00footnotetext: Support of the research of both authors by the Austrian
Science Fund (FWF), project I 4579-N, and the Czech Science Foundation (GAČR),
project 20-09869L, entitled “The many facets of orthomodularity”, is
gratefully acknowledged.

# Extensions and congruences of partial lattices

Ivan Chajda and Helmut Länger

###### Abstract

For a partial lattice $\mathbf{L}$ the so-called two-point extension is
defined in order to extend $\mathbf{L}$ to a lattice. We are motivated by the
fact that the one-point extension broadly used for partial algebras does not
work in this case, i.e. the one-point extension of a partial lattice need not
be a lattice. We describe these two-point extensions and prove several
properties of them. We introduce the concept of a congruence on a partial
lattice and show its relationship to the notion of a homomorphism and its
connections with congruences on the corresponding two-point extension. In
particular we prove that the quotient $\mathbf{L}/E$ of a partial lattice
$\mathbf{L}$ by a congruence $E$ on $\mathbf{L}$ is again a partial lattice
and that the two-point extension of $\mathbf{L}/E$ is isomorphic to the
quotient lattice of the two-point extension $\mathbf{L}^{*}$ of $\mathbf{L}$
by the congruence on $\mathbf{L}^{*}$ generated by $E$. Several illustrative
examples are enclosed.

AMS Subject Classification: 06B75, 06A06, 06B05, 06B10, 08A55

Keywords: Partial lattice, partial sublattice, congruence, two-point extension

##  1 Introduction

Although our paper is devoted to extensions of partial lattices to lattices as
well as to congruences on partial lattices, we assume that it should be
helpful to get a brief introduction to partial algebras in general and then
apply these concepts to partial lattices. The source of the following concepts
are the monographs [2] and [3] by P. Burmeister and, concerning partial
lattices, the paper [4] by the first author and Z. Seidl. It is worth noticing
that the closedness of classes of partial algebras with respect to quotients,
subalgebras etc. was profoundly investigated by K. Denecke in [5] and by B.
Staruch and B. Staruch in [6].

In particular, we use the following concepts concerning various types of
identities in partial algebras (see [3]).

Let $\tau=(n_{i};i\in I)$ be a similarity type, $\mathbf{A}=(A,F)$ and
$\mathbf{B}=(B,F)$ with $F=(f_{i};i\in I)$ partial algebras of type $\tau$ and
$p,q$ terms of type $\tau$. Then

$p\stackrel{{\scriptstyle e}}{{=}}q$

means: $p$ and $q$ are defined and they are equal. We say that $\mathbf{A}$
satisfies the weak identity

$p\stackrel{{\scriptstyle w}}{{\approx}}q$

if the following holds: If $a_{1},\ldots,a_{n}\in A$ and
$p(a_{1},\ldots,a_{n})$ and $q(a_{1},\ldots,a_{n})$ are defined then
$p(a_{1},\ldots,a_{n})=q(a_{1},\ldots,a_{n})$. We say that $\mathbf{A}$
satisfies the strong identity

$p\stackrel{{\scriptstyle s}}{{\approx}}q$

if the following holds: For $a_{1},\ldots,a_{n}\in A$ the expression
$p(a_{1},\ldots,a_{n})$ is defined if and only if $q(a_{1},\ldots,a_{n})$ is
defined and in this case $p(a_{1},\ldots,a_{n})=q(a_{1},\ldots,a_{n})$. The
identity $p\approx q$ is called regular if $p$ and $q$ contain the same
variables.

The partial algebra $\mathbf{A}$ is called a weak subalgebra of $\mathbf{B}$
if $A\subseteq B$ and if for all $i\in I$ and all $a_{1},\ldots,a_{n_{i}}\in
A$, if $f_{i}(a_{1},\ldots,a_{n_{i}})$ is defined in $\mathbf{A}$ then it is
defined in $\mathbf{B}$ and has the same value. The weak subalgebra
$\mathbf{A}$ of $\mathbf{B}$ is called a subalgebra of $\mathbf{B}$ if for all
$i\in I$ and all $a_{1},\ldots,a_{n_{i}}\in A$, if
$f_{i}(a_{1},\ldots,a_{n_{i}})$ is defined in $\mathbf{B}$ then it is defined
in $\mathbf{A}$.

A homomorphism from $\mathbf{A}$ to $\mathbf{B}$ is a mapping $h\colon
A\rightarrow B$ such that for all $i\in I$ and all
$a_{1},\ldots,a_{n_{i}},a\in A$,
$f_{i}(a_{1},\ldots,a_{n_{i}})\stackrel{{\scriptstyle e}}{{=}}a$ implies
$f_{i}\big{(}h(a_{1}),\ldots,h(a_{n_{i}})\big{)}\stackrel{{\scriptstyle
e}}{{=}}h(a)$. A homomorphism from $\mathbf{A}$ to $\mathbf{B}$ is called
closed if for all $i\in I$ and all $a_{1},\ldots,a_{n_{i}}\in A$, if
$f_{i}\big{(}h(a_{1}),\ldots,h(a_{n_{i}})\big{)}$ is defined in $\mathbf{B}$
then $f_{i}(a_{1},\ldots,a_{n_{i}})$ is defined in $\mathbf{A}$. It is easy to
see that $\mathbf{A}$ is a weak subalgebra of $\mathbf{B}$ if and only if
$A\subseteq B$ and $\operatorname{id}_{A}$ is a homomorphism from $\mathbf{A}$
to $\mathbf{B}$ and that $\mathbf{A}$ is a subalgebra of $\mathbf{B}$ if and
only if $A\subseteq B$ and $\operatorname{id}_{A}$ is a closed homomorphism
from $\mathbf{A}$ to $\mathbf{B}$. An isomorphism is a bijective closed
homomorphism.

Let $\mathbf{A}=\big{(}A,(f_{i};i\in I)\big{)}$ be a partial algebra and
assume $c\notin A$. Then the one-point extension $\bar{\mathbf{A}}$ of
$\mathbf{A}$ is defined as follows:

  1. (i)

If all $f_{i}$ are defined everywhere then $\bar{\mathbf{A}}:=\mathbf{A}$.

  2. (ii)

if at least one $f_{i}$ is not defined everywhere then
$\bar{\mathbf{A}}:=\big{(}A\cup\\{c\\},(g_{i};i\in I)\big{)}$ where for all
$i\in I$ and all $a_{1},\ldots,a_{n_{i}}\in A\cup\\{c\\}$

$g_{i}(a_{1},\ldots,a_{n_{i}}):=\left\\{\begin{array}[]{ll}f_{i}(a_{1},\ldots,a_{n_{i}})&\text{if
}f_{i}(a_{1},\ldots,a_{n_{i}})\text{ is defined in }\mathbf{A},\\\
c&\text{otherwise}.\end{array}\right.$

It was proved in [5] and [6] that a class of partial algebras can be described
by a set of strong and regular identities if and only if it is closed under
the formation of subalgebras, closed homomorphic images, direct products,
initial segments and one-point extensions.

However, not every interesting class of partial algebras can be described by
strong and regular identities. For example, partial lattices treated in [4]
are described by identities containing those for absorption that are neither
regular nor strong. This is in accordance with the fact that the one-point
extension of a partial lattice is not a lattice in general. We want to present
another construction for partial lattices, the so-called two-point extension,
which preserves also weak identities like the absorption identities.

##  2 Partial lattices

There exist several definitions of a partial lattice. For our purposes we
adopt the following one from [4].

######  Definition 2.1.

A partial lattice is a partial algebra $(L,\vee,\wedge)$ of type $(2,2)$
satisfying the following identities:

$\displaystyle x\vee x\stackrel{{\scriptstyle s}}{{\approx}}x$
$\displaystyle\text{ and }x\wedge x\stackrel{{\scriptstyle s}}{{\approx}}x,$
$\displaystyle x\vee y\stackrel{{\scriptstyle s}}{{\approx}}y\vee x$
$\displaystyle\text{ and }x\wedge y\stackrel{{\scriptstyle
s}}{{\approx}}y\wedge x,$ $\displaystyle(x\vee y)\vee z\stackrel{{\scriptstyle
s}}{{\approx}}x\vee(y\vee z)$ $\displaystyle\text{ and }(x\wedge y)\wedge
z\stackrel{{\scriptstyle s}}{{\approx}}x\wedge(y\wedge z)$

as well as the following duality conditions:

$\displaystyle x\vee y\stackrel{{\scriptstyle e}}{{=}}x$ $\displaystyle\text{
implies }x\wedge y\stackrel{{\scriptstyle e}}{{=}}y,$ $\displaystyle x\wedge
y\stackrel{{\scriptstyle e}}{{=}}x$ $\displaystyle\text{ implies }x\vee
y\stackrel{{\scriptstyle e}}{{=}}y.$

We call weak subalgebras of a partial lattice also partial sublattices.

The following lemma was proved in [4]. For the sake of completeness we repeat
the proof.

######  Lemma 2.2.

Let $\mathbf{L}$ be a partial lattice. Then the following hold:

  1. (i)

$\mathbf{L}$ satisfies $(x\vee y)\wedge x\stackrel{{\scriptstyle
w}}{{\approx}}x$ and $(x\wedge y)\vee x\stackrel{{\scriptstyle
w}}{{\approx}}x$.

  2. (ii)

If $\mathbf{L}$ satisfies $(x\vee y)\wedge x\stackrel{{\scriptstyle
s}}{{\approx}}x$ and $(x\wedge y)\vee x\stackrel{{\scriptstyle
s}}{{\approx}}x$ then $\mathbf{L}$ is a lattice.

###### Proof.

  1. (i)

Let $a,b\in L$ and assume $(a\vee b)\wedge a$ to be defined. Then $a\vee b$ is
defined. Since $a\vee a\stackrel{{\scriptstyle e}}{{=}}a$ we conclude that
$(a\vee a)\vee b$ is defined. Hence $a\vee(a\vee b)$ and $(a\vee b)\vee a$ are
defined and we obtain

$(a\vee b)\vee a\stackrel{{\scriptstyle e}}{{=}}a\vee(a\vee
b)\stackrel{{\scriptstyle e}}{{=}}(a\vee a)\vee b\stackrel{{\scriptstyle
e}}{{=}}a\vee b.$

Applying the duality conditions yields $(a\vee b)\wedge
a\stackrel{{\scriptstyle e}}{{=}}a$. The second weak identity follows
analogously.

  2. (ii)

is clear.

∎

Let $\mathbf{P}=(P,\leq)$ be a poset and $a,b\in P$. We define

$\displaystyle L(a,b)$ $\displaystyle:=\\{x\in P\mid x\leq a,b\\},$
$\displaystyle U(a,b)$ $\displaystyle:=\\{x\in P\mid a,b\leq x\\}.$

The following notions as well as the following result are taken from [4]:

The poset $\mathbf{P}$ is said to satisfy the

  * •

lower bound property (LBP) if for all $x,y\in L$ the set $L(x,y)$ is either
empty or possesses a greatest element,

  * •

upper bound property (UBP) if for all $x,y\in L$ the set $U(x,y)$ is either
empty or possesses a smallest element.

Let $\mathbf{L}=(L,\vee,\wedge)$ be a partial lattice. On $L$ we define the
binary relations $\leq_{\vee}$ and $\leq_{\wedge}$ as follows:

$\displaystyle x\leq_{\vee}y$ $\displaystyle\text{ if and only if }x\vee
y\stackrel{{\scriptstyle e}}{{=}}y,$ $\displaystyle x\leq_{\wedge}y$
$\displaystyle\text{ if and only if }x\wedge y\stackrel{{\scriptstyle
e}}{{=}}x.$

Due to the duality conditions, the relations $\leq_{\vee}$ and $\leq_{\wedge}$
coincide and form a partial order relation on $L$. We call it the induced
order of $\mathbf{L}$ and denote them by $\leq$.

We call a poset satisfying the (LBP) and the (UBP) a partially lattice-ordered
set.

######  Example 2.3.

The poset depicted in Figure 1

$0$$a$$c$$b$$d$$1$Fig. 1

is not a partially lattice-ordered set since e.g.
$U(a,b)=\\{c,d,1\\}\neq\emptyset$, but $U(a,b)$ has no smallest element.

The following theorem was partly proved in [4]. For the sake of completeness
we provide a complete proof.

######  Theorem 2.4.

  1. (i)

Let $\mathbf{L}=(L,\vee,\wedge)$ be a partial lattice and $\leq$ its induced
order. Then $\mathbb{P}(\mathbf{L}):=(L,\leq)$ is a partially lattice-ordered
set.

  2. (ii)

Let $\mathbf{P}=(P,\leq)$ be a partially lattice-ordered set and define
partial binary operations $\vee$ and $\wedge$ on $P$ as follows:

$\displaystyle x\vee y$
$\displaystyle\left\\{\begin{array}[]{ll}:=\sup(x,y)&\text{if
}U(x,y)\neq\emptyset,\\\ \text{is
undefined}&\text{otherwise},\end{array}\right.$ $\displaystyle x\wedge y$
$\displaystyle\left\\{\begin{array}[]{ll}:=\inf(x,y)&\text{if
}L(x,y)\neq\emptyset,\\\ \text{is
undefined}&\text{otherwise}\end{array}\right.$

($x,y\in P$). Then $\mathbb{L}(\mathbf{P}):=(P,\vee,\wedge)$ is a partial
lattice.

  3. (iii)

Let $\mathbf{L}=(L,\vee,\wedge)$ be a partial lattice. Then
$\mathbb{L}\big{(}\mathbb{P}(\mathbf{L})\big{)}=\mathbf{L}$.

  4. (iv)

Let $\mathbf{P}=(P,\leq)$ be a partially lattice-ordered set. Then
$\mathbb{P}\big{(}\mathbb{L}(\mathbf{P})\big{)}=\mathbf{P}$.

###### Proof.

  1. (i)

Let $a,b,c\in L$. Then $a\vee a\stackrel{{\scriptstyle e}}{{=}}a$ and hence
$a\leq a$. If $a\leq b\leq a$ then $a\vee b\stackrel{{\scriptstyle e}}{{=}}b$
and $b\vee a\stackrel{{\scriptstyle e}}{{=}}a$ and hence
$a\stackrel{{\scriptstyle e}}{{=}}b\vee a\stackrel{{\scriptstyle e}}{{=}}a\vee
b\stackrel{{\scriptstyle e}}{{=}}b$. If $a\leq b\leq c$ then

$c\stackrel{{\scriptstyle e}}{{=}}b\vee c\stackrel{{\scriptstyle
e}}{{=}}(a\vee b)\vee c\stackrel{{\scriptstyle e}}{{=}}a\vee(b\vee
c)\stackrel{{\scriptstyle e}}{{=}}a\vee c,$

i.e. $a\leq c$. Now assume $c\in U(a,b)$. Then $a\vee c\stackrel{{\scriptstyle
e}}{{=}}c$ and $b\vee c\stackrel{{\scriptstyle e}}{{=}}c$ and hence

$c\stackrel{{\scriptstyle e}}{{=}}a\vee c\stackrel{{\scriptstyle
e}}{{=}}a\vee(b\vee c)\stackrel{{\scriptstyle e}}{{=}}(a\vee b)\vee c,$

i.e. $a\vee b$ is defined and $a\vee b\leq c$. Because of

$\displaystyle a\vee(a\vee b)$ $\displaystyle\stackrel{{\scriptstyle
e}}{{=}}(a\vee a)\vee b\stackrel{{\scriptstyle e}}{{=}}a\vee b,$
$\displaystyle b\vee(a\vee b)$ $\displaystyle\stackrel{{\scriptstyle
e}}{{=}}(a\vee b)\vee b\stackrel{{\scriptstyle e}}{{=}}a\vee(b\vee
b)\stackrel{{\scriptstyle e}}{{=}}a\vee b$

we have $a,b\leq a\vee b$. Together we obtain that in case
$U(a,b)\neq\emptyset$ the element $a\vee b$ is defined and coincides with
$\sup(a,b)$. The dual assertion can be proved analogously.

  2. (ii)

Let $a,b,c\in P$. Clearly, $a\vee a\stackrel{{\scriptstyle s}}{{=}}a$,
$a\wedge a\stackrel{{\scriptstyle s}}{{=}}a$, $a\vee b\stackrel{{\scriptstyle
s}}{{=}}b\vee a$ and $a\wedge b\stackrel{{\scriptstyle s}}{{=}}b\wedge a$.
Assume $(a\vee b)\vee c$ to be defined. Since $(a\vee b)\vee c\in U(b,c)$ the
element $b\vee c$ is defined and since $(a\vee b)\vee c\in U(a,b\vee c)$ the
element $a\vee(b\vee c)$ is defined and we obtain $(a\vee b)\vee
c=\sup(a,b,c)=a\vee(b\vee c)$. If, conversely, $a\vee(b\vee c)$ is defined
then

$a\vee(b\vee c)\stackrel{{\scriptstyle e}}{{=}}(b\vee c)\vee
a\stackrel{{\scriptstyle e}}{{=}}(c\vee b)\vee a\stackrel{{\scriptstyle
e}}{{=}}c\vee(b\vee a)\stackrel{{\scriptstyle e}}{{=}}(b\vee a)\vee
c\stackrel{{\scriptstyle e}}{{=}}(a\vee b)\vee c.$

The strong identity $(x\wedge y)\wedge z\stackrel{{\scriptstyle
s}}{{\approx}}x\wedge(y\wedge z)$ can be proved analogously. Finally, the
duality conditions can be easily verified.

  3. (iii)

Let $\mathbb{P}(\mathbf{L})=(L,\leq)$ and
$\mathbb{L}\big{(}\mathbb{P}(\mathbf{L})\big{)}=(L,\sqcup,\sqcap)$ and $a,b\in
L$. According to the proof of (i), if $U(a,b)\neq\emptyset$ then $a\vee b$ is
defined and $a\vee b=\sup(a,b)$. Conversely, if $a\vee b$ is defined then

$\displaystyle a\vee(a\vee b)$ $\displaystyle\stackrel{{\scriptstyle
e}}{{=}}(a\vee a)\vee b\stackrel{{\scriptstyle e}}{{=}}a\vee b,$
$\displaystyle b\vee(a\vee b)$ $\displaystyle\stackrel{{\scriptstyle
e}}{{=}}(a\vee b)\vee b\stackrel{{\scriptstyle e}}{{=}}a\vee(b\vee
b)\stackrel{{\scriptstyle e}}{{=}}a\vee b$

and hence $a\vee b\in U(a,b)$ showing $U(a,b)\neq\emptyset$. Hence the
following are equivalent: $a\vee b$ is defined; $U(a,b)\neq\emptyset$;
$a\sqcup b$ is defined. In this case we have $a\sqcup b=\sup(a,b)=a\vee b$.
Analogously, one can prove that $a\sqcap b$ is defined if and only if $a\wedge
b$ is defined and in this case $a\sqcap b=a\wedge b$.

  4. (iv)

If $\mathbb{L}(\mathbf{P})=(P,\vee,\wedge)$,
$\mathbb{P}\big{(}\mathbb{L}(\mathbf{P})\big{)}=(P,\sqsubseteq)$ and $a,b\in
P$ then the following are equivalent: $a\sqsubseteq b$; $a\vee
b\stackrel{{\scriptstyle e}}{{=}}b$; $\sup(a,b)\stackrel{{\scriptstyle
e}}{{=}}b$; $a\leq b$. This shows
$\mathbb{P}\big{(}\mathbb{L}(\mathbf{P})\big{)}=\mathbf{P}$.

∎

##  3 Extensions

As mentioned above, the one-point extension of a partial lattice need not be a
lattice. In order to avoid this difficulty, we introduce the two-point
extension of a partial lattice as follows.

######  Definition 3.1.

Let $\mathbf{L}=(L,\vee,\wedge)$ be a partial lattice and $\leq$ its induced
order and assume $0,1\notin L$. Then the two-point extension
$\mathbf{L}^{*}=(L^{*},\leq^{*})$ of $\mathbf{L}$ is defined as follows:

  1. (i)

If $\vee$ and $\wedge$ are defined everywhere then

$\mathbf{L}^{*}:=(L,\leq),$

  2. (ii)

if $\wedge$ is defined everywhere, but $\vee$ is not then

$\mathbf{L}^{*}:=(L\cup\\{1\\},\leq\cup(L^{*}\times\\{1\\}),$

  3. (iii)

if $\vee$ is defined everywhere, but $\wedge$ is not then

$\mathbf{L}^{*}:=(L\cup\\{0\\},\leq\cup(\\{0\\}\times L^{*}),$

  4. (iv)

if neither $\vee$ nor $\wedge$ is defined everywhere then

$\mathbf{L}^{*}:=(L\cup\\{0,1\\},\leq\cup(\\{0\\}\times
L^{*})\cup(L^{*}\times\\{1\\}).$

Clearly, $\mathbf{L}$ is a partial sublattice of $\mathbf{L}^{*}$. If
$\mathbf{L}$ is an infinite lattice having neither a smallest nor a greatest
element then its two-point extension coincides with $\mathbf{L}$. Hence, one
should have in mind that $\mathbf{L}^{*}$ neither need have a smallest nor a
greatest element.

The following lemma is obvious.

######  Lemma 3.2.

Let $\mathbf{L}=(L,\vee,\wedge)$ be a partial lattice and $a,b,c,d\in L$. Then

  1. (i)

$\mathbf{L}^{*}$ is a lattice with lattice operations

$x\vee^{*}y:=\sup_{\leq^{*}}(x,y)\text{ and
}x\wedge^{*}y:=\inf_{\leq^{*}}(x,y)$

($x,y\in L^{*}$),

  2. (ii)

$a\vee b\stackrel{{\scriptstyle e}}{{=}}c$ if and only if $a\vee^{*}b=c$, and
$a\wedge b\stackrel{{\scriptstyle e}}{{=}}d$ if and only if $a\wedge^{*}b=d$,

  3. (iii)

$a\vee^{*}b=\left\\{\begin{array}[]{ll}a\vee b&\text{ if
}U(a,b)\neq\emptyset,\\\ 1&\text{ otherwise},\end{array}\right.$ and
$a\wedge^{*}b=\left\\{\begin{array}[]{ll}a\wedge b&\text{ if
}L(a,b)\neq\emptyset,\\\ 0&\text{ otherwise}.\end{array}\right.$

For homomorphisms of partial lattices and their two-point extensions, we can
prove the following result.

######  Theorem 3.3.

Let $\mathbf{L}_{i}=(L_{i},\vee,\wedge),i=1,2,$ be partial lattices.

  1. (i)

Let $h^{*}$ be a homomorphism from $\mathbf{L}_{1}^{*}$ to
$\mathbf{L}_{2}^{*}$ with $h^{*}(L_{1})\subseteq L_{2}$. Then $h^{*}|L_{1}$ is
a homomorphism from $\mathbf{L}_{1}$ to $\mathbf{L}_{2}$.

  2. (ii)

Let $h$ be a closed homomorphism from $\mathbf{L}_{1}$ to $\mathbf{L}_{2}$.
Then there exists some homomorphism $h^{*}$ from $\mathbf{L}_{1}^{*}$ to
$\mathbf{L}_{2}^{*}$ with $h^{*}|L_{1}=h$.

###### Proof.

  1. (i)

Put $h:=h^{*}|L_{1}$ and let $a,b\in L_{1}$. First assume $a\vee b$ to be
defined. If $h(a)\vee h(b)$ would not be defined then we would conclude

$1_{\mathbf{L}_{2}}=h(a)\vee^{*}h(b)=h^{*}(a)\vee^{*}h^{*}(b)=h^{*}(a\vee^{*}b)=h(a\vee
b)\in h(L_{1})\subseteq L_{2},$

a contradiction. Hence $h(a)\vee h(b)$ is defined and

$h(a\vee b)=h^{*}(a\vee^{*}b)=h^{*}(a)\vee^{*}h^{*}(b)=h(a)\vee h(b).$

Analogously, one can prove that whenever $a\wedge b$ is defined, also
$h(a)\wedge h(b)$ is defined and $h(a\wedge b)=h(a)\wedge h(b)$.

  2. (ii)

Define $h^{*}\colon L_{1}^{*}\rightarrow L_{2}^{*}$ by

$\displaystyle h^{*}(x)$ $\displaystyle:=h(x)\text{ for all }x\in L_{1},$
$\displaystyle h^{*}(0_{\mathbf{L}_{1}^{*}})$
$\displaystyle:=0_{\mathbf{L}_{2}^{*}}\text{ if }0_{\mathbf{L}_{1}^{*}}\in
L_{1}^{*},$ $\displaystyle h^{*}(1_{\mathbf{L}_{1}^{*}})$
$\displaystyle:=1_{\mathbf{L}_{2}^{*}}\text{ if }1_{\mathbf{L}_{1}^{*}}\in
L_{1}^{*}.$

$\bullet$ $h^{*}$ is well-defined.
If $0_{\mathbf{L}_{2}^{*}}\notin L_{2}^{*}$ then $h(x)\wedge h(y)$ is defined
for all $x,y\in L_{1}$ and hence, since $h$ is closed, $x\wedge y$ is defined
for all $x,y\in L_{1}$ showing $0_{\mathbf{L}_{1}^{*}}\notin L_{1}^{*}$.
Analogously, one can prove that $1_{\mathbf{L}_{2}^{*}}\notin L_{2}^{*}$
implies $1_{\mathbf{L}_{1}^{*}}\notin L_{1}^{*}$.
$\bullet$ $h^{*}$ is a homomorphism from $\mathbf{L}_{1}^{*}$ to
$\mathbf{L}_{2}^{*}$.
Let $a,b\in L_{1}$. If $a\vee b$ is defined then $h(a)\vee h(b)$ is defined
and

$h^{*}(a\vee^{*}b)=h(a\vee b)=h(a)\vee h(b)=h^{*}(a)\vee^{*}h(b).$

If $a\vee b$ is not defined then, since $h$ is closed, $h(a)\vee h(b)$ is not
defined, too, and

$h^{*}(a\vee^{*}b)=h^{*}(1_{\mathbf{L}_{1}^{*}})=1_{\mathbf{L}_{2}^{*}}=h(a)\vee^{*}h(b)=h^{*}(a)\vee^{*}h^{*}(b).$

Analogously, one can show $h^{*}(a\wedge^{*}b)=h(a)\wedge^{*}h(b)$. It is not
hard to prove

$\displaystyle h^{*}(x\vee^{*}y)$ $\displaystyle=h^{*}(x)\vee^{*}h(y),$
$\displaystyle h^{*}(x\wedge^{*}y)$ $\displaystyle=h^{*}(x)\wedge^{*}h(y)$

for all $(x,y)\in(L_{1}^{*})^{2}\setminus L_{1}^{2}$.
$\bullet$ $h^{*}|L_{1}=h$.
This is clear.

∎

The following example shows that $\mathbf{L}^{*}$ need not be the smallest
lattice including $\mathbf{L}$ as a partial sublattice. Moreover, we show that
in general the operator ∗ does not preserve inclusion.

######  Example 3.4.

The partial lattice $\mathbf{L}_{1}=(L_{1},\vee,\wedge)$ visualized in Figure
2:

Fig. 2

is a partial sublattice of the partial lattice
$\mathbf{L}_{2}=(L_{2},\vee,\wedge)$ depicted in Figure 3:

Fig. 3

and one can see that $\operatorname{id}_{L_{1}}$ is a homomorphism from
$\mathbf{L}_{1}$ to $\mathbf{L}_{2}$ which is not closed. Although
$L_{1}\subseteq L_{2}$, we do not have $L_{1}^{*}\subseteq L_{2}^{*}$ since
$\mathbf{L}_{1}^{*}\cong\mathbf{N}_{5}$ whereas
$\mathbf{L}_{2}^{*}=\mathbf{L}_{2}$. Moreover, $\mathbf{L}_{1}^{*}$ is not the
smallest lattice including $\mathbf{L}_{1}$ as a partial sublattice.

##  4 Congruences and quotient partial lattices

We are going to introduce the concept of congruence on a partial lattice and
show how to define the partial lattice operations for congruence classes in a
natural way.

######  Definition 4.1.

Let $\mathbf{L}=(L,\vee,\wedge)$ be a partial lattice. By a congruence on
$\mathbf{L}$ is meant an equivalence relation $E$ on $L$ satisfying
$\Theta(E)\cap L^{2}=E$ where $\Theta(E)$ denotes the congruence on the
lattice $\mathbf{L}^{*}$ generated by $E$. Let $\operatorname{Con}\mathbf{L}$
denote the set of all congruences on $\mathbf{L}$. For
$E\in\operatorname{Con}\mathbf{L}$ we define two partial operation $\vee$ and
$\wedge$ on $L/E$ by

$\displaystyle[x]E\vee[y]E$
$\displaystyle\left\\{\begin{array}[]{ll}:=[x\vee^{*}y]\big{(}\Theta(E)\big{)}\cap
L&\text{if this set is non-empty},\\\ \text{is
undefined}&\text{otherwise},\end{array}\right.$ $\displaystyle[x]E\wedge[y]E$
$\displaystyle\left\\{\begin{array}[]{ll}:=[x\wedge^{*}y]\big{(}\Theta(E)\big{)}\cap
L&\text{if this set is non-empty},\\\ \text{is
undefined}&\text{otherwise}.\end{array}\right.$

It is easy to see that the operations $\vee$ and $\wedge$ on $L/E$ are well-
defined.

The next lemma shows that congruences as defined above have the expected
properties.

######  Lemma 4.2.

Let $\mathbf{L}=(L,\vee,\wedge)$ be a partial lattice. Then

  1. (i)

$\operatorname{Con}\mathbf{L}=\\{\Theta\cap
L^{2}\mid\Theta\in\operatorname{Con}\mathbf{L}^{*}\\}$,

  2. (ii)

$(\operatorname{Con}\mathbf{L},\subseteq)$ is a complete lattice.

###### Proof.

  1. (i)

Let $E$ be an equivalence relation on $L$. If
$E\in\operatorname{Con}\mathbf{L}$ then
$\Theta(E)\in\operatorname{Con}\mathbf{L}^{*}$ and $E=\Theta(E)\cap L^{2}$.
If, conversely, $\Theta\in\operatorname{Con}\mathbf{L}^{*}$ and $\Theta\cap
L^{2}=E$ then $E\subseteq\Theta$ and hence $\Theta(E)\subseteq\Theta$ whence

$E\subseteq\Theta(E)\cap L^{2}\subseteq\Theta\cap L^{2}=E$

which implies $\Theta(E)\cap L^{2}=E$ and hence
$E\in\operatorname{Con}\mathbf{L}$.

  2. (ii)

If $E_{i}\in\operatorname{Con}\mathbf{L}$ for all $i\in I$ then for every
$i\in I$ there exists some $\Theta_{i}\in\operatorname{Con}\mathbf{L}^{*}$
satisfying $\Theta_{i}\cap L^{2}=E_{i}$ and hence $\bigcap\limits_{i\in
I}\Theta_{i}\in\operatorname{Con}\mathbf{L}^{*}$ and

$\bigcap_{i\in I}E_{i}=\bigcap_{i\in I}(\Theta_{i}\cap L^{2})=(\bigcap_{i\in
I}\Theta_{i})\cap L^{2}\in\operatorname{Con}\mathbf{L}$

according to (i).

∎

The following lemma describes the partial lattice operations in quotients of
partial lattices.

######  Lemma 4.3.

Let $\mathbf{L}=(L,\vee,\wedge)$ be a partial lattice, $a,b\in L$ and
$E\in\operatorname{Con}\mathbf{L}$. Then

$[a]E\vee[b]E\left\\{\begin{array}[]{ll}=[a\vee b]E&\text{if
}U(a,b)\neq\emptyset,\\\ \text{is undefined}&\text{if }U(a,b)=\emptyset\text{
and }[1]\big{(}\Theta(E)\big{)}=\\{1\\},\\\ =[\alpha]E&\text{if
}U(a,b)=\emptyset\text{ and
}[1]\big{(}\Theta(E)\big{)}\neq\\{1\\}\end{array}\right.$

where $\alpha\in[1]\big{(}\Theta(E)\big{)}\setminus\\{1\\}$.

###### Proof.

If $U(a,b)\neq\emptyset$ then

$[a\vee^{*}b]\big{(}\Theta(E)\big{)}\cap L=[a\vee
b]\big{(}\Theta(E)\big{)}\cap L=[a\vee b]E\neq\emptyset$

and hence $[a]E\vee[b]E=[a\vee b]E$. If $U(a,b)=\emptyset$ and
$[1]\big{(}\Theta(E)\big{)}=\\{1\\}$ then

$[a\vee^{*}b]\big{(}\Theta(E)\big{)}\cap L=[1]\big{(}\Theta(E)\big{)}\cap
L=\\{1\\}\cap L=\emptyset$

and hence $[a]E\vee[b]E$ is not defined. Assume, finally, $U(a,b)=\emptyset$
and $[1]\big{(}\Theta(E)\big{)}\neq\\{1\\}$. Then
$[1]\big{(}\Theta(E)\big{)}\cap L=\emptyset$ would imply
$[1]\big{(}\Theta(E)\big{)}=\\{0,1\\}$ and hence $\Theta(E)=(L^{*})^{2}$
whence $L=L^{*}\cap L=[1]\big{(}\Theta(E)\big{)}\cap L=\emptyset$, a
contradiction. This shows that there exists some
$\alpha\in[1]\big{(}\Theta(E)\big{)}\cap L$. Now

$[a\vee^{*}b]\big{(}\Theta(E)\big{)}\cap L=[1]\big{(}\Theta(E)\big{)}\cap
L=[\alpha]\big{(}\Theta(E)\big{)}\cap L=[\alpha]E\neq\emptyset$

and hence $[a]E\vee[b]E=[\alpha]E$. ∎

######  Remark 4.4.

Due to duality, the dual version of Lemma 4.3 holds, too.

The main point is to show that the quotient partial algebra
$(L/E,\vee,\wedge)$ as defined above is again a partial lattice, the so-called
quotient partial lattice by the congruence $E$.

######  Theorem 4.5.

Let $\mathbf{L}=(L,\vee,\wedge)$ be a partial lattice and
$E\in\operatorname{Con}\mathbf{L}$. Then $\mathbf{L}/E:=(L/E,\vee,\wedge)$ is
a partial lattice.

###### Proof.

In this proof we often use Lemma 4.3. We prove only a part of the conditions
mentioned in Definition 2.1, the remaining conditions follow by duality. Let
$a,b,c\in L$. It is evident directly by Definition 4.1 that the partial
operations $\vee$ and $\wedge$ on $L/E$ satisfy strong idempotency and strong
commutativity. We need to prove also strong associativity. Assume
$([a]E\vee[b]E)\vee[c]E$ to be defined. We have

$U(a,b),U(a\vee b,c)\neq\emptyset\text{ if and only if }U(b,c),U(a,b\vee
c)\neq\emptyset$

and in this case we conclude

$(a\vee b)\vee c=\sup(a,b,c)=a\vee(b\vee c).$

$\bullet$ Case 1. $U(a,b),U(a\vee b,c)\neq\emptyset$.
Then $[a]E\vee([b]E\vee[c]E)$ is defined and

$\displaystyle([a]E\vee[b]E)\vee[c]E$ $\displaystyle=[a\vee
b]E\vee[c]E=[(a\vee b)\vee c]E=[a\vee(b\vee c)]E=$
$\displaystyle=[a]E\vee[b\vee c]E=[a]E\vee([b]E\vee[c]E).$

$\bullet$ Case 2. $U(a,b)=\emptyset$ or ($U(a,b)\neq\emptyset$ and $U(a\vee
b,c)=\emptyset$).
Then $1\in L^{*}$, $[1]\big{(}\Theta(E)\big{)}\neq\\{1\\}$ and
$[a]E\vee([b]E\vee[c]E)$ is defined. According to the proof of Lemma 4.3 there
exists some $\alpha\in[1]\big{(}\Theta(E)\big{)}\cap L$ and

$\displaystyle[x]E\vee[\alpha]E$
$\displaystyle=[x\vee^{*}\alpha]\big{(}\Theta(E)\big{)}\cap
L=[x\vee^{*}1]\big{(}\Theta(E)\big{)}\cap L=[1]\big{(}\Theta(E)\big{)}\cap L=$
$\displaystyle=[\alpha]\big{(}\Theta(E)\big{)}\cap L=[\alpha]E$

for all $x\in L$.
$\bullet$ Case 2a. $U(a,b)=U(b,c)=\emptyset$.
Then

$([a]E\vee[b]E)\vee[c]E=[\alpha]E\vee[c]E=[\alpha]E=[a]E\vee[\alpha]E=[a]E\vee([b]E\vee[c]E).$

Hence $([a]E\vee[b]E)\vee[c]E=[a]E\vee([b]E\vee[c]E)$.
$\bullet$ Case 2b. $U(a,b)=\emptyset$ and $U(b,c)\neq\emptyset$.
Then $U(a,b\vee c)=\emptyset$ and

$([a]E\vee[b]E)\vee[c]E=[\alpha]E\vee[c]E=[\alpha]E=[a]E\vee[b\vee
c]E=[a]E\vee([b]E\vee[c]E).$

$\bullet$ Case 2c. $U(a,b)\neq\emptyset$ and $U(b,c)=\emptyset$.
Then $U(a\vee b,c)=\emptyset$ and

$([a]E\vee[b]E)\vee[c]E=[a\vee
b]E\vee[c]E=[\alpha]E=[a]E\vee[\alpha]E=[a]E\vee([b]E\vee[c]E).$

$\bullet$ Case 2d. $U(a,b),U(b,c)\neq\emptyset$ and $U(a\vee b,c)=U(a,b\vee
c)=\emptyset$.
Then

$([a]E\vee[b]E)\vee[c]E=[a\vee b]E\vee[c]E=[\alpha]E=[a]E\vee[b\vee
c]E=[a]E\vee([b]E\vee[c]E).$

If, conversely $[a]E\vee([b]E\vee[c]E)$ is defined then

$\displaystyle[a]E\vee([b]E\vee[c]E)$ $\displaystyle\stackrel{{\scriptstyle
e}}{{=}}([b]E\vee[c]E)\vee[a]E\stackrel{{\scriptstyle
e}}{{=}}([c]E\vee[b]E)\vee[a]E\stackrel{{\scriptstyle e}}{{=}}$
$\displaystyle\stackrel{{\scriptstyle
e}}{{=}}[c]E\vee([b]E\vee[a]E)\stackrel{{\scriptstyle
e}}{{=}}([b]E\vee[a]E)\vee[c]E\stackrel{{\scriptstyle
e}}{{=}}([a]E\vee[b]E)\vee[c]E.$

Finally, the duality conditions remain to be proved. If
$[a]E\vee[b]E\stackrel{{\scriptstyle e}}{{=}}[a]E$ then
$a\in[a]E\subseteq[a\vee^{*}b]\big{(}\Theta(E)\big{)}$, i.e.
$a\mathrel{\Theta(E)}(a\vee^{*}b)$ and hence
$(a\wedge^{*}b)\mathrel{\Theta(E)}\big{(}(a\vee^{*}b)\wedge^{*}b\big{)}=b$
whence $[a\wedge^{*}b]\big{(}\Theta(E)\big{)}\cap
L=[b]\big{(}\Theta(E)\big{)}\cap L=[b]E\neq\emptyset$, i.e.
$[a]E\wedge[b]E\stackrel{{\scriptstyle e}}{{=}}[b]E$. ∎

Similar to the situation for algebras there exist natural relationships
between congruences and homomorphisms also for partial lattices.

######  Lemma 4.6.

Let $\mathbf{L}=(L,\vee,\wedge)$ be a partial lattice and
$E\in\operatorname{Con}\mathbf{L}$ and define $h\colon L\rightarrow L/E$ by
$h(x):=[x]E$ for all $x\in L$. Then $h$ is a homomorphism from $\mathbf{L}$ to
$\mathbf{L}/E$ and $\ker h=E$. If, in addition,
$[0_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}=\\{0_{\mathbf{L}^{*}}\\}$
provided $0_{\mathbf{L}^{*}}\in L^{*}$ and
$[1_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}=\\{1_{\mathbf{L}^{*}}\\}$
provided $1_{\mathbf{L}^{*}}\in L^{*}$ then $h$ is a closed homomorphism from
$\mathbf{L}$ to $\mathbf{L}/E$.

###### Proof.

This easily follows from Lemma 4.3. ∎

For partial lattices we can now prove the following version of the
Homomorphism Theorem.

######  Theorem 4.7.

For $i=1,2$ let $\mathbf{L}_{i}=(L_{i},\vee,\wedge)$ be partial lattices and
$h$ a closed homomorphism from $\mathbf{L}_{1}$ to $\mathbf{L}_{2}$. Then
$\ker h\in\operatorname{Con}\mathbf{L}_{1}$. If, in addition,
$[0_{\mathbf{L}_{1}^{*}}]\big{(}\Theta(\ker
h)\big{)}=\\{0_{\mathbf{L}_{1}^{*}}\\}$ provided $0_{\mathbf{L}_{1}^{*}}\in
L_{1}^{*}$ and $[1_{\mathbf{L}_{1}^{*}}]\big{(}\Theta(\ker
h)\big{)}=\\{1_{\mathbf{L}_{1}^{*}}\\}$ provided $1_{\mathbf{L}_{1}^{*}}\in
L_{1}^{*}$ then $\big{(}h(L_{1}),\vee,\wedge\big{)}$ (with partial operations
defined as in $\mathbf{L}_{2}$) is a partial lattice which is isomorphic to
$\mathbf{L}_{1}/(\ker h)$.

###### Proof.

According to Theorem 3.3 there exists some homomorphism $h^{*}$ from
$\mathbf{L}_{1}^{*}$ to $\mathbf{L}_{2}^{*}$ satisfying $h^{*}|L_{1}=h$. Now
$\ker h^{*}\in\operatorname{Con}\mathbf{L}_{1}$ and hence $\ker h=\ker
h^{*}\cap L_{1}^{2}\in\operatorname{Con}\mathbf{L}_{1}$. If the additional
condition holds then, according to Lemma 4.3, the mapping $h(x)\mapsto[x]\ker
h$ is a well-defined isomorphism from $\big{(}h(L_{1}),\vee,\wedge\big{)}$ to
$\mathbf{L}_{1}/(\ker h)$ since for $a,b\in L_{1}$ the following are
equivalent: $h(a)\vee h(b)$ exists in $\mathbf{L}_{2}$; $a\vee b$ exists in
$\mathbf{L}_{1}$; $[a]\ker h\vee[b]\ker h$ exists in $\mathbf{L}_{1}/(\ker
h)$. Analogous statements hold for $\wedge$ instead of $\vee$. ∎

Now we show that our concept of a quotient partial lattice $\mathbf{L}/E$ is
sound, i.e. its two-point extension is isomorphic to the quotient lattice of
the two-point extension of $\mathbf{L}$ with respect to $\Theta(E)$.

######  Theorem 4.8.

Let $\mathbf{L}=(L,\vee,\wedge)$ be a partial lattice and
$E\in\operatorname{Con}\mathbf{L}$. Then
$(\mathbf{L}/E)^{*}\cong\mathbf{L}^{*}/\big{(}\Theta(E)\big{)}$.

###### Proof.

Define $f\colon(L/E)^{*}\rightarrow L^{*}/\big{(}\Theta(E)\big{)}$ in the
following way:

$\displaystyle f([x]E)$ $\displaystyle:=[x]\big{(}\Theta(E)\big{)}\text{ for
all }x\in L,$ $\displaystyle f(0_{(\mathbf{L}/E)^{*}})$
$\displaystyle:=[0_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}\text{ if
}0_{(\mathbf{L}/E)^{*}}\in(L/E)^{*},$ $\displaystyle
f(1_{(\mathbf{L}/E)^{*}})$
$\displaystyle:=[1_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}\text{ if
}1_{(\mathbf{L}/E)^{*}}\in(L/E)^{*}.$

We will prove that $f$ is an isomorphism from $(\mathbf{L}/E)^{*}$ to
$\mathbf{L}^{*}/\big{(}\Theta(E)\big{)}$. Let $a,b\in L$.
$\bullet$ $f$ is well-defined.
If $[a]E=[b]E$ then $[a]\big{(}\Theta(E)\big{)}=[b]\big{(}\Theta(E)\big{)}$.
We prove that $U([a]E,[b]E)=\emptyset$ implies $U(a,b)=\emptyset$. Assume
$U(a,b)\neq\emptyset$, say $c\in U(a,b)$. Then $a\vee c\stackrel{{\scriptstyle
e}}{{=}}c$ and $b\vee c\stackrel{{\scriptstyle e}}{{=}}c$. Hence

$[a]E\vee[c]E=[a\vee^{*}c]\big{(}\Theta(E)\big{)}\cap
L=[c]\big{(}\Theta(E)\big{)}\cap L=[c]E$

and, analogously, $[b]E\vee[c]E=[c]E$, i.e. $[c]E\in
U([a]E,[b]E)\neq\emptyset$. Hence, if $1_{(\mathbf{L}/E)^{*}}\in(L/E)^{*}$
then there exist $d,e\in L$ with $U([d]E,[e]E)=\emptyset$. Therefore
$U(d,e)=\emptyset$ which shows $1_{\mathbf{L}^{*}}\in L^{*}$. Analogously, one
can prove that $0_{(\mathbf{L}/E)^{*}}\in(L/E)^{*}$ implies
$0_{\mathbf{L}^{*}}\in L^{*}$.
$\bullet$ $f$ is injective.
If $[a]\big{(}\Theta(E)\big{)}=[b]\big{(}\Theta(E)\big{)}$ then
$(a,b)\in\Theta(E)\cap L^{2}=E$, i.e. $[a]E=[b]E$. Now assume
$1_{(\mathbf{L}/E)^{*}}\in(L/E)^{*}$ and
$[a]\big{(}\Theta(E)\big{)}=[1_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}$. Then
there exist $c,d\in L$ with $U([c]E,[d]E)=\emptyset$. Now we have

$\displaystyle[c]E\vee[a]E$
$\displaystyle=[c\vee^{*}a]\big{(}\Theta(E)\big{)}\cap
L=[c\vee^{*}1_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}\cap
L=[1_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}\cap L=$
$\displaystyle=[a]\big{(}\Theta(E)\big{)}\cap L=[a]E$

and, analogously, $[d]E\vee[a]E=[a]E$. This shows $[a]E\in U([c]E,[d]E)$, a
contradiction. Hence
$[a]\big{(}\Theta(E)\big{)}\neq[1_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}$.
Analogously, one obtains that in case $0_{(\mathbf{L}/E)^{*}}\in(L/E)^{*}$ we
have
$[a]\big{(}\Theta(E)\big{)}\neq[0_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}$.
Now assume $0_{(\mathbf{L}/E)^{*}},1_{(\mathbf{L}/E)^{*}}\in(L/E)^{*}$ and
$[0_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}=[1_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}$.
Then $0_{\mathbf{L}^{*}},1_{\mathbf{L}^{*}}\in L^{*}$ and
$(0_{\mathbf{L}^{*}},1_{\mathbf{L}^{*}})\in\Theta(E)$ and hence
$\Theta(E)=(L^{*})^{2}$ whence

$E=\Theta(E)\cap L^{2}=(L^{*})^{2}\cap L^{2}=L^{2}$

which implies $|L/E|=1$ and hence
$0_{(\mathbf{L}/E)^{*}},1_{(\mathbf{L}/E)^{*}}\notin(L/E)^{*}$, a
contradiction.
$\bullet$ $f$ is surjective.
Assume $1_{\mathbf{L}^{*}}\in L^{*}$. If $1_{(\mathbf{L}/E)^{*}}\in(L/E)^{*}$
then $f(1_{(\mathbf{L}/E)^{*}})=[1_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}$.
Now assume $1_{(\mathbf{L}/E)^{*}}\notin(L/E)^{*}$. Then $[x]E\vee[y]E$ is
defined in $L/E$ for every $x,y\in L$. Suppose
$[1_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}=\\{1_{\mathbf{L}^{*}}\\}$. Then
$U(x,y)\neq\emptyset$ for all $x,y\in L$ and hence $1_{\mathbf{L}^{*}}\notin
L^{*}$, a contradiction. Therefore
$[1_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}\neq\\{1_{\mathbf{L}^{*}}\\}$.
According to the proof of Lemma 4.3 there exists some
$\alpha\in[1_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}\cap L$ and we obtain
$f([\alpha]E)=[\alpha]\big{(}\Theta(E)\big{)}=[1_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}$.
Analogously, one can show that in case $0_{\mathbf{L}^{*}}\in L^{*}$ there
exists some $x\in(L/E)^{*}$ with
$f(x)=[0_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}$.
$\bullet$ $f$ is a homomorphism from $(\mathbf{L}/E)^{*}$ to
$\mathbf{L}^{*}/\big{(}\Theta(E)\big{)}$.
First assume $U(a,b)\neq\emptyset$. Then

$\displaystyle f([a]E\vee[b]E)$
$\displaystyle=f\Big{(}[a\vee^{*}b]\big{(}\Theta(E)\big{)}\cap
L\Big{)}=f\Big{(}[a\vee b]\big{(}\Theta(E)\big{)}\cap L\Big{)}=f([a\vee b]E)=$
$\displaystyle=[a\vee
b]\big{(}\Theta(E)\big{)}=[a]\big{(}\Theta(E)\big{)}\vee[b]\big{(}\Theta(E)\big{)}=f([a]E)\vee
f([b]E).$

Now assume $U(a,b)=\emptyset$. Then $1_{\mathbf{L}^{*}}\in L^{*}$. First
assume $[1_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}=\\{1_{\mathbf{L}^{*}}\\}$.
Then $1_{(\mathbf{L}/E)^{*}}\in(L/E)^{*}$ and

$\displaystyle f([a]E\vee[b]E)$
$\displaystyle=f(1_{(\mathbf{L}/E)^{*}})=[1_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}=[a\vee
b]\big{(}\Theta(E)\big{)}=[a]\big{(}\Theta(E)\big{)}\vee[b]\big{(}\Theta(E)\big{)}=$
$\displaystyle=f([a]E)\vee f([b]E).$

Finally, assume
$[1_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}\neq\\{1_{\mathbf{L}^{*}}\\}$.
According to the proof of Lemma 4.3 there exists some
$\alpha\in[1_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}\cap L$ and we have

$\displaystyle f([a]E\vee[b]E)$
$\displaystyle=f\Big{(}[a\vee^{*}b]\big{(}\Theta(E)\big{)}\cap
L\Big{)}=f\Big{(}[1_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}\cap
L\Big{)}=f\Big{(}[\alpha]\big{(}\Theta(E)\big{)}\cap L\Big{)}=$
$\displaystyle=f([\alpha]E)=[\alpha]\big{(}\Theta(E)\big{)}=[1_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}=[a\vee
b]\big{(}\Theta(E)\big{)}=$
$\displaystyle=[a]\big{(}\Theta(E)\big{)}\vee[b]\big{(}\Theta(E)\big{)}=f([a]E)\vee
f([b]E).$

It is easy to see that $f(x\vee y)=f(x)\vee f(y)$ for all
$(x,y)\in\big{(}(L/E)^{*}\big{)}^{2}\setminus(L/E)^{2}$. Analogously, one can
prove $f(x\wedge y)=f(x)\wedge f(y)$ for all $x,y\in(L/E)^{*}$. ∎

##  5 Examples

In the following we present several examples showing different situations
concerning two-point extensions and quotient partial lattices.

######  Example 5.1.

Let $\mathbf{L}$ denote the partial lattice visualized in Figure 4:

$a$$c$$b$Fig. 4

The lattice $\mathbf{L}^{*}$ is depicted in Figure 5:

$0_{\mathbf{L}^{*}}$$a$$c$$b$$1_{\mathbf{L}^{*}}$Fig. 5

Here $0_{\mathbf{L}^{*}},1_{\mathbf{L}^{*}}\in L^{*}$. Put
$E:=\\{a,c\\}^{2}\cup\\{b\\}^{2}$ Then
$\Theta(E)=\\{0_{\mathbf{L}^{*}}\\}^{2}\cup\\{a,c\\}^{2}\cup\\{b\\}^{2}\cup\\{1_{\mathbf{L}^{*}}\\}^{2}$
and hence $\Theta(E)$ is a congruence on $\mathbf{L}^{*}$ satisfying
$\Theta(E)\cap L^{2}=E$ and $\mathbf{L}/E$ is visualized in Figure 6:

$[a]E$$[b]E$Fig. 6

The lattice $(\mathbf{L}/E)^{*}$ is depicted in Figure 7:

$0_{(\mathbf{L}/E)^{*}}$$[a]E$$[b]E$$1_{(\mathbf{L}/E)^{*}}$Fig. 7

and we have $0_{(\mathbf{L}/E)^{*}},1_{(\mathbf{L}/E)^{*}}\in(L/E)^{*}$.
Finally, $\mathbf{L}^{*}/\big{(}\Theta(E)\big{)}$ is visualized in Figure 8:

$[0_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}$$[a]\big{(}\Theta(E)\big{)}$$[b]\big{(}\Theta(E)\big{)}$$[1_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}$Fig.
8

In accordance with Theorem 4.8,
$(\mathbf{L}/E)^{*}\cong\mathbf{L}^{*}/\big{(}\Theta(E)\big{)}$.

######  Example 5.2.

Let $\mathbf{L}$ be the partial lattice depicted in Figure 9:

$a$$b$$c$$d$Fig. 9

The lattice $\mathbf{L}^{*}$ is visualized in Figure 10:

$0_{\mathbf{L}^{*}}$$a$$b$$c$$d$$1_{\mathbf{L}^{*}}$Fig. 10

Here $0_{\mathbf{L}^{*}},1_{\mathbf{L}^{*}}\in L^{*}$. Put
$E:=\\{a\\}^{2}\cup\\{b,d\\}^{2}\cup\\{c\\}^{2}$. Let us mention that
$(b,d)\in E$ and $a\vee b=c$, but $a\vee d$ does not exist in $\mathbf{L}$. We
have
$\Theta(E)=\\{0_{\mathbf{L}^{*}}\\}^{2}\cup\\{a\\}^{2}\cup\\{b,d\\}^{2}\cup\\{c,1_{\mathbf{L}^{*}}\\}^{2}$
and hence $\Theta(E)\cap L^{2}=E$ and $\mathbf{L}/E$ is depicted in Figure 11:

$[a]E$$[d]E$$[c]E$Fig. 11

One can see that although $[a]E\vee[d]E$ exists in $\mathbf{L}/E$, $a\vee d$
does not exist in $\mathbf{L}$. The lattice $(\mathbf{L}/E)^{*}$ is visualized
in Figure 12:

$0_{(\mathbf{L}/E)^{*}}$$[a]E$$[d]E$$[c]E$Fig. 12

and we have $0_{(\mathbf{L}/E)^{*}}\in(L/E)^{*}$ and
$1_{(\mathbf{L}/E)^{*}}\notin(L/E)^{*}$. Finally,
$\mathbf{L}^{*}/\big{(}\Theta(E)\big{)}$ is depicted in Figure 13:

$[0_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}$$[a]\big{(}\Theta(E)\big{)}$$[b]\big{(}\Theta(E)\big{)}$$[c]\big{(}\Theta(E)\big{)}$Fig.
13

######  Example 5.3.

Let $\mathbf{L}$ denote the partial lattice from Example 5.2. Then
$0_{\mathbf{L}^{*}},1_{\mathbf{L}^{*}}\in L^{*}$. Put
$E:=\\{a,c\\}^{2}\cup\\{b\\}^{2}\cup\\{d\\}^{2}$. Here $(a,c)\in E$ and
$c\wedge d=b$, but $a\wedge d$ does not exist in $\mathbf{L}$. We have
$\Theta(E)=\\{0_{\mathbf{L}^{*}},b\\}^{2}\cup\\{a,c\\}^{2}\cup\\{d\\}^{2}\cup\\{1_{\mathbf{L}^{*}}\\}^{2}$
and hence $\Theta(E)\cap L^{2}=E$ and $\mathbf{L}/E$ is visualized in Figure
14:

$[b]E$$[a]E$$[d]E$Fig. 14

One can see that although $[a]E\wedge[b]E$ exists in $\mathbf{L}/E$, $a\wedge
b$ does not exists in $\mathbf{L}$. The lattice $(\mathbf{L}/E)^{*}$ is
depicted in Figure 15:

$[b]E$$[a]E$$[d]E$$[1_{(\mathbf{L}/E)^{*}}]E$Fig. 15

and we have $0_{(\mathbf{L}/E)^{*}}\notin(L/E)^{*}$ and
$1_{(\mathbf{L}/E)^{*}}\in(L/E)^{*}$. Finally,
$\mathbf{L}^{*}/\big{(}\Theta(E)\big{)}$ is visualized in Figure 16:

$[0_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}$$[a]\big{(}\Theta(E)\big{)}$$[d]\big{(}\Theta(E)\big{)}E$$[1_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}$Fig.
16

The next example shows that $\mathbf{L}/E$ may be a lattice even if
$\mathbf{L}$ is only a partial lattice.

######  Example 5.4.

Let $\mathbf{L}$ denote the partial lattice from Example 5.2. Then
$0_{\mathbf{L}^{*}},1_{\mathbf{L}^{*}}\in L^{*}$. Put
$E:=\\{a\\}^{2}\cup\\{b,c\\}^{2}\cup\\{d\\}^{2}$. Then
$\Theta(E)=\\{0_{\mathbf{L}^{*}},a\\}^{2}\cup\\{b,c\\}^{2}\cup\\{d,1_{\mathbf{L}^{*}}\\}^{2}$
and hence $\Theta(E)\cap L^{2}=E$ and $\mathbf{L}/E$ is depicted in Figure 17:

$[a]E$$[b]E$$[d]E$Fig. 17

Hence $(\mathbf{L}/E)^{*}=\mathbf{L}/E$ and
$0_{(\mathbf{L}/E)^{*}},1_{(\mathbf{L}/E)^{*}}\notin(L/E)^{*}$. Finally,
$\mathbf{L}^{*}/\big{(}\Theta(E)\big{)}$ is visualized in Figure 18:

$[0_{\mathbf{L}^{*}}]\big{(}\Theta(E)\big{)}$$[b]\big{(}\Theta(E)\big{)}$$[d]\big{(}\Theta(E)\big{)}$Fig.
18

##  6 Conclusion

We introduced the so-called two-point extension of a partial lattice which
extends it to a lattice with everywhere defined operations. This was not
possible by using the one-point extension which was intensively used for
partial algebras in general because the one-point extension of a partial
lattice need not be a lattice. However, this is not a final step concerning
this research. Namely lattice distributivity or modularity can be defined for
partial lattices by strong and regular identities but these are not preserved
by the two-point extension. For example, if a partial lattice
$\mathbf{L}=(L,\vee,\wedge)$ is a finite antichain containing $n\geq 3$
elements then the join, respectively meet of two distinct elements of $L$ is
not defined and hence $\mathbf{L}$ trivially satisfies the strong distributive
identity. However, its two-point extension $\mathbf{L}^{*}$ is isomorphic to
the non-modular lattice $\mathbf{M}_{n}$. Hence, this research may continue
with finding other tools which avoid this difficulty.

## References

  * [1] 9
  * [2] P. Burmeister, A Model Theoretic Oriented Approach to Partial Algebras. Math. Research 32, Akademie-Verlag, Berlin 1986.
  * [3] P. Burmeister, Partial algebras – an introductory survey. Algebras and Orders (Montreal, 1991), 1–70, NATO Adv. Sci. Inst. Ser. C: Math. Phys. Sci. 389, Kluwer, Dordrecht 1993.
  * [4] I. Chajda and Z. Seidl, An algebraic approach to partial lattices. Demonstratio Math. 30 (1997), 485–494.
  * [5] K. Denecke, Strong regular varieties of partial algebras. I. Beiträge Algebra Geom. 31 (1991), 163–177.
  * [6] B. Staruch and B. Staruch, Strong regular varieties of partial algebras. Algebra Universalis 31 (1994), 157–176.

Authors’ addresses:

Ivan Chajda
Palacký University Olomouc
Faculty of Science
Department of Algebra and Geometry
17\. listopadu 12
771 46 Olomouc
Czech Republic
<EMAIL_ADDRESS>
Helmut Länger
TU Wien
Faculty of Mathematics and Geoinformation
Institute of Discrete Mathematics and Geometry
Wiedner Hauptstraße 8-10
1040 Vienna
Austria, and
Palacký University Olomouc
Faculty of Science
Department of Algebra and Geometry
17\. listopadu 12
771 46 Olomouc
Czech Republic
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