1 Introduction

There are several ways how to embed an ordered algebra into a complete ordered algebra of the same type. One such possibility is given in [15], where it is shown that certain injective hulls of ordered algebras have properties similar to those of Dedekind–MacNeille completions. In this paper we will follow a different approach—constructing completions with the help of admissible ideals.

In [5], Bruns and Lakser introduced admissible subsets and so-called D-ideals in semilattices. They proved that the set of all D-ideals (which is a complete lattice) is the injective hull of the semilattice. In [3], Bishop studied the completion by complete ideals of a lattice and proved a universal property for it. Krishnan [7] contstructed a completion for pomonoids which is compatible with joins. Rasouli [9] used a similar approach to construct a completion for S-posets where S is a pomonoid. In a recent paper [16], completions of marked quantales are considered. Such structures (semilattices, lattices, pomonoids, S-posets, posemigroups) can be considered as special cases of ordered universal algebras of different types. It is natural to ask if the results in the mentioned papers have a common generalization to the ordered algebras.

In this article we will give a construction that assigns the sup-algebra of \(\mathscr {D}\)-ideals (denoted by \(\mathscr {D}(A)\)) to each ordered algebra \(\mathcal {A}\). We will prove that \(\mathscr {D}(A)\) is a join-completion for \(\mathcal {A}\) and prove a universal property of this construction. As a consequence, we will obtain a reflector functor \(\mathscr {D}\) to the category of sup-algebras. The source category of this functor has ordered algebras as objects, but the morphisms are not all homomorphisms, but those which preserve admissible joins. We note that also in [12] different sup-algebra completions are considered, one of them being \(\mathcal {D}(A)\). But the definitions of \(\mathscr {D}(A)\) and \(\mathcal {D}(A)\) differ a litt’le bit and the universal property is not considered in [12].

2 Preliminaries

We recall some definitions that will be needed in this paper.

Definition 2.1

([1, Definition 4.16]). A subcategory \(\mathcal {A}\) of a category \(\mathcal {B}\) is called a reflective subcategory if for every \(\mathcal {B}\)-object B there is a \(\mathcal {B}\)-morphism \(r: B \rightarrow A\) from B to an \(\mathcal {A}\)-object A with the following universal property: for any \(\mathcal {B}\)-morphism \(f: B \rightarrow A'\) from B to an \(\mathcal {A}\)-object \(A'\), there exists a unique \(\mathcal {A}\)-morphism \(f': A \rightarrow A'\) such that \(f'r=f\). In other words, \(\mathcal {A}\) is a reflective subcategory of \(\mathcal {B}\) if the inclusion functor \(\mathcal {A}\rightarrow \mathcal {B}\) has a left adjoint functor \(\mathcal {B}\rightarrow \mathcal {A}\) (see [8], page 91), which is usually called a reflector.

In this paper we will show how to construct a reflector from a certain category of ordered algebras to the category of sup-algebras of the same type.

Let \(\Omega \) be a type. An ordered \(\Omega \)-algebra is a triplet \(\mathcal {A}= (A,\Omega _{A},\leqslant _{A})\) comprising a poset \((A,\leqslant _{A})\) and a set \(\Omega _{A}\) of operations on A (for every k-ary operation symbol \(\omega \in \Omega _{k}\) there is a k-ary operation \(\omega _{A} \in \Omega _{A}\) on A) such that all the operations \(\omega _{A}\) are monotone mappings ([4]).

Let \(\mathcal {A}\) and \(\mathcal {B}\) be ordered \(\Omega \)-algebras. We say that a monotone mapping \(f: \mathcal {A}\rightarrow \mathcal {B}\) is a lax morphism, if

$$\begin{aligned} \omega _B( f(a_1),\dots ,f(a_n))\leqslant f (\omega _A(a_1,\dots ,a_n)) \end{aligned}$$
(2.1)

for every \(n \in \mathbb {N},\) \(\omega \in \Omega _n\), \( a_1, \dots ,a_n\in A\), and

$$\begin{aligned} \omega _B \leqslant f (\omega _A) \end{aligned}$$
(2.2)

for every \(\omega \in \Omega _0\).

If \(f: \mathcal {A}\rightarrow \mathcal {B}\) is monotone and operation-preserving, i.e., the inequalities in (2.1) and (2.2) turn out to be equalities, then f is a homomorphism of ordered algebras.

Throughout this text, a type \(\Omega \) is fixed, all algebras that we consider will be \(\Omega \)-algebras and all homomorphisms will be homomorphisms of \(\Omega \)-algebras, even if \(\Omega \) is not explicitly mentioned.

Linear functions on an ordered algebra \(\mathcal {A}\) are defined as follows (see [14]).

  1. (1)

    The identity mapping \(A\rightarrow A, x\mapsto x,\) is a linear function.

  2. (2)

    If \(n\in \mathbb {N}\), \(\omega \in \Omega _n\), \(i\in \{1,\ldots ,n\}\), \(a_1,\ldots ,a_{i-1},a_{i+1},\ldots ,a_n\in A\) and \(p:A\rightarrow A\) is a linear function, then the mapping

    $$\begin{aligned} A\rightarrow A, \;\; x\mapsto \omega (a_1,\ldots ,a_{i-1},p(x),a_{i+1},\ldots ,a_n) \end{aligned}$$

    is a linear function.

Linear functions obtained by step (1) or by step (2) with \(p=id_A\) are called elementary translations of \(\mathcal {A}\). We write \(L_\mathcal {A}\) for the set of all linear functions on an ordered algebra \(\mathcal {A}\). Linear functions are the composites of elementary translations.

Definition 2.2

A subset M of an ordered algebra \(\mathcal {A}\) is called admissible, if

  • \(\bigvee p(M)\) exists for all \(p \in L_\mathcal {A}\) (in particular \(\bigvee M = \bigvee id_A(M)\) exists),

  • for all \(p \in L_\mathcal {A}\),

    $$\begin{aligned} p\left( \bigvee M\right) = \bigvee p(M). \end{aligned}$$

Our definition generalizes the definition of an admissible subset of a semilattice which was introduced in [5]. Note that [12] also defines admissible subsets of an ordered algebra, but that definition differs from ours: instead of linear functions, unary polynomial functions are used.

We write \(S\mathord \downarrow = \{a\in P\mid a\leqslant s \text{ for } \text{ some } s\in S\}\) if S is a subset of a poset P, and we call S a lower subset of P if \(S\mathord \downarrow =S\).

Definition 2.3

A lower subset S of an ordered algebra \(\mathcal {A}\) is called a \(\mathscr {D}\)-ideal (cf. [5, p. 116] or [12, Definition3.6]) if for any admissible subset M of S, one has that \(\bigvee M \in S\).

Let us denote by \(\mathscr {D}(A)\) the set of all \(\mathscr {D}\)-ideals of A. Note that \(a\mathord \downarrow \) is a \(\mathscr {D}\)-ideal for every \(a \in A\) (see Remark 3.7 in [12]).

Definition 2.4

We say that an ordered algebra homomorphism \(f:\mathcal {A}\rightarrow \mathcal {B}\) preserves admissible joins if, for any admissible subset M of A, the subset \(f(M)\subseteq B\) is admissible and

$$\begin{aligned} f\left( \bigvee M\right) = \bigvee f(M). \end{aligned}$$

All ordered algebras with admissible joins preserving homomorphisms constitute a category, which we denote by \({\textsf{OAlg}^*}\). This is a (non-full) subcategory of \(\textsf{OAlg}\), the category of ordered algebras with all homomorphisms as morphisms.

3 Sup-algebras

An ordered \(\Omega \)-algebra \(\mathcal {Q}\!= (Q,\Omega _{Q},\leqslant _{Q})\) is called a sup-algebra if the poset \((Q,\leqslant _{Q})\) is a complete lattice and all elementary translations preserve joins. Sup-algebras were introduced by Pedro Resende in [10] as a common generalization for many quantale-like structures. All sup-algebras with join-preserving homomorphisms form a category denoted by \(\textsf{SupAlg}\).

Lemma 3.1

Let \(\mathcal {A}\) be an ordered algebra such that \((A,\leqslant )\) is a complete lattice. Then the following assertions are equivalent.

  1. (1)

    \(\mathcal {A}\) is a sup-algebra.

  2. (2)

    All linear functions on \(\mathcal {A}\) preserve joins.

  3. (3)

    All subsets of \(\mathcal {A}\) are admissible.

Proof

(1) \(\Leftrightarrow \) (2). This is true because linear functions are the composites of elementary translations.

(2) \(\Leftrightarrow \) (3). This holds because all subsets of A have joins. \(\square \)

Proposition 3.2

\(\textsf{SupAlg}\) is a full subcategory of \({\textsf{OAlg}^*}\).

Proof

If \(\mathcal {Q}\) and \(\mathcal {R}\) are sup-algebras then all subsets of Q and R are admissible by Lemma 3.1. Hence a homomorphism \(f: \mathcal {Q}\rightarrow \mathcal {R}\) preserves admissible joins if and only if it preserves all joins. \(\square \)

If \(\mathcal {Q}\) is a sup-algebra, then every elementary translation preserves joins, hence it has a right adjoint.

A closure operator j on a sup-algebra \(\mathcal {Q}\) is a nucleus if it is a lax endomorphism of \(\mathcal {Q}\). The subset \(Q_j = \{q\in Q\mid j(q)=q\}\) can be made into a sup-algebra called a quantic quotient of \(\mathcal {Q}\) (see [10, Theorem 2.2.7] or [13, Proposition 16]).

Lemma 3.3

[13, Proposition 15] If \(\mathcal {Q}= (Q,\Omega _{Q},\leqslant _{Q})\) is a sup-algebra and \(S \subseteq Q\), then \(S = Q_j\) for some nucleus j on \(\mathcal {Q}\) if and only if S is closed under meets and under right adjoints of elementary translations. In this case, the nucleus j is defined by

$$\begin{aligned} j(q):= \bigwedge \{ s \in S \mid q \leqslant s \} \end{aligned}$$
(3.1)

for \(q\in Q\).

For an ordered algebra \(\mathcal {A}= (A,\Omega _{A},\leqslant _{A})\), let \(\mathscr {P}(A)\) be the set of all lower subsets of A. Then \(\mathscr {P}(A)\) is a sup-algebra equipped with the inclusion as ordering and

$$\begin{aligned} \omega _{\mathscr {P}(A)} (D_1, \dots , D_{n}) = \{\omega _{A} (d_1, \dots , d_{n}) \mid d_i \in D_i, i=1,\dots ,n \} \mathord \downarrow \end{aligned}$$

as n-ary operations for every \(\omega \in \Omega _{n}\), \(n \in \mathbb {N}\). If \(\omega \in \Omega _0\), then \(\omega _{\mathscr {P}(A)} = \omega _A\mathord \downarrow \).

A nucleus j on \(\mathscr {P}(A)\) is called principal closed, if \(j(a\mathord \downarrow ) = a\mathord \downarrow \) for all \(a\in A\).

Proposition 3.4

Let \(\mathcal {A}=(A,\Omega _{A},\leqslant _{A})\) be an ordered algebra. Then \(\mathscr {D}(A)\) is a quantic quotient of the sup-algebra \(\mathscr {P}(A)\).

Although our point of view is a little bit different from that of [12] (namely \(\mathscr {P}(A)\) denotes the set of all subsets of A in [12]), Proposition 3.4 can still be proved precisely as Theorem 3.9 in [12].

Now, given an ordered algebra \(\mathcal {A}\), Lemma 3.3 provides a nucleus \(j: \mathscr {P}(A)\rightarrow \mathscr {P}(A)\), defined by (3.1), that is,

$$\begin{aligned} j(C) = \bigcap \{ S \in \mathscr {D}(A) \mid C \subseteq S \} \end{aligned}$$
(3.2)

for every \(C \in \mathscr {P}(A)\), such that \(\mathscr {D}(A) = \mathscr {P}(A)_j = \{C\in \mathscr {P}(A)\mid j(C)=C\}\). It can be shown that j is a principal closed nucleus. In the sup-algebra \(\mathscr {D}(A)\) joins and meets are calculated as

$$\begin{aligned} \bigvee _{i \in I} S_i{} & {} = j\Bigl ( \bigcup _{i \in I}S_i\Bigr ) = \bigcap \Biggl \{ S \in \mathscr {D}(A) \Bigm | \bigcup _{i \in I} S_i \subseteq S \Biggr \}, \\{} & {} \quad \bigwedge _{i \in I} S_i = \bigcap _{i \in I} S_i, \end{aligned}$$

and operations are

$$\begin{aligned} \begin{aligned} \omega _{\mathscr {D}(A)} (S_1, \dots , S_{n})&= j(\omega _{\mathscr {P}(A)} (S_1, \dots , S_{n})) \\&= \bigcap \{ S \in \mathscr {D}(A) \mid \omega _{\mathscr {P}(A)} (S_1, \dots , S_{n}) \subseteq S \}, \end{aligned} \end{aligned}$$
(3.3)

for every \(n \in \mathbb {N}\), \(\omega \in \Omega _{n}\). If \(\omega \in \Omega _0\), then \(\omega _{\mathscr {D}(A)} = j(\omega _{\mathscr {P}(A)}) = j(\omega _A\mathord \downarrow )\). Nullary operations of \(\mathscr {D}(A)\) can be described as follows.

Lemma 3.5

If \(\mathcal {A}\) is an ordered-algebra and \(\omega \in \Omega _0\), then \(\omega _{\mathscr {D}(A)} = \omega _A\mathord \downarrow \).

Proof

Clearly,

$$\begin{aligned} \omega _A\in \bigcap \{S\in \mathscr {D}(A)\mid \omega _A\in S\} = \bigcap \{S\in \mathscr {D}(A)\mid \omega _A\mathord \downarrow \subseteq S\} = \omega _{\mathscr {D}(A)} \end{aligned}$$

and hence \(\omega _A\mathord \downarrow \subseteq \omega _{\mathscr {D}(A)}\). Take arbitrary \(a\in \omega _{\mathscr {D}(A)}\). Then \(a\in S\) for all \(S\in \mathscr {D}(A)\) containing \(\omega _A\). In particular, \(a\in \omega _A\mathord \downarrow \), and thus \(\omega _{\mathscr {D}(A)} \subseteq \omega _A\mathord \downarrow \). \(\square \)

Proposition 3.6

[15]. Let \(\mathcal {A}\) be an ordered algebra and let j be a principal closed nucleus on \(\mathscr {P}(\mathcal {A})\). Then the mapping \(\eta :\mathcal {A}\rightarrow \mathscr {P}(\mathcal {A})_j, a\mapsto a\mathord \downarrow \), has the following properties:

  1. (1)

    \(\eta \) is a homomorphism of ordered algebras which is an order-embedding,

  2. (2)

    \(\eta (A)\) is join-dense in the lattice \(\mathscr {P}(\mathcal {A})_j\),

  3. (3)

    \(\eta \) preserves all existing meets in \(\mathcal {A}\).

4 Completions by \(\mathscr {D}\)-ideals

Generalizing the notion of join-completion of a poset (cf. [2, 11], or [6]) we say that a join-completion of an ordered algebra \(\mathcal {A}\) is a pair \((\eta , \mathcal {R}(A))\), where

  1. (1)

    \(\mathcal {R}(A)\) is a sup-algebra,

  2. (2)

    \(\eta : \mathcal {A}\rightarrow \mathcal {R}(A)\) is a homomorphism of ordered algebras which is an order-embedding,

  3. (3)

    the set \(\eta (A)\) is join-dense in \(\mathcal {R}(A)\).

Meet-completions are defined dually.

It turns out that \(\mathscr {D}(A)\), the sup-algebra of \(\mathscr {D}\)-ideals, is a join-completion of \(\mathcal {A}\).

Proposition 4.1

If \(\mathcal {A}\) is an ordered algebra, then the mapping \(r: A \rightarrow \mathscr {D}(A)\), \(a\mapsto a \mathord \downarrow \), is a homomorphism of ordered algebras that is an order-embedding and preserves admissible joins. Moreover, r(A) is join-dense in \(\mathscr {D}(A)\) and r preserves all meets that exist in A.

Proof

Let M be an admissible subset of A. Since \(\mathscr {D}(A)\) is a sup-algebra, \(p(\bigvee r(M)) = \bigvee p(r(M))\) for every \(p\in L_{\mathscr {D}(A)}\) by Lemma 3.1. Thus r(M) is an admissible subset of \(\mathscr {D}(A)\).

It remains to prove that \(r\left( \bigvee M \right) = \bigvee r(M)\) (all other claims follow from Proposition 3.4 and Proposition 3.6). Since \(\mathscr {D}(A)\) is a sup-algebra, the join of r(M) exists and, in fact,

$$\begin{aligned} \bigvee r(M) = \bigvee \limits _{m\in M} r(m)= \bigcap \left\{ S \in \mathscr {D}(A) \ \Bigm | \ \bigcup \limits _{m\in M} m \mathord \downarrow \subseteq S \right\} . \end{aligned}$$

If S is a \(\mathscr {D}\)-ideal with \(\bigcup \nolimits _{m\in M} m \mathord \downarrow \subseteq S\), then \(M \subseteq S\) and hence \(\bigvee M\in S\) by the definition of a \(\mathscr {D}\)-ideal. This yields \(\left( \bigvee M\right) \mathord \downarrow \subseteq S\), and thus \(r\left( \bigvee M \right) \subseteq \bigvee r(M)\). The inclusion \( \bigvee r(M) \subseteq r\left( \bigvee M \right) \) is clear. \(\square \)

Example 4.2

In general, r need not preserve existing joins. Let \(A = \{a,b,c\}\) be a commutative posemigroup with the multiplication and order given in Figure 1.

Fig. 1
figure 1

The multiplication and order of A

Full size image

Its quantale of \(\mathscr {D}\)-ideals is \(\mathscr {D}(A)=\{a\mathord \downarrow , b\mathord \downarrow ,c\mathord \downarrow ,\{b,c\},\emptyset \}\). We see that \(r(\bigvee \{b,c\}) = r(a) = \{a,b,c\}\), but

$$\begin{aligned} \bigvee r(\{b,c\}) = \{b\} \vee \{c\} = \bigcap \{S\in \mathscr {D}(A)\mid \{b,c\} \subseteq S\} = \{b,c\}. \end{aligned}$$

Our main result is the following. It generalizes, for example, [3, Theorem 3] about lattices and [9, Theorem 3.2] about S-posets.

Theorem 4.3

Let \(\mathcal {A}\) be an ordered algebra. Then, for every sup-algebra \(\mathcal {Q}\) and every \({\textsf{OAlg}^*}\)-morphism \(f: \mathcal {A}\rightarrow \mathcal {Q}\), there exists a unique \(\textsf{SupAlg}\)-morphism \(g: \mathscr {D}(A) \rightarrow \mathcal {Q}\) such that the diagram

commutes.

Proof

Proposition 4.1 has established that r is an \({\textsf{OAlg}^*}\)-morphism. Given a sup-algebra \(\mathcal {Q}\) and a morphism \(f: \mathcal {A}\rightarrow \mathcal {Q}\) in the category \({\textsf{OAlg}^*}\), define \(g: \mathscr {D}(A) \rightarrow Q\) by

$$\begin{aligned} g(D)= \bigvee \limits _{d\in D} f(d) \end{aligned}$$

for every \(\mathscr {D}\)-ideal D of A. We need to prove that g is a join-preserving sup-algebra homomorphism. For every \(\omega \in \Omega _0\), using Lemma 3.5, we have

$$\begin{aligned} g\left( \omega _{\mathscr {D}(A)}\right) = \bigvee \limits _{d\in {\omega _{A} \mathord \downarrow }} f(d) = f(\omega _{A}) = \omega _{Q}. \end{aligned}$$

We also need to prove the equality

$$\begin{aligned} \omega _{Q} (g(D_1), \dots , g(D_{n}))=g(\omega _{\mathscr {D}(A)}(D_1, \dots , D_{n})) \end{aligned}$$

for any \(n \in \mathbb {N}, \omega \in \Omega _{n}\) and \(\mathscr {D}\)-ideals \(D_1, \dots , D_{n}\) of A. By the definition of g, we have

$$\begin{aligned} g(\omega _{\mathscr {D}(A)}(D_1, \dots , D_{n})) =\bigvee \limits _{d\in \omega _{\mathscr {D}(A)}(D_1, \dots , D_{n})} f(d), \end{aligned}$$

where, according to (3.3),

$$\begin{aligned} \omega _{\mathscr {D}(A)}(D_1, \dots , D_{n}) =\bigcap \{ S \in \mathscr {D}(A) \mid \omega _{\mathscr {P}(A)}(D_1, \dots , D_{n}) \subseteq S \}. \end{aligned}$$
(4.1)

We compute

$$\begin{aligned} \omega _{Q}(g(D_1), \dots , g(D_{n})) = \omega _{Q}\Biggl (\bigvee \limits _{d_1\in D_1} f(d_1), \dots , \bigvee \limits _{d_{n}\in D_{n}} f(d_{n})\Biggr ) \end{aligned}$$
figure a
figure b
figure c

By (4.1) we have \(\omega _{\mathscr {P}(A)}(D_1, \dots , D_{n}) \subseteq \omega _{\mathscr {D}(A)}(D_1, \dots , D_{n})\), so in \(\mathcal {Q}\) we obtain the inequality \( \omega _{Q}(g(D_1), \dots , g(D_{n})) \leqslant g(\omega _{\mathscr {D}(A)}(D_1, \dots , D_{n})). \)

To prove the opposite inequality, we first show that, for every \(q_0\in Q\), the subset

$$\begin{aligned} \mathscr {K}= \left\{ a \in A \mid f(a)\leqslant q_0\right\} \end{aligned}$$

of A is a \(\mathscr {D}\)-ideal. Since f is monotone, \(\mathscr {K}\) is a lower subset of \(\mathcal {A}\). Assume that \(M \subseteq \mathscr {K}\) is an admissible subset. By the definition of \(\mathscr {K}\), \(f(m) \leqslant q_0\) for every \(m \in M\). As f preserves admissible joins, we obtain

$$\begin{aligned} f\Bigl (\bigvee M\Bigr ) = \bigvee \limits _{m\in M} f(m) \leqslant q_0, \end{aligned}$$

which means that \(\bigvee M \in \mathscr {K}\). Thus \(\mathscr {K}\) is a \(\mathscr {D}\)-ideal.

Now consider \(\mathscr {K}\) corresponding to the element

$$\begin{aligned} q_0:= \bigvee \limits _{d\in \omega _{\mathscr {P}(A)}(D_1, \dots , D_{n})} f(d) = \omega _Q(g(D_1),\ldots ,g(D_n)). \end{aligned}$$

Then we have \(\omega _{\mathscr {P}(A)}(D_1, \dots , D_{n}) \subseteq \mathscr {K}\). Since \(\mathscr {K}\) is a \(\mathscr {D}\)-ideal of \(\mathcal {A}\), it belongs to the set \(\{S\in \mathscr {D}(A)\mid \omega _{\mathscr {P}(A)}(S_1,\ldots ,S_n) \subseteq S\}\) and hence (4.1) implies that \(\omega _{\mathscr {D}(A)}(D_1, \dots , D_{n})\subseteq \mathscr {K}\). We conclude that

$$\begin{aligned} g(\omega _{\mathscr {D}(A)}(D_1, \dots , D_{n}))&=\bigvee \limits _{d\in \omega _{\mathscr {D}(A)}(D_1, \dots , D_{n})} f(d)\leqslant \bigvee \limits _{d\in \mathscr {K}} f(d) \\&\leqslant q_0 = \omega _{Q}(g(D_1), \dots , g(D_{n})). \end{aligned}$$

The equality \(g(\omega _{\mathscr {D}(A)}(D_1, \dots , D_{n})) = \omega _{Q}(g(D_1), \dots , g(D_{n}))\) follows.

Next we verify that g preserves joins. Assume that \(\{ D_i \mid i \in I\}\) is a set of \(\mathscr {D}\)-ideals of \(\mathcal {A}\). Write \({\widetilde{D}}=\bigvee \limits _{i \in I} D_i\) for the join in \(\mathscr {D}(A)\). Then

$$\begin{aligned} {\widetilde{D}} =\bigcap \Bigl \{ S \in \mathscr {D}(A)\ \Bigm | \ D \subseteq S \Bigr \}, \end{aligned}$$

where \(D=\bigcup \nolimits _{i \in I} D_i\). The inequality \(\bigvee \nolimits _{i \in I} g(D_i) \leqslant g\Bigl (\bigvee \nolimits _{i \in I} D_i\Bigr )\) is clear. We put \(q_0:= \bigvee _{d \in D} f(d)\) and consider the set

$$\begin{aligned} {\widetilde{\mathscr {K}}}= \Bigl \{ a \in A \ \Bigm | \ f(a) \leqslant q_0 \Bigr \}. \end{aligned}$$

Then \({\widetilde{\mathscr {K}}}\) is also a \(\mathscr {D}\)-ideal of \(\mathcal {A}\) by the argument that we used for \(\mathscr {K}\) above. Since \({\widetilde{\mathscr {K}}}\) is a \(\mathscr {D}\)-ideal and \(D\subseteq {\widetilde{\mathscr {K}}}\), we have \({\widetilde{D}}\subseteq {\widetilde{\mathscr {K}}}\). So \(\bigvee \nolimits _{d\in {\widetilde{D}}} f(d)\leqslant \bigvee \nolimits _{d\in {\widetilde{\mathscr {K}}}} f(d)\). For every \(a\in {\widetilde{\mathscr {K}}}\), the inequality \(f(a)\leqslant q_0\) holds, therefore \(\bigvee \nolimits _{d\in {\widetilde{D}}} f(d)\leqslant q_0\). Hence we have

$$\begin{aligned} g\Bigl (\bigvee \limits _{i \in I} D_i\Bigr ) =\,&g\left( {\widetilde{D}}\right) = \bigvee \limits _{d\in {\widetilde{D}}} f(d) \\ \leqslant \,&q_0 =\bigvee _{i\in I} \bigvee _{d\in D_i} f(d)\\ =\,&\bigvee \limits _{i \in I} g(D_i) \leqslant g\Bigl (\bigvee \limits _{i \in I} D_i\Bigr ) \end{aligned}$$

yielding \(\bigvee \nolimits _{i \in I} g(D_i) = g\Bigl (\bigvee \nolimits _{i \in I} D_i\Bigr )\).

It is straightforward to check that \(gr=f\). It remains to show that g is unique with that property. Suppose that \(h: \mathscr {D}(A) \rightarrow \mathcal {Q}\) is a \(\textsf{SupAlg}\)-morphism such that \(hr = f\). Then \(h(a \mathord \downarrow ) =f(a)\) for every \(a \in A\). For any \(D\in \mathscr {D}(A)\),

$$\begin{aligned} \begin{aligned} \bigvee \limits _{d \in D} (d \mathord \downarrow )&= \bigcap \Biggl \{S\in \mathscr {D}(A) \ \Bigm | \ \bigcup _{d\in D}d\mathord \downarrow \subseteq S\Biggr \} \\&= \bigcap \{S\in \mathscr {D}(A)\mid D\subseteq S\} = D, \end{aligned} \end{aligned}$$
(4.2)

so

$$\begin{aligned} g(D) = \bigvee \limits _{d \in D} f(d) = \bigvee \limits _{d \in D} h(d \mathord \downarrow ) = h \Biggl (\bigvee \limits _{d \in D} (d \mathord \downarrow )\Biggr ) = h(D). \end{aligned}$$

This completes the proof. \(\square \)

Example 4.4

In general, g is not an order-embedding when f is an order-embedding in Theorem 4.3.

Let \(S = \{a,b,c\}\) be the posemigroup considered in Example 4.2. Then \(Q=S^0\) with externally adjoined zero element 0 being the bottom element is also a posemigroup. If \(f:S\longrightarrow Q\) is the inclusion mapping then

$$\begin{aligned} g(\{b,c\})=f(b)\vee f(c)=f(a)=g(a\mathord \downarrow ) = g(\{a,b,c\}) \end{aligned}$$

shows that g is not an order-embedding.

Proposition 4.5

\(\textsf{SupAlg}\) is a reflective subcategory of \({\textsf{OAlg}^*}\) with the reflector functor \(\mathscr {D}:{\textsf{OAlg}^*}\rightarrow \textsf{SupAlg}\) defined by the assignment

where \(\mathscr {D}(f)(D) = j(f(D)\mathord \downarrow )\) for every \(D \in \mathscr {D}(A)\) and \(j:\mathscr {P}(B)\rightarrow \mathscr {P}(B)\) is defined as in (3.2).

Proof

By Theorem 4.3, we know that \(\textsf{SupAlg}\) is a reflective subcategory of \({\textsf{OAlg}^*}\). Hence the inclusion functor \(\textsf{SupAlg}\rightarrow {\textsf{OAlg}^*}\) has a left adjoint functor \(\mathscr {D}: {\textsf{OAlg}^*}\rightarrow \textsf{SupAlg}\) which, by [1, Proposition 4.22], can be described explicitly as follows. It maps an object \(\mathcal {A}\) of the category \({\textsf{OAlg}^*}\) to an object \(\mathscr {D}(A)\) of \(\textsf{SupAlg}\) and a morphism \(f: \mathcal {A}\rightarrow \mathcal {B}\) in \({\textsf{OAlg}^*}\) to the unique morphism \(\mathscr {D}(f): \mathscr {D}(A) \rightarrow \mathscr {D}(B)\) in \(\textsf{SupAlg}\) such that the square

commutes. Using (4.2) we conclude that

$$\begin{aligned} \mathscr {D}(f)(D)&= \mathscr {D}(f) \Biggl ( \bigvee _{d\in D} r_A(d) \Biggr ) = \bigvee _{d\in D} \mathscr {D}(f) \left( r_A(d) \right) = \bigvee _{d\in D} r_B \left( f(d) \right) \\&= \bigvee _{d \in D} f(d)\mathord \downarrow = j\Biggl (\bigcup _{d \in D} f(d)\mathord \downarrow \Biggr ) = j(f(D)\mathord \downarrow ), \end{aligned}$$

for each \(D\in \mathscr {D}(A)\). \(\square \)