#  Projective Fraïssé limits of trees with confluent epimorphisms

Włodzimierz J. Charatonik  W. J. Charatonik and R. P. Roe, Department of
Mathematics and Statistics
Missouri University of Science and Technology
400 W 12th St
Rolla MO 65409-0020<EMAIL_ADDRESS>,  Aleksandra Kwiatkowska  A. Kwiatkowska,
Institut für Mathematische Logik und Grundlagenforschung, Universität Münster,
Einsteinstrasse 62, 48149 Münster, Germany and Instytut Matematyczny,
Uniwersytet Wrocławski, pl. Grunwaldzki 2/4, 50-384 Wrocław, Poland,
<EMAIL_ADDRESS>,  Robert P. Roe  and  Shujie Yang  S. Yang, Institut
für Mathematische Logik und Grundlagenforschung, Universität Münster,
Einsteinstrasse 62, 48149 Münster, Germany<EMAIL_ADDRESS>
###### Abstract.

We continue the study of projective Fraïssé limits developed by Irwin-Solecki
and Panagiotopoulos-Solecki by investigating families of epimorphisms between
finite trees and finite rooted trees. Ideas of monotone, confluent, and light
mappings from continuum theory as well as several properties of continua are
modified so as to apply them to topological graphs. As the topological
realizations of the projective Fraïssé limits we obtain the dendrite $D_{3}$,
the Mohler-Nikiel universal dendroid, as well as new, interesting continua for
which we do not yet have topological characterizations.

###### Key words and phrases:

Fraïssé limit, topological graph, confluent, monotone, light

###### 2020 Mathematics Subject Classification:

03C98, 54D80, 54E40, 54F15, 54F50

Funded by the Deutsche Forschungsgemeinschaft (DFG, German Research
Foundation) under Germany’s Excellence Strategy EXC 2044–390685587,
Mathematics Münster: Dynamics–Geometry–Structure and by CRC 1442 Geometry:
Deformations and Rigidity.

##  1\. Introduction and Definitions

###  1.1. Introduction

In [15], T. Irwin and S. Solecki introduced the idea of a projective Fraïssé
limit as a dualization of the injective Fraïssé limit from model theory. In
that paper they constructed the pseudo-arc as the topological realization of a
projective Fraïssé limit of a certain family of finite graphs and epimorphisms
between members of the class. Subsequently, D. Bartošová and A. Kwiatkowska,
[3], W. Kubiś and A. Kwiatkowska, [18] and A. Panagiotopoulos and S. Solecki,
[23], extended these ideas to study, repectively, the Lelek fan, the Poulsen
simplex, and the Menger curve as projective Fraïssé limits. Further, recent
results on the applications of the Fraïssé theory to continuum theory are in
[2], [4], [11], [16], [24].

A continuum is a compact connected (metric) space. In continuum theory, a
continuous map $f\colon X\to Y$ between continua is monotone if preimages of
subcontinua are subcontinua. It is confluent if for each subcontinuum $Q$ of
$Y$ each component of $f^{-1}(Q)$ is mapped onto $Q$ by $f$. It is light if
the preimage of each point is discrete. We adapt those ideas of monotone,
confluent, and light mappings from continuum theory as well as several
properties of continua (such as hereditary indecomposability, dendroids,
dendrites) so as to apply them to topological graphs.

In the article we undertake a systematic study of graphs, trees, and rooted
trees with monotone, light, and confluent epimorphisms. We develop tools for
studying projective Fraïssé limits of families of such epimorphisms and
continua obtained as their topological realizations.

In Section 3, we first notice that all finite trees with monotone maps do not
amalgamate (Example 3.12). However, when we put restrictions on the order of
vertices we obtain the well-known Ważewski dendrite $D_{3}$. The main results
of Section 3 are Theorems 3.13 and 3.15.

######  Theorem 1.1.

The topological realization of the projective Fraïssé limit of finite trees
having all vertices of order less than or equal to 3 with monotone
epimorphisms is the Ważewski dendrite $D_{3}$

The main result of Section 4 is:

######  Theorem 1.2.

Finite connected graphs with confluent epimorphisms form a projective Fraïssé
family.

The topological realization of the obtained projective Fraïssé limit as far as
we know has not been investigated in continuum theory, before this work. A
detailed study of that continuum is in the continuation of this article [9].
In contrast, finite connected graphs with monotone epimorphisms also form a
projective Fraïssé family, and the topological realization of the limit is the
well known Menger universal curve [23].

In Sections 5-10, we investigate, in detail, rooted trees with confluent
epimorphism. Inverse limits of continua that are trees with confluent bonding
maps were studied, for example, in [6], [22], [5]. Our investigations are most
importantly motivated by the universal Molher-Nikiel dendroid [21] constructed
as the inverse limits of continua that are rooted trees with confluent bonding
maps, see Section 9. The Mohler-Nikiel dendroid is universal for the class of
smooth dendroids. It has a closed set of end points and is such that each
point is either an end point or a ramification point.

Rooted trees with monotone epimorphisms, unlike connected graphs with monotone
maps, do not amalgamate (see Example 8.1). Any confluent epimorphism between
rooted trees can be written as a composition of monotone and light confluent
epimorphisms (see Theorem 4.14 and Remark 5.14). Furthermore, any light
confluent epimorphism can be further written as a composition of elementary
light confluent epimorphisms (see Definition 5.17). Light confluent
epimorphisms between rooted trees do amalgamate, but inverse limits in this
class will have trivial topological realizations (the transitive closure of
the edge relation will consist of a single equivalence class). It turns out
that by taking all light confluent epimorphisms, but only some monotone ones,
we obtain two interesting Fraïssé families. One of them has the Mohler-Nikiel
continuum as the topological realization, and the other one we believe is
different from any known continua.

An epimorphism $f\colon S\to T$ is simple confluent if the rooted tree $S$ is
obtained from the rooted tree $T$ by repeating the following 2 operations:
take an edge and split it into two edges (simple-monotone epimorphism) or take
a vertex and a component above it and double it (elementary light confluent
epimorphism). An epimorphism $f\colon S\to T$ is simple*-confluent if $S$ is
obtained from $T$ by repeating the following 3 operations: the two operation
as above, and additionally we can pick a vertex and attach a new edge to it
(see Definitions 8.2 and 8.3).

We summarize the main results of Sections 5-10 below.

######  Theorem 1.3.

  1. (1)

If $A,B,C$ are rooted trees, $f\colon B\to A$ is light confluent, and $g\colon
C\to A$ is confluent, then there is a rooted tree $D$, confluent $f_{0}\colon
D\to B$, and light confluent $g_{0}\colon D\to C$ with $f\circ f_{0}=g\circ
g_{0}$ (Theorem 5.16).

  2. (2)

Simple confluent epimorphisms form a projective Fraïssé family (denoted later
as $\mathcal{T}_{\mathcal{CE}}$, see Theorem 8.7).

  3. (3)

Simple*-confluent epimorphisms form a projective Fraïssé family (denoted later
as $\mathcal{T}_{\mathcal{C}}$, see Theorem 8.18).

In Theorems 8.14 and 8.24 we obtain intrinsic characterisations of simple
confluent and simple*-confluent epimorphisms. They are interesting on their
own, but we also will need them to show that the projective Fraïssé limits of
$\mathcal{T}_{\mathcal{CE}}$ and $\mathcal{T}_{\mathcal{C}}$ are well defined.

Main continuum theory consequences are as follows.

######  Theorem 1.4.

  1. (1)

The topological realization of the projective Fraïssé limit of
$\mathcal{T}_{\mathcal{CE}}$ is the Mohler-Nikiel universal dendroid (Theorem
10.1)

  2. (2)

The topological realization of the projective Fraïssé limit of
$\mathcal{T}_{\mathcal{C}}$ is a dendroid with a dense set of end points and
such that each point is either an end point or a ramification point (Theorem
8.30).

###  1.2. Background

We start with some basic definitions and background about projective Fraïssé
families and limits.

A graph is an ordered pair $A=(V(A),E(A))$, where $E(A)\subseteq V(A)^{2}$ is
a reflexive and symmetric relation on $V(A)$. The elements of $V(A)$ are
called vertices of graph $A$ and elements of $E(A)$ are called edges. Given
vertices $a$ and $b$ in a graph $A$, where $a\not=b$, we will use $\langle
a,b\rangle$ to denote an edge in $A$. Note that since $E(A)$ is reflexive
there is always an edge from a vertex in $A$ to itself, such an edge will be
called a degenerate edge. When it is clear from context that $x$ is a vertex
we will use $x\in A$ for $x\in V(A)$.

Given two graphs $A$ and $B$ a function $f\colon V(A)\to V(B)$ is a
homomorphism if it maps edges to edges, i.e. $\langle a,b\rangle\in E(A)$
implies $\langle f(a),f(b)\rangle\in E(B)$. A homomorphism $f$ is an
epimorphism between graphs if it is moreover surjective on both vertices and
edges. An isomorphism is an injective epimorphism.

By a rooted graph we mean a graph $G$ with a distinguished vertex $r(G)$. An
epimorphism $f\colon G\to H$, between rooted graphs, is an epimorphism between
corresponding (unrooted) graphs, such that $f(r(G))=r(H)$.

######  Definition 1.5.

A topological graph (rooted topological graph) $K$ is a graph $(V(K),E(K))$
(respectively, rooted graph $(V(K),E(K),r(K))$, where $r(K)$ is the
distinguished vertex), whose domain $V(K)$ is a 0-dimensional, compact,
second-countable (thus metrizable) space and $E(K)$ is a closed, reflexive and
symmetric subset of $V(K)^{2}$.

An epimorphism between topological graphs is an epimorphism between graphs,
which is additionally continuous.

A topological graph and a rooted topological graph are examples of topological
$\mathcal{L}$-structures. For a general definition of a topological
$\mathcal{L}$-structure see [15].

######  Definition 1.6.

A finite graph $T$ is a tree if for every two distinct vertices $a,b\in T$
there is a unique finite sequence $v_{0}=a,v_{1},\dots,v_{n}=b$ of vertices in
$T$ such that for every $i\in\\{0,1,\dots,n-1\\}$ we have $\langle
v_{i},v_{i+1}\rangle\in E(T)$ and $v_{j}\neq v_{i}$ for $j\neq i$. Let $n$ be
a natural number, a vertex $p\in T$ has order $n$ (${\rm ord}(p)=n$) if there
are $n$ non-degenerate edges in $T$ that contain $p$. If ${\rm ord}(p)=1$ then
$p$ is an end vertex, ${\rm ord}(p)=2$ then $p$ is an ordinary vertex, and if
${\rm ord}(p)\geq 3$ then $p$ a ramification vertex.

The theory of projective Fraïssé limits was developed in [15] and further
refined in [23]. We literally recall their definitions here. We will restrict
to the case of finite (rooted) graphs.

######  Definition 1.7.

Let $\mathcal{F}$ be a family of finite graphs or a family of finite rooted
graphs with a fixed family of epimorphisms among the structures in
$\mathcal{F}$. We say that $\mathcal{F}$ is a projective Fraïssé family if

  1. (1)

$\mathcal{F}$ is countable up to isomorphism, that is, any sub-collection of
pairwise non-isomorphic structures of $\mathcal{F}$ is countable;

  2. (2)

epimorphisms in $\mathcal{F}$ are closed under composition and each identity
map is in $\mathcal{F}$;

  3. (3)

for $B,C\in\mathcal{F}$ there exist $D\in\mathcal{F}$ and epimorphisms $f$ and
$g$ in $\mathcal{F}$ such that $f\colon D\to B$ and $g\colon D\to C$; and

  4. (4)

for every two epimorphisms $f\colon B\to A$ and $g\colon C\to A$ in
$\mathcal{F}$ there exist epimorphisms $f_{0}\colon D\to B$ and $g_{0}\colon
D\to C$ in $\mathcal{F}$ such that $f\circ f_{0}=g\circ g_{0}$, i.e. the
diagram (D1) commutes.

(D1)
${B}$${A}$${D}$${C}$$\scriptstyle{f}$$\scriptstyle{f_{0}}$$\scriptstyle{g_{0}}$$\scriptstyle{g}$

We will refer to property (3) as the joint projection property (JPP) and
property (4) as the projective amalgamation property (AP).

######  Notation 1.8.

Given a sequences $F_{n}$ of finite graphs and epimorphisms $\alpha_{n}\colon
F_{n+1}\to F_{n}$, which are called bonding maps, we denote the inverse
sequence by $\\{F_{n},\alpha_{n}\\}$. For $m>n$ we let
$\alpha_{n}^{m}=\alpha_{n}\circ\dots\circ\alpha_{m-1}$ and note that
$\alpha_{n}^{n+1}=\alpha_{n}$. Given an inverse sequence
$\\{F_{n},\alpha_{n}\\}$ the associated inverse limit space is the subspace of
the product space $\Pi F_{i}$ determined by $\\{(x_{1},x_{2},\ldots):x_{i}\in
F_{i}\text{ and }x_{i}=\alpha_{i}(x_{i+1})\\}$ and is denoted as ${\bf
F}=\varprojlim\\{F_{n},\alpha_{n}\\}$. Further, we denote the canonical
projection from the inverse limit space ${\bf F}$ to the $n$th factor space
$F_{n}$ by $\alpha_{n}^{\infty}$. Note that if
$x,y\in\varprojlim\\{F_{n},\alpha_{n}\\}$ then the metric
$d(x,y)=\sum_{i=1}^{\infty}d_{i}(x_{i},y_{i})/2^{i}$, where $d_{i}$ is the
discrete metric on $F_{i}$, is compatible with the topology of the product
space. So, in particular, if $a\in F_{n}$ then
$\operatorname{diam}((\alpha^{\infty}_{n})^{-1}(a))\leq 1/2^{n-1}$.

The family $\mathcal{F}$ of finite graphs or finite rooted graphs and
epimorphisms is enlarged to a family $\mathcal{F}^{\omega}$ which includes all
topological graphs obtained as inverse limits of graphs in $\mathcal{F}$ with
bonding maps from the family of epimorphisms. If ${\bf
G}=\varprojlim\\{G_{n},\alpha_{n}\\}\in\mathcal{F}^{\omega}$ and $a=(a_{n})$
and $b=(b_{n})$ are elements of ${\bf G}$ then $\langle a,b\rangle$ is an edge
in ${\bf G}$ if and only if for each $n$, $\langle a_{n},b_{n}\rangle$ is an
edge in $G_{n}$. An epimorphism $h$ between a (rooted) topological graph ${\bf
G}=\varprojlim\\{G_{n},\alpha_{n}\\}$ in $\mathcal{F}^{\omega}$ and
$A\in\mathcal{F}$ is in the family $\mathcal{F}^{\omega}$ if and only if there
is an $m$ and an epimorphism $h^{\prime}\colon G_{m}\to A$,
$h^{\prime}\in\mathcal{F}$, such that $h=h^{\prime}\circ\alpha_{m}^{\infty}$.
Finally, if $K$ and $L$ are in $\mathcal{F}^{\omega}$ an epimorphism $h\colon
L\to K$ is in the family $\mathcal{F}^{\omega}$ if and only if for any
$A\in\mathcal{F}$ and any epimorphism $g\colon K\to A$ in
$\mathcal{F}^{\omega}$, $g\circ h\in\mathcal{F}^{\omega}$.

######  Lemma 1.9.

Let $\\{F_{n},\alpha_{n}\\}$ be an inverse sequence of finite (rooted) graphs,
where each $\alpha_{n}$ is an epimorphism. Let ${\bf F}$ denote its inverse
limit.

  1. (1)

If $A$ is a finite (rooted) graph and $f\colon{\bf F}\to A$ is an epimorphism,
then there is $n$ and an epimorphism $\gamma\colon F_{n}\to A$ such that
$f=\gamma\circ\alpha^{\infty}_{n}$.

  2. (2)

Suppose that $\\{G_{m},\beta_{m}\\}$ is another inverse sequence of finite
(rooted) graphs, where each $\beta_{m}$ is an epimorphism, such that ${\bf F}$
is its inverse limit. Then there are subsequences $(n_{i})$ and $(m_{j})$ and
there are epimorphisms $g_{i}\colon F_{n_{i}}\to G_{m_{i}}$ and $h_{i}\colon
G_{m_{i+1}}\to F_{n_{i}}$ such that $\alpha^{n_{i+1}}_{n_{i}}=h_{i}g_{i+1}$
and $\beta^{m_{i+1}}_{m_{i}}=g_{i}h_{i}$.

###### Proof.

For (1), take any $n$ such that any element of the partition
$\\{(\alpha^{\infty}_{n})^{-1}(x)\colon x\in F_{n}\\}$ is contained in some
element of the partition $\\{f^{-1}(x)\colon x\in A\\}$. For $z\in F_{n}$ we
let $\gamma(z)=f(t)$ for any $t\in(\alpha^{\infty}_{n})^{-1}(z)$. Then
$f=\gamma\circ\alpha^{\infty}_{n}$. Since $f$ and $\alpha^{\infty}_{n}$ are
epimorphism, by Lemma 2.1 in [15], $\gamma$ is an epimorphism. For (2), if we
already constructed $(g_{i})$ and $(h_{i})$ for $i<i_{0}$, to get $g_{i_{0}}$
apply part (1) to $f_{1}=\beta^{\infty}_{m_{i_{0}}}$ and get $n_{i_{0}}$ and
$g_{i_{0}}$ such that $f_{1}=g_{i_{0}}\circ\alpha^{\infty}_{n_{i_{0}}}$. Next,
we apply part (1) to $f_{2}=\alpha^{\infty}_{n_{i_{0}}}$ and get
$m_{{i_{0}+1}}$ and $h_{i_{0}}$ such that
$f_{2}=h_{i_{0}}\circ\beta^{\infty}_{m_{i_{0}+1}}$. Those $g_{i}$ and $h_{i}$
are as required. ∎

We need conditions that imply whether or not an epimorphism $h\colon{\bf G}\to
A$ belongs to $\mathcal{F}^{\omega}$ does not depend on the choice of the
inverse limit representation of ${\bf G}$. This is needed for the proof of
Theorem 1.14 below, i.e., for the existence and uniqueness of a projective
Fraïssé limit.

Note that if ${\bf G}=\varprojlim\\{G_{n},\alpha_{n}\\}$, then each
$\alpha^{\infty}_{n}$ is in $\mathcal{F}^{\omega}$.

######  Corollary 1.10.

Let $\mathcal{F}$ be a projective Fraïssé family of finite (rooted) graphs.
Suppose that for all epimorphisms between finite (rooted) graphs $f\colon B\to
A$, $g\colon C\to A$, $\phi\colon C\to B$ such that $f\circ\phi=g$, we have
that if $g\in\mathcal{F}$, then $f\in\mathcal{F}$. Then the definition of
$\mathcal{F}^{\omega}$ does not depend on choices of the inverse limit
representations.

###### Proof.

Let ${\bf
F}=\varprojlim\\{F_{n},\alpha_{n}\\}=\varprojlim\\{G_{n},\beta_{n}\\}$. It
suffices to check that maps $g_{i}$ and $h_{i}$ constructed in Lemma 1.9 are
in $\mathcal{F}$. By our assumption, if
$\alpha^{n_{i+1}}_{n_{i}}=h_{i}g_{i+1}$, then since $\alpha^{n_{i+1}}_{n_{i}}$
is in $\mathcal{F}$, we get that $h_{i}$ is in $\mathcal{F}$. Similarly, if
$\beta^{m_{i+1}}_{m_{i}}=g_{i}h_{i}$, then since $\beta^{m_{i+1}}_{m_{i}}$ is
in $\mathcal{F}$, we have that $g_{i}$ is in $\mathcal{F}$.

Suppose now that $h\colon\varprojlim\\{F_{n},\alpha_{n}\\}\to A$, $m$, and
$h^{\prime}\colon F_{m}\to A$ with $h^{\prime}\in\mathcal{F}$, are such that
$h=h^{\prime}\circ\alpha_{m}^{\infty}$. Find $n_{i}\geq m$ and consider
$h_{i}\colon G_{m_{i+1}}\to F_{n_{i}}$. Then
$h=h^{\prime}\circ\alpha^{n_{i}}_{m}\circ h_{i}\circ\beta^{\infty}_{m_{i+1}}$
and $h^{\prime}\circ\alpha^{n_{i}}_{m}\circ h_{i}\in\mathcal{F}$.

∎

In Section 8, we will need a strengthening of Corollary 1.10.

######  Corollary 1.11.

Let $\mathcal{F}$ be a projective Fraïssé family of finite (rooted) graphs.
Let $\mathcal{F}_{0}\supseteq\mathcal{F}$ be a family of epimorphisms such
that for all epimorphisms $f\colon B\to A$, $g\colon C\to A$, $\phi\colon C\to
B$ if $f\circ\phi=g$, we have: (1) if $g\in\mathcal{F}_{0}$, then
$f\in\mathcal{F}_{0}$, and (2) if $\phi\in\mathcal{F}_{0}$ and
$g\in\mathcal{F}$, then $f\in\mathcal{F}$. Then the definition of
$\mathcal{F}^{\omega}$ does not depend on choices of the inverse limit
representations.

###### Proof.

In the proof of Corollary 1.10, if all $\alpha_{n}$ and $\beta_{n}$ are in
$\mathcal{F}$, then by (1) all $g_{i}$, $h_{i}$ are in $\mathcal{F}_{0}$.
Since $\alpha^{n_{i+1}}_{n_{i}}=h_{i}g_{i+1}$ and
$\beta^{m_{i+1}}_{m_{i}}=g_{i}h_{i}$, for all $i$, by (2) we get
$g_{i},h_{i}\in\mathcal{F}$. Then proceed as in the second paragraph of the
proof of Corollary 1.10.

∎

In the proof of [15, Theorem 2.4] an inverse sequence satisfying certain
properties, see definition below, was introduced and used to show the
existence and uniqueness of the projective Fraïssé limit.

######  Definition 1.12.

Given a projective Fraïssé family $\mathcal{F}$ an inverse sequence
$\\{F_{n},\alpha_{n}\\}$ where $F_{n}\in{\mathcal{F}}$ and $\alpha_{n}\colon
F_{n+1}\to F_{n}$ are epimorphisms in $\mathcal{F}$ is said to be a Fraïssé
sequence for $\mathcal{F}$ if the following two conditions hold.

  1. (1)

For any $G\in{\mathcal{F}}$ there is an $n$ and an epimorphism from $F_{n}$
onto $G$;

  2. (2)

For any $n$, any pair $G,H\in{\mathcal{F}}$, and any epimorphisms $g\colon
H\to G$ and $f\colon F_{n}\to G$ there exists $m>n$ and an epimorphism
$h\colon F_{m}\to H$ such that $g\circ h=f\circ\alpha_{n}^{m}$ i.e the diagram
(D2) commutes.

(D2)
${F_{n}}$${F_{m}}$${G}$${H}$$\scriptstyle{f}$$\scriptstyle{h}$$\scriptstyle{\alpha^{m}_{n}}$$\scriptstyle{g}$

Note that the name Fraïssé sequence has not been standardized. Other names
that have been used include generic sequence and fundamental sequence. Let us
also remark, and we will use it in later sections, that we have the following
characterization of a Fraïssé sequence.

######  Proposition 1.13.

Let $\mathcal{F}$ be a projective Fraïssé family. An inverse sequence
$\\{F_{n},\alpha_{n}\\}$, where $F_{n}\in{\mathcal{F}}$ and $\alpha_{n}\colon
F_{n+1}\to F_{n}$ are epimorphisms, is a Fraïssé sequence if and only if for
each $m$, $A\in\mathcal{F}$ and an epimorphism $\phi\colon A\to F_{m}$ there
exist $n$ and an epimorphism $\psi\colon F_{n}\to A$ such that
$\phi\circ\psi=\alpha^{n}_{m}$.

###### Proof.

Clearly a Fraïssé sequence has the property stated in the proposition; just
take $f={\rm Id}$. The converse is proved in [4, Proposition 2.3]. ∎

An epimorphism $f\colon A\to B$ between (rooted) topological graphs is said to
be an $\varepsilon$-map if for every $x\in B$,
$\operatorname{diam}(f^{-1}(x))<\varepsilon$.

The following theorem was proved in [15] for the case when we take all
epimorphisms, and then refined to the version below in [23].

######  Theorem 1.14.

Let $\mathcal{F}$ be a projective Fraïssé family with a fixed family of
epimorphisms among the structures of $\mathcal{F}$. There exists a unique
topological graph (rooted topological graph) $\mathbb{F}$ in
$\mathcal{F}^{\omega}$ such that

  1. (1)

for each $A\in\mathcal{F}$, there exists an epimorphism from $\mathbb{F}$ to
$A$;

  2. (2)

for $A,B\in\mathcal{F}$ and epimorphisms $f\colon\mathbb{F}\to A$ and $g\colon
B\to A$ there exists an epimorphism $h\colon\mathbb{F}\to B$ such that
$f=g\circ h$.

  3. (3)

For every $\varepsilon>0$ there is $G\in\mathcal{F}$ and an epimorphism
$f\colon\mathbb{F}\to G$ such that $f$ is an $\varepsilon$-map.

This topological graph (rooted topological graph) $\mathbb{F}$ is called the
projective Fraïssé limit of $\mathcal{F}$.

###### Proof.

The uniquness of $\mathbb{F}$ along with the first two conditions are
precisely [23, Theorem 3.1]. For the third condition note that in the proof of
[23, Theorem 3.1] it is shown that the projective Fraïssé limit $\mathbb{F}$
is the inverse limit of a Fraïssé sequence of $\mathcal{F}$. Condition (3)
then follows from the definition of a metric on the inverse limit space. ∎

######  Definition 1.15.

Given a (rooted) topological graph $K$, if $E(K)$ is also transitive then it
is an equivalence relation and $K$ is known as a prespace. The quotient space
$K/E(K)$ is called the topological realization of $K$ and is denoted by $|K|$.

######  Theorem 1.16.

Each compact metric space is a topological realization of a topological graph.

###### Proof.

Let $X$ be a compact metric space and let $f\colon C\to X$ be a surjective
mapping from the Cantor set $C$. Define a topological graph $K$ by putting
$V(K)=C$ and $\langle a,b\rangle\in E(K)$ if and only if $f(a)=f(b)$. Then $K$
is a compact topological graph, $E(K)$ is transitive, and $|K|$ is
homeomorphic to $X$. ∎

##  2\. Connectedness properties of topological graphs

In this section we propose definitions of some connectedness properties of
topological graphs analogous to respective definitions for continua. We have
decided to keep the terminology original to continuum theory. Let us start
with the definitions of connected and locally connected topological graphs as
in [23].

######  Definition 2.1.

Given a topological graph $G$, a subset $S$ of $V(G)$ is disconnected if there
are two nonempty closed subsets $P$ and $Q$ of $S$ such that $P\cup Q=S$ and
if $a\in P$ and $b\in Q$, then $\langle a,b\rangle\notin E(G)$. A subset $S$
of $V(G)$ is connected if it is not disconnected. A graph $G$ is connected if
$V(G)$ is connected.

######  Definition 2.2.

A topological graph $G$ is locally connected if it admits a basis of its
topology consisting of connected sets in the above sense.

######  Proposition 2.3.

If $f\colon G\to H$ is an epimorphism between topological graphs and $G$ is
connected, then $H$ is connected.

###### Proof.

Suppose $H$ is disconnected. Then there are two disjoint nonempty closed
subsets $A$ and $B$ of $V(H)$ such that $V(H)=A\cup B$ and there are no edges
between $A$ and $B$. So $V(G)=f^{-1}(A)\cup f^{-1}(B)$ and $f^{-1}(A)\cap
f^{-1}(B)=\emptyset$. Since $G$ is connected, there are vertices $a\in
f^{-1}(A)$ and $b\in f^{-1}(B)$ such that $\langle a,b\rangle\in E(G)$, and
thus $\langle f(a),f(b)\rangle\in E(H)$, a contradiction. ∎

######  Definition 2.4.

Given a topological graph $G$, a subset $S$ of $V(G)$, and a vertex $a\in S$,
the component of $S$ containing $a$ is the largest connected subset $C$ of $S$
that contains $a$; in other words $C=\bigcup\\{P\subseteq S\colon a\in P\
\text{ and }P\text{ is connected}\\}$.

######  Definition 2.5.

A continuum $K$ is hereditarily unicoherent if for every two non-empty
subcontinua $P$ and $Q$ of $K$ the intersection $P\cap Q$ is connected. The
continuum $K$ is unicoherent if in addition $P\cup Q=K$.

A topological graph $G$ is hereditarily unicoherent if for every two non-empty
closed connected topological graphs, $P$ and $Q$, with $V(P)\subseteq V(G)$,
$V(Q)\subseteq V(G)$, $E(P)\subseteq E(G)$, and $E(Q)\subseteq E(G)$, the
intersection $P\cap Q=(V(P)\cap V(Q),E(P)\cap E(Q))$ is connected. The
topological graph $G$ is unicoherent if in addition $V(P)\cup V(Q)=V(G)$.

The reason why we allow $E(P)\neq E(G)\cap V(P)^{2}$, and $E(Q)\neq E(G)\cap
V(Q)^{2}$ in the definition of hereditary unicoherence, is that, for example,
we do not want complete topological graphs to be dendrites (see Definition
2.10).

######  Notation 2.6.

If $G$ is a topological graph and $S\subseteq V(G)$ by $G\backslash S$ we mean
a topological graph $F$ such that $V(F)=V(G)\backslash S$ and if $e=\langle
a,b\rangle\in E(G)$ then $e\in E(F)$ if and only if $\\{a,b\\}\subseteq V(F)$

######  Definition 2.7.

We say that a topological graph $G$ is an arc if it is connected and there are
two vertices $a,b\in V(G)$ such that for every $x\in V(G)\setminus\\{a,b\\}$
the graph $G\setminus\\{x\\}$ is not connected. The vertices $a$ and $b$ are
called end vertices of the arc and we say that $G$ joins $a$ and $b$.

In topology, there is only one, up to homeomorphism, metric arc, while for
topological graphs we have finite arcs, countable arcs as well as uncountable
arcs.

######  Example 2.8.

Let $G$ be the topological graph whose set of vertices is the Cantor ternary
set, and the edges are defined by $\langle a,b\rangle\in E(G)$ if and only if
$a=b$ or $a$ and $b$ are end vertices of the same deleted interval. Then $G$
is an uncountable arc, whose topological realization is homeomorphic to
$[0,1]$.

######  Definition 2.9.

A topological graph $H$ is an induced subgraph of a topological graph $G$ if
$V(H)\subseteq V(G)$ and for any $x,y\in V(H)$ we have $\langle x,y\rangle\in
E(H)$ iff $\langle x,y\rangle\in E(G)$.

A topological graph $G$ is called arcwise connected if for every two vertices
$a,b\in V(G)$ there is an induced subgraph $A$ of $G$ that is an arc and
contains $a$ and $b$.

######  Definition 2.10.

A hereditarily unicoherent and arcwise connected continuum is a dendroid. A
locally connected dendroid is a dendrite.

A hereditarily unicoherent and arcwise connected topological graph is a
dendroid. A locally connected topological graph, which is a dendroid, is a
dendrite.

######  Proposition 2.11.

For a connected finite graph $G$ the following conditions are equivalent:

  1. (1)

$G$ is a tree;

  2. (2)

$G$ is unicoherent;

  3. (3)

$G$ is a dendroid;

  4. (4)

$G$ is a dendrite;

###### Proof.

The implications $(1)\implies(2)\implies(3)\implies(4)$ follow from
definitions. To see $(4)\implies(1)$ note that a finite dendrite cannot
contain a cycle, so it is a tree. ∎

The following observations use the fact that if a subset of a topological
graph $G$ is connected then its topological realization, viewed as a subspace
of $|G|$, is connected. Moreover, if $\pi\colon G\to|G|$ is the quotient map
and $A\subseteq|G|$ is connected, then $\pi^{-1}(A)$ is connected.

######  Observation 2.12.

If $G$ is a hereditarily unicoherent topological graph and $E(G)$ is
transitive, then its topological realization $|G|$ is a hereditarily
unicoherent continuum.

The quotient map takes arcs to arcs, therefore we have the following
observation. Compare with Lemma 3.7.

######  Observation 2.13.

If a topological graph $G$ is an arc and $E(G)$ is transitive, then its
topological realization $|G|$ is an arc or a point.

######  Observation 2.14.

If a topological graph $G$ is arcwise connected and $E(G)$ is transitive, then
its topological realization $|G|$ is arcwise connected.

######  Observation 2.15.

If a topological graph $G$ is a dendroid and $E(G)$ is transitive, then its
topological realization $|G|$ is a dendroid.

######  Observation 2.16.

If a topological graph $G$ is a dendrite and $E(G)$ is transitive, then its
topological realization $|G|$ is a dendrite.

###### Proof.

We show that if $G$ is locally connected, then so is $|G|$. Let $p\in
U\subseteq|G|$, where $U$ is open, and let $\pi\colon G\to|G|$ be the quotient
map. We claim there is an open connected $V\subseteq G$ such that
$\pi^{-1}(p)\subseteq V\subseteq\pi^{-1}(U)$. To obtain such a $V$ simply take
the union of open connected neighborhoods of vertices of $\pi^{-1}(p)$,
contained in $\pi^{-1}(U)$. This union is connected as $\pi^{-1}(p)$ is
connected. Then $\pi(V)$ is a connected neighborhood of $p$. Indeed, it
contains an open set $U^{\prime}=\\{x\in|G|\colon\pi^{-1}(x)\subseteq V\\}$
and $p\in U^{\prime}$. This implies local connectedness of $|G|$ (see for
example Nadler [20], Exercise 5.22).

∎

The following Theorem provides a sufficient condition in order for a
projective Fraïssé family to have a transitive set of edges in the projective
Fraïssé limit.

######  Theorem 2.17.

Suppose that $\mathcal{G}$ is a projective Fraïssé family of graphs and for
every $G\in\mathcal{G}$, for every $a,b,c\in V(G)$ such that $\langle
a,b\rangle\in E(G)$ and $\langle b,c\rangle\in E(G)$ there is a graph $H$ and
an epimorphism $f^{H}_{G}\colon H\to G$ such that for every vertices $p,q,r\in
V(H)$ such that $f^{H}_{G}(p)=a$, $f^{H}_{G}(q)=b$, and $f^{H}_{G}(r)=c$ we
have $\langle p,q\rangle\notin E(H)$ or $\langle q,r\rangle\notin E(H)$. Then
there are no distinct vertices $a,b,c$ in the projective Fraïssé limit
$\mathbb{G}$ such that the edges $\langle a,b\rangle$ and $\langle b,c\rangle$
are in $\mathbb{G}$ hence $\mathcal{G}$ has a transitive set of edges.
Moreover, for any $a\in\mathbb{G}$ there is at most one $b\in\mathbb{G}$,
$b\neq a$, such that $\langle a,b\rangle\in E(\mathbb{G})$.

###### Proof.

Let $\mathcal{G}$ be a family that satisfies the assumptions of the theorem.
Suppose on the contrary that there are three vertices $a,b,c\in\mathbb{G}$
such that $\langle a,b\rangle,\langle b,c\rangle\in E(\mathbb{G})$. Let a
graph $G\in\mathcal{G}$ and an epimorphism $f_{G}\colon\mathbb{G}\to G$ be
such that $f_{G}(a),f_{G}(b),\text{ and }f_{G}(c)$ are three distinct vertices
of $G$. Then $\langle f_{G}(a),f_{G}(b)\rangle,\langle
f_{G}(b),f_{G}(c)\rangle\in E(G)$, and thus, by our assumption, there is a
graph $H$ and an epimorphism $f^{H}_{G}\colon H\to G$ such that for an
epimorphism $f_{H}\colon\mathbb{G}\to H$ satisfying $f_{G}=f^{H}_{G}\circ
f_{H}$ we have $\langle f_{H}(a),f_{H}(b)\rangle\notin E(H)$ or $\langle
f_{H}(b),f_{H}(c)\rangle\notin E(H)$. This contradicts the fact that $f_{H}$
maps edges to edges. ∎

######  Theorem 2.18.

If $\mathcal{T}$ is a projective Fraïssé family of trees, and $\mathbb{T}$ is
a projective Fraïssé limit of $\mathcal{T}$, then $\mathbb{T}$ is hereditarily
unicoherent.

###### Proof.

Suppose $\mathbb{T}$ is not hereditarily unicoherent and fix a metric $d$ on
$\mathbb{T}$. Then there exist closed connected subsets $P$ and $Q$ of
$V(\mathbb{T})$ such that $P\cap Q$ is not connected. Let $p\in P\backslash Q$
and $q\in Q\backslash P$. Let $A$ and $B$ be disjoint closed sets in
$\mathbb{T}$ such that $p,q\notin A\cup B$, $P\cap Q\subseteq A\cup B$,
$A\cap(P\cap Q)\neq\emptyset$, $B\cap(P\cap Q)\neq\emptyset$, and
$d(P\setminus(A\cup B),Q\setminus(A\cup B))>0$ (i.e.
$\overline{P\setminus(A\cup B)}$ and $\overline{Q\setminus(A\cup B)}$ are
disjoint). Let $\varepsilon>0$ be such that $d(A,B)>\varepsilon$ and
$d(P\setminus(A\cup B),Q\setminus(A\cup B))>\varepsilon$.

By Condition 3 of Theorem 1.14, there exists a tree $G$ and an
$\varepsilon$-map, $f_{G}\colon\mathbb{T}\to G$, such that $f_{G}(p)\not\in
f_{G}(Q)$ and $f_{G}(q)\not\in f_{G}(P)$. In particular, none of the
$f_{G}(P)$ and $f_{G}(Q)$ is contained in the other. The sets $f_{G}(P)$ and
$f_{G}(Q)$ are clearly closed and connected. Moreover, $f_{G}(P)\cap
f_{G}(Q)\subseteq f_{G}(A)\cup f_{G}(B)$. Indeed, if $x\in P$ and $y\in Q$ are
such that $f_{G}(x)=f_{G}(y)\notin f_{G}(A)\cup f_{G}(B)$, then $x\in
P\setminus(A\cup B)$, $y\in Q\setminus(A\cup B)$, and hence
$d(x,y)>\varepsilon$, which is impossible as $f_{G}$ is an $\varepsilon$-map.
As $f_{G}(A)$ and $f_{G}(B)$ are disjoint closed sets satisfying
$f_{G}(A)\cap(f_{G}(P)\cap f_{G}(Q))\supseteq f_{G}(A\cap P\cap
Q)\neq\emptyset$ and $f_{G}(B)\cap(f_{G}(P)\cap f_{G}(Q))\neq\emptyset$, this
last containment implies that $f_{G}(P)\cap f_{G}(Q)$ is not connected. This
contradicts the fact that $G$ is a tree.

∎

##  3\. Monotone epimorphisms

In this section we investigate a family of finite trees with monotone
epimorphisms. We show that this family, which we denote by
$\mathcal{T}_{\mathcal{M}3}$, is a projective Fraïssé family (Theorem 3.13)
and that the topological realization of $\mathcal{T}_{\mathcal{M}3}$ is
homeomorphic to $D_{3}$, the standard universal denrite of order 3, also known
as the Ważewski denrite of order 3 (Theorem 3.15).

Kwiatkowska [17] uses an injective Fraïssé construction of $D_{3}$ and other
Ważewski dendrites to study their groups of homeomorphisms. Further results on
the homeomorphism groups of Ważewski dendrites are in Duchesne [13].

First, we want to show an example that the family of all epimorphisms between
trees, even trees with all vertices having order less than or equal to three,
do not have amalgams that are connected graphs.

######  Example 3.1.

There exist a triod $T$, arcs $I$ and $J$ for which there is no connected
graph $G$ and epimorphisms $f_{0}$ and $g_{0}$ such that the diagram below
commutes.

(D2)
${I}$${T}$${G}$${J}$$\scriptstyle{f}$$\scriptstyle{f_{0}}$$\scriptstyle{g_{0}}$$\scriptstyle{g}$

The triod $T$ has the center $b$ and the end vertices $a$, $c$, and $d$; the
arc $I$ has vertices $p_{1},p_{2},p_{3},p_{4}$, and $p_{5}$, similarly the arc
$J$ has vertices $q_{1},q_{2},q_{3},q_{4}$, and $q_{5}$. The epimorphisms $f$
and $g$ are pictured below. Precisely, we have $f(p_{1})=d$, $f(p_{2})=b$,
$f(p_{3})=c$, $f(p_{4})=b$, and $f(p_{5})=a$; similarly, $g(q_{1})=d$,
$g(q_{2})=b$, $g(q_{3})=a$, $g(q_{4})=b$, and $g(q_{5})=c$.

$d$$a$$b$$c$$p_{1}$$p_{2}$$p_{3}$$p_{4}$$p_{5}$$d$$a$$b$$c$$q_{1}$$q_{2}$$q_{3}$$q_{4}$$q_{5}$$f\colon
I\to T$$g\colon J\to T$

Suppose that there is a connected graph $G$ and epimorphisms $f_{0}\colon G\to
I$ and $g_{0}\colon G\to J$ such that the diagram (D2) commutes. Let
$x_{0}\in(f_{0})^{-1}(p_{1})$ and let $x_{0},x_{1},\dots x_{n}$ be a sequence
of vertices of $G$ such that:

  1. (1)

for each $i\in\\{0,1,\dots,n-1\\}$ we have $\langle x_{i},x_{i+1}\rangle\in
E(G)$;

  2. (2)

$x_{0},x_{1},\dots x_{n-1}\in(f\circ f_{0})^{-1}(\\{b,d\\})$;

  3. (3)

$x_{n}\notin(f\circ f_{0})^{-1}(\\{b,d\\})$.

Then we have

  1. (1)

$f_{0}(x_{0})=p_{1}$,

  2. (2)

$f_{0}(x_{0}),f_{0}(x_{1}),\dots f_{0}(x_{n-1})\in\\{p_{1},p_{2}\\}$;

  3. (3)

$f_{0}(x_{n})=p_{3}$;

  4. (4)

$g_{0}(x_{0})=q_{1}$,

  5. (5)

$g_{0}(x_{0}),g_{0}(x_{1}),\dots g_{0}(x_{n-1})\in\\{q_{1},q_{2}\\}$;

  6. (6)

$g_{0}(x_{n})=q_{3}$.

By conditions (3) and (6) we have that $f(f_{0}(x_{n}))=c$, while
$g(g_{0}(x_{n}))=a$, so the diagram (D2) does not commute, a contradiction.

######  Definition 3.2.

Given two topological graphs $G$ and $H$ an epimorphism $f\colon G\to H$ is
called monotone if the preimage of a closed connected subset of $H$ is a
connected subset of $G$.

The following is an immediate consequence of the definition.

######  Remark 3.3.

If $f\colon G\to H$ is a monotone epimorphism between topological graphs,
$U,V\subseteq G$ are connected and disjoint, then $f(U)$ and $f(V)$ are
disjoint.

######  Lemma 3.4.

Let $G$ and $H$ be topological graphs and let $f\colon G\to H$ be an
epimorphism. If the preimage of every vertex in $H$ is connected, then $f$ is
monotone.

###### Proof.

Suppose towards the contradiction that the preimage of a closed connected
subset $C$ of $V(H)$ is not connected. Let $D,E\subseteq G$ be disjoint
nonempty closed sets such that $f^{-1}(C)=D\cup E$. By the assumption, note
that for every $c\in C$, we have $f^{-1}(c)\subseteq D$ or $f^{-1}(c)\subseteq
E$. Therefore $f(D)$ and $f(E)$ are disjoint. Since they are closed nonempty,
and $C=f(D)\cup f(E)$, we get a contradiction with connectedness of $C$. ∎

######  Lemma 3.5.

Consider the epimorphisms $f\colon F\to G$ and $g\colon G\to H$ between
topological graphs.

  * (a)

If $f$ and $g$ are monotone, then so is $g\circ f$.

  * (b)

If the composition $g\circ f\colon F\to H$ is monotone, then $g$ is monotone.

###### Proof.

Part (a) is clear. For part (b), write $h=g\circ f$ and let $c\in V(H)$. Then
$g^{-1}(c)=f(h^{-1}(c))$. By the assumption, $h^{-1}(c)$ is connected. Since
images by epimorphisms of connected sets are connected, $g^{-1}(c)$ is
connected. ∎

The following example shows that the concept of monotone epimorphism for
graphs is not exactly how continuum theory sees monotone maps between
continua.

######  Example 3.6.

There is a monotone epimorphism from a cyclic graph onto an arc.

Let $G$ be a complete graph with three vertices $a,b$, and $c$, i.e. a graph
where $E(G)=V(G)^{2}$, and let $H$ be an arc with two vertices $p$ and $q$,
and $E(H)=V(H)^{2}$. Define $f\colon G\to H$ by $f(a)=p$ and $f(b)=f(c)=q$.
The reader can verify that $f$ is a monotone epimorphism.

The following result, which is well known in continuum theory, also holds in
the setting of topological graphs and is used in Theorem 3.15.

######  Lemma 3.7.

If $f\colon G\to H$ is a monotone epimorphism between topological graphs and
$G$ is an arc, then $H$ is an arc and the images of end vertices of $G$ are
end vertices of $H$.

###### Proof.

Denote the end vertices of $G$ by $a$ and $b$. We need to show that every
vertex in $H\setminus\\{f(a),f(b)\\}$ disconnects $H$. Let $y\in
H\setminus\\{f(a),f(b)\\}$; then, since $G$ is an arc the graph $G\setminus
f^{-1}(y)$ is disconnected. Let $G\setminus f^{-1}(y)$ be the union of two
disjoint graphs $G\setminus f^{-1}(y)=U\cup V$. Thus
$H\setminus\\{y\\}=f(U)\cup f(V)$ and $f(U)\cap f(V)=\emptyset$, so
$H\setminus\\{y\\}$ is disconnected as needed. ∎

It is well known in continuum theory that the inverse limit of arcs with
monotone maps is an arc. It is easy to see that the same result holds for
graphs.

######  Lemma 3.8.

The inverse limit of finite arcs with monotone bonding epimorphisms is an arc.

######  Proposition 3.9.

If $\mathcal{G}$ is a projective Fraïssé family of graphs with monotone
epimorphisms and $\mathbb{G}$ is a projective Fraïssé limit of $\mathcal{G}$,
then for every $G\in\mathcal{G}$ any epimorphism $f_{G}\colon\mathbb{G}\to G$
is monotone.

###### Proof.

Let $\\{F_{n},\alpha_{n}\\}$ be a Fraïssé sequence for $\mathcal{F}$. Fix $m$
and let $x\in F_{m}$. We show that $T=(\alpha^{\infty}_{m})^{-1}(x)$ is
connected. By monotonicity of each $\alpha_{m}^{n}$, we have that for every
$n>m$, $(\alpha^{n}_{m})^{-1}(x)$ is connected. Suppose towards the
contradiction that $T$ is not connected. Then there exist nonempty closed
disjoint sets $R$ and $S$ such that $R\cup S=T$. Choose $\varepsilon>0$ such
that $d(a,b)>\varepsilon$ for every $a\in R$ and $b\in S$, where $d$ is a
metric on $\mathbb{G}$. Let $n_{0}$ be large enough so that
$\alpha^{\infty}_{n_{0}}$ is an $\varepsilon$-map. Then
$\alpha^{\infty}_{n_{0}}(R)$ and $\alpha^{\infty}_{n_{0}}(S)$ are two disjoint
nonempty closed subsets of $F_{n_{0}}$ such that
$(\alpha^{n_{0}}_{m})^{-1}(x)=\alpha^{\infty}_{n_{0}}(R)\cup\alpha^{\infty}_{n_{0}}(S)$,
contrary to monotonicity of $\alpha_{m}^{n_{0}}$. This finishes the proof of
monotonicity of each $\alpha^{\infty}_{n}$.

Finally, there is an $m_{0}$ and a monotone epimorphism $f\colon F_{m_{0}}\to
G$ such that $f_{G}=f\circ\alpha_{m_{0}}^{\infty}$. As
$\alpha_{m_{0}}^{\infty}$ is monotone, so is $f_{G}$.

∎

######  Proposition 3.10.

If $\mathcal{T}$ is a projective Fraïssé family of finite trees with monotone
epimorphisms, and $\mathbb{T}$ is a projective Fraïssé limit of $\mathcal{T}$,
then $\mathbb{T}$ is arcwise connected.

###### Proof.

Let $\\{T_{n},\alpha_{n}\\}$ be a Fraïssé sequence for $\mathbb{T}$ and take
$a=(a_{n}),b=(b_{n})\in\mathbb{T}$. For each $n$ let $I_{n}$ be the arc
joining $a_{n}$ and $b_{n}$. Since each tree $T_{n}$ is uniquely arcwise
connected, we have $\alpha_{n}(I_{n+1})=I_{n}$. Therefore
$\varprojlim\\{I_{n},\alpha_{n}|I_{n}\\}$ is an arc joining $a$ and $b$.

∎

######  Corollary 3.11.

If $\mathcal{T}$ is a projective Fraïssé family of trees with monotone
epimorphisms, and $\mathbb{T}$ is a projective Fraïssé limit of $\mathcal{T}$,
then $\mathbb{T}$ is a dendrite.

###### Proof.

The limit $\mathbb{T}$ is arcwise connected by Proposition 3.10, it is
hereditarily unicoherent by Theorem 2.18, and it is locally connected by [23,
Theorem 2.1]. ∎

The next example shows that it is not always possible to find an amalgamation
of trees with monotone epimorphisms when the order of some of the ramification
vertices is greater than 3.

######  Example 3.12.

Let $A$ be the 4-od with end vertices $a$, $b$, $c$, $d$ and order 4
ramification vertex $x$. Let $B$ be the tree having end vertices $a_{B}$,
$b_{B}$, $c_{B}$, $d_{B}$ and order 3 ramification vertices $x_{1}$ and
$x_{2}$ with edges $\langle a_{B},x_{1}\rangle$, $\langle b_{B},x_{1}\rangle$,
$\langle x_{1},x_{2}\rangle$, $\langle c_{B},x_{2}\rangle$, and $\langle
d_{B},x_{2}\rangle$ and let $C$ be the tree having end vertices $a_{C}$,
$b_{C}$, $c_{C}$, $d_{C}$ and order 3 ramification vertices $x^{\prime}_{1}$
and $x^{\prime}_{2}$ with edges $\langle a_{C},x^{\prime}_{1}\rangle$,
$\langle c_{C},x^{\prime}_{1}\rangle$, $\langle
x^{\prime}_{1},x^{\prime}_{2}\rangle$, $\langle b_{C},x^{\prime}_{2}\rangle$,
and $\langle d_{C},x^{\prime}_{2}\rangle$. Let $f\colon B\to A$ be the
monotone epimorphism such that $f(x_{1})=f(x_{2})=x$ and otherwise maps
corresponding letters. Finally, let $g\colon C\to A$ be the monotone
epimorphism such that $g(x_{1}^{\prime})=g(x_{2}^{\prime})=x$ and otherwise
maps corresponding letters.

$x$$a$$b$$c$$d$$A$$a_{B}$$x_{1}$$x_{2}$$c_{B}$$b_{B}$$d_{B}$$B$$b_{C}$$x^{\prime}_{2}$$x^{\prime}_{1}$$a_{C}$$d_{C}$$c_{C}$$C$$f$$g$$D$$f_{0}$$g_{0}$

Assume towards the contradiction that there is a tree $D$ and monotone
epimorphisms $f_{0}\colon D\to B$ and $g_{0}\colon D\to C$ such that $f\circ
f_{0}=g\circ g_{0}$. If $a_{D},b_{D},c_{D},d_{D}$ are vertices in $D$ mapped
by $f\circ f_{0}$ to $a,b,c,d$, respectively, then pick arcs: $I_{ab}$ joining
$a_{D}$ and $b_{D}$, $I_{bc}$ joining $b_{D}$ and $c_{D}$, $I_{cd}$ joining
$c_{D}$ and $d_{D}$, and $I_{da}$ joining $d_{D}$ and $a_{D}$. Then, by the
monotonicity of $f_{0}$ and $g_{0}$, we have
$f_{0}(I_{ab})=\\{a_{B},x_{1},b_{B}\\}$,
$f_{0}(I_{cd})=\\{c_{B},x_{2},d_{B}\\}$,
$g_{0}(I_{bc})=\\{a_{C},x^{\prime}_{1},c_{C}\\}$, and
$g_{0}(I_{da})=\\{b_{C},x^{\prime}_{2},d_{C}\\}$. Therefore $I_{ab}\cap
I_{cd}=\emptyset$ and $I_{bc}\cap I_{da}=\emptyset$. It is not hard to see
that $I_{ab}\cup I_{bc}\cup I_{cd}\cup I_{da}$ contains a cycle, which
contradicts that $D$ is a tree.

Let $\mathcal{T}_{\mathcal{M}\it 3}$ be the family of trees having at least
two vertices, such that all vertices have order $\leq 3$, with monotone
epimorphisms.

######  Theorem 3.13.

$\mathcal{T}_{\mathcal{M}\it 3}$ has the JPP and AP.

###### Proof.

For the proof of the JPP, let $B,C\in\mathcal{T}_{\mathcal{M}\it 3}$ and let
$e_{B}$ and $e_{C}$ be fixed end vertices of $B$ and $C$ respectively. Take a
tree $D$ to be the disjoint union of $B$ and $C$ with the vertices $e_{B}$ and
$e_{C}$ identified. Let $f\colon D\to B$ be the epimorphism that is the
identity on $B$ and takes all vertices of $C$ to $e_{B}$. Similarly define
$g\colon D\to C$.

For the proof of the AP, let $A,B,C$ be trees such that each vertex has order
$\leq 3$, and let $f\colon B\to A$ and $g\colon C\to A$ be monotone.

Observe first that if $x\in A$ is such that ${\rm ord}(x)=3$ and $a,b,c$ are
all vertices adjacent to $x$, then for any $a_{1}\in f^{-1}(a)$, $b_{1}\in
f^{-1}(b)$ and $c_{1}\in f^{-1}(c)$, there is a unique $x_{1}\in f^{-1}(x)$
such that any two of the arcs: $[a_{1},x_{1}],[b_{1},x_{1}],[c_{1},x_{1}]$
intersect exactly in $x_{1}$. Indeed, let $x_{a},x_{b},x_{c}\in f^{-1}(x)$,
$a^{\prime}\in f^{-1}(a)$, $b^{\prime}\in f^{-1}(b)$, and $c^{\prime}\in
f^{-1}(c)$ be such that $\langle x_{a},a^{\prime}\rangle,\langle
x_{b},b^{\prime}\rangle,\langle x_{c},b^{\prime}\rangle\in E(B)$. By the
monotonicity of $f$ there is a unique choice of such vertices. In the case
that two or three of $x_{a},x_{b},x_{c}$ are equal, we take $x_{1}$ to be that
common vertex. In the case when $x_{a},x_{b},x_{c}$ are pairwise different, we
let $p$ to be the arc joining $x_{a}$ and $x_{b}$. Since $B$ is a tree there
is a unique arc $q$ joining $x_{c}$ and the arc $p$. The common vertex of $p$
and $q$ is the required $x_{1}$. Call such a point $x_{1}$ a center for $f$
and $x$.

We first focus on $f\colon B\to A$. Now let $i_{f}\colon A\to B$ be the
following injection. Given $x\in A$, if ${\rm ord}(x)=3$ we let $i_{f}(x)$ to
be the center for $f$ and $x$, if ${\rm ord}(x)=2$, let $i_{f}(x)$ be any
vertex in $f^{-1}(x)$, and if ${\rm ord}(x)=1$, let $i_{f}(x)$ be any end
vertex in $f^{-1}(x)$. We analogously define $i_{g}$ for $g$.

Fix an edge $e=\langle a,b\rangle\in E(A)$ and let

$i_{f}(a)=a^{B}_{0},a^{B}_{1},a^{B}_{2},\ldots,a^{B}_{m_{B}},b^{B}_{1},b^{B}_{2},\ldots,b^{B}_{n_{B}},b^{B}_{n_{B}+1}=i_{f}(b)$

be the arc joining $i_{f}(a)$ and $i_{f}(b)$, where $f(a^{B}_{i})=a$,
$i=1,2,\ldots,m_{B}$, and $f(b^{B}_{j})=b$, $j=1,2,\ldots,n_{B}$. Then, for
each $i\in\\{1,\ldots,m_{B}\\}$, $B\setminus\\{a^{B}_{i}\\}$ has a component
containing $i_{f}(a)$, a component containing $i_{f}(b)$, and possibly a third
component. If there is a third component denote it by $\hat{a}^{B}_{i}$, and
note that $f(\hat{a}^{B}_{i})=a$. Furthermore, it can happen that ${\rm
ord}(a)=2$ but ${\rm ord}(i_{f}(a))=3$. In that case there is a unique
component $\hat{a}^{B}$ of $B\setminus\\{i_{f}(a)\\}$, not containing vertices
from $i_{f}(A)$ such that $f(\hat{a}^{B})=a$. Let
$x^{B}_{i}\in\hat{a}^{B}_{i}$ be such that $\langle
a^{B}_{i},x^{B}_{i}\rangle\in E(B)$, $i=1,2,\ldots,m_{B}$, in case
$\hat{a}^{B}_{i}\neq\emptyset$. Let $a^{B}\in\hat{a}^{B}$ be such that
$\langle a^{B},i_{f}(a)\rangle\in E(B)$, in case $\hat{a}^{B}\neq\emptyset$.
In a similar manner define components $\hat{b}^{B}_{j}$ for each $b_{j}$,
$j=1,2,\ldots,n_{B}$. Some of those components can be empty. This time we have
that $f(\hat{b}_{j}^{B})=b$. Let $y^{B}_{j}\in\hat{b}^{B}_{j}$ be such that
$\langle b^{B}_{j},y^{B}_{j}\rangle\in E(B)$, $j=1,2,\ldots,n_{B}$, in case
$\hat{b}^{B}_{j}\neq\emptyset$.

We do a similar analysis for $g$ and $C$ and an edge $e=\langle a,b\rangle\in
E(A)$. If
$i_{g}(a)=a^{C}_{0},a^{C}_{1},a^{C}_{2},\ldots,a^{C}_{m_{C}},b^{C}_{1},b^{C}_{2},\ldots,b^{C}_{n_{C}},b^{C}_{n_{C}+1}=i_{g}(b)$
is the arc joining $i_{g}(a)$ and $i_{g}(b)$, we define components
$\hat{a}^{C}_{i}$, $\hat{b}^{C}_{j}$, and vertices $x^{C}_{i}$, $y^{C}_{j}$,
analogously as above. Define also $\hat{a}^{C}$ and $a^{C}$.

To get $D$ we just have to put together all those new components. Here is how
we build $D$. To simplify the notation, below we will often write $a$ for
$i_{f}(a)$ and for $i_{g}(a)$, where $a$ is a vertex in $A$. For every edge
$e=\langle a,b\rangle\in E(A)$ we do the following. Replace $e$ by edges:

$\langle a,a^{B}_{1}\rangle,\ldots,\langle
a^{B}_{m_{B}},a^{C}_{1}\rangle,\ldots,\langle
a^{C}_{m_{C}},b^{B}_{1}\rangle,\ldots,\langle
b^{B}_{n_{B}},b^{C}_{1}\rangle,\ldots,\langle b^{C}_{n_{C}},b\rangle.$

Then we attach $\hat{a}^{B}_{i}$, $i=1,\ldots,m_{B}$, $\hat{a}^{C}_{i}$,
$i=1,\ldots,m_{C}$, as well as $\hat{b}^{B}_{j}$, $j=1,2,\ldots,n_{B}$,
$\hat{b}^{C}_{j}$, $j=1,2,\ldots,n_{C}$, in appropriate places. Specifically,
take the disjoint union of all nonempty $\hat{a}^{B}_{i}$, $\hat{a}^{C}_{i}$,
$\hat{b}^{B}_{j}$, $\hat{b}^{C}_{j}$, as well as the path
$a,a^{B}_{1},\ldots,a^{B}_{m_{B}},a^{C}_{1},\ldots,a^{C}_{m_{C}},b^{B}_{1},\ldots,b^{B}_{n_{B}},b^{C}_{1},\ldots,b^{C}_{n_{C}},b$
and add appropriate edges $\langle x^{B}_{i},a^{B}_{i}\rangle$, $\langle
x^{C}_{i},a^{C}_{i}\rangle$, $\langle y^{B}_{j},b^{B}_{j}\rangle$, $\langle
y^{C}_{j},b^{C}_{j}\rangle$. Next, for every $a\in A$ such that ${\rm
ord}(a)=2$, take a vertex $a^{\prime}$ and an edge $\langle
a,a^{\prime}\rangle$. In case $\hat{a}^{B}$ was nonempty take it together with
the edge $\langle a^{\prime},a^{B}\rangle$. Similarly, if $\hat{a}^{C}$ was
nonempty take it together with the edge $\langle a^{\prime},a^{C}\rangle$.
This finishes the definition of $D$.

We now have to specify monotone epimorphisms $f_{0}\colon D\to B$ and
$g_{0}\colon D\to C$ such that $f\circ f_{0}=g\circ g_{0}$. Let us first
define $f_{0}$. For every edge $e=\langle a,b\rangle\in E(A)$, let
$f_{0}(\hat{a}_{i}^{C})=f_{0}(a_{i}^{C})=a_{m_{B}}^{B}$,
$f_{0}(\hat{b}_{j}^{C})=f_{0}(b_{j}^{C})=b$, and for every $a\in V(A)$, let
$f_{0}(\hat{a}^{C})=f_{0}(a^{\prime})=a$. We let $f_{0}$ to be the identity
otherwise. We now define $g_{0}$. For every edge $e=\langle a,b\rangle\in
E(A)$, let $g_{0}(\hat{a}_{i}^{B})=g_{0}(a_{i}^{B})=a$,
$g_{0}(\hat{b}_{j}^{B})=g_{0}(b_{j}^{B})=b_{1}^{C}$, and for every $a\in
V(A)$, let $g_{0}(\hat{a}^{B})=g_{0}(a^{\prime})=a$. We let $g_{0}$ to be the
identity otherwise. The maps $f_{0},g_{0}$ are as required.

∎

The Ważewski denrite $D_{3}$, see for example [7], is a dendrite characterized
by the following conditions:

  1. (1)

each ramification point is of order 3;

  2. (2)

the set of ramification points is dense.

We show that the topological realization of the projective Fraïssé limit of
the family $\mathcal{T}_{\mathcal{M}\it 3}$ is homeomorphic to the dendrite
$D_{3}$. We divide the proof into two steps.

######  Theorem 3.14.

The projective Fraïssé limit of $\mathcal{T}_{\mathcal{M}\it 3}$ has
transitive set of edges and the topological realization of the projective
Fraïssé limit of $\mathcal{T}_{\mathcal{M}\it 3}$ is a dendrite with a dense
set of ramification points.

###### Proof.

First, we show that the family $\mathcal{T}_{\mathcal{M}\it 3}$ satisfies the
hypothesis of Theorem 2.17 and thus has a transitive set of edges. Let
$G\in\mathcal{T}_{\mathcal{M}\it 3}$ and $a,b,c\in V(G)$ such that $\langle
a,b\rangle,\langle b,c\rangle\in E(G)$. Define
$H\in\mathcal{T}_{\mathcal{M}\it 3}$ by
$V(H)=V(G)\cup\\{a^{\prime},b^{\prime}\\}$, $E(H)=E(G)\cup\\{\langle
a,a^{\prime}\rangle,\langle a^{\prime},b^{\prime}\rangle,\langle
b^{\prime},b\rangle\\}\setminus\\{\langle a,b\rangle\\}$, and the epimorphism
$f^{H}_{G}$ where $f^{H}_{G}(a^{\prime})=a$, $f^{H}_{G}(b^{\prime})=b$ and
$f^{H}_{G}$ is the identity otherwise. Then, for $p\in\\{a,a^{\prime}\\}$,
$q\in\\{b,b^{\prime}\\}$, and $r=c$ we have either $\langle p,q\rangle\not\in
E(H)$ or $\langle q,r\rangle\not\in E(H)$.

The projective Fraïssé limit $\mathbb{D}$ of $\mathcal{T}_{\mathcal{M}\it 3}$
is a dendrite by Corollary 3.11, so its topological realization $|\mathbb{D}|$
is a (topological) dendrite by Observation 2.16. It remains to show that the
set of ramification points of $|\mathbb{D}|$ is dense.

Let us introduce the necessary notation. Let
$\varphi\colon\mathbb{D}\to|\mathbb{D}|$ be the quotient mapping. To prove the
density of the set of ramification points of $|\mathbb{D}|$ suppose on the
contrary that $U$ is an open connected subset of $|\mathbb{D}|$ that contains
no ramification point, i.e $U$ is homeomorphic to $(0,1)$. Then
$\varphi^{-1}(U)$ is an open subset of $\mathbb{D}$. By Property (3) of
Theorem 1.14, there is a tree $G$, an epimorphism $f_{G}\colon\mathbb{D}\to G$
in $\mathcal{T}_{\mathcal{M}\it 3}$, and $a\in V(G)$ such that
$f_{G}^{-1}(a)\subseteq\varphi^{-1}(U)$. Define a graph $H$ by putting
$V(H)=V(G)\cup\\{b,c,d,a^{\prime},c^{\prime},d^{\prime}\\}$ and
$E(H)=E(G)\cup\\{\langle a^{\prime},b\rangle,\langle
b,c^{\prime}\rangle,\langle b,d^{\prime}\rangle,\langle
a^{\prime},a\rangle,\langle c,c^{\prime}\rangle,\langle
d,d^{\prime}\rangle\\}$. Finally, define the monotone epimorphism
$f^{H}_{G}\colon H\to G$ by

$f^{H}_{G}(x)=\begin{cases}x\text{ if }x\in G\\\ a\text{ if
}x\in\\{b,c,d,a^{\prime},c^{\prime},d^{\prime}\\}.\end{cases}$

Let $f_{H}\colon\mathbb{D}\to H$ be an epimorphism such that
$f_{G}=f^{H}_{G}\circ f_{H}$; note that by Proposition 3.9 the epimorphism
$f_{H}$ is monotone. Thus, the three sets
$A=\varphi(f_{H}^{-1}(\\{a,a^{\prime},b\\}))$,
$B=\varphi(f_{H}^{-1}(\\{b,c,c^{\prime}\\}))$, and
$C=\varphi(f_{H}^{-1}(\\{b,d,d^{\prime}\\}))$ are continua that satisfy $A\cup
B\cup C\subseteq U$, $A\not\subseteq B\cup C$, $B\not\subseteq A\cup C$,
$C\not\subseteq A\cup B$, and $A\cap B\cap C\not=\emptyset$. For example,
$A\not\subseteq B\cup C$ follows from the fact that
$\phi(f_{H}^{-1}(a))\subseteq A\setminus(B\cup C)$. This contradicts the fact
that $U$ is homeomorphic to $(0,1)$.

∎

######  Theorem 3.15.

The topological realization of the projective Fraïssé limit of
$\mathcal{T}_{\mathcal{M}\it 3}$ is homeomorphic to the dendrite $D_{3}$ .

###### Proof.

By Theorem 3.14 the topological realization $|\mathbb{D}|$ of the projective
Fraïssé limit of $\mathcal{T}_{\mathcal{M}\it 3}$ is a dendrite with a dense
set of ramification points; it remains to show that every ramification point
of $|\mathbb{D}|$ has order 3. Suppose on the contrary that $p\in|\mathbb{D}|$
is a point of order 4 or more. Then there are four arcs $A,B,C,D$ in
$|\mathbb{D}|$ such that the intersection of any two of them is $\\{p\\}$.

Denote $\varphi^{-1}(p)=\\{p_{1},p_{2}\\}$ with a possibility that
$p_{1}=p_{2}$ and note that the sets $\varphi^{-1}(A)$, $\varphi^{-1}(B)$,
$\varphi^{-1}(C)$, and $\varphi^{-1}(D)$ contains arcs
$\alpha,\beta,\gamma,\delta$ in $\mathbb{D}$ and that the intersection of any
two of these arcs is $\\{p_{1},p_{2}\\}$.

Let $G$ be a tree and $f_{G}\colon\mathbb{D}\to G$ be a monotone epimorphism
such that:

  1. (1)

$f_{G}(\alpha)$, $f_{G}(\beta)$, $f_{G}(\gamma)$, and $f_{G}(\delta)$ are non-
degenerate subsets of $V(G)$;

  2. (2)

for any two adjacent vertices of $G$ at most one of them has order 3.

We can achieve (2) since for any $G\in\mathcal{T}_{\mathcal{M}\it 3}$, if we
split each edge of $G$ into two, that is, replace each edge $\langle
a,b\rangle$ of $G$ by the edges $\langle a,x\rangle$ and $\langle x,b\rangle$
where $x$ is a “new” vertex, the resulting graph will have the required
property. By Lemma 3.7 the sets $f_{G}(\alpha)$, $f_{G}(\beta)$,
$f_{G}(\gamma)$, and $f_{G}(\delta)$ are arcs. We claim that the intersection
of any two of them is $\\{f_{G}(p_{1}),f_{G}(p_{2})\\}$. Suppose on the
contrary that $r\in f_{G}(\alpha)\cap
f_{G}(\beta)\setminus\\{f_{G}(p_{1}),f_{G}(p_{2})\\}$; then $f_{G}^{-1}(r)$ is
a connected subset of $V(\mathbb{D})$ that intersects both
$\alpha\setminus\\{p_{1},p_{2}\\}$ and $\beta\setminus\\{p_{1},p_{2}\\}$
contrary to hereditary unicoherence of $\mathbb{D}$. The claim contradicts the
condition that both $f_{G}(p_{1})$ and $f_{G}(p_{2})$ are points of order at
most 3 and not both of them are of order 3. ∎

######  Remark 3.16.

The family of all finite trees is cofinal in $\mathcal{T}_{\mathcal{M}\it 3}$.
Indeed, for a finite tree $T$ and a vertex $v$ of order $n>3$ we can find a
finite tree $T^{\prime}$ obtained from $T$ by replacing $v$ with two vertices
joined by an edge, such that one of the vertices has order 3 and the other has
order $n-1$.

Codenotti-Kwiatkowska [11] generalized Theorems 3.13 and 3.15 and constructed
all generalized Ważewski dendrites in a Fraïssé theoretic framework. On the
one hand, this work was an inspiration for the article [11]. On the other
hand, the proof of Theorem 3.13 presented above is a simplification and
correction of the one presented in an earlier version of this article and is
inspired by the work in [11].

##  4\. Confluent epimorphisms

Motivated by the definition of a confluent map between continua (see [8]) we
give an analogous definition for confluent epimorphisms between graphs and
develop various tools which will be used in studying these families.

######  Definition 4.1.

Given two topological graphs $G$ and $H$, an epimorphism $f\colon G\to H$ is
called confluent if for every closed connected subset $Q$ of $V(H)$ and every
component $C$ of $f^{-1}(Q)$ we have $f(C)=Q$. Equivalently, if for every
closed connected subset $Q$ of $V(H)$ and every vertex $a\in V(G)$ such that
$f(a)\in Q$ there is a connected set $C$ of $V(G)$ such that $a\in C$ and
$f(C)=Q$. Clearly, every monotone epimorphism is confluent.

######  Observation 4.2.

If $f\colon X\to Y$ and $g\colon Y\to Z$ are confluent epimorphisms between
topological graphs, then $g\circ f\colon X\to Z$ is a confluent epimorphism.

######  Proposition 4.3.

Given two finite graphs $G$ and $H$ the following conditions are equivalent
for an epimorphism $f\colon G\to H$:

  1. (1)

$f$ is confluent;

  2. (2)

for every edge $P\in E(H)$ and for every vertex $a\in V(G)$ such that $f(a)\in
P$, there is an edge $E\in E(G)$ and a connected set $R\subseteq V(G)$ such
that $E\cap R\neq\emptyset$, $f(E)=P$, $a\in V(R)$, and $f(R)=\\{f(a)\\}$.

  3. (3)

for every edge $P\in E(H)$ and every component $C$ of $f^{-1}(P)$ there is an
edge $E$ in $C$ such that $f(E)=P$.

###### Proof.

(1) $\Rightarrow$ (2): Let $P\in E(H)$ and $a\in V(G)$ such that $f(a)\in P$.
Then there exists a connected subset $C\subseteq G$ such that $a\in C$ and
$f(C)=P$. Since $C$ is connected there is an edge $E\in C$ such that $f(E)=P$.
If $a\in E$ let $R=\\{a\\}$ otherwise let $R$ be the component of
$f^{-1}(f(a))$ containing $a$. In either case (2) is satisfied.

(2) $\Rightarrow$ (3): This is immediate.

(3) $\Rightarrow$ (1): Given a connected set $Q$ in $H$ and $a\in V(G)$ with
$f(a)\in Q$ start with an edge in $Q$ containing $f(a)$ and use condition (3)
edge-by-edge to build a connected set in $G$ containing $a$.

∎

######  Proposition 4.4.

Consider the epimorphisms $f\colon F\to G$ and $g\colon G\to H$ between
topological graphs. If the composition $g\circ f\colon F\to H$ is confluent,
then $g$ is confluent.

###### Proof.

Let $E$ be a closed connected subset of $H$, and let $a\in G$ be a vertex such
that $g(a)\in E$. We need to find a closed connected subset of $S$ of $G$ that
contains $a$ and such that $g(S)=E$. Since $g\circ f$ is confluent, there is a
closed connected subset $C$ of $F$ that contains a vertex in $f^{-1}(a)$ and
such that $(g\circ f)(C)=E$. Then $S=f(C)$ satisfies the requirements. ∎

######  Proposition 4.5.

If $\mathcal{F}$ is a projective Fraïssé family of graphs with confluent
epimorphisms and $\mathbb{F}$ is a projective Fraïssé limit of $\mathcal{F}$,
then for every graph $G\in\mathcal{F}$ every epimorphism
$f_{G}\colon\mathbb{F}\to G$ is confluent.

###### Proof.

We first show that the canonical projections
$\alpha^{\infty}_{m}\colon\varprojlim\\{F_{n},\alpha_{n}\\}\to F_{m}$ are
confluent. Let $m$ and $\langle a_{m},b_{m}\rangle\in E(F_{m})$ be given. Fix
$a=(a_{n})\in\mathbb{F}$, so we have $\alpha^{\infty}_{m}(a)=a_{m}$. We will
find $b\in\mathbb{F}$ with $\alpha^{\infty}_{m}(b)=b_{m}$, such that there is
an arc $J$ joining $a$ and $b$, for which
$\alpha^{\infty}_{m}(J)=\\{a_{m},b_{m}\\}$. We construct $J$ as the inverse
limit of finite arcs. Let $J_{m}=\\{a_{m},b_{m}\\}$, and for $k<m$, let
$J_{k}=\alpha^{m}_{k}(J_{m})$. By the confluence of $\alpha_{m}$, there is an
arc $J_{m+1}\subseteq F_{m+1}$ such that $a_{m+1}$ is an end vertex of
$J_{m+1}$, $\alpha_{m}(J_{m+1})=J_{m}$, and $\alpha_{m}(b_{m+1})=b_{m}$, where
$b_{m+1}$ is the other end vertex of $J_{m+1}$. Suppose that we constructed an
arc $J_{n}$, for some $n>m$, such that $\alpha_{m}^{n}(J_{n})=J_{m}$, $a_{n}$
is an end vertex of $J_{n}$, and $\alpha_{m}^{n}(b_{n})=b_{m}$, where $b_{n}$
is the other end vertex of $J_{n}$. By the confluence of $\alpha_{n}$, there
is an arc $J_{n+1}\subseteq F_{n+1}$ such that $a_{n+1}$ is an end vertex of
$J_{n+1}$, $\alpha_{n}(J_{n+1})=J_{n}$, and $\alpha_{n}(b_{n+1})=b_{n}$, where
$b_{n+1}$ is the other end vertex of $J_{n+1}$. This finishes the construction
of $\\{J_{n}\\}$ and $(b_{n})$. Finally, we let
$J=\varprojlim\\{J_{n},\alpha_{n}|J_{n}\\}$ and $b=(b_{n})$.

Finally, there is an $m_{0}$ and a confluent epimorphism $f\colon F_{m_{0}}\to
G$ such that $f_{G}=f\circ\alpha_{m_{0}}^{\infty}$. As
$\alpha_{m_{0}}^{\infty}$ is confluent, so is $f_{G}$.

∎

######  Definition 4.6.

An epimorphism $f\colon G\to H$ between graphs is called light if for any
vertex $h\in V(H)$ and for any $a,b\in f^{-1}(h)$ we have $\langle
a,b\rangle\notin E(G)$.

######  Definition 4.7.

Recall the diagram (D1).

(D1)
${B}$${A}$${D}$${C}$$\scriptstyle{f}$$\scriptstyle{f_{0}}$$\scriptstyle{g_{0}}$$\scriptstyle{g}$

For given graphs $A,B,C$ and epimorphisms $f\colon B\to A$, $g\colon C\to A$
by standard amalgamation procedure we mean the graph $D$ defined by
$V(D)=\\{(b,c)\in B\times C:f(b)=g(c)\\}$;
$E(D)=\\{\langle(b,c),(b^{\prime},c^{\prime})\rangle:\langle
b,b^{\prime}\rangle\in E(B)\text{ and }\langle c,c^{\prime}\rangle\in
E(C)\\}$; and $f_{0}((b,c))=b$, $g_{0}((b,c))=c$. Note that $f_{0}$ and
$g_{0}$ are epimorphisms and $f\circ f_{0}=g\circ g_{0}$.

######  Proposition 4.8.

Using the standard amalgamation procedure and the notation of diagram (D1), if
the epimorphism $f$ is light, then the epimorphism $g_{0}$ is light.

###### Proof.

Suppose $g_{0}$ is not light. Then there is a vertex $h\in V(C)$ such that
$(a_{1},h),(a_{2},h)\in g_{0}^{-1}(h)$, and
$\langle(a_{1},h),(a_{2},h)\rangle\in E(D)$. Then, $\langle
a_{1},a_{2}\rangle\in E(B)$ and $a_{1},a_{2}\in f^{-1}(g(h))$, contradicting
$f$ being light. ∎

######  Proposition 4.9.

Using the standard amalgamation procedure and the notation of the diagram
(D1), if the epimorphism $f$ is monotone, then the epimorphism $g_{0}$ is
monotone.

###### Proof.

Fix a vertex $c\in V(C)$ and consider $g_{0}^{-1}(c)=\\{\langle
x,c\rangle:f(x)=g(c)\\}$. This set is isomorphic to $f^{-1}(g(c))$ which is
connected by monotonicity of $f$. ∎

######  Proposition 4.10.

Using the standard amalgamation procedure and the notation of diagram (D1), if
the epimorphism $f$ is confluent, then the epimorphism $g_{0}$ is confluent.

###### Proof.

Let $\langle c,c^{\prime}\rangle\in E(C)$ and $(b,c)\in V(D)$. By Proposition
4.3 we need to find an edge $\langle d_{1},d_{2}\rangle\in E(D)$ that is
mapped by $g_{0}$ onto $\langle c,c^{\prime}\rangle$ and a connected set $R$
such that $(b,c),d_{1}\in R$, and $g_{0}(R)=\\{c\\}$. Since $f$ is confluent,
there are vertices $b_{1},b_{2}\in B$ and a connected set $R_{B}$ such that
$\langle b_{1},b_{2}\rangle\in E(B)$, $b,b_{1}\in R_{B}$ and $f(\langle
b_{1},b_{2}\rangle)=\langle g(c),g(c^{\prime})\rangle$. It is enough to put
$d_{1}=(b_{1},c)$, $d_{2}=(b_{2},c^{\prime})$, and $R=R_{B}\times\\{c\\}$.

∎

######  Lemma 4.11.

Let $f\colon B\to A$ be a confluent epimorphism between (not necessarily
connected) graphs. Let $X\subseteq A$ be connected and let $Y$ be a component
of $f^{-1}(X)$. Then $f|_{Y}$ is confluent.

###### Proof.

Let $P\in E(X)$ and let $a\in Y$ be such that $f(a)\in V(P)$. Since $f$ is
confluent, by Proposition 4.3(2), there is an edge $E\subseteq E(B)$ and a
connected set $R\subseteq V(B)$ such that $E\cap R\not=\emptyset$, $f(E)=P$,
$a\in V(R)$, and $f(R)=\\{f(a)\\}$. Since $R$ is connected,
$f(R)=\\{f(a)\\}\subseteq V(P)\subseteq V(X)$, and $Y$ is a component of
$f^{-1}(X)$, it follows that $R\subseteq Y$ and the edge $E$ is in $Y$. Then
the connected set $R$ and the edge $E$ satisfy Condition (2) of Proposition
4.3 for $f_{0}$.

∎

Proposition 4.10 and Lemma 4.11 yields the following corollary.

######  Corollary 4.12.

Using the standard amalgamation procedure and the notation of diagram (D1), if
the graphs $A,B,C$ are connected and the epimorphisms $f$ and $g$ are
confluent, then for each component $P$ of the graph $D$ we have $f_{0}(P)=B$
and $g_{0}(P)=C$. Moreover, $f_{0}|_{P}$ and $g_{0}|_{P}$ are confluent.

######  Corollary 4.13.

The family of finite connected graphs with confluent epimorphisms is a
projective Fraïssé family.

###### Proof.

The AP follows from Corollary 4.12. The JPP for graphs $B$ and $C$ follows
from AP if $A$, in the notation of diagram (D1), is a graph with only one
vertex. ∎

We finish this section with the following decomposition results.

######  Proposition 4.14.

Given an epimorphism $f\colon G\to H$ between graphs there is a graph $M$ and
epimorphisms $m\colon G\to M$, $l\colon M\to H$ such that $f=l\circ m$, $m$ is
monotone, and $l$ is light.

###### Proof.

Define an equivalence relation $\sim$ on $V(G)$ by $x\sim y$ if and only if
$f(x)=f(y)$ and $x$ and $y$ are in the same component of $f^{-1}(f(x))$.
Putting $V(M)=V(G)/\\!\\!\\!\sim$, $m\colon V(G)\to V(M)$ as the projection,
$\langle[x]_{\sim},[y]_{\sim}\rangle\in E(M)$ if and only if there are
$x^{\prime}\in[x]_{\sim},y^{\prime}\in[y]_{\sim}$ such that $\langle
x^{\prime},y^{\prime}\rangle\in E(G)$, and $l([x]_{\sim})=f(x)$ one can verify
that the conclusions of the proposition are satisfied. ∎

######  Corollary 4.15.

Under the hypothesis of Proposition 4.14 if the epimorphism $f$ is confluent,
then both $m$ and $l$ are confluent.

###### Proof.

The epimorphism $m$ is confluent since it is monotone, while $l$ is confluent
by Proposition 4.4. ∎

##  5\. Rooted trees

The following example shows that confluent epimorphisms on (unrooted) trees do
not behave well. Because of this we concentrate on families of rooted trees
and develop several results that will be used later.

######  Example 5.1.

Let $A$ consist of just one edge $\langle 0,1\rangle$, $B$ be an arc with
three vertices $\\{a,b,c\\}$ and two edges $\langle a,c\rangle$, $\langle
c,b\rangle$, and $C$ be again an arc with three vertices $\\{p,q,r\\}$ and two
edges $\langle p,r\rangle$, $\langle r,q\rangle$. Let $f\colon B\to A$ be such
that $f(a)=f(b)=0$ and $f(c)=1$ and let $g\colon C\to A$ be such that
$g(p)=g(q)=1$ and $g(r)=0$, see figure below. It is easy to see that $f$ and
$g$ are confluent epimorphisms.

10abcD$f$$g$$f_{0}$$g_{0}$pqr

Note that the standard amalgamation procedure gives the graph $D$ to be the
cycle $(c,p)$, $(a,r)$, $(c,q)$, $(b,r)$, $(c,p)$ hence not a tree.

In fact, no amalgamation of the trees $A$, $B$, and $C$ yields a tree. To see
this assume there is an amalgamation of $A$, $B$, and $C$, that is, there is
finite tree $D$ and confluent epimorphisms $f_{0}\colon D\to B$ and
$g_{0}\colon D\to C$ such that $g\circ g_{0}=f\circ f_{0}$.

Without loss of generality, $f_{0}$ and $g_{0}$ are light confluent. Indeed,
let $\sim$ be the equivalence relation on $D$ given by $p\sim q$ if and only
if the arc joining $p$ and $q$ is contained in a component of $(f\circ
f_{0})^{-1}(a)$, for some $a\in A$. Take $D^{\prime}=D/\\!\\!\sim$,
$f^{\prime}_{0}=f_{0}/\\!\\!\sim\colon D^{\prime}\to B$ and
$g^{\prime}_{0}=g_{0}/\\!\\!\sim\colon D^{\prime}\to C$ induced from $f_{0}$,
$g_{0}$, and $\sim$. As $f$ and $g$ are light, $f^{\prime}_{0}$ and
$g^{\prime}_{0}$ are well defined and light. Moreover, $D^{\prime}$ is still a
tree, $f^{\prime}_{0}$, $g^{\prime}_{0}$ are confluent, and $g\circ
g^{\prime}_{0}=f\circ f^{\prime}_{0}$.

But now, by confluence of $f$ and $g$, the tree $D$ is such that any vertex is
adjacent to at least two other vertices. Therefore $D$ is an infinite tree. A
contradiction.

We will see that rooted trees with simple-confluent epimorphisms as well as
with simple*-confluent epimorphisms form projective Fraïssé families (see
Section 8).

######  Definition 5.2.

By a rooted tree we mean a rooted graph which is a tree. Given a rooted tree
$T$, we define an order $\leq$ by $x\leq y$ if the arc containing the root of
$T$, $r(T)$, and $y$ contains $x$. In the remainder of this article whenever
we write an edge as $\langle a,b\rangle$ we will mean that $a<b$.

Assumption. We require epimorphisms $f\colon T\to S$ between rooted trees to
be order-preserving, that is, if $a,b\in T$ with $a\leq b$ then $f(a)\leq
f(b)$.

Epimorphisms between rooted trees need not be confluent as the following
example shows.

######  Example 5.3.

Let $S$ be rooted tree which is a triod with vertices $\\{A,B,C,D\\}$, edges
$\\{\langle A,B\rangle,\langle B,C\rangle,\langle B,D\rangle\\}$, and $A$ is
the root of $T$. Let $T$ be another rooted tree with vertices
$\\{a,b_{1},b_{2},c,d_{1},d_{2}\\}$, edges $\\{\langle a,b_{1}\rangle,\langle
a,b_{2}\rangle,\langle b_{1},c\rangle,\langle b_{1},d_{1}\rangle,\langle
b_{2},d_{2}\rangle\\}$, and $a$ is the root of $S$. Let $f\colon T\to S$ be
given by $f(a)=A$, $f(b_{1})=f(b_{2})=B$, $f(c)=C$, and $f(d_{1})=f(d_{2})=D$.
Then $f$ is an order-preserving epimorphism on rooted trees but is not
confluent.

$c$$d_{1}$$a$$b_{1}$$b_{2}$$d_{2}$$A$$B$$C$$D$$T$$S$

The following example shows that the standard amalgamation procedure, when
applied to rooted trees, does not necessarily yield a tree.

######  Example 5.4.

Let $A$ consist of a single vertex, $B$ and $C$ be edges, where a vertex in
each edge is designated as the root. Then the standard amalgamation procedure
gives a complete graph on four vertices.

######  Definition 5.5.

By a chain in an rooted tree $G$ we mean sequence $a_{1},a_{2},\dots a_{n}$ of
vertices of $G$ such that:

  1. (1)

$a_{1}=r(G)$;

  2. (2)

$a_{1}\leq a_{2}\leq\dots\leq a_{n}$.

A chain is proper if moreover, it satisfies the condition

  1. (3)

$a_{1}\neq a_{2}\neq\dots\neq a_{n}$.

A branch is a maximal proper chain.

######  Observation 5.6.

If $f\colon G\to H$ is an epimorphism between rooted trees and $C$ is a chain
in $G$, then $f|_{C}$ is monotone.

######  Proposition 5.7.

If $\mathcal{T}$ is a projective Fraïssé family of rooted trees, and
$\mathbb{T}$ is a projective Fraïssé limit of $\mathcal{T}$, then $\mathbb{T}$
is arcwise connected.

###### Proof.

Let $\\{T_{n},\alpha_{n}\\}$ be a Fraïssé sequence for $\mathbb{T}$ and take
$x=(x_{n})\in\mathbb{T}$. For any $n$ let $I_{n}$ be the proper chain in
$T_{n}$ from $r(T_{n})$ to $x_{n}$. By Observation 5.6, we have that
$\mathbb{I}=\varprojlim\\{I_{n},\alpha_{n}|_{I_{n}}\\}$ is an arc in
$\mathbb{T}$ joining $x$ and $r(\mathbb{T})$.

∎

By Theorem 2.18 and Proposition 5.7 we get the following corollary.

######  Corollary 5.8.

If $\mathcal{T}$ is a projective Fraïssé family of rooted trees, and
$\mathbb{T}$ is a projective Fraïssé limit of $\mathcal{T}$, then $\mathbb{T}$
is a dendroid.

Recall the following definition for continua.

######  Definition 5.9.

A continuum $X$ which is a dendroid is said to be a smooth dendroid if there
exists a point $p\in X$ such that if $x_{n}\in X$ is a sequence of points that
converges to a point $x\in X$ then the sequence of arcs $px_{n}$ converges to
the arc $px$. This is equivalent to saying the order $\leq$, defined by $x\leq
y$ if every arc joining $p$ and $y$ contains $x$, is closed.

We now give an analogous definition for a rooted topological graph to be a
smooth dendroid.

######  Definition 5.10.

A rooted topological graph $X$ with the root $r(X)$ is called a smooth
dendroid if $X$ is a dendroid according to Definition 2.10 and the order
$\leq$, defined by $x\leq y$ if every arc joining $r(X)$ and $y$ contains $x$,
is closed.

######  Observation 5.11.

If a rooted topological graph $G$ is a smooth dendroid and $E(G)$ is
transitive, then its topological realization $|G|$ is a smooth dendroid in the
topological sense.

######  Proposition 5.12.

If $\mathcal{T}$ is a projective Fraïssé family of rooted trees, and
$\mathbb{T}$ is a projective Fraïssé limit of $\mathcal{T}$, then $\mathbb{T}$
is a smooth dendroid.

###### Proof.

First, note that by Corollary 5.8, $\mathbb{T}$ is a dendroid.

The order $\leq$ is closed on $\mathbb{T}$. Indeed, let
$\\{T_{n},\alpha_{n}\\}$ be a Fraïssé sequence for $\mathbb{T}$ and take
$x,y\in\mathbb{T}$. Then $x\leq y$ iff for every $n$, $x_{n}\leq y_{n}$. By
the definition of the product topology, $\leq$ is closed on $\mathbb{T}$.
Therefore the dendroid $\mathbb{T}$ is smooth. ∎

The Lelek fan can be characterized as a smooth fan (smooth dendroid having
exactly one ramification point) with a dense set of end points, [10]. In [3],
it is shown that the topological realization of the projective Fraïssé limit
for the family of all rooted trees with order-preserving epimorphisms is the
Lelek fan.

######  Definition 5.13.

We say that an epimorphism $f\colon S\to T$ between rooted trees is end vertex
preserving if, besides preserving order, we have that an image of an end
vertex in $S$ is an end vertex in $T$.

######  Remark 5.14.

If in Theorem 4.14, $G$ and $H$ are rooted trees and $f$ is order-preserving,
then $M$ is a rooted tree and $l$, $m$ are order-preserving. If additionally
$f$ is end vertex preserving, then so are $l$ and $m$.

The following will be used in Section 8.

######  Lemma 5.15.

Consider epimorphisms $f\colon F\to G$ and $g\colon G\to H$ between rooted
trees. If the composition $g\circ f\colon F\to H$ is end vertex preserving
then $g$ is end vertex preserving.

###### Proof.

Let $e$ be an end vertex in $G$. Because $f$ is order-preserving there is
$e^{\prime}\in f^{-1}(e)$ which is an end vertex of $H$. Since $h$ is end
vertex preserving, $g(e)=h(e^{\prime})$ is an end vertex. ∎

Recall the standard amalgamation procedure (Definition 4.7).

######  Theorem 5.16.

Let $A,B,C$ be rooted trees. Let $f\colon B\to A$ be light confluent and let
$g\colon C\to A$ be confluent. Then $D$ obtained via the standard amalgamation
procedure is a rooted tree. The obtained $f_{0}\colon D\to B$ is confluent and
$g_{0}\colon D\to C$ is light confluent. Moreover, in case $g$ is light then
so is $f_{0}$. In case both $f,g$ are end vertex preserving, then so are
$f_{0},g_{0}$.

###### Proof.

Let $r_{B}$ be the root of $B$ and $r_{C}$ be the root of $C$. Since $A$ is a
tree, $D$ is connected. Indeed, let $(a,b),(c,d)\in D$. Since $f(a)=g(b)$,
$f,g$ are order-preserving, and since there is a unique path in $A$ from
$f(a)=g(b)$ to $r_{A}$, we have that the $f|_{[r_{B},a]}$ and $g|_{[r_{C},b]}$
are monotone and onto $[r_{A},f(a)=g(b)]$. Therefore we can find a path from
$(r_{B},r_{C})$ to $(a,b)$ in $D$. Similarly, we can find a path from
$(r_{B},r_{C})$ to $(c,d)$ in $D$. Combine the two paths to get a path from
$(a,b)$ to $(c,d)$.

Moreover, $D$ is a tree. Indeed, suppose towards the contradiction that this
is not the case. Then there is a cycle
$(a_{1},b_{1}),(a_{2},b_{2}),\ldots,(a_{n},b_{n})$, $n\geq 3$, with
$\langle(a_{i},b_{i}),(a_{i+1},b_{i+1})\rangle\in E(D)$, $i=1,2,\ldots,n-1$,
and $\langle(a_{n},b_{n}),(a_{1},b_{1})\rangle\in E(D)$. Therefore $\langle
a_{i},a_{i+1}\rangle\in E(B)$ and $\langle a_{n},a_{1}\rangle\in E(B)$, as
well as $\langle b_{i},b_{i+1}\rangle\in E(C)$ and $\langle
b_{n},b_{1}\rangle\in E(C)$. Since $B$ is a tree, either all $a_{i}$’s are
equal or there are exactly two which are distinct, and similarly for $C$. We
have to conclude that all $(a_{i},b_{i})$ are equal except possibly two (as
edges are allowed). The only non-trivial case is when there are
$a^{\prime}\neq a^{\prime\prime}\in B$ and $b^{\prime}\neq b^{\prime\prime}\in
C$ such that $\\{a_{1},\ldots,a_{n}\\}=\\{a^{\prime},a^{\prime\prime}\\}$ and
$\\{b_{1},\ldots,b_{n}\\}=\\{b^{\prime},b^{\prime\prime}\\}$. Since $f$ is
light, if $a_{i}\neq a_{i+1}$, we must have $f(a_{i})\neq f(a_{i+1})$. As
$f(a_{i})=g(b_{i})\neq g(b_{i+1})=f(a_{i+1})$, this implies $b_{i}\neq
b_{i+1}$ (the $i+1$ is taken modulo $n$). Hence in the set of pairs
$\\{(a_{1},b_{1}),(a_{2},b_{2}),\ldots,(a_{n},b_{n})\\}$, we cannot have both
$(a^{\prime},b^{\prime})$ and $(a^{\prime\prime},b^{\prime})$, and similarly,
we cannot have both $(a^{\prime},b^{\prime\prime})$ and
$(a^{\prime\prime},b^{\prime\prime})$. Therefore
$|\\{(a_{1},b_{1}),(a_{2},b_{2}),\ldots,(a_{n},b_{n})\\}|\leq 2$, as we
wanted.

Declare $r=(r_{B},r_{C})$ to be the root of $D$.

(1) The $f_{0}$ and $g_{0}$ are confluent surjections. (see Proposition 4.10
and use that $D$ is connected)

(2) The $g_{0}$ is light (see Proposition 4.8). For the same reason, if $g$ is
light then so is $f_{0}$.

(3) We have the order $\leq_{D}$ on the rooted tree $D$ given by
$(a,b)\leq_{D}(c,d)$ iff the path joining $r$ and $(c,d)$ contains $(a,b)$. We
have to show that $(a_{1},b_{1})\leq_{D}(a_{2},b_{2})$ iff
$a_{1}\leq_{B}a_{2}$ and $b_{1}\leq_{C}b_{2}$. In particular, $f_{0}$ and
$g_{0}$ are order-preserving.

Suppose first that we have $a_{1}\leq_{B}a_{2}$, $b_{1}\leq_{C}b_{2}$,
$f(a_{1})=g(b_{1})$, and $f(a_{2})=g(b_{2})$. Denote $q=f(a_{2})$. Note that
$f|_{[r_{B},a_{2}]}$ is monotone and onto $[r_{A},q]$. Similarly,
$g|_{[r_{C},b_{2}]}$ is monotone and onto $[r_{A},q]$. Therefore we can find a
path from $(r_{B},r_{C})$ to $(a_{2},b_{2})$, which contains $(a_{1},b_{1})$
but does not contain $(a_{2},b_{2})$ or $(r_{B},r_{C})$ in its interior. This
shows that if $a_{1}\leq_{B}a_{2}$, $b_{1}\leq_{C}b_{2}$, then
$(a_{1},b_{1})\leq_{D}(a_{2},b_{2})$.

Suppose now that $(a_{1},b_{1})<_{D}(a_{2},b_{2})$ and that $(a_{1},b_{1})$
and $(a_{2},b_{2})$ are adjacent. Denote $p=f(a_{1})=g(b_{1})$ and
$q=f(a_{2})=g(b_{2})$. Since $f$ is light, we have $p\neq q$. Since $a_{1}$
and $a_{2}$ are adjacent, they are comparable wrt $\leq_{B}$, and analogously,
$b_{1}$ and $b_{2}$ are comparable wrt $\leq_{C}$. Therefore $p$ and $q$ are
comparable wrt $\leq_{A}$. Thus $p<q$ or $q<p$. In the first case, since $f$
and $g$ are order-preserving, we get $a_{1}<_{B}a_{2}$ and $b_{1}<_{C}b_{2}$.
In the second case we would get $a_{2}<_{B}a_{1}$ and $b_{2}<_{C}b_{1}$.
However, the second case is impossible, since it implies
$(a_{2},b_{2})<_{D}(a_{1},b_{1})$, as we have just seen.

In the general case, where $(a_{1},b_{1})\leq_{D}(a_{2},b_{2})$ but
$(a_{1},b_{1})$ and $(a_{2},b_{2})$ are not necessarily adjacent, we can
reduce to the one just discussed by taking the shortest path from
$(a_{1},b_{1})$ to $(a_{2},b_{2})$ and considering consecutive pair in it one
by one.

(4) Let $(e_{1},e_{2})\in D$. We have to show that if $(e_{1},e_{2})$ is an
end vertex of $D$, then $e_{1}$ is an end vertex of $B$ and $e_{2}$ is an end
vertex of $C$, i.e. that $f_{0}$ and $g_{0}$ are end vertex preserving.

Let $(a,b)\in D$ be an end vertex, i.e. a maximal with respect to $\leq_{D}$
vertex. Using that $f$ is light we show first that $b$ is an end vertex.
Suppose towards a contradiction that $b$ is not maximal with respect to
$\leq_{C}$. Then there is $b<_{C}t$ with $\langle b,t\rangle\in E(C)$. In case
$g(b)=g(t)$, we have $(a,b)<_{D}(a,t)$, which contradicts the maximality of
$(a,b)$, so suppose that $g(b)<_{A}g(t)$. Since $f$ is light and confluent
there is $a<_{B}s$ with $\langle a,s\rangle\in E(B)$, such that $f(s)=g(t)$.
But then $(a,b)<_{D}(s,t)$, which again contradicts the maximality of $(a,b)$.
This concludes the proof that $b$ is maximal wrt $\leq_{C}$, i.e. it is an end
vertex of $C$. Now as $g$ is end vertex preserving, we have that $g(b)$ is an
end vertex of $A$. In case there is $a<_{B}s$ with $\langle a,s\rangle\in
E(B)$, since $f$ is order-preserving, we must have $f(s)=f(a)=g(b)$, which is
impossible as $f$ is light. This shows that $a$ is an end vertex of $B$.

∎

######  Definition 5.17.

An epimorphism $f:G\to H$ between rooted trees is called elementary light
confluent if there is a vertex $v\in G$ and two components $C_{1}$ and $C_{2}$
of $G\setminus\\{v\\}$ such that:

  1. (1)

$f|_{C_{1}}$ and $f|_{C_{2}}$ are one-to one;

  2. (2)

$f(C_{1})=f(C_{2})$ is a component of $H\setminus f(v)$;

  3. (3)

$f|_{(G\setminus(C_{1}\cup C_{2}))}$ is one-to-one.

A typical elementary light confluent epimorphism is pictured in Figure 1.

Figure 1. Elementary light confluent epimorphism

It will follow from Corollary 8.15 that any light confluent epimorphism
decomposes into elementary light confluent epimorphisms.

##  6\. Kelley topological graphs

We first recall definitions concerning Kelley continua and adapt these to the
setting of topological graphs. Results of this section will be applied in
Theorem 8.30.

######  Notation 6.1.

For a given metric space $X$ with a metric $d$, the $r$-neighborhood of a
closed set $A\subseteq X$ is $N(A,r)=\\{x\in X:\text{ there is }a\in
A:d(x,a)<r\\}$. The Hausdorff distance $H$ between closed subsets of $X$ is
defined by $H(A,B)=\inf\\{r>0:A\subseteq N(A,r)\text{ and }B\subseteq
N(A,r)\\}$.

######  Definition 6.2.

A continuum $X$ is said to be Kelley continuum if for every subcontinuum $K$
of $X$, every $p\in K$, and every sequence $p_{n}\to p$ in $X$ there are
subcontinua $K_{n}$ of $X$ such that $p_{n}\in K_{n}$ and $\lim K_{n}=K$. By
compactness, this definition is equivalent to saying that for every
$\varepsilon>0$ there is $\delta>0$ such that for each two points $p,q\in X$
satisfying $d(p,q)<\delta$, and for each subcontinuum $K\subseteq X$ such that
$p\in K$ there is a subcontinuum $L\subseteq X$ satisfying $q\in L$ and
$H(K,L)<\varepsilon$.

It is well known that every Kelley dendroid is smooth, see [12].

######  Definition 6.3.

A topological graph $X$ is called Kelley if $X$ is connected and for every
closed and connected set $K\subseteq X$, every vertex $p\in K$, and every
sequence $p_{n}\to p$ of vertices in $X$ there are closed and connected sets
$K_{n}$ such that $p_{n}\in K_{n}$ and $\lim K_{n}=K$. By compactness, this
definition is equivalent to saying that for every $\varepsilon>0$ there is
$\delta>0$ such that for each two vertices $p,q\in X$ satisfying
$d(p,q)<\delta$, and for each closed connected set $K\subseteq X$ such that
$p\in K$ there is a closed connected set $L\subseteq X$ satisfying $q\in L$
and $H(K,L)<\varepsilon$.

Note that every finite graph is Kelley. Indeed, if we take the discrete metric
on finite graphs to have the value of $1$ when vertices are not equal then
taking $\delta=1/2$ in the definition implies $p=q$ and so we can take $K_{n}$
to equal $K$.

######  Observation 6.4.

If a topological graph $G$ is Kelley and $E(G)$ is transitive, then its
topological realization $|G|$ is a Kelley continuum.

In [14, Theorem 3.1] it is shown that if $\\{X_{i},f_{i}\\}$ is an inverse
sequence where the $X_{i}$’s are Kelley continua and the $f_{i}$’s are
confluent bonding maps then the inverse limit space
$\varprojlim\\{X_{i},f_{i}\\}$ is a Kelley continuum. The proof of the
following theorem is essentially the proof given in Theorem 3.1 adapted for
the present setting.

######  Proposition 6.5.

If $\mathcal{G}$ is a projective Fraïssé family of connected finite graphs
with confluent epimorphisms, then the projective Fraïssé limit is Kelley.

###### Proof.

Let $\\{G_{n},\alpha_{n}\\}$ be a Fraïssé sequence for $\mathcal{G}$ so
${\mathbb{G}}=\varprojlim\\{G_{n},\alpha_{n}\\}$. Let the metric on
${\mathbb{G}}$ be given by
$d(x,y)=\sum_{i=1}^{\infty}d_{i}(x_{i},y_{i})/2^{i}$ where $d_{i}$ is the
discrete metric on the finite graphs $G_{i}$. Note that if
$x,y\in{\mathbb{G}}$ and $N$ is a positive integer then $d(x,y)\leq 1/2^{N}$
if and only if $\alpha^{\infty}_{i}(x)=\alpha^{\infty}_{i}(y)$ for all $i\leq
N$. Thus, if $K$ and $L$ are closed connected graphs in ${\mathbb{G}}$ such
that $\alpha_{N}(K)=\alpha_{N}(L)$ then $H(K,L)\leq 1/2^{N}$.

Let $\varepsilon>0$ be given and let $N$ be such that $1/2^{N}<\varepsilon$.
Suppose $K$ is a closed connected graph in ${\mathbb{G}}$, $p$ is a vertex in
$K$, $q$ is a vertex in ${\mathbb{G}}$, and $d(p,q)<1/2^{N}$. Then
$\alpha^{\infty}_{N}(q)\in\alpha^{\infty}_{N}(K)$. For each $m>N$ let $L_{m}$
be the component of $(\alpha_{N}^{m})^{-1}(\alpha_{N}^{\infty}(K))$ that
contains $\alpha_{m}^{\infty}(q)$ and for $m\leq N$ let
$L_{m}=\alpha_{m}^{N}(\alpha_{N}^{\infty}(K))$. Then
$L=\varprojlim\\{L_{i},\alpha_{i}|_{L_{i}}\\}$ being the inverse limit of
closed connected graphs is a closed connected topological graph in
${\mathbb{G}}$ containing $q$ and $H(K,L)<\varepsilon$.

∎

##  7\. End vertices and ramification vertices in topological graphs

In this section we define ramification and end vertices for topological graphs
and study how they project onto the topological realization of a projective
Fraïssé limit.

######  Definition 7.1.

For a topological graph $D$ that is a dendroid, we say that a vertex $v$ has
order at least $n$, in symbols ${\rm ord}(v)\geq n$, if there are arcs
$A_{1},A_{2},\dots A_{n}$ such that $A_{i}\cap A_{j}=\\{v\\}$ for
$i,j\in\\{1,2,\dots n\\}$ satisfying $i\neq j$. We define ${\rm ord}(v)=n$ if
${\rm ord}(v)\geq n$ and ${\rm ord}(v)\geq n+1$ is not true. If ${\rm
ord}(v)\geq n$ for each positive integer $n$, then we say that $v$ has
infinite order, in symbols ${\rm ord}(v)=\infty$. Vertices of order 1 are
called end vertices and vertices of order $\geq 3$ are called ramification
vertices. Note that these definitions agree with Definitions 1.6 in the case
of trees.

We call an end vertex isolated if it belongs to a non-degenerate edge. Note
that, according to this definition, every end vertex in a finite graph is
isolated.

Let us recall a continuum theory definition.

######  Definition 7.2.

For a continuum $X$ a point $p\in X$ is called a ramification point of $X$ in
the classical sense if there are three arcs $A,B,C$ in $X$ such that $A\cap
B=A\cap C=B\cap C=\\{p\\}$. Similarly, a point $p\in X$ is called an end point
of $X$ in the classical sense if it is an end point of every arc in $X$ that
contains $p$.

Suppose that $D$ is a dendroid, $e$ is an isolated end vertex, and $r$ is a
ramification vertex such that $\langle r,e\rangle\in E(D)$. If $D$ is
transitive, $\varphi\colon D\to|D|$ is the quotient map, then, since
$\varphi(r)=\varphi(e)$, $\varphi(e)$ is not an end point. The proposition
below captures that a converse to this holds.

######  Proposition 7.3.

Suppose $D$ is a topological graph that is a dendroid, the set of edges of $D$
is transitive and $e$ is a non-isolated end vertex of $D$. Let $\varphi$ be
the quotient map from $D$ onto its topological realization $|D|$. Then
$\varphi(e)$ is an end point of $|D|$ in the classical sense.

###### Proof.

Assume the conclusion is not true. Then there exists an arc $A\subseteq|D|$
such that $A\setminus\\{\varphi(e)\\}=H\cup K$ where $H$ and $K$ are nonempty
sets such that $\overline{H}\cap K=H\cap\overline{K}=\emptyset$. The set
$\varphi^{-1}(H\cup K)$ is not connected since the image of a connected set is
connected. However the sets $\varphi^{-1}(H)$ and $\varphi^{-1}(K)$ are each
connected. Let $a$ and $b$ be vertices in $\varphi^{-1}(H)$ and
$\varphi^{-1}(K)$ respectively and $A_{H}$ and $A_{K}$ be arcs in $D$ from $a$
to $e$ and $b$ to $e$. The arcs $A_{H}$ and $A_{K}$ then lie in
$\varphi^{-1}(H)\cup\\{e\\}$ and $\varphi^{-1}(K)\cup\\{e\\}$ respectively.
The intersection of $A_{H}$ and $A_{K}$ is $\varphi^{-1}(\varphi(e))=\\{e\\}$
since $e$ is a non-isolated end vertex of $D$. So
$\varphi^{-1}(H)\cup\varphi^{-1}(K)\cup\\{e\\}$ is an arc containing $e$ but
$e$ is not the end vertex of this arc. Thus $e$ is not an end vertex of $D$
contrary to the hypothesis. ∎

######  Proposition 7.4.

Let $f\colon X\to Y$ be an order-preserving epimorphism between topological
graphs, which are smooth dendroids. Then, for every end vertex $y\in Y$ there
is an end vertex $x\in X$ such that $f(x)=y$.

###### Proof.

Let $z$ be any vertex in $X$ such that $f(z)=y$, and let $x$ be an end vertex
of $X$ satisfying $z\leq x$. Since $f$ is order-preserving we have $f(z)=y\leq
f(x)$, but $y$ is an end point, so $f(z)=f(x)=y$. ∎

######  Proposition 7.5.

Suppose $D$ is a topological graph that is a dendroid, the set of edges of $D$
is transitive and $r$ is a ramification vertex of $D$. Let $\varphi$ be the
quotient map from $D$ onto its topological realization $|D|$. If
$\varphi^{-1}(\varphi(r))$ contains no isolated end vertices, then
$\varphi(r)$ is a ramification point of $|D|$ in the classical sense.

###### Proof.

Suppose $r$ is a ramification vertex of $D$ and let $A$, $B$, and $C$ be arcs
in $D$ such that $A\cap B=B\cap C=C\cap A=\\{r\\}$. The images $\varphi(A)$,
$\varphi(B)$, and $\varphi(C)$ are arcs or degenerate with pairwise
intersections equaling $\varphi(r)$. Suppose $\varphi(A)$ is degenerate then
$A$ consists of a single edge $\langle r,e\rangle$ and
$A=\varphi^{-1}(\varphi(r))$ contains the isolated end vertex $e$ contrary to
the hypothesis. Thus $\varphi(r)$ is a ramification point of $|D|$.

∎

######  Theorem 7.6.

Suppose that $\mathcal{T}$ is a projective Fraïssé family of rooted trees and
for every $T\in\mathcal{T}$, for every end vertex $e\in T\setminus\\{r(T)\\}$
and every $a\in T$ such that $\langle a,e\rangle\in E(T)$ there is a tree $S$,
an epimorphism $f\colon S\to T$ such that for every $p\in f^{-1}(a)$ there are
vertices $q,r\in S$ satisfying $p\leq q\leq r$, $q\neq r$, and $f(q)=f(r)=e$.
Then the projective Fraïssé limit of $\mathcal{T}$ has no isolated end
vertices.

###### Proof.

Let $\mathcal{T}$ be a family that satisfies the assumptions of the theorem,
and let $\mathbb{T}$ be the projective Fraïssé limit of $\mathcal{T}$. Suppose
$e$ is an isolated end vertex of $\mathbb{T}$ then there exists
$a\in\mathbb{T}$ such that $\langle a,e\rangle\in E(\mathbb{T})$. Let
$T\in\mathcal{T}$ and $f_{T}:\mathbb{T}\to T$ be an epimorphism such that
$f_{T}(a)\not=f_{T}(e)$ then $\langle f_{T}(a),f_{T}(e)\rangle\in E(T)$. From
the hypothesis we know there is a tree $S$ and an epimorphism $f\colon S\to T$
such that for every $p\in f^{-1}(f_{T}(a))$ there are vertices $q,r\in S$
satisfying $p\leq q\leq r$, $q\neq r$, and $f(q)=f(r)=f_{T}(e)$. Let
$f_{S}\colon\mathbb{T}\to S$ be an epimorphism such that $f_{T}=f\circ f_{S}$.
Then $f_{S}(a)\inf^{-1}(f_{T}(a))$ and $r=f_{S}(e)$, since $f_{S}$ is an
order-preserving epimorphism, but these vertices are not members of an edge in
$S$ contrary to the fact that $f_{S}$ is an epimorphism. ∎

##  8\. Simple- and simple*-confluent epimorphisms and rooted trees

In this section, we use the results obtained earlier to investigate families
of rooted trees with confluent order-preserving epimorphisms.

###  8.1. Epimorphisms between rooted trees

Families of rooted trees with all confluent or even all monotone epimorphisms
do not form a projective Fraïssé family, they do not amalgamate, as the
following example shows. Therefore, we introduce subfamilies of monotone
epimorphisms consisting of simple-monotone or simple*-monotone epimorphisms.
Composing these epimorphisms with light confluent epimorphisms we obtain
simple-confluent and simple*-confluent epimorphisms, which have the
amalgamation property.

######  Example 8.1.

Let $A$ be the tree with vertices $V(A)=\\{a_{A},b_{A},c_{A},x_{A}\\}$ and
edges $E(A)=\\{\langle x_{A},a_{A}\rangle,\langle x_{A},b_{A}\rangle,\langle
x_{A},c_{A}\rangle\\}$ with $x_{A}$ being the root of $A$, $B$ be the tree
with vertices $V(B)=\\{a_{B},b_{B},c_{B},x_{B},y_{B}\\}$ and edges
$E(B)=\\{\langle y_{B},a_{B}\rangle,\langle y_{B},b_{B}\rangle,\langle
x_{B},y_{B}\rangle,\langle x_{B},c_{B}\rangle\\}$ with $x_{B}$ being the root
of $B$, and $C$ be the tree with vertices
$V(C)=\\{a_{C},b_{C},c_{C},x_{C},y_{C}\\}$ and edges $E(C)=\\{\langle
x_{C},a_{C}\rangle,\langle y_{C},b_{C}\rangle,\langle
y_{C},c_{C}\rangle,\langle x_{C},y_{C}\rangle\\}$ with $x_{C}$ being the root
of $C$. For epimorphisms $f\colon B\to A$ and $g\colon C\to A$ let
$f(x_{B})=f(y_{B})=x_{A}=g(x_{C})=g(y_{C})$ and otherwise map letters
correspondingly; see Figure 2.

$x_{A}$$A$$a_{A}$$b_{A}$$c_{A}$$x_{B}$$a_{B}$$b_{B}$$B$$c_{B}$$y_{B}$$x_{C}$$C$$a_{C}$$b_{C}$$c_{C}$$y_{C}$$D^{\prime}$$f$$g$$f_{0}$$g_{0}$
Figure 2. Failure of amalgamation for monotone epimorphisms and rooted trees

Clearly, $f$ and $g$ are monotone epimorphisms. By Theorem 3.13, when $A$,
$B$, and $C$ are viewed as unrooted trees, there is an amalgamation. In
particular, the tree $D$ having vertices
$\\{a_{D},y_{D},b_{D},x_{D},z_{D},c_{D}\\}$ and edges $\\{\langle
y_{D},a_{D}\rangle,\langle x_{D},y_{D}\rangle,\langle
x_{D},b_{D}\rangle,\langle x_{D},z_{D}\rangle,\langle z_{D},c_{D}\rangle\\}$
is an amalgamation with $f_{0}(x_{D})=y_{B}$, $f_{0}(z_{D})=x_{B}$,
$g_{0}(y_{D})=x_{C}$, and $g_{0}(z_{D})=x_{C}$. No matter which vertex is
selected as the root of $D$, either $f_{0}$ or $g_{0}$ does not map it to the
root of $B$ or $C$. So this is not an amalgamation of rooted trees as one of
$f_{0}$ or $g_{0}$ is not order-preserving.

In fact, there is no amalgamation of these rooted trees. Suppose, towards a
contradiction, that there is a rooted tree $D^{\prime}$ and confluent
epimorphisms $f_{0}\colon D^{\prime}\to B$, $g_{0}\colon D^{\prime}\to C$
satisfying $f\circ f_{0}=g\circ g_{0}$. Since $f\circ f_{0}$ is an epimorphism
there exists $t,s\in D^{\prime}$ be such that $\langle t,s\rangle\in
E(D^{\prime})$, $f\circ f_{0}(t)=x_{A}$, and $f\circ f_{0}(s)=b_{A}$. Then
$f_{0}(t)=y_{B}$ and $g_{0}(t)=y_{C}$. Therefore, the set $Y=\\{t\in
D^{\prime}\colon f_{0}(t)=y_{B},\ g_{0}(t)=y_{C}\\}$ is nonempty. Let $t\in Y$
and consider the arc $r(D^{\prime})t$ from $r(D^{\prime})$ to $t$. Let $u$ be
the largest vertex on this arc that is not in $Y$. Then there is vertex
$t^{\prime}\in Y\cap r(D^{\prime})t$ such that $\langle u,t^{\prime}\rangle$
is an edge in $D^{\prime}$. Since $u\not\in Y$ and $u<t^{\prime}$,
$f_{0}(u)=x_{B}$ or $g_{0}(u)=x_{C}$; without loss of generality,
$g_{0}(u)=x_{C}$.

Let $C$ be the component of $g_{0}^{-1}(y_{C})$ that contains $t^{\prime}$.
Then $t^{\prime}\in C\subseteq Y$ and $t^{\prime}$ is the least vertex of $C$.
Therefore any vertex $v\in D^{\prime}$ adjacent to $C$, $v\neq u$, satisfies
$f\circ f_{0}(v)\in\\{a_{A},b_{A}\\}$ and $g\circ
g_{0}(v)\in\\{b_{A},c_{A}\\}$, which implies $f\circ f_{0}(v)=g\circ
g_{0}(v)=b_{A}$. This gives a contradiction with confluence of $g_{0}$
(applied to the edge $\langle y_{C},c_{C}\rangle$ and the component $C$ of
$g_{0}^{-1}(y_{C})$).

To obtain a projective Fraïssé families of finite rooted trees we start with
the following definitions.

######  Definition 8.2.

An epimorphism $f\colon S\to T$ will be called a splitting edge map if $S$ is
obtained from $T$ as follows. Let $V(S)=V(T)\cup\\{x\\}$, where $x$ is a “new”
vertex, and for some edge $\langle a,b\rangle\in E(T)$, we have
$E(S)=(E(T)\setminus\langle a,b\rangle)\cup\\{\langle a,x\rangle,\langle
x,b\rangle\\}$. The map $f$ is the identity on $T$ and it maps $x$ either to
$a$ or to $b$. We will call an epimorphism simple-monotone if it is a
composition of splitting edge maps. We call an epimorphism a simple-confluent
if it is a composition of light confluent epimorphisms and simple-monotone
epimorphisms. Specifically, $f$ is simple-confluent if there are
$f_{1},f_{2},\ldots,f_{n}$, for some $n$, such that $f=f_{1}\circ
f_{2}\circ\ldots\circ f_{n}$, where each $f_{i}$ is light confluent or simple-
monotone.

######  Definition 8.3.

An epimorphism $f\colon S\to T$ will be called an adding edge map if $S$ is
obtained from $T$ as follows. Let $V(S)=V(T)\cup\\{y\\}$, where $y$ is a “new”
vertex, and for some vertex $v\in T$ we have $E(S)=E(T)\cup\\{\langle
v,y\rangle\\}$. The map $f\colon S\to T$ takes $y$ to $v$ and is equal to the
identity otherwise. Define simple*-monotone epimorphisms to be compositions of
splitting edge maps and adding edge maps. Let simple*-confluent epimorphisms
be compositions of simple*-monotone maps and light confluent maps.

Note that simple-confluent epimorphisms are end vertex preserving. Also, the
epimorphisms in Example 8.1 are neither simple-confluent nor
simple*-confluent.

######  Notation 8.4.

The family of all rooted trees and simple-confluent order-preserving
epimorphisms is denoted by $\mathcal{T}_{\mathcal{CE}}$ and the family of all
rooted trees and simple*-confluent order-preserving epimorphisms is denoted by
$\mathcal{T}_{\mathcal{C}}$. We assume that every rooted tree has at least two
vertices.

###  8.2. Amalgamation of simple-confluent epimorphisms

The goal of this subsection is to prove that the family of all rooted trees
with simple-confluent epimorphisms is a projective Fraïssé family, see Theorem
8.7.

######  Definition 8.5.

Let $T$ be a rooted tree. We define recursively the height of each vertex in
$T$. If $r$ is the root of $T$, we let ${\rm ht}(r)=0$. If $a$ is such that
${\rm ht}(a)=n$ and $b$ is an immediate successor of $a$, i.e. $\langle
a,b\rangle\in E(T)$ and $a<_{T}b$, then we let ${\rm ht}(b)=n+1$. It is not
hard to see that for an $a\in T$, ${\rm ht}(a)$ is just the number of edges in
the arc joining the root and $a$. Finally, let ${\rm ht}(T)=\max\\{{\rm
ht}(a)\colon a\in T\\}$, be the height of $T$.

For a vertex $a\in T$ we denote the set of immediate successors of $a$ by
${\rm im}(a)$ and the successor order of $a$, denoted ${\rm sord}(a)$, will
simply be $|{\rm im}(a)|$. Let ${\rm sord}(T)=\max\\{{\rm sord}(a)\colon a\in
T\\}$.

A rooted tree $T$ is regular if for each vertex $a\in T$ that is not an end
vertex sord(a)= sord(T) and for each end vertex $e\in T$, ${\rm ht}(e)={\rm
ht}(T)$.

If $T$ is a rooted tree and $a\in T$, let $C_{a}=\\{x\in T\colon x>a\\}$ be
the set of vertices above $a$ with the edge relation induced from $T$. We also
let $\overline{C}_{a}=\\{x\in T\colon x\geq a\\}$.

######  Theorem 8.6.

The family $\mathcal{T}_{\mathcal{CE}}$ has the JPP.

###### Proof.

Let $A$ and $B$ be rooted trees with $k=\max\\{{\rm ht}(A),{\rm ht}(B)\\}$ and
$m=\max\\{{\rm sord}(A),{\rm sord}(B)\\}$. It suffices to show that if $T$ is
a rooted tree, $k\geq{\rm ht}(T)$, and $m\geq{\rm sord}(T)$, then there is a
regular rooted tree $S$, with ${\rm ht}(S)=k$ and ${\rm sord}(S)=m$, and a
simple-confluent epimorphism from $S$ to $T$.

First, we get a rooted tree $R$ such that there is a simple-monotone
epimorphism from $R$ onto $T$ and each end vertex of $R$ has height $k$. For
that, we proceed with the following algorithm. For any end vertex $e\in T$, if
$a_{e}$ is the immediate predecessor of $e$, and ${\rm ht}(e)=k^{\prime}<k$,
replace the edge $\langle a_{e},e\rangle$ by $k-k^{\prime}+1$ many edges. This
gives the required $R$. Mapping the added vertices to the respective end
vertices gives a simple-monotone epimorphism from $R$ to $T$.

To get from $R$ to the required tree $S$, we will proceed recursively as
follows. Suppose that we have already constructed a tree $S_{i}$ with $k={\rm
ht}(S_{i})$, $m\geq{\rm sord}(S_{i})$, and such that for some vertex $b\in
S_{i}$, ${\rm sord}(b)<m$ and for every vertex $a\in S_{i}$ with $a>b$, ${\rm
sord}(a)=m$, as well as a light confluent epimorphism $f_{i}\colon S_{i}\to
R$. Suppose ${\rm sord}(b)=m^{\prime}$. Pick a component $C$ of $C_{b}$ and
replace $C$ by $m-m^{\prime}+1$ many copies of $C$. This gives a tree
$S_{i+1}$ with one fewer vertex having successor order less than $m$. Mapping
all of the new components in $S_{i+1}$ to $C$ gives a light confluent
epimorphism from $S_{i+1}$ to $S_{i}$.

∎

######  Theorem 8.7.

The family $\mathcal{T}_{\mathcal{CE}}$ of all rooted trees with simple-
confluent epimorphisms is a projective Fraïssé family.

We need to verify condition (4) of Definition 1.7. Theorem 8.7 will follow
from Theorem 5.16 and the two lemmas below. Let $f\colon B\to A$ and $g\colon
C\to A$ be simple-confluent epimorphisms. We find
$D\in\mathcal{T}_{\mathcal{CE}}$ and simple-confluent epimorphisms
$f_{0}\colon D\to B$ and $g_{0}\colon D\to C$ such that $f\circ f_{0}=g\circ
g_{0}$. We will consider the following three cases:

  * (i)

$f,g$ are light confluent. Then we find $f_{0},g_{0}$ light confluent.
(Theorem 5.16)

  * (ii)

$f,g$ are simple-monotone. Then we find $f_{0},g_{0}$ simple-monotone. (Lemma
8.9)

  * (iii)

$f$ is simple-monotone and $g$ is light confluent. Then we find $f_{0}$ light
confluent and $g_{0}$ simple-monotone. (Lemma 8.8)

This will suffice as every simple-confluent epimorphism is a composition of
light confluent and simple-monotone epimorphisms. Diagram (D3) illustrates the
scheme of the general proof for the case when each $f$ and $g$ is a
composition of a single simple-monotone and a single light confluent
epimorphism.

(D3)
${B}$${*}$${*}$${A}$${*}$${D}$${*}$${*}$${C}$$\scriptstyle{m}$$\scriptstyle{l}$$\scriptstyle{m}$$\scriptstyle{l}$$\scriptstyle{l}$$\scriptstyle{l}$$\scriptstyle{m}$$\scriptstyle{m}$$\scriptstyle{l}$$\scriptstyle{m}$$\scriptstyle{l}$$\scriptstyle{m}$

######  Lemma 8.8.

Let $f\colon B\to A$ be a simple-monotone epimorphism and let $g\colon C\to A$
be a light confluent epimorphism. Let $D$ be obtained via the standard
amalgamation procedure and let $f_{0}$ and $g_{0}$ be the corresponding
projections. Then $f_{0}$ is light confluent and $g_{0}$ is simple-monotone.

###### Proof.

By Theorem 5.16, all we are left to show is that $g_{0}$ is simple-monotone.

Since any simple-monotone epimorphism is a composition of splitting edge maps,
we can assume without loss of generality that $f$ is a splitting edge map.
Therefore we may assume that $f$ replaces an edge $\langle a,b\rangle\in E(A)$
by edges $\langle a,x\rangle$ and $\langle x,b\rangle$, and that $f(x)=a$. Let
$\langle a_{1},b_{1}\rangle,\langle a_{2},b_{2}\rangle,\ldots,\langle
a_{n},b_{n}\rangle$ be all edges mapped to $\langle a,b\rangle$ by $g$. Note
that $D$ is obtained from $C$ by splitting each of $\langle
a_{i},b_{i}\rangle$, $i=1,2,\ldots,n$, by some vertex $x_{i}$ that is mapped
by $g_{0}$ to $a_{i}$. Therefore $g_{0}$ is a composition of $n$ many
splitting edge maps, as needed.

∎

######  Lemma 8.9.

Let $f\colon B\to A$ and $g\colon C\to A$ be simple-monotone epimorphisms.
Then there is a rooted tree $D$ and simple-monotone epimorphisms $f_{0}\colon
D\to B$ and $g_{0}\colon D\to C$ such that $f\circ f_{0}=g\circ g_{0}$.

###### Proof.

It suffices to show that if $f\colon B\to A$ and $g\colon C\to A$ are
splitting edge maps, then we can find a rooted tree $D$ and splitting edge
maps $f_{0}\colon D\to B$ and $g_{0}\colon D\to C$ such that $f\circ
f_{0}=g\circ g_{0}$.

The $B$ be obtained from $A$ by splitting some edge $e_{B}$ of $A$. The new
vertex $x$ is mapped by $f$ to one of the end vertices of $e_{B}$. Similarly,
$C$ is obtained from $A$ by splitting some edge $e_{C}$ of $A$. The new vertex
$y$ is mapped by $g$ to one of the end vertices of $e_{C}$.

Let $D$ be obtained from $A$ as follows. In case $e_{B}\neq e_{C}$, then we
obtain $D$ by simultaneously splitting $e_{B}$ and $e_{C}$. If
$e_{B}=e_{C}=e$, then we split $e$ twice. It is clear how the required $f_{0}$
and $g_{0}$ are defined and that they are splitting edge maps.

∎

######  Notation 8.10.

Let $\mathbb{T}_{\mathcal{CE}}$ be the projective Fraïssé limit of the family
$\mathcal{T}_{\mathcal{CE}}$.

###  8.3. More on simple-confluent epimorphisms

We introduce a family of epimorphisms between rooted trees that we call
special. It is then shown that special epimorphisms are precisely simple-
confluent epimorphisms, see Theorem 8.14. The definition of a special
epimorphism is intrinsic, unlike that of a simple-confluent epimorphism, which
allows us to more easily work with the family of simple-confluent
epimorphisms. We conclude this subsection by showing that
$(\mathbb{T}_{\mathcal{CE}})^{\omega}$ is independent of the inverse limit
representations of elements of the family, see Corollary 8.17.

Recall that if $T$ is a rooted tree and $a\in T$, then $C_{a}=\\{x\in T\colon
x>a\\}$ is the set of vertices above $a$ with the edge relation induced from
$T$. Let $f\colon S\to T$ be a confluent epimorphism between finite rooted
trees, let $p\in T$, and $q\in S$ be such that $f(q)=p$. Note that if $E$ is a
component of $C_{q}$ and $D$ is a component of $C_{p}$ such that $f(E)\cap
D\neq\emptyset$ then, by confluence of $f$, we get $D\subseteq f(E)$.
Therefore to each component $E$ of $C_{q}$ we can assign the collection of all
components of $C_{p}$ contained in $f(E)$. The union of those components is
equal to $f(E)\setminus\\{p\\}$.

######  Definition 8.11.

Let $f\colon S\to T$ be a confluent epimorphism between finite rooted trees,
and let $p$ be a ramification vertex in $T$. Call a vertex $q\in S$ with
$f(q)=p$ a special vertex for $p$ (wrt $f)$ if the following hold:

  1. (1)

for any component $E$ of $C_{q}$ there is exactly one component $D$ of $C_{p}$
such that $D\subseteq f(E)$;

  2. (2)

for any component $D$ of $C_{p}$ there is a component $E$ of $C_{q}$ such that
$D\subseteq f(E)$.

If $f$ is understood from the context, and $q$ is special for $p=f(q)$ (wrt
$f$), we simply say that $q$ is special. If $q_{1}\in S$ is such that there
exists $q_{1}<q_{2}$ with $f(q_{1})=f(q_{2})$, and $q_{2}$ is special, we say
that $q_{1}$ is below a special vertex. Similarly, if $q_{2}\in S$ is such
that there exists $q_{1}<q_{2}$ with $f(q_{1})=f(q_{2})$, and $q_{1}$ is
special, we say that $q_{2}$ is above a special vertex.

We denote by $P_{f}(p)$ the set of vertices in $S$, which are special for $p$.

Note that we can rephrase (1) in the definition into: for any component $E$ of
$C_{q}$, $f(E)\setminus\\{p\\}$ is a component of $C_{p}$. Note also that any
two different vertices in $P_{f}(p)$ are incomparable. If $q$ is special for
$p$, we let $\alpha_{f}(q,p)$ to be the induced surjection from the set of
components of $C_{q}$ to the set of components of $C_{p}$, that is,
$\alpha_{f}(q,p)(E)=D$ if and only if $D=f(E)\setminus\\{p\\}$.

######  Definition 8.12.

A confluent end vertex preserving epimorphism $f\colon S\to T$ between rooted
trees will be called a special epimorphism if for every ramification vertex
$x\in T$ and $b\in S$ such that $f(b)>x$, there exists $x^{\prime}<b$, which
is special for $x$ (wrt $f$).

Note that if $f\colon S\to T$ is a special epimorphism then any vertex in $S$,
which is mapped to a ramification vertex, either is special, or it is above a
special vertex, or else it is below a special vertex.

######  Lemma 8.13.

Let $f\colon S\to T$ be a special epimorphism between finite rooted trees, let
$x\in T$ be a ramification vertex, and let $y\in S$ be such that $f(y)=x$.
Then:

  1. (1)

$y$ is special, or

  2. (2)

there is a component $D$ of $C_{x}$ such that for any component $E$ of $C_{y}$
it holds $f(E)\setminus\\{x\\}=D$, or else

  3. (3)

for any component $E$ of $C_{y}$ it holds $f(E)=C_{x}\cup\\{x\\}$.

###### Proof.

Indeed, if $y$ is above a special vertex, then (2) holds. If $y$ is below a
special vertex, then (3) holds. In this last case, we additionally notice,
using just the definition of a special epimorphism, that any component of
$C_{y}$ contains a vertex, which is special for $x$. ∎

######  Theorem 8.14.

An epimorphism between finite rooted trees is simple-confluent if and only if
it is special.

###### Proof.

First, we argue that any simple-confluent epimorphism is special. Indeed, in
case $f$ is simple-monotone and $p$ is a ramification vertex, then $q$ is
special for $p$ iff it is the unique ramification vertex with $f(q)=p$. In
case $f$ is light confluent, then $P_{f}(p)=f^{-1}(p)$. In both cases $f$ is a
special epimorphism. Then proceed along a decomposition into light confluent
and simple-monotone epimorphisms, noticing that if $f=f_{2}\circ f_{1}$, where
$f_{1}$ is simple-monotone or light confluent, $f_{2}$ is simple-confluent,
and $p$ is a ramification vertex, then $P_{f}(p)=\bigcup_{x\in
P_{f_{2}}(p)}P_{f_{1}}(x)$.

For the opposite direction, suppose that the conclusion holds for any special
epimorphism whose domain has $<n$ vertices. Let now $h\colon H\to F$ be
special and such that $|V(H)|=n$. We show that if $h$ is not an isomorphism,
then there are $f\colon G\to F$ special and $g\colon H\to G$ simple-confluent,
which is not an isomorphism, such that $h=f\circ g$. Once we show this last
statement, we are done with the proof. Indeed, since $|V(G)|<n$, by the
inductive assumption $f$ is simple-confluent, and hence $h=f\circ g$ must be
simple-confluent.

Case 1. There is a vertex $a\in H$ of order 2 and $a^{\prime}\neq a$ adjacent
to $a$ such that $h(a)=h(a^{\prime})$. In that case we let $G$ to be $H$ with
$a$ and $a^{\prime}$ identified, and let $g,f$ be the appropriately induced
maps. Specifically, let $\sim$ be the equivalence relation on $H$, which
identifies $a$ and $a^{\prime}$. Define $G$ as the graph $H/\\!\\!\sim$ and
let $g\colon H\to G$ be the quotient epimorphism. Then $g$ is a simple-
monotone epimorphism. Define the epimorphism $f\colon G\to F$ by the condition
$h=f\circ g$. Moreover for any ramification vertex $p\in F$,
$P_{f}(p)=g(P_{h}(p))$ and $f$ is special.

Case 2. Case 1 does not hold. Consider the subset $S$ of $H$ defined by the
conditions $x\in S$ if and only if there are $y,z\in H$ such that:

  1. (1)

$y$ and $z$ are incomparable;

  2. (2)

$x<y$ and $x<z$;

  3. (3)

$h(y)=h(z)$.

Since $h$ is not an isomorphism, $S\neq\emptyset$. Indeed, this follows from
the following claim.

Claim 1. Suppose that there are $a<b$ are such that $h(a)=h(b)$. Then $a\in S$
or $b\in S$.

###### Proof of Claim 1.

Because $f$ is order-preserving we may assume that $b$ is adjacent to $a$.
Since we are in Case 2, $a$ is a ramification vertex, and $b$ is either a
ramification vertex or an end vertex. Denote $c=h(a)=h(b)$. In case $c$ is an
end vertex or an ordinary vertex, we get $a\in S$. Indeed, in the case that
$c$ is an end vertex, then all vertices above $a$ get mapped to $c$, so $a\in
S$. In the case that $c$ is an ordinary vertex, let $d$ be the vertex that is
adjacent and above $c$. No component of $C_{a}$ is mapped by $h$ onto $c$ as
$h$ is end vertex preserving. Thus there are vertices in different components
of $C_{a}$ that are mapped to $d$, again making $a\in S$.

Now consider the case that $c$ is a ramification vertex. Then either $a$ is
special, or it is above a special vertex, or it is below a special vertex. In
the first two cases $b$ is above a special vertex. Then, by Lemma 8.13(2), we
get $b\in S$, as there is a component $D$ of $C_{c}$ such that for any
component $E$ of $C_{b}$ it holds $f(E)\setminus\\{c\\}=D$. In case $a$ is
below a special vertex, we get, by Lemma 8.13(3), $a\in S$, as any component
of $C_{a}$ is mapped onto $C_{c}\cup\\{c\\}$. ∎

Let $v$ be a maximal element in $S$ with respect to the order $\leq$, let
$y,z$ be two vertices as in the definition of $S$, and denote $u=h(v)$. Let
$C_{1}$ and $C_{2}$ be components of $C_{v}$ that contain $y$ and $z$
respectively.

By confluence of $h$, each of $h(C_{1})$, $h(C_{2})$ contains one or more
components of $C_{u}$.

Claim 2. $h(C_{1})=h(C_{2})$.

###### Proof of Claim 2.

(a) Suppose that $u$ is an end vertex. Then $h(C_{1})=h(C_{2})=\\{u\\}$.

(b) Suppose that $u$ is an ordinary vertex or $u$ is a ramification vertex and
$v$ is special or $v$ is above a special vertex.

In case $u$ is an ordinary vertex, this is clear. In the case that $u$ is a
ramification vertex it follows from Lemma 8.13(1),(2) that for each component
$E$ of $C_{v}$, $h(E)$ contains exactly one component of $C_{u}$.

We have to verify that both $h(C_{1})$ and $h(C_{2})$ contain the same
component, and that $h(C_{1})$ or $h(C_{2})$ does not contain $u$. For that,
we will show that there is no $v^{\prime}>v$ with $h(v^{\prime})=u$. Indeed,
if there is such a $v^{\prime}$, since we are not in Case 1 and not in (a) of
this proof, such a $v^{\prime}$ would have to be a ramification vertex.
Furthermore, there is a component $D$ of $C_{u}$ such that for any component
$E$ of $C_{v^{\prime}}$, it holds $f(E)\setminus\\{u\\}=D$, which implies
$v^{\prime}\in S$, resulting in a contradiction of the maximality of $v$.
Therefore $h(y)=h(z)>u$, and $h(C_{1})=h(C_{2})$ is a single component of
$C_{u}$, .

(c) Suppose that $u$ is a ramification vertex and $v$ is below a special
vertex. Then, by Lemma 8.13(3), $h(C_{1})=h(C_{2})=C_{u}\cup~{}\\{u\\}$.

∎

Claim 3. Each of the restrictions $h|_{C_{1}}$ and $h|_{C_{2}}$ is injective.

###### Proof of Claim 3.

Fix one of the $C_{1}$ or $C_{2}$ and call it $C$. In case there are
incomparable $y^{\prime},z^{\prime}\in C$ such that
$h(y^{\prime})=h(z^{\prime})$, then there is $x\in C$ such that $x<y^{\prime}$
and $x<z^{\prime}$ (we can, for example, take $x$ to be the meet of
$y^{\prime}$ and $z^{\prime}$), making $x$ in $S$, and we have a contradiction
with maximality of $v$.

Now suppose there are $a,b\in C$ such that $v<a<b$ and $h(a)=h(b)$. Apply
Claim 1 and get $a\in S$ or $b\in S$, which is a contradiction with the
maximality of $v$.

∎

To finish the proof of Theorem 8.14 define the equivalence relation $\sim$ by
$p\sim q$ if and only if $h(p)=h(q)$ and $p,q\in C_{1}\cup C_{2}$. Define $G$
as the graph $H/\\!\\!\sim$ and let $g\colon H\to G$ be the quotient
epimorphism. Then $g$ is an elementary light confluent epimorphism. Define the
epimorphism $f\colon G\to F$ by the condition $h=f\circ g$ and note that by
Proposition 4.4, $f$ is confluent. Since $h$ is end vertex preserving, Lemma
5.15 implies that $f$ is end vertex preserving. Moreover for any ramification
vertex $p\in F$, $P_{f}(p)=g(P_{h}(p))$ and $f$ is special. ∎

As a by-product of Theorem 8.14, we obtain the following.

######  Corollary 8.15.

Light confluent epimorphisms between rooted trees are compositions of
elementary light confluent epimorphisms.

Our goal now is to prove that the definition of
$(\mathcal{T}_{\mathcal{CE}})^{\omega}$ does not depend on choices of the
inverse limit representations (Corollary 8.17). We first need the following
theorem and lemma.

######  Theorem 8.16.

Let $f\colon H\to G$ and $g\colon G\to F$ be confluent epimorphisms between
finite rooted trees. If the composition $g\circ f\colon H\to F$ is simple-
confluent, then $g$ is simple-confluent.

###### Proof.

Because of Theorem 8.14 we will show that $g$ is special assuming that the
composition $h=g\circ f$ is special. We have that $g$ is end vertex preserving
by Lemma 5.15.

Let $x$ be a ramification vertex in $F$ and $u\in G$ be such that $g(u)>x$.
Take $v\in H$ such that $f(v)=u$. Let $x^{\prime\prime}<v$ be special for $x$
(wrt $h$). We claim that $x^{\prime}=f(x^{\prime\prime})$ is special for $x$
(wrt $g$). We know that $g$ is an epimorphism between rooted trees, in
particular, it is order-preserving, therefore $x^{\prime}<u$. We have to
verify (1) and (2) in the definition of a special vertex.

To prove (2), let $A$ be a component of $C_{x}$ and let $D$ be a component of
$C_{x^{\prime\prime}}$ such that
$h(D\cup\\{x^{\prime\prime}\\})=A\cup\\{x\\}$; it exists since $h$ special.
Let $B$ be a component of $C_{x^{\prime}}$ intersecting $f(D)$. Then, using
confluence of $g$, it holds $g(B\cup\\{x^{\prime}\\})\supseteq A\cup\\{x\\}$.

To prove (1), suppose towards the contradiction that there are components
$A_{1},A_{2}$ of $C_{x}$ such that for some component $B$ of $C_{x^{\prime}}$,
we have $g^{-1}(A_{1}),g^{-1}(A_{2})\subseteq B$. We find a component $D$ of
$f^{-1}(B)$, $y\in H$ which is special for $x$ (wrt $h$) and satisfies
$f(y)=x^{\prime}$, and $E$, a component of $C_{y}$, such that $E\cap
D\neq\emptyset$. To find such a $y$ first take any $b\in B$ with $g(b)\neq x$.
Take $c,d\in H$ such that $f(c)=x^{\prime}$, $f(d)=b$, and $c<d$, which exist
as $f$ is an epimorphism. Now let $y<d$ be special for $x$ (wrt $h$). Let $E$
be a component of $C_{y}$ containing $d$ and let $D$ be a component of
$f^{-1}(B)$ intersecting $E$. Clearly $D\subseteq E$ and by confluence of $f$,
$f(E\cup\\{y\\})\supseteq B\cup\\{x^{\prime}\\}$. Hence
$h(E\cup\\{y\\})\supseteq A_{1}\cup A_{2}$, which contradicts that $y$ is
special for $x$ (wrt $h$). ∎

######  Corollary 8.17.

The definition of $(\mathcal{T}_{\mathcal{CE}})^{\omega}$ does not depend on
choices of the inverse limit representations.

###### Proof.

Apply Corollary 1.11 with $\mathcal{F}_{0}$ equal to the family of rooted
trees with confluent epimorphisms and $\mathcal{F}$ equal to the family of
rooted trees with simple-confluent epimorphisms. The assumption (1) in its
statement follows from Proposition 4.4 and (2) follows from Theorem 8.16. ∎

###  8.4. Simple*-confluent epimorphisms and the family
$\mathcal{T}_{\mathcal{C}}$

A family of epimorphisms, which we call special*, is introduced and is shown
to be equal to the family of simple*-confluent epimorphisms. Then we prove
results parallel to the ones in the the previous subsection. We conclude with
Theorem 8.30 which summarizes the properties of $|\mathbb{T}_{\mathcal{C}}|$,
the topological realization of the projective Fraïssé limit of the family all
rooted trees with simple*-confluent epimorphisms.

######  Theorem 8.18.

The family $\mathcal{T}_{\mathcal{C}}$ of all rooted trees with
simple*-confluent epimorphisms is a projective Fraïssé family.

It is easy to see that Theorem 8.6 also shows that the family
$\mathcal{T}_{\mathcal{C}}$ has the JPP. The amalgamation of
$\mathcal{T}_{\mathcal{C}}$ follows from Theorem 5.16 and the two lemmas
below.

######  Lemma 8.19.

Let $f\colon B\to A$ be a simple*-monotone epimorphism and let $g\colon C\to
A$ be a light confluent epimorphism. Let $D$ be obtained via the standard
amalgamation procedure and let $f_{0}$ and $g_{0}$ be the corresponding
projections. Then $f_{0}$ is light confluent and $g_{0}$ is simple*-monotone.

###### Proof.

By Theorem 5.16, all we are left to show is that $g_{0}$ is simple*-monotone.

Since a simple*-monotone epimorphism is a composition of splitting edge maps
and adding edge maps, and by Lemma 8.8, we can assume without loss of
generality that $f$ is an adding edge map. Then the proof is analogous to the
proof of Lemma 8.8. ∎

######  Lemma 8.20.

Let $f\colon B\to A$ and $g\colon C\to A$ be simple*-monotone epimorphisms.
Then there is a rooted tree $D$ and simple*-monotone epimorphisms $f_{0}\colon
D\to B$ and $g_{0}\colon D\to C$ such that $f\circ f_{0}=g\circ g_{0}$.

###### Proof.

It suffices to show that if each of $f\colon B\to A$ and $g\colon C\to A$ is a
splitting edge map or an adding edge map, then we can find a rooted tree $D$,
$f_{0}\colon D\to B$, and $g_{0}\colon D\to C$ such that each of $f_{0}$ and
$g_{0}$ is a splitting edge map or an adding edge map, and $f\circ
f_{0}=g\circ g_{0}$.

The proof is analogous to the proof of Lemma 8.9. ∎

######  Definition 8.21.

Let $f\colon S\to T$ be a confluent epimorphism between finite rooted trees,
and let $p$ be a ramification vertex. Call a vertex $q\in S$ with $f(q)=p$ a
special* for $p$ (wrt $f$) if the following hold:

  1. (1)

for any component $E$ of $C_{q}$ either $f(E)=\\{p\\}$, or there is exactly
one component $D$ of $C_{p}$ such that $D\subseteq f(E)$;

  2. (2)

for any component $D$ of $C_{p}$ there is a component $E$ of $C_{q}$ such that
$D\subseteq f(E)$.

######  Definition 8.22.

A confluent epimorphism $f\colon S\to T$ between rooted trees will be called a
special* epimorphism if for every ramification vertex $x\in T$ and $b\in S$
such that $f(b)>x$, there exists $x^{\prime}<b$, which is special* for $x$
(wrt to $f$).

Then we have the following analogs of Lemma 8.13, Theorem 8.14, and Theorem
8.16.

######  Lemma 8.23.

Let the epimorphism $f\colon S\to T$ between finite rooted trees be special*,
let $x\in T$ be a ramification vertex, and let $y\in S$ be such that $f(y)=x$.
Then:

  1. (1)

$y$ is special*, or

  2. (2)

there is a component $D$ of $C_{x}$ such that for any component $E$ of $C_{y}$
it holds $f(E)\setminus\\{x\\}=D$ or $f(E)=\\{x\\}$, or else

  3. (3)

for any component $E$ of $C_{y}$ it holds $f(E)=C_{x}\cup\\{x\\}$ or
$f(E)=\\{x\\}$.

Note that unless $y$ and $x$ in the statement of Lemma 8.23 are such that
$f(C_{y})=\\{x\\}$, then (2) corresponds to the case when $y$ is above a
special vertex, and (3) corresponds to the case when $y$ is below a special
vertex.

######  Theorem 8.24.

An epimorphism between finite rooted trees is simple*-confluent if and only if
it is special*.

###### Proof.

A proof that every simple*-confluent epimorphism is special* is essentially
the same as that every simple-confluent epimorphism is special.

The opposite direction will follow from the following. We prove that a
special* epimorphism can be written as a composition of a special epimorphism
and a simple*-confluent epimorphism. Then the conclusion will follow from
Theorem 8.14 since the composition of simple*-confluent and simple-confluent
is simple*-confluent.

Claim. Let $h\colon H\to F$ be a special* epimorphism. Suppose that there is a
vertex $a\in H$ and a component $E$ of $C_{a}$ such that $h(E)=\\{h(a)\\}$.
Then there is a rooted tree $G$, a special* epimorphism $f\colon G\to F$, and
a simple*-confluent epimorphism $g\colon H\to G$ with $g(E)=\\{g(a)\\}$ and
$h=f\circ g$. In fact, $g=g_{2}\circ g_{1}$, where $g_{2}$ is an edge adding
map, and $g_{1}$ is a simple-confluent epimorphism.

###### Proof of Claim.

We let $\sim$ to be the equivalence relation on $H$, which identifies all
vertices in $E\cup\\{a\\}$. Define $G$ as the graph $H/\\!\\!\sim$ and let
$g\colon H\to G$ be the quotient epimorphism. We will see in a moment that $g$
is a simple*-confluent epimorphism. Define the epimorphism $f\colon G\to F$ by
the condition $h=f\circ g$. Then it is easy to see that $f$ is special*.

We let $g_{2}\colon G^{\prime}\to G$ be the map that adds an edge with one of
its end vertices being $g(a)$. Denote $a^{\prime}=g(a)$ and let $b^{\prime}$
be the other end vertex of the added edge. Let $g^{\prime}_{1}\colon
E\cup\\{a\\}\to(\\{a^{\prime},b^{\prime}\\},\langle
a^{\prime},b^{\prime}\rangle)$ be any simple-confluent epimorphism, and
finally, let $g_{1}\colon H\to G^{\prime}$ be equal to $g^{\prime}_{1}$ on $E$
and to the identity otherwise. Then $g_{1}$ is simple-confluent as well. This
in particular shows that $g$ is simple*-confluent.

∎

Let $h\colon H\to F$ be a special* epimorphism. Apply Claim as many times as
it is possible and obtain a decomposition $h=h_{2}\circ h_{1}$, where
$h_{1}\colon H\to G$ is simple*-confluent, $h_{2}\colon G\to F$ is special*,
and for no ordinary or ramification vertex $a$ there is a component $E$ of
$C_{a}$ with $h_{2}(E)=\\{h_{2}(a)\\}$. Then $h_{2}$ is special. Indeed, all
that is left to show is that $h_{2}$ is end vertex preserving. Suppose that it
is not the case and let $e$ be an end vertex of $G$ such that $x=h_{2}(e)$ is
not an end vertex. Let $y$ be the immediate predecessor of $e$ and let $E$ be
the component of $C_{y}$ that contains $e$.

If $x=h_{2}(y)=h_{2}(e)$, then $h_{2}(E)=\\{x\\}$, which is impossible. On the
other hand, if $h_{2}(y)\neq h_{2}(e)$, then by the confluence of $h_{2}$, we
have $h_{2}(E)\supseteq E^{\prime}$, where $E^{\prime}$ is the component of
$h_{2}(y)$ containing $x$. But this is also impossible as $h_{2}(E)=\\{x\\}$
and $x$ is not an end vertex.

∎

The proof of the next theorem is identical to that of Theorem 8.16.

######  Theorem 8.25.

Let $f\colon H\to G$ and $g\colon G\to F$ be confluent epimorphisms between
finite rooted trees. If the composition $g\circ f\colon H\to F$ is
simple*-confluent, then $g$ is simple*-confluent.

Therefore by Theorem 8.25 and Corollary 1.11, we have:

######  Corollary 8.26.

The definition of $(\mathcal{T}_{\mathcal{C}})^{\omega}$ does not depend on
choices of the inverse limit representations.

######  Notation 8.27.

Let $\mathbb{T}_{\mathcal{C}}$ be the projective Fraïssé limit of the family
$\mathcal{T}_{\mathcal{C}}$.

###  8.5. The continuum $|\mathbb{T}_{\mathcal{C}}|$

We exhibit several properties of the continuum $|\mathbb{T}_{\mathcal{C}}|$.
We do not know a characterization of the continuum
$|\mathbb{T}_{\mathcal{C}}|$ and believe it has not been previously studied.
On the other hand, in the remaining sections, we show that
$|\mathbb{T}_{\mathcal{CE}}|$ is homeomorphic to the Mohler-Nikiel universal
dendroid.

By Proposition 5.12, $|\mathbb{T}_{\mathcal{C}}|$ is a smooth dendroid.

######  Lemma 8.28.

A vertex $r\in\mathbb{T}_{\mathcal{C}}$ is either an end vertex or a
ramification vertex of $\mathbb{T}_{\mathcal{C}}$.

###### Proof.

Suppose a vertex $r\in\mathbb{T}_{\mathcal{C}}$ is not an end vertex. Then
there are vertices $a,b\in\mathbb{T}_{\mathcal{C}}$ such that $a<r<b$. Let
$0<\varepsilon<\min\\{d(a,r),d(r,b),d(a,b)\\}$,
$Z\in\mathcal{T}_{\mathcal{C}}$, and epimorphism
$f_{Z}\colon\mathbb{T}_{\mathcal{C}}\to Z$ be an $\varepsilon$-map. Define a
tree $W$ by $V(W)=V(Z)\cup\\{d\\}$ and $E(W)=E(Z)\cup\\{\langle
f_{Z}(r),d\rangle\\}$, and an epimorphism $f_{Z}^{W}\colon W\to Z$ by
$f_{Z}^{W}(x)=x$ if $x\in Z$ and $f_{Z}^{W}(d)=f_{Z}(r)$. Let
$f_{W}\colon\mathbb{T}_{\mathcal{C}}\to W$ be an epimorphism such that
$f_{Z}=f_{Z}^{W}\circ f_{W}$, and let $C$ be the component of
$(f_{W})^{-1}(\\{f_{W}(r),d\\})$ containing $r$. By confluence of $f_{W}$, see
Proposition 4.5, $f_{W}(C)=\\{f_{W}(r),d\\}$. Therefore, there is an arc $A$
in $\mathbb{T}_{\mathcal{C}}$ with $r$ as one of its end vertices, such that
the three arcs: $A$, $ra$, $rb$ witness that $r$ is a ramification vertex of
$\mathbb{T}_{\mathcal{C}}$. ∎

######  Lemma 8.29.

Let $\mathbb{T}_{\mathcal{C}}$ be the projective Fraïssé limit of the family
$\mathcal{T}_{\mathcal{C}}$. Then the set of end vertices is dense in
$\mathbb{T}_{\mathcal{C}}$.

###### Proof.

Let $p\in\mathbb{T}_{\mathcal{C}}$ and let $\varepsilon>0$ be given. We want
to find an end vertex $e\in\mathbb{T}_{\mathcal{C}}$ such that
$d(p,e)<\varepsilon$. Let $G$ be a tree and
$f_{G}\colon\mathbb{T}_{\mathcal{C}}\to G$ be an $\varepsilon$-map. Define a
tree $H$ such that $V(H)=V(G)\cup\\{q\\}$ and $E(H)=E(G)\cup\\{\langle
f_{G}(p),q\rangle\\}$, and let $f^{H}_{G}$ be an epimorphism defined by
${f^{H}_{G}}|_{G}$ is the identity and $f^{H}_{G}(q)=f_{G}(p)$. Let
$f_{H}\colon\mathbb{T}_{\mathcal{C}}\to H$ be an epimorphism such that
$f_{G}^{H}\circ f_{H}=f_{G}$. By Proposition 7.4 there is
$e\in\mathbb{T}_{\mathcal{C}}$ which is an end vertex and $f_{H}(e)=q$. Then
$d(p,e)<\varepsilon$ since $f_{G}(p)=f_{G}(e)$ and $f_{G}$ is an
$\varepsilon$-map.

∎

We summarize the results of this section in the following theorem.

######  Theorem 8.30.

The projective Fraïssé limit $\mathbb{T}_{\mathcal{C}}$ of the family
$\mathcal{T}_{\mathcal{C}}$ has transitive set of edges and its topological
realization $|\mathbb{T}_{\mathcal{C}}|$ is a Kelley dendroid (thus smooth)
with a dense set of end points and such that each point is either an end point
or a ramification point in the classical sense.

###### Proof.

Proceeding as in the proof of Theorem 3.14 we may show that the family
$\mathcal{T}_{\mathcal{C}}$ satisfies the hypothesis of Theorem 2.17. Hence
the set of edges is transitive. Knowing that $E(\mathbb{T}_{\mathcal{C}})$ is
transitive we see from Proposition 5.12, Proposition 6.5 and Observation 6.4
that the topological realization $|\mathbb{T}_{\mathcal{C}}|$ is a Kelley
dendroid.

To see that the set of end points is dense in $|\mathbb{T}_{\mathcal{C}}|$,
take a nonempty open set $U$ in $|\mathbb{T}_{\mathcal{C}}|$. The set
$\varphi^{-1}(U)$, where $\varphi$ is the quotient map, is open in
$\mathbb{T}_{\mathcal{C}}$. So by Lemma 8.29 there is an end vertex
$e\in\varphi^{-1}(U)$. It follows from Theorem 7.6 that
$\mathbb{T}_{\mathcal{C}}$ has no isolated end vertices. Then by Proposition