

For $c$ odd or $t$ odd, it can immediately be concluded that the system (8)
has no solution. If both integers are even, consider
$f^{\prime}:\mathbb{T}^{3}\to\mathbb{T}^{3}$ defined by
$f^{\prime}(x,y,z)=(ax+by+cz+\epsilon(z),rx+sy+tz+\delta(z),ux+vy+wz).$ Note
that $f^{\prime}$ is homotopic to $f$ and
$\operatorname{Coin}(f^{\prime},f^{\prime}\circ h_{2i+ij+1})=\emptyset$,
because

$f^{\prime}(x,y,z)=(f^{\prime}\circ
h_{2i+ij+1})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}\epsilon(z)=\epsilon(z+\frac{1}{2})\\\
\delta(z)=\delta(z+\frac{1}{2})\\\
C_{ij}y=\frac{w}{2}\end{array}\right.\Leftrightarrow\left\\{\begin{array}[]{l}z=0,\frac{1}{2}\\\
z=\frac{1}{4},\frac{3}{4}\\\ C_{ij}y=\frac{w}{2}\end{array}.\right.$ (9)

  3. (3)

Only one of the integers $A_{ij}$, $B_{ij}$ and $C_{ij}$ is null. Without loss
of generality, suppose $A_{ij}=0$, $B_{ij}\neq 0$ and $C_{ij}\neq 0$. So,

$f(x,y,z)=(f\circ
h_{2i+ij+1})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}0=\frac{c}{2}\\\
B_{ij}y=\frac{t}{2}\\\ C_{ij}y=\frac{w}{2}\end{array}.\right.$

Obviously, if $c$ is odd then $\operatorname{Coin}(f,f\circ
h_{2i+ij+1})=\emptyset$. Now, for $c$ even, consider
$f^{\prime}:\mathbb{T}^{3}\to\mathbb{T}^{3}$ defined by
$f^{\prime}(x,y,z)=(ax+by+cz+\delta(z),rx+sy+tz+\epsilon(z),ux+vy+wz)$. Then,

$f^{\prime}(x,y,z)=(f^{\prime}\circ
h_{2i+ij+1})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}\delta(z)=\delta(z+\frac{1}{2})\\\
B_{ij}y+\epsilon(z)=\frac{t}{2}+\epsilon(z+\frac{1}{2})\\\
C_{ij}y=\frac{w}{2}\end{array}\right.$ (10)

Let $(x,y,z)\in\mathbb{T}^{3}$ be such that $y$ is the solution to the
equation $C_{ij}y=\frac{w}{2}$. Like $C_{ij}\neq 0$, there exists a finite
amount of $y$ satisfying this equation. If there exists $y$ such that
$Cy=\frac{w}{2}$ and $By=\frac{t}{2}$, then the system (10) coincides with the
system (9). Therefore, such values of $y$ are not solutions to the system
$\eqref{4sisth4}$. For the values of $y$ such that $C_{ij}y=\frac{w}{2}$ and
$B_{ij}y\neq\frac{t}{2}$, we have
$By+\epsilon(z)\neq\frac{t}{2}+\epsilon(z+\frac{1}{2})$ for $n_{0}$ big
enough. This demonstrates that all values of $y$ which satisfy the equation
$Cy=\frac{w}{2}$ are not solutions to the system (10). Therefore,
$\operatorname{Coin}(f^{\prime},f^{\prime}\circ h_{2i+ij+1})=\emptyset$.

  4. (4)

If none of the integers $A_{ij}$, $B_{ij}$ and $C_{ij}$ are null. Consider
$f^{\prime}:\mathbb{T}^{3}\to\mathbb{T}^{3}$ defined by

$f^{\prime}(x,y,z)=\left(ax+by+cz,rx+sy+tz+\epsilon(z),ux+vy+wz+\delta(z)\right).$

Note that the map $f^{\prime}$ is homotopic to $f$. Now,

$f^{\prime}(x,y,z)=(f^{\prime}\circ
h_{2i+ij+1})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}A_{ij}y=\frac{c}{2}\\\
B_{ij}y+\epsilon(z)=\frac{t}{2}+\epsilon(z+\frac{1}{2})\\\
C_{ij}y+\delta(z)=\frac{w}{2}+\delta(z+\frac{1}{2})\end{array}.\right.$ (11)

Suppose that $A_{ij}y=\frac{c}{2}$ (otherwise
$\operatorname{Coin}(f^{\prime},f^{\prime}\circ h_{2i+ij+1})=\emptyset$). The
$\operatorname{Coin}(f^{\prime},f^{\prime}\circ h_{2i+ij+1})$ calculation will
be divided into two cases:

    1. (a)

$B_{ij}y=\frac{t}{2}$ and $C_{ij}y=\frac{w}{2}$;
Thus

$f^{\prime}(x,y,z)=(f^{\prime}\circ
h_{2i+ij+1})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}0=0\\\
\epsilon(z)=\epsilon(z+\frac{1}{2})\\\
\delta(z)=\delta(z+\frac{1}{2})\end{array}\right.\Leftrightarrow\left\\{\begin{array}[]{l}0=0\\\
z=0,\frac{1}{2}\\\ z=\frac{1}{4},\frac{3}{4}\end{array},\right.$

concluding that the system (11) has no solution.

    2. (b)

$B_{ij}y\neq\frac{t}{2}$ or $C_{ij}y\neq\frac{w}{2}$;
Without loss of generality, suppose $B_{ij}y\neq\frac{t}{2}$. Like $A_{ij}\neq
0$ and $B_{ij}\neq 0$, there exists a finite amount of $y$ satisfying the
conditions $A_{ij}y=\frac{c}{2}$ and $B_{ij}y\neq\frac{t}{2}$. Choosing
$n_{0}$ big enough we have
$B_{ij}y+\epsilon(z)\neq\frac{t}{2}+\epsilon(z+\frac{1}{2})$, concluding that
the system (11) has no solution.

For all possible cases, $\operatorname{Coin}(f^{\prime},f^{\prime}\circ
h_{2i+ij+1})=\emptyset$. Therefore,

$\operatorname{NBU}(f,h_{2i+ij+1})=0.$

###  5.2. $h_{2}(x,y,z)=\left(-x,-y,z+\frac{1}{2}\right)$

Consider the integer values

$\displaystyle p=\det\begin{pmatrix}r&s\\\ u&v\end{pmatrix}=rv-su,\qquad
q=\det\begin{pmatrix}a&b\\\ u&v\end{pmatrix}=av-bu$ and $\displaystyle
o=\det\begin{pmatrix}a&b\\\ r&s\end{pmatrix}=as-br$

The proof for the free involution $h_{2}$ will be divided as follows: first
$(a,r,u)=(0,0,0)$ or $(b,s,v)=(0,0,0)$, the second case $(a,r,u)\neq(0,0,0)$
and $(b,s,v)\neq(0,0,0)$ will be divided into four new cases: $c,t$ and $w$
are even integers; the set $\\{c,t,w\\}$ has an odd integer and two even
numbers; the set $\\{c,t,w\\}$ has two odd numbers and an even integer; and
lastly, $c,t$ and $w$ are odd integers. In these subcases, they will be
analyzed when $p$ and $q$ are zeros or not.

First, if $(a,r,u)=(0,0,0)$ or $(b,s,v)=(0,0,0)$ then the equation
$f(x,y,z)=(f\circ h_{2})(x,y,z)$ corresponds to a system which is similar to
the system studied in the previous section. So, applying the same technique
used previously we can conclude that
$\operatorname{Coin}(f^{\prime},f^{\prime}\circ h_{2})=\emptyset$ for some
$f^{\prime}$ is homotopic to $f$. Therefore, $\operatorname{NBU}(f,h_{2})=0$.

In the second case $(a,r,u)\neq(0,0,0)$ and $(b,s,v)\neq(0,0,0)$. Without loss
of generality, suppose $a\neq 0$. The computation of
$\operatorname{NBU}(f,h_{2})$ will be done in four sub-cases:

First subcase, suppose that $c$, $t$ and $w$ are even integers. Thus,

$\displaystyle f(x,y,z)=(f\circ h_{2})(x,y,z)$
$\displaystyle\Leftrightarrow\left\\{\begin{array}[]{l}ax+by+cz=-ax-
by+cz+\frac{c}{2}\\\ rx+sy+tz=-rx-sy+tz+\frac{t}{2}\\\ ux+vy+wz=-ux-
vy+wz+\frac{w}{2}\end{array}\right.$ (15)
$\displaystyle\Leftrightarrow\left\\{\begin{array}[]{l}2ax+2by=k_{1}\\\
2rx+2sy=k_{2}\\\ 2ux+2vy=k_{3}\end{array},\right.$ (19)

with $k_{1}$, $k_{2}$, $k_{3}\in\mathbb{Z}$. Note that $(0,0,z)$,
$\left(0,\frac{1}{2},z\right)$, $\left(\frac{1}{2},0,z\right)$ and
$\left(\frac{1}{2},\frac{1}{2},z\right)\in\mathbb{T}^{3}$ are solutions of the
system (19), for all $z$. Thereby, $\operatorname{Coin}(f,f\circ
h_{2})\neq\emptyset$.

To resolve the system $\eqref{1estrelah2}$ in $\mathbb{T}^{3}$, i.e., find the
points of coincidences between the maps $f$ and $f\circ h_{2}$, a different
technique will be used. This technique consists of finding possible solutions
to the system $\eqref{1estrelah2}$ in $\mathbb{R}^{3}$ and projecting said
solutions in $\mathbb{T}^{3}$, through natural projection
$\pi:\mathbb{R}^{3}\to\mathbb{T}^{3}$ given by
$\pi(x,y,z)=[(x,y,z)]_{\mathbb{T}^{3}}$, getting the points of
$\operatorname{Coin}(f,f\circ h_{2})$. In other words, the objective is to
find points of coincidence between pairs of coverings
$(\widetilde{f},\widetilde{f\circ h_{2}})$ and project them by $\pi$ in
$\mathbb{T}^{3}$, in order to find what the system $\eqref{1estrelah2}$
solutions in $\mathbb{T}^{3}$ are.

Note that changing the covering class of the pair
$(\widetilde{f},\widetilde{f\circ h_{2}})$, the values $k_{1}$, $k_{2}$ and
$k_{3}$ may vary, and so we can suppose all such numbers to be non-nulls.

Studying the system $\eqref{1estrelah2}$ in $\mathbb{R}^{3}$, one obtains from
the first and third equations that the triples $(x,y,z)\in\mathbb{R}^{3}$ with

$x=\frac{qk_{1}-abk_{3}+buk_{1}}{2qa}\qquad\text{and}\qquad
y=\frac{ak_{3}-uk_{1}}{2q}$

are possible solutions, for $q\neq 0$. Projecting these possible solutions in
$\mathbb{T}^{3}$, like $(qk_{1}-abk_{3}+buk_{1})\in\mathbb{Z}$ and
$(ak_{3}-uk_{1})\in\mathbb{Z}$, it follows that

$\operatorname{Coin}(f,f\circ h_{2})\subseteq S=\\{(x,y,z)\in\mathbb{T}^{3}\
|\ 2qax=0\quad\text{and}\quad 2qy=0\\},$

since any coincidence point between the maps $f$ and $f\circ h_{2}$ in
$\mathbb{T}^{3}$ is the image of a coincidence point of some pair of coverings
$(\widetilde{f},\widetilde{f\circ h_{2}})$.

Consider the triples $(x,y,z)\in S\cap\operatorname{Coin}(f,f\circ h_{2})$,
then

$f(x,y,z)=(f\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}x=\frac{i}{2qa}\\\
y=\frac{j}{2q}\\\ 0=0\end{array},\right.$

where $i\in\\{0,1,2,\ldots,2qa-1\\}$ and $j\in\\{0,1,\ldots,2q-1\\}$.

Let $f^{\prime}:\mathbb{T}^{3}\to\mathbb{T}^{3}$ be defined by
$f^{\prime}(x,y,z)=(ax+by+cz+\epsilon(y),rx+sy+tz+\epsilon(z),ux+vy+wz+\epsilon(x))$,
where $\epsilon:\mathbb{T}^{1}\to\mathbb{T}^{1}$ is given by
$\epsilon(z)=\frac{1}{n_{0}}\operatorname{sen}(2\pi z)$ with
$n_{0}\in\mathbb{N}$ conveniently chosen. Then,

$\displaystyle f^{\prime}(x,y,z)=(f^{\prime}\circ h_{2})(x,y,z)$
$\displaystyle\Leftrightarrow\left\\{\begin{array}[]{l}x+\epsilon(y)=\frac{i}{2qa}+\epsilon(-y)\\\
y+\epsilon(z)=\frac{j}{2q}+\epsilon(z+\frac{1}{2})\\\
\epsilon(x)=\epsilon(-x)\end{array}\right.$
$\displaystyle\Leftrightarrow\left\\{\begin{array}[]{l}y=0,\frac{1}{2}\\\
z=0,\frac{1}{2}\\\ x=0,\frac{1}{2}\end{array}.\right.$

Therewith, $\operatorname{Coin}(f^{\prime},f^{\prime}\circ
h_{2})=\left\\{\begin{array}[]{l}(0,0,0),(0,0,\frac{1}{2}),(0,\frac{1}{2},0),(0,\frac{1}{2},\frac{1}{2})\\\
(\frac{1}{2},0,0),(\frac{1}{2},0,\frac{1}{2}),(\frac{1}{2},\frac{1}{2},0),(\frac{1}{2},\frac{1}{2},\frac{1}{2})\end{array}\right\\}$
and

$\operatorname{BUCoin}(f^{\prime},h_{2})=\left\\{\begin{array}[]{l}\left[\left(0,0,0\right),\left(0,0,\frac{1}{2}\right)\right],\left[\left(0,\frac{1}{2},0\right),\left(0,\frac{1}{2},\frac{1}{2}\right)\right]\\\
\left[\left(\frac{1}{2},0,0\right),\left(\frac{1}{2},0,\frac{1}{2}\right)\right],\left[\left(\frac{1}{2},\frac{1}{2},0\right),\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)\right]\end{array}\right\\}.$

There are four single Borsuk-Ulam coincidences classes. The pseudo-index shows
that all four are essential. Therefore, $\operatorname{NBU}(f,h_{2})=4$.

Now, if $q=0$ we will at first assume $p=0$. Thus, $su=rv$ and $bu=av$. For
$u\neq 0$ we have $s=\frac{rv}{u}$ and $b=\frac{av}{u}$. Then,

$\displaystyle f(x,y,z)=(f\circ h_{2})(x,y,z)$
$\displaystyle\Leftrightarrow\left\\{\begin{array}[]{l}2ax+2\left(\frac{av}{u}\right)y=0\\\
2rx+2\left(\frac{rv}{u}\right)y=0\\\ 2vy=-2ux\end{array}\right.$
$\displaystyle\Leftrightarrow\left\\{\begin{array}[]{l}2ax-\frac{a}{u}2ux=0\\\
2rx-\frac{r}{u}2ux=0\\\ 2vy=-2ux\end{array}\right.$
$\displaystyle\Leftrightarrow\left\\{\begin{array}[]{l}0=0\\\ 0=0\\\
2vy=-2ux\end{array}.\right.$

Let $f^{\prime}:\mathbb{T}^{3}\to\mathbb{T}^{3}$ be defined by
$f^{\prime}(x,y,z)=(ax+by+cz+\delta(z),rx+sy+tz+\epsilon(z),ux+vy+wz)$.

$f^{\prime}(x,y,z)=(f^{\prime}\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}\delta(z)=\delta(z+\frac{1}{2})\\\
\epsilon(z)=\epsilon(z+\frac{1}{2})\\\ 2vy=-2ux\end{array}\right.\\\
\Leftrightarrow\left\\{\begin{array}[]{l}z=\frac{1}{4},\frac{3}{4}\\\
z=0,\frac{1}{2}\\\ 2vy=-2ux\end{array}.\right.$

So, the system has no solution in $\mathbb{T}^{3}$, that is,
$\operatorname{Coin}(f^{\prime},f^{\prime}\circ h_{2})=\emptyset$. Therefore,
$\operatorname{NBU}(f,h_{2})=0$.

For $u=0$ we have $rv=0$ and $av=0$. Since $(a,r,u)\neq(0,0,0)$ we have $v=0$.
Thus,

$f(x,y,z)=(f\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}2ax+2by=0\\\
2rx+2sy=0\\\ 0=0\end{array}.\right.$

Let $f^{\prime}:\mathbb{T}^{3}\to\mathbb{T}^{3}$ be defined by
$f^{\prime}(x,y,z)=(ax+by+cz,rx+sy+tz,wz+\epsilon(x))$.

$f^{\prime}(x,y,z)=(f^{\prime}\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}2ax+2by=0\\\
2rx+2sy=0\\\
\epsilon(x)=\epsilon(-x)\end{array}\right.\Leftrightarrow\left\\{\begin{array}[]{l}2by=0\\\
2sy=0\\\ x=0,\frac{1}{2}\end{array}.\right.$ (20)

The case $(b,s,v)=(0,0,0)$ has already been analyzed at the beginning of this
section. So $(b,s)\neq(0,0)$, since $v=0$. Without loss of generality, assume
$b\neq 0$.

Consider $(x,y,z)\in\mathbb{T}^{3}$ with $y=\frac{i}{2b}$ satisfying the first
two equations of (20) with $i\in\\{0,1,2,\ldots,2b-1\\}$. Note that $y=0$ and
$y=\frac{1}{2}$ are examples of $y=\frac{i}{2b}$ satisfying the desired
equations (for $i=0$ and $i=b$, respectively). Thus,

$f^{\prime}(x,y,z)=(f^{\prime}\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}y=\frac{i}{2b}\\\ 0=0\\\
x=0,\frac{1}{2}\end{array}.\right.$

Let $f^{\prime\prime}:\mathbb{T}^{3}\to\mathbb{T}^{3}$ be given by

$f^{\prime\prime}(x,y,z)=(ax+by+cz+\epsilon(z),rx+sy+tz+\epsilon(y),wz+\epsilon(x)).$

Then, $f^{\prime\prime}$ is homotopic to $f$ the

$\displaystyle f^{\prime\prime}(x,y,z)=(f^{\prime\prime}\circ h_{2})(x,y,z)$
$\displaystyle\Leftrightarrow\left\\{\begin{array}[]{l}y+\epsilon(z)=\frac{i_{\theta}}{2b}+\epsilon(z+\frac{1}{2})\\\
y=0,\frac{1}{2}\\\ x=0,\frac{1}{2}\end{array}\right.$
$\displaystyle\Leftrightarrow\left\\{\begin{array}[]{l}z=0,\frac{1}{2}\\\
y=0,\frac{1}{2}\\\ x=0,\frac{1}{2}\end{array}\right.$

Therefore,

$\operatorname{BUCoin}(f^{\prime\prime},h_{2})=\left\\{\begin{array}[]{l}\left[\left(0,0,0\right),\left(0,0,\frac{1}{2}\right)\right],\left[\left(0,\frac{1}{2},0\right),\left(0,\frac{1}{2},\frac{1}{2}\right)\right]\\\
\left[\left(\frac{1}{2},0,0\right),\left(\frac{1}{2},0,\frac{1}{2}\right)\right],\left[\left(\frac{1}{2},\frac{1}{2},0\right),\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)\right]\end{array}\right\\}$

with all four Borsuk-Ulam coincidence classes essential. Therefore,

$\operatorname{NBU}(f,h_{2})=4.$

If $q=0$ and $p\neq 0$, we have $bu=av$ and $su\neq rv$. If $u\neq 0$ then
$b=\frac{av}{u}$. That way,

$f(x,y,z)=(f\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}2ax+2\left(\frac{av}{u}\right)y=0\\\
2rx+2sy=0\\\
2vy=-2ux\end{array}\right.\Leftrightarrow\left\\{\begin{array}[]{l}0=0\\\
2rx+2sy=0\\\ 2vy=-2ux\end{array}.\right.$

Consider $f^{\prime}:\mathbb{T}^{3}\to\mathbb{T}^{3}$ given by

$f^{\prime}(x,y,z)=(ax+by+cz+\epsilon(x),rx+sy+tz,ux+vy+wz).$
$f^{\prime}(x,y,z)=(f^{\prime}\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}\epsilon(x)=\epsilon(-x)\\\
2rx+2sy=0\\\
2vy=-2ux\end{array}\right.\Leftrightarrow\left\\{\begin{array}[]{l}x=0,\frac{1}{2}\\\
2sy=0\\\ 2vy=0\end{array}.\right.$

We know $(s,v)\neq(0,0)$ since $p\neq 0$. Moreover, $y=0$ and $y=\frac{1}{2}$
satisfy the above system. Using the same technique applied to the system (20),
for $f^{\prime\prime}:\mathbb{T}^{3}\to\mathbb{T}^{3}$ defined by
$f^{\prime\prime}(x,y,z)=(ax+by+cz+\epsilon(x),rx+sy+tz+\epsilon(z),ux+vy+wz+\epsilon(y))$,
we conclude that $\operatorname{NBU}(f,h_{2})=4$.

If $u=0$ we have $av=0$. From the beginning $a\neq 0$, so $v=0$, which
contradicts the fact that $p\neq 0$. Therefore, $u\neq 0$.

In the second subcase, the set $\\{c,t,w\\}$ has an odd integer and two even
numbers. Without loss of generality, suppose $c$ odd. So,

$f(x,y,z)=(f\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}2ax+2by=k_{1}+\frac{1}{2}\\\
2rx+2sy=k_{2}\\\ 2ux+2vy=k_{3}\end{array},\right.$ (21)

with $k_{1}$, $k_{2}$, $k_{3}\in\mathbb{Z}$. We will find possible system
solutions (21) in $\mathbb{R}^{3}$ and then project them in $\mathbb{T}^{3}$.
Studying the system (21), from the first and third equations, we have that
$(x,y,z)\in\mathbb{R}^{3}$ with

$x=\frac{2qk_{1}+q-2abk_{3}+2buk_{1}+bu}{4qa}\qquad\text{and}\qquad
y=\frac{2ak_{3}-2uk_{1}-u}{4q}$

being possible solutions, for $q\neq 0$.

Projecting these possible solutions in $\mathbb{T}^{3}$, we have

$\operatorname{Coin}(f,f\circ h_{2})\subseteq S=\\{(x,y,z)\in\mathbb{T}^{3}\
|\ 4qax=0\quad\text{and}\quad 4qy=0\\},$

since $(2qk_{1}+q-2abk_{3}+2buk_{1}+bu)\in\mathbb{Z}$ and
$(2ak_{3}-2uk_{1}-u)\in\mathbb{Z}$.

Consider $(x,y,z)\in S\cap\operatorname{Coin}(f,f\circ h_{2})$, then

$f(x,y,z)=(f\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}x=\frac{i}{4qa}\\\
y=\frac{j}{4q}\\\ 0=0\end{array},\right.$

where $i\in\\{0,1,2,\ldots,4qa-1\\}$ and $j\in\\{0,1,\ldots,4q-1\\}$.

Define $f^{\prime}:\mathbb{T}^{3}\to\mathbb{T}^{3}$ by
$f^{\prime}(x,y,z)=(ax+by+cz+,rx+sy+tz,ux+vy+wz+\epsilon(x))$.

$f^{\prime}(x,y,z)=(f^{\prime}\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}x=\frac{i}{4qa}\\\
y=\frac{j}{4q}\\\ x=0,\frac{1}{2}\end{array}.\right.$ (22)

If $(0,y,z)$ and $\left(\frac{1}{2},y,z\right)$ are not in
$\operatorname{Coin}(f,f\circ h_{2})$, then the system (22) has no solution.
So, $\operatorname{Coin}(f^{\prime},f^{\prime}\circ h_{2})=\emptyset$.
Therefore, $\operatorname{NBU}(f,h_{2})=0$. If $(0,y,z)$ and
$\left(\frac{1}{2},y,z\right)$ are in $\operatorname{Coin}(f,f\circ h_{2})$,
for some $y$ and $z$, so considering
$f^{\prime\prime}:\mathbb{T}^{3}\to\mathbb{T}^{3}$ given by
$f^{\prime\prime}(x,y,z)=(ax+by+cz+\epsilon(y),rx+sy+tz,ux+vy+wz+\epsilon(x))$,
then we have that $f^{\prime\prime}$ is homotopic to $f$ and

$f^{\prime\prime}(x,y,z)=(f^{\prime\prime}\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}\epsilon(y)=\epsilon(-y)\\\
y=\frac{j}{4q}\\\
x=0,\frac{1}{2}\end{array}\right.\Leftrightarrow\left\\{\begin{array}[]{l}y=0,\frac{1}{2}\\\
y=\frac{j}{4q}\\\ x=0,\frac{1}{2}\end{array}.\right.$

However $(0,0,z)$, $\left(\frac{1}{2},0,z\right)$,
$\left(0,\frac{1}{2},z\right)$,
$\left(\frac{1}{2},\frac{1}{2},z\right)\in\mathbb{T}^{3}$ are not in
$\operatorname{Coin}(f,f\circ h_{2})$, since they do not satisfy the first
equation of the system (21). Therewith, $\frac{j}{4q}\neq 0$ and
$\frac{j}{4q}\neq\frac{1}{2}$. Therefore
$\operatorname{Coin}(f^{\prime\prime},f^{\prime\prime}\circ h_{2})=\emptyset$
and $\operatorname{NBU}(f,h_{2})=0$.

For $q=0$ suppose, firstly, that $p=0$. Thus, $su=rv$ and $bu=av$. Considering
$u\neq 0$ we have $s=\frac{rv}{u}$ and $b=\frac{av}{u}$. Therewith,

$\displaystyle f(x,y,z)=(f\circ h_{2})(x,y,z)$
$\displaystyle\Leftrightarrow\left\\{\begin{array}[]{l}2ax+2\left(\frac{av}{u}\right)y=\frac{1}{2}\\\
2rx+2\left(\frac{rv}{u}\right)y=0\\\ 2vy=-2ux\end{array}\right.$
$\displaystyle\Leftrightarrow\left\\{\begin{array}[]{l}2ax-\frac{a}{u}2ux=\frac{1}{2}\\\
2rx-\frac{r}{u}2ux=0\\\ 2vy=-2ux\end{array}\right.$
$\displaystyle\Leftrightarrow\left\\{\begin{array}[]{l}0=\frac{1}{2}\\\ 0=0\\\
2vy=-2ux\end{array}.\right.$

So, $\operatorname{Coin}(f,f\circ h_{2})=\emptyset$. Therefore,
$\operatorname{NBU}(f,h_{2})=0$.

Considering $u=0$ we have $rv=0$ and $av=0$. Since $(a,r,u)\neq(0,0,0)$, it
follows that $v=0$. Thus,

$f(x,y,z)=(f\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}2ax+2by=\frac{1}{2}\\\
2rx+2sy=0\\\ 0=0\end{array}.\right.$

Let $f^{\prime}:\mathbb{T}^{3}\to\mathbb{T}^{3}$ be defined by
$f^{\prime}(x,y,z)=(ax+by+cz,rx+sy+tz,wz+\epsilon(x))$.

$f^{\prime}(x,y,z)=(f^{\prime}\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}2ax+2by=\frac{1}{2}\\\
2rx+2sy=0\\\
\epsilon(x)=\epsilon(-x)\end{array}\right.\Leftrightarrow\left\\{\begin{array}[]{l}2by=\frac{1}{2}\\\
2sy=0\\\ x=0,\frac{1}{2}\end{array}.\right.$ (23)

If $b=0$ then $\operatorname{NBU}(f,h_{2})=0$. For $b\neq 0$ consider
$y=\frac{i}{4b}$ satisfying the first two equations of (23), with
$i\in\\{1,3,5,\ldots,4b-1\\}$, note that $y=0$ and $y=\frac{1}{2}$ do not
satisfy such condition. Thus,

$f^{\prime}(x,y,z)=(f^{\prime}\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}y=\frac{i}{4b}\\\ 0=0\\\
x=0,\frac{1}{2}\end{array}.\right.$

Let $f^{\prime\prime}:\mathbb{T}^{3}\to\mathbb{T}^{3}$ be given by
$f^{\prime\prime}(x,y,z)=(ax+by+cz,rx+sy+tz+\epsilon(y),wz+\epsilon(x))$.

$f^{\prime\prime}(x,y,z)=(f^{\prime\prime}\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}y=\frac{i}{4b}\\\
\epsilon(y)=\epsilon(-y)\\\
x=0,\frac{1}{2}\end{array}\right.\Leftrightarrow\left\\{\begin{array}[]{l}y=\frac{i}{4b}\\\
y=0,\frac{1}{2}\\\ x=0,\frac{1}{2}\end{array}.\right.$

Therefore, $\operatorname{Coin}(f^{\prime\prime},f^{\prime\prime}\circ
h_{2})=\emptyset$ and $\operatorname{NBU}(f,h_{2})=0$.

Now, if $q=0$ and $p\neq 0$ then $bu=av$ and $su\neq rv$. If $u\neq 0$ then
$b=\frac{av}{u}$. That way,

$f(x,y,z)=(f\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}2ax+2\left(\frac{av}{u}\right)y=\frac{1}{2}\\\
2rx+2sy=0\\\
2vy=-2ux\end{array}\right.\Leftrightarrow\left\\{\begin{array}[]{l}0=\frac{1}{2}\\\
2rx+2sy=0\\\ 2vy=-2ux\end{array}.\right.$

Therefore, $\operatorname{Coin}(f,f\circ h_{2})=\emptyset$ and
$\operatorname{NBU}(f,h_{2})=0$.

If $u=0$ we have $av=0$. From the beginning $a\neq 0$, so $v=0$, contradicting
the fact that $p\neq 0$. Therefore, $u\neq 0$.

The third subcase, in which the set $\\{c,t,w\\}$ has two odd numbers and an
even number, is similar to the second case, and the Nielsen-Borsuk-Ulam number
is also zero.

In the fourth and last subcase, $c,t,w$ are odd. For $q\neq 0$ we conclude,
similarly to the previous cases, that $\operatorname{NBU}(f,h_{2})=0$.

If $q=0$, $p=0$ and $u\neq 0$, we have

$\displaystyle f(x,y,z)=(f\circ h_{2})(x,y,z)$
$\displaystyle\Leftrightarrow\left\\{\begin{array}[]{l}2ax+2\left(\frac{av}{u}\right)y=\frac{1}{2}\\\
2rx+2\left(\frac{rv}{u}\right)y=\frac{1}{2}\\\
2vy=\frac{1}{2}-2ux\end{array}\right.$
$\displaystyle\Leftrightarrow\left\\{\begin{array}[]{l}2ax+\frac{a}{2u}-\frac{a}{u}2uy=\frac{1}{2}\\\
2rx+\frac{r}{2u}-\frac{r}{u}2uy=\frac{1}{2}\\\
2vy=\frac{1}{2}-2ux\end{array}\right.$
$\displaystyle\Leftrightarrow\left\\{\begin{array}[]{l}\frac{a}{2u}=\frac{1}{2}\\\
\frac{r}{2u}=\frac{1}{2}\\\ 2vy=\frac{1}{2}-2ux\end{array}.\right.$

If $\frac{a}{2u}\neq\frac{1}{2}$ or $\frac{r}{2u}\neq\frac{1}{2}$, it follows
that $\operatorname{NBU}(f,h_{2})=0$, because the system above has no solution
in $\mathbb{T}^{3}$. Otherwise, if $\frac{a}{2u}=\frac{1}{2}$ and
$\frac{r}{2u}=\frac{1}{2}$ then $\operatorname{NBU}(f,h_{2})=0$, just consider
$f^{\prime}:\mathbb{T}^{3}\to\mathbb{T}^{3}$ defined by
$f^{\prime}(x,y,z)=(ax+by+cz+\delta(z),rx+sy+tz+\epsilon(z),ux+vy+wz)$.

If $q=0$, $p=0$ and $u=0$, we have $v=0$. Then,

$f(x,y,z)=(f\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}2ax+2by=\frac{1}{2}\\\
2rx+2sy=\frac{1}{2}\\\ 0=0\end{array}.\right.$

Define $f^{\prime}:\mathbb{T}^{3}\to\mathbb{T}^{3}$ by
$f^{\prime}(x,y,z)=(ax+by+cz,rx+sy+tz,wz+\epsilon(x))$. So, $f^{\prime}$ is
homotopic to $f$ and

$f^{\prime}(x,y,z)=(f^{\prime}\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}2ax+2by=\frac{1}{2}\\\
2rx+2sy=\frac{1}{2}\\\
x=0,\frac{1}{2}\end{array}\right.\Leftrightarrow\left\\{\begin{array}[]{l}2by=\frac{1}{2}\\\
2sy=\frac{1}{2}\\\ x=0,\frac{1}{2}\end{array}.\right.$

We know $(b,s)\neq(0,0)$, because $(b,s,v)\neq(0,0,0)$. Clearly the
$\operatorname{NBU}(f,h_{2})=0$ if $b=0$ or $s=0$. For $b\neq 0$ and $s\neq
0$, we can use the same technique applied to the system
$\eqref{sistemabsimpar}$, and conclude that $\operatorname{NBU}(f,h_{2})=0.$

Lastly, for $q=0$ and $p\neq 0$, we have $u\neq 0$, $b=\frac{av}{u}$ and
$su\neq rv$. Thus,

$f(x,y,z)=(f\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}\frac{a}{2u}=\frac{1}{2}\\\
2rx+2sy=\frac{1}{2}\\\ 2vy=\frac{1}{2}-2ux\end{array}.\right.$

If $\frac{a}{2u}\neq\frac{1}{2}$, it follows that
$\operatorname{Coin}(f,f\circ h_{2})=\emptyset$. Otherwise, let
$f^{\prime}:\mathbb{T}^{3}\to\mathbb{T}^{3}$ be given by
$f^{\prime}(x,y,z)=(ax+by+cz+\epsilon(x),rx+sy+tz,ux+vy+wz)$. Then,

$f^{\prime}(x,y,z)=(f^{\prime}\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}x=0,\frac{1}{2}\\\
2sy=\frac{1}{2}\\\ 2vy=\frac{1}{2}\end{array}.\right.$ (24)

Note that $(s,v)\neq(0,0)$ (because $p\neq 0$) and $(x,y,z)\in\mathbb{T}^{3}$
with $y=0$ or $y=\frac{1}{2}$ are not solutions to the system above. If $s=0$
or $v=0$ then $\operatorname{Coin}(f^{\prime},f^{\prime}\circ
h_{2})=\emptyset$. If $s\neq 0$ and $v\neq 0$, consider
$(x,y,z)\in\mathbb{T}^{3}$ such that $y=\frac{i}{4s}$, satisfies the last two
equations of the system (24), where $i\in\\{1,3,\ldots,4s-1\\}$. So,

$f^{\prime}(x,y,z)=(f^{\prime}\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}x=0,\frac{1}{2}\\\
y=\frac{i}{4s}\\\ 0=0\end{array}.\right.$

For $f^{\prime\prime}:\mathbb{T}^{3}\to\mathbb{T}^{3}$ defined by
$f^{\prime\prime}(x,y,z)=(ax+by+cz+\epsilon(x),rx+sy+tz,ux+vy+wz+\epsilon(y))$.

$f^{\prime\prime}(x,y,z)=(f^{\prime\prime}\circ
h_{2})(x,y,z)\Leftrightarrow\left\\{\begin{array}[]{l}x=0,\frac{1}{2}\\\
y=\frac{i}{4s}\\\ y=0,\frac{1}{2}\end{array}.\right.$ (25)

Since $y=0$ and $y=\frac{1}{2}$ do not satisfy the second equation of (25), it
follows that $\operatorname{Coin}(f^{\prime\prime},f^{\prime\prime}\circ
h_{2})=\emptyset$, and $\operatorname{NBU}(f,h_{2})=0$. ∎

##  6\. Nielsen-Borsuk-Ulam number in $\mathbb{T}^{n}$, $n>3$

For the $n-$torus, with $n>3$, there is no classification of free involutions
in the literature. That way, the study of the Nielsen-Borsuk-Ulam number in
this particular space cannot be made in general. What we can do is consider a
free involution $\tau$ and calculate the Borsuk-Ulam number
$\operatorname{NBU}(f,\tau)$ for any map $f:\mathbb{T}^{n}\to\mathbb{T}^{n}$.

Consider the following free involutions in $\mathbb{T}^{n}$:

$\displaystyle\tau_{1}(x_{1},x_{2},\ldots,x_{n-1},x_{n})$
$\displaystyle=\left(x_{1},x_{2},\ldots,x_{n-1},x_{n}+\frac{1}{2}\right)$
$\displaystyle\tau_{3}(x_{1},x_{2},\ldots,x_{n-1},x_{n})$
$\displaystyle=\left(x_{1},x_{2},\ldots,x_{n-2},-x_{n-1},x_{n}+\frac{1}{2}\right)$
$\displaystyle\tau_{4}(x_{1},x_{2},\ldots,x_{n-1},x_{n})$
$\displaystyle=\left(x_{1}+x_{2},-x_{2},x_{3},\ldots,x_{n-1},x_{n}+\frac{1}{2}\right).$

Observe that the mentioned involutions are generalizations to the $n-$torus of
the involutions $h_{1}$, $h_{3}$ and $h_{4}$ of $3-$torus. Applying the same
method used previously in $\mathbb{T}^{3}$ to these involutions, we can
demonstrate that the Nielsen-Borsuk-Ulam number is zero for those involutions,
i.e., for any map $f:\mathbb{T}^{n}\to\mathbb{T}^{n}$ we have

$\operatorname{NBU}(f,\tau_{1})=0,\qquad\operatorname{NBU}(f,\tau_{3})=0\quad\text{and}\quad\operatorname{NBU}(f,\tau_{4})=0.$

For the free involution

$\tau_{2}(x_{1},x_{2},\ldots,x_{n-1},x_{n})=\left(-x_{1},-x_{2},\ldots,-x_{n-1},x_{n}+\frac{1}{2}\right)$

in $\mathbb{T}^{n}$, we have $\operatorname{NBU}(f,\tau_{2})\neq 0$ for some
map $f:\mathbb{T}^{n}\to\mathbb{T}^{n}$. Indeed, let
$g:\mathbb{T}^{n}\to\mathbb{T}^{n}$ be a map such that

$g_{\\#}=\begin{pmatrix}1&0&0&\cdots&0&2b_{1}\\\ 0&1&0&\cdots&0&2b_{2}\\\
0&0&1&\cdots&0&2b_{3}\\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
0&0&0&\cdots&1&2b_{n-1}\\\ 0&0&0&\cdots&0&2b_{n}\end{pmatrix}$

where $b_{i}\in\mathbb{Z}$, i.e.,

$g(x_{1},\ldots,x_{n-1},x_{n})=\left(x_{1},\ldots,x_{n-1},2b_{1}x_{1}+2b_{2}x_{2}+\ldots+2b_{n}x_{n}\right).$

Let $g^{\prime}:\mathbb{T}^{n}\to\mathbb{T}^{n}$ be defined by

$g^{\prime}(x_{1},\ldots,x_{n-1},x_{n})=\left(x_{1},\ldots,x_{n-1},2b_{1}x_{1}+2b_{2}x_{2}+\ldots+2b_{n}x_{n}+\epsilon(x_{n})\right),$

where $\epsilon:\mathbb{T}^{1}\to\mathbb{T}^{1}$ is given by
$\epsilon(x)=\frac{1}{n_{0}}\operatorname{sen}(2\pi x)$, with
$n_{0}\in\mathbb{N}$ conveniently chosen. Note that $g^{\prime}$ is homotopic
to $g$ and

$g^{\prime}(x_{1},\ldots,x_{n})=(g^{\prime}\circ\tau_{2})(x_{1},\ldots,x_{n})\Leftrightarrow\left\\{\begin{array}[]{c}x_{1}=0,\frac{1}{2}\\\
x_{2}=0,\frac{1}{2}\\\ \vdots\\\ x_{n}=0,\frac{1}{2}\end{array}.\right.$

Then, we have that the cardinality of the coincidence set of pair
$(g^{\prime},g^{\prime}\circ\tau_{2})$ is equal to $2^{n}$,
$\\#\operatorname{Coin}(g^{\prime},g^{\prime}\circ\tau_{2})=2^{n}$, and the
cardinality of the Borsuk-Ulam coincidence set of pair $(g^{\prime},\tau_{2})$
is $2^{n-1}$, $\\#\operatorname{BUCoin}(g^{\prime},\tau_{2})=2^{n-1}$.
Therefore, there exists $2^{n-1}$ essential Borsuk-Ulam coincidence classes.
Thus, we can conclude that $\operatorname{NBU}(g,\tau_{2})=2^{n-1}$.

The results obtained here for the Nielsen-Borsuk-Ulam number in low dimension
$n-$torus, $n=1,2,3$, and the example of the map $g$ in $\mathbb{T}^{n}$
above, induces the formulation of the following conjecture:

######  Conjecture 6.1.

Let $f:\mathbb{T}^{n}\to\mathbb{T}^{n}$ be a map and $\tau$ a free involution
in $\mathbb{T}^{n}$. Then

$\operatorname{NBU}(f,\tau)=\left\\{\begin{array}[]{cl}2^{n-1}&\text{or }\\\
0.&\end{array}\right.$

## References

  * [1] A. Bauval, D. L. Gonçalves, and C. Hayat, The Borsuk-Ulam theorem for the Seifert manifolds having flat geometry. https://arxiv.org/pdf/1807.00159.pdf, 2018.
  * [2] F. S. Cotrim and D. Vendrúscolo.  Nielsen coincidence theory applied to Borsuk-Ulam geometric problems. Topology Appl., 159(18) (2012), 3738–3745.
  * [3] F. S. Cotrim and D. Vendrúscolo. The Nielsen Borsuk-Ulam number. Bull. Belg. Math. Soc. Simon Stevin,24(4) (2017), 613–619.
  * [4] D. L. Gonçalves. The Borsuk-Ulam theorem for surfaces. Quaest. Math., 29(1) (2006), 117–123.
  * [5] D. L. Gonçalves and J. Guaschi. The Borsuk-Ulam theorem for maps into a surface. Topology Appl., 157(10-11) (2010), 1742–1759.
  * [6] D. L. Gonçalves, J. Guaschi, and V. C. Laass. The Borsuk-Ulam property for homotopy classes of self-maps of surfaces of Euler characteristic zero. J. Fixed Point Theory Appl., 21(2) (2019) Art. 65.
  * [7] J. Hempel. Free cyclic actions on $S^{1}\times S^{1}\times S^{1}$. Proc. Amer. Math. Soc., 48 (1975), 221–227.
  * [8] J. Jezierski and W. Marzantowicz.  Homotopy methods in topological fixed and periodic points theory, volume 3 of Topological Fixed Point Theory and Its Applications. Springer, Dordrecht, 2006.
  * [9] V. C. Laass. A propriedade de Borsuk-Ulam para funções entre superfícies. PhD thesis, Instituto de Matemática e Estatística da Universidade de São Paulo, 2015.
  * [10] J. Matoušek. Using the Borsuk-Ulam theorem. Universitext. Springer-Verlag, Berlin, 2003.