Monotone and cone preserving mappings on posets
aa r X i v : . [ m a t h . C O ] F e b Monotone and cone preserving mappings on posets
Ivan Chajda and Helmut L¨anger
Abstract
We define several sorts of mappings on a poset like monotone, strictly mono-tone, upper cone preserving and variants of these. Our aim is to characterize posetsin which some of these mappings coincide. We define special mappings determinedby two elements and investigate when these are strictly monotone or upper conepreserving. If the considered poset is a semilattice then its monotone mappingscoincide with semilattice homomorphisms if and only if the poset is a chain. Simi-larly, we study posets which need not be semilattices but whose upper cones havea minimal element. We extend this investigation to posets that are direct productsof chains or an ordinal sum of an antichain and a finite chain. We characterizeequivalence relations induced by strongly monotone mappings and show that thequotient set of a poset by such an equivalence relation is a poset again.
AMS Subject Classification:
Keywords:
Poset, directed poset, semilattice, chain, monotone, strictly monotone, up-per cone preserving, strictly upper cone preserving, strongly upper cone preserving, or-dinal sum, induced equivalence relation
Partially ordered sets, shortly posets, are relational structures which occur frequentlyboth in various areas of mathematics and in applications. Posets were studied fromnumerous points of view depending on their application. One possible approach is toconsider various mappings on a given poset and check when they coincide. Examplesof such mappings are monotone mappings, cone preserving mappings, filter preservingmappings, etc. If the poset in question is of a particular form, e.g. if it is a semilatticeor lattice, we can consider also homomorphisms. If a poset is directed then it can beconverted into a so-called directoid, i.e. a groupoid with one binary operation. Homo-morphisms of such directed posets were already investigated by the first author in [2].For a bit more general relational structures, so-called quasiordered sets, cone preserv-ing mappings were studied in [4]. Homomorphisms of semilattices were investigated byL. R. Berrone in [1]. Support of the research by the Austrian Science Fund (FWF), project I 4579-N, and the CzechScience Foundation (GA ˇCR), project 20-09869L, entitled “The many facets of orthomodularity”, as wellas by ¨OAD, project CZ 02/2019, entitled “Function algebras and ordered structures related to logic anddata fusion”, and, concerning the first author, by IGA, project PˇrF 2020 014, is gratefully acknowledged.
Concerning the concepts used here, numerous of them are familiarly known and theremaining ones are introduced or recalled below.Let P := ( P, ≤ ) be a poset, A, B ⊆ P and a, b ∈ P . Then A ≤ B should mean x ≤ y forall ( x, y ) ∈ A × B . Instead of A ≤ { b } , { a } ≤ B and { a } ≤ { b } we simply write A ≤ b , a ≤ B and a ≤ b , respectively. The sets L ( A ) := { x ∈ P | x ≤ A } ,U ( A ) := { x ∈ P | A ≤ x } are called the lower and upper cone of A , respectively. Instead of L ( A ∪ B ), L ( A ∪ { b } ), L ( { a } ∪ B ), L ( { a, b } ) and L ( { a } ) we simply write L ( A, B ), L ( A, b ), L ( a, B ), L ( a, b ) and L ( a ), respectively. In a similar way we proceed for U and in analogous cases. Moreover,put L ∗ ( a ) := ( L ( a )) \ { a } and U ∗ ( a ) := ( U ( a )) \ { a } . P is called up-directed if U ( x, y ) = ∅ for all x, y ∈ P . The subset A of P is called a filter of P if x ∈ A and x ≤ y imply y ∈ A .Let Fil P denote the set of all filters of P . For each a ∈ P , the set [ a ) := { x ∈ P | a ≤ x } is a filter of P , the so-called principal filter generated by a . Remark 2.1. If P is a poset then (Fil P , ⊆ ) is a complete lattice with smallest element ∅ and greatest element P and _ i ∈ I F i = [ i ∈ I F i , ^ i ∈ I F i = \ i ∈ I F i for every family F i , i ∈ I, of filters of P . A mapping f : P → P is called(i) monotone if x ≤ y implies f ( x ) ≤ f ( y ),(ii) strictly monotone if x < y implies f ( x ) < f ( y ),(iii) upper cone preserving if f ( U ( x, y )) = U ( f ( x ) , f ( y )) for all x, y ∈ P ,(iv) strictly upper cone preserving if f ( U ( x, y )) = U ( f ( x ) , f ( y )) for all x, y ∈ P with x = y , 2v) strongly upper cone preserving if f ( U ( x, y )) = U ( f ( x ) , f ( y )) for all x, y ∈ P with f ( x ) = f ( y ).Observe that for monotone f we have f ( L ( x, y )) ⊆ L ( f ( x ) , f ( y )) and f ( U ( x, y )) ⊆ U ( f ( x ) , f ( y )) for all x, y ∈ P .Throughout the paper, we consider only non-void posets.In the following, for every poset ( P, ≤ ) and every element a ∈ P let f a denote the constantmapping from P to P with value a .Using of the mapping f a which is evidently monotone, we can characterize up-directedposets having a maximal element as follows. Lemma 2.2.
Let P = ( P, ≤ ) be a poset and a ∈ P . Then P is up-directed and a is amaximal element of P if and only if f a is upper cone preserving.Proof. Let b, c ∈ P . If P is up-directed and a maximal then f a ( U ( b, c )) = { a } = U ( a ) = U ( a, a ) = U ( f a ( b ) , f a ( c ))showing that f a is upper cone preserving. Conversely, if f a is upper cone preserving then f a ( U ( b, c )) = U ( f a ( b ) , f a ( c )) = U ( a, a ) = U ( a ) ⊇ { a } 6 = ∅ ,U ( a ) = U ( a, a )) = U ( f a ( a ) , f a ( a )) = f a ( U ( a, a )) = { a } showing that P is up-directed and that a is maximal.Several elementary facts on cone preserving mappings are stated in the next lemma. Lemma 2.3.
Let P = ( P, ≤ ) be a poset, f : P → P and A ⊆ P . Then the following hold: (i) f is monotone if and only if f ( U ( x )) ⊆ U ( f ( x )) for all x ∈ P , (ii) if f is upper cone preserving then it is monotone and f ( F ) ∈ Fil P for all F ∈ Fil P , (iii) if every monotone mapping from P to P is upper cone preserving then | P | = 1 , (iv) if f is monotone then f ( L ( A )) ⊆ L ( f ( A )) and f ( U ( A )) ⊆ U ( f ( A )) .Proof. (i) This is obvious.(ii) Assume f to be upper cone preserving. Then f ( U ( x )) = f ( U ( x, x )) = U ( f ( x ) , f ( x )) = U ( f ( x )) for all x ∈ P and hence f is monotone according to (i). Moreover, if F ∈ Fil P then f ( F ) = f ( [ x ∈ F U ( x )) = [ x ∈ F f ( U ( x )) = [ x ∈ F f ( U ( x, x )) = [ x ∈ F U ( f ( x ) , f ( x )) == [ x ∈ F U ( f ( x )) ∈ Fil P . f is monotone and a ∈ f ( L ( A )) then there exists some b ∈ L ( A ) with f ( b ) = a and since f is monotone we have a = f ( b ) ∈ L ( f ( A )). The statement for U followsby duality. Example 2.4.
Consider the poset depicted in Figure 1: ✉ ✉✉ ✉✉ (cid:0)(cid:0)(cid:0)(cid:0)❅❅❅❅ ✁✁✁✁ ❆❆❆❆ a bc d
Then f : P → P defined by f ( x ) := (cid:26) c if x ∈ { a, b, c } , otherwiseis upper cone preserving and hence monotone according to Lemma 2.3 (ii) . Since theabove poset is not a singleton there must exist some monotone mapping which is notupper cone preserving according to Lemma 2.3 (iii) . The mapping g : P → P defined by g ( x ) := (cid:26) b if x = a,x otherwiseis monotone, but not upper cone preserving since g ( U ( a, b )) = g ( { c, d, } ) = { c, d, } 6 = { b, c, d, } = U ( b ) = U ( b, b ) = U ( g ( a ) , g ( b )) . For injective mappings, we can show that they are upper cone preserving provided theypreserve principal filters.
Proposition 2.5.
Let ( P, ≤ ) be a poset. Then every injective mapping f : P → P satisfying f ([ x )) = [ f ( x )) for all x ∈ P is upper cone preserving.Proof. If a, b ∈ P and f : P → P is injective and satisfies f ([ x )) = [ f ( x )) for all x ∈ P then f ( U ( a )) = U ( f ( a )) and f ( U ( a, b )) = f ( U ( a ) ∩ U ( b )) = f ( U ( a )) ∩ f ( U ( b )) = U ( f ( a )) ∩ U ( f ( b )) = U ( f ( a ) , f ( b )) . Mappings determined by two elements
In the following, for every poset ( P, ≤ ) and every a, b ∈ P with a = b let f ab denote themapping from P to P defined by f ab ( x ) := (cid:26) b if x = a,x otherwise . The question when the mapping f ab is strictly monotone is answered in the next propo-sition. Proposition 3.1.
Let ( P, ≤ ) be a poset and a, b ∈ P with a = b . Then f ab is strictlymonotone if and only if a k b , L ∗ ( a ) ⊆ L ∗ ( b ) and U ∗ ( a ) ⊆ U ∗ ( b ) .Proof. Obviously, f ab is strictly monotone if a k b (since b < b is impossible) and for all x ∈ P the following hold: a < x implies b < x, (1) x < a implies x < b. (2)Now (1) and (2) are equivalent to U ∗ ( a ) ⊆ U ∗ ( b ) and L ∗ ( a ) ⊆ L ∗ ( b ), respectively.Similarly, we can ask when the mapping f ab is upper cone preserving. The answer is asfollows. Theorem 3.2.
Let P = ( P, ≤ ) be a poset and a, b ∈ P with a = b . Then f ab is uppercone preserving if and only if a is a minimal element of P and U ∗ ( a ) = U ( b ) .Proof. Let c, d ∈ P \ { a } . First assume f ab to be upper cone preserving. Then b ≤ a would imply a ∈ U ( b ) = U ( b, b ) = U ( f ab ( a ) , f ab ( a )) = f ab ( U ( a, a )) = f ab ( U ( a )) = U ∗ ( a ) ∪ { b } and hence a = b , a contradiction. Therefore b a . Now b ∈ U ( b ) = U ( b, b ) = U ( f ab ( a ) , f ab ( b )) = f ab ( U ( a, b )) = U ( a, b ) ⊆ U ( a )since a / ∈ U ( a, b ) and hence a ≤ b , i.e. a < b . Now c < a would imply a ∈ U ( c ) = U ( c, c ) = U ( f ab ( c ) , f ab ( c )) = f ab ( U ( c, c )) = f ab ( U ( c )) = (( U ( c )) \ { a } ) ∪ { b } and hence a = b , a contradiction. This shows that a is minimal. Moreover, U ∗ ( a ) = f ab ( U ( a )) = f ab ( U ( a, a )) = U ( f ab ( a ) , f ab ( a )) = U ( b, b ) = U ( b )since b ∈ U ∗ ( a ). Conversely, assume a to be minimal and U ∗ ( a ) = U ( b ). Then a < b and f ab ( U ( a, a )) = f ab ( U ( a )) = U ∗ ( a ) = U ( b ) = U ( b, b ) = U ( f ab ( a ) , f ab ( a )) ,f ab ( U ( a, c )) = U ( a, c ) = U ( b, c ) = U ( f ab ( a ) , f ab ( c )) ,f ab ( U ( c, d )) = U ( c, d ) = U ( f ab ( c ) , f ab ( d ))and hence f ab is upper cone preserving. Observe that c, d a because of c, d = a and theminimality of a . 5t should be remarked that U ∗ ( a ) = U ( b ) implies a ≺ b . Namely, from b ∈ U ( b ) = U ∗ ( a )we conclude a < b . If there would exist some c ∈ P with a < c < b then c ∈ U ∗ ( a ) = U ( b ),a contradiction. This shows a ≺ b .By Lemma 2.3 (ii), every upper cone preserving mapping is monotone. The questionis for which posets not every strictly monotone mapping is upper cone preserving. Theanswer is as follows. Remark 3.3. If ( P, ≤ ) is a poset containing two elements a and b with a k b satisfying L ∗ ( a ) ⊆ L ∗ ( b ) and U ∗ ( a ) ⊆ U ∗ ( b ) then not every strictly monotone mapping from P to P is upper cone preserving. Such a poset is depicted in Fig. 1.Proof. Let ( P, ≤ ) be a poset having two elements a and b with a k b satisfying L ∗ ( a ) ⊆ L ∗ ( b ) and U ∗ ( a ) ⊆ U ∗ ( b ). Then f ab is strictly monotone by Proposition 3.1, but notupper cone preserving by Theorem 3.2.In Theorem 3.2 we characterized when the mapping f ab is upper cone preserving. Nowwe show when this mapping is strictly upper cone preserving. Theorem 3.4.
Let ( P, ≤ ) be a poset and a, b ∈ P with a = b . Then f ab is strictly uppercone preserving if and only if | L ( a ) | ≤ and U ∗ ( a ) = U ( b ) .Proof. First assume f ab to be strictly upper cone preserving. Then b ≤ a would imply a ∈ U ( b ) = U ( b, b ) = U ( f ab ( a ) , f ab ( b )) = f ab ( U ( a, b )) = f ab ( U ( a )) = U ∗ ( a ) ∪ { b } and hence a = b , a contradiction. Therefore b a . Now b ∈ U ( b ) = U ( b, b ) = U ( f ab ( a ) , f ab ( b )) = f ab ( U ( a, b )) = U ( a, b ) ⊆ U ( a )since a / ∈ U ( a, b ) and hence a ≤ b , i.e. a < b and therefore U ( b ) ⊆ U ∗ ( a ). If c ∈ U ∗ ( a )then c ∈ U ( c ) = f ab ( U ( c )) = f ab ( U ( a, c )) = U ( f ab ( a ) , f ab ( c )) = U ( b, c ) ⊆ U ( b )showing U ∗ ( a ) ⊆ U ( b ). Altogether, we obtain U ∗ ( a ) = U ( b ). Now | L ( a ) | > d, e ∈ P with d = e and d, e < a and hence f ab ( U ( d, e )) = ( U ( d, e )) \ { a } 6 = U ( d, e ) = U ( f ab ( d ) , f ab ( e ))contradicting the fact that f ab is strongly upper cone preserving. Hence | L ( a ) | ≤
2. If,conversely, | L ( a ) | ≤ U ∗ ( a ) = U ( b ) and g, h ∈ P \ { a } then f ab ( U ( a, g )) = (cid:26) f ab ( U ( a )) = U ∗ ( a ) = U ( b ) = U ( b, g ) = U ( f ab ( a ) , f ab ( g )) if g ≤ a,U ( a, g ) = U ( b, g ) = U ( f ab ( a ) , f ab ( g )) otherwise ,f ab ( U ( g, h )) = U ( g, h ) = U ( f ab ( g ) , f ab ( h )) if g = h and hence f ab is strictly upper cone preserving.6 Chains
Chains are relatively simple posets. We derive an easy condition under which everystrictly monotone mapping on a chain is upper cone preserving.
Proposition 4.1.
Let P = ( P, ≤ ) be a chain and f a strictly monotone mapping from P to P . Then f is upper cone preserving if and only if f ( P ) ∈ Fil P .Proof. First assume f ( P ) ∈ Fil P . Let a, b ∈ P with a ≤ b and c ∈ U ( f ( a ) , f ( b )). Then c ≥ f ( b ). Since f ( P ) ∈ Fil P , there exists some d ∈ P with c = f ( d ). Now d < b would imply c = f ( d ) < f ( b ), a contradiction. Hence d ≥ b and c = f ( d ) ∈ f ( U ( b )) = f ( U ( a, b )). This shows U ( f ( a ) , f ( b )) ⊆ f ( U ( a, b )). The opposite inclusion follows fromLemma 2.3 (iv). Hence f is upper cone preserving. The rest of the proof follows fromLemma 2.3 (ii).Another interesting question concerns posets which are semilattices. Because every semi-lattice homomorphism is a monotone mapping, we can ask when every monotone map-ping of a given semilattice into itself is a homomorphism. Using the method developedby Berrone ([1]), we can prove the following result. Theorem 4.2.
A join-semilattice ( P, ∨ ) is a chain if and only if every monotone mappingfrom P to P is a homomorphism.Proof. Let P = ( P, ∨ ) be a join-semilattice and a, b ∈ P . If P is a chain and f a monotonemapping from P to P then f ( a ∨ b ) = (cid:26) f ( b ) = f ( a ) ∨ f ( b ) if a ≤ b,f ( a ) = f ( a ) ∨ f ( b ) otherwiseand hence f is a homomorphism. Now assume P not to be a chain. Then there exist c, d ∈ P with c k d . Define g : P → P by g ( x ) := (cid:26) c if x < c ∨ d,c ∨ d otherwise . Assume a ≤ b . If a < c ∨ d then g ( a ) = c ≤ g ( b ). If a < c ∨ d then b < c ∨ d and hence g ( a ) = c ∨ d = g ( b ). This shows that g is monotone. But g is not a homomorphism since g ( c ∨ d ) = c ∨ d = c = c ∨ c = g ( c ) ∨ g ( d ) . We have proved that there exists a monotone mapping from P to P that is not a homo-morphism .By duality, Theorem 4.2 also holds for meet-semilattices and hence also for lattices. Onthe other hand, the result of Theorem 4.2 can be extended to direct products of chains.For this, let us recall the following concepts.For i = 1 , f i : A i → B i . Then f × f denotes the mapping from A × A to B × B defined by ( f × f )( x , x ) := ( f ( x ) , f ( x )) for all ( x , x ) ∈ A × A . g : A × A → B × B is called directly decomposable if there exist g : A → B and g : A → B with g × g = g .Let C be a chain and P := C × C . As proved in [3], every lattice homomorphism from P to P is directly decomposable since P is a lattice and the variety of lattices is congruencedistributive. We can ask if monotone directly decomposable mappings from P to P arelattice homomorphisms. The following corollary of Theorem 4.2 gives a positive answer. Corollary 4.3.
Let ( C , ≤ ) , ( C , ≤ ) be chains, ( P, ≤ ) := ( C , ≤ ) × ( C , ≤ ) and f a mono-tone directly decomposable mapping from P to P . Then f is a lattice homomorphism.Proof. If f = f × f with f i : C i → C i for i = 1 , f , f are a monotone and, byTheorem 4.2, also (semi-)lattice homomorphisms which implies that f is (semi-)latticehomomorphism, too.Direct decomposability of homomorphisms was investigated by the authors and M. Gold-stern in [3]. For mappings which need not be homomorphisms we cannot use methodsinvolved in congruence distributive varieties. A simple characterization of directly de-composable mappings is formulated in the following lemma. Lemma 4.4.
Let A , A , B , B be non-void sets and f : A × A → B × B and for i = 1 , let p i denote the projection of B × B onto B i . Then the following are equivalent: (i) f is decomposable, (ii) p ( f ( x , x )) = p ( f ( x , y )) and p ( f ( x , x )) = p ( f ( y , x )) for all x , y ∈ A and x , y ∈ A .Proof. (i) ⇒ (ii):If f = f × f then p ( f ( x , x )) = p ( f ( x ) , f ( x )) = f ( x ) = p ( f ( x ) , f ( y )) = p ( f ( x , y )) ,p ( f ( x , x )) = p ( f ( x ) , f ( x )) = f ( x ) = p ( f ( y ) , f ( x )) = p ( f ( y , x ))for all x , y ∈ A and x , y ∈ A .(ii) ⇒ (i):Let a ∈ A and a ∈ A and for i = 1 , f i : A i → B i by f ( x ) := p ( f ( x , a )) for all x ∈ A ,f ( x ) := p ( f ( a , x )) for all x ∈ A . Because of (ii), f and f are well-defined and f ( x , x ) = ( p ( f ( x , x )) , p ( f ( x , x ))) = ( p ( f ( x , a )) , p ( f ( a , x ))) = ( f ( x ) , f ( x ))for all ( x , x ) ∈ A × A , i.e. f = f × f .Instead of join-semilattices we can investigate posets whose upper cones U ( x, y ) have aminimal element. Of course, every join-semilattice has this property, but there manyother examples of such posets, e.g. all finite up-directed posets.8 heorem 4.5. If ( P, ≤ ) is a poset, a, b ∈ P , a k b and U ( a, b ) has a minimal elementthen there exists a monotone mapping f from P to P with f ( U ( a, b )) = U ( f ( a ) , f ( b )) and hence there exists a monotone mapping from P to P which is not strictly upper conepreserving.Proof. Let ( P, ≤ ) be a poset and a, b, c ∈ P and assume a k b and that c is a minimalelement of U ( a, b ). Define f : P → P by f ( x ) := (cid:26) a if x < c,c otherwise . Let d, e ∈ P with d ≤ e . If d < c then f ( d ) = a ≤ f ( e ). If d < c then e < c andhence f ( d ) = c = f ( e ). This shows that f is monotone. We have a, b ≤ c . Since a = c would imply b ≤ c = a and b = c would imply a ≤ c = b , we have a, b < c andtherefore a ∈ U ( a ) = U ( a, a ) = U ( f ( a ) , f ( b )). Now assume f ( U ( a, b )) = U ( f ( a ) , f ( b )).Then a ∈ f ( U ( a, b )) and hence there exists some d ∈ U ( a, b ) with f ( d ) = a . Since c isa minimal element of U ( a, b ) we have d < c and hence a = f ( d ) = c , a contradiction.Therefore f ( U ( a, b )) = U ( f ( a ) , f ( b )). Corollary 4.6. If ( P, ≤ ) is a poset which is not a chain and which satisfies the DescendingChain Condition then there exists a monotone mapping from P to P which is not strictlyupper cone preserving and hence not upper cone preserving. On the other hand, if a poset in question is a chain, we can give a necessary and sufficientcondition for a monotone mapping to be upper cone preserving.
Proposition 4.7.
Let ( C, ≤ ) be a chain and f : C → C monotone. Then f is upper conepreserving if and only if U ( f ( x )) ⊆ f ( C ) for all x ∈ C .Proof. Let a ∈ C . If f is upper cone preserving then U ( f ( a )) = U ( f ( a ) , f ( a )) = f ( U ( a, a )) ⊆ f ( C ) . Conversely, assume U ( f ( x )) ⊆ f ( C ) for all x ∈ C . Since f is monotone, we have f ( U ( a )) ⊆ U ( f ( a )) according to Lemma 2.3 (i). Now let b ∈ U ( f ( a )). If b = f ( a ) then b ∈ f ( U ( a )). Now assume b > f ( a ). Since b ∈ U ( f ( a )) ⊆ f ( C ), there exists some c ∈ C with f ( c ) = b . Now c ≤ a would imply b = f ( c ) ≤ f ( a ), a contradiction. Hence c ∈ U ( a ) and therefore b = f ( c ) ∈ f ( U ( a )). This shows U ( f ( a )) ⊆ f ( U ( a )) and hence f ( U ( a )) = U ( f ( a )). Now, for x, y ∈ C we have U ( f ( x ) , f ( y )) = (cid:26) U ( f ( y )) = f ( U ( y )) = f ( U ( x, y )) if x ≤ y,U ( f ( x )) = f ( U ( x )) = f ( U ( x, y )) otherwise , i.e. f is upper cone preserving. We have seen that the Descending Chain Condition together with the property thatevery monotone mapping is strictly upper cone preserving forces a poset to be a chain. It9eems that our conditions are too restrictive. In fact, if we replace monotone mappingsby strictly monotone ones, we can obtain a richer structure of posets in which strictlymonotone mappings are strongly upper cone preserving.The ordinal sum of two posets ( A, ≤ ) and ( B, ≤ ) with A ∩ B = ∅ is the poset with baseset A ∪ B where the order inside A and inside B coincides with the original one and A < B , i.e. every element of A is below every element of B . Now, we can state thefollowing result. Proposition 5.1.
Every strictly monotone mapping on the ordinal sum of an antichainand a finite chain is strongly upper cone preserving.Proof. If f is a strictly monotone mapping on the ordinal sum ( P, ≤ ) of an antichain( A, ≤ ) and a finite chain ( C, ≤ ), a, b ∈ P and f ( a ) = f ( b ) then f ( A ) ⊆ A , f ( x ) = x forall x ∈ C and f ( U ( a, b )) = f ( C ) = C = U ( f ( a ) , f ( b )) if a, b ∈ A,f ( U ( b )) = U ( b ) = U ( f ( a ) , b ) = U ( f ( a ) , f ( b )) if a ∈ A and b ∈ C,U ( a, b ) = U ( f ( a ) , f ( b )) if a, b ∈ C. Example 5.2.
Examples of such ordinal sums are visualized in the Figure 2: ✉ ✉ ✉ ✉ . . . ✉ . . . ✉ . . . ✉ . . . ✉ ✉✉ ✉✉✉✉ ✁✁✁✁ ❆❆❆❆ ✁✁✁✁ ❆❆❆❆ (cid:0)(cid:0)(cid:0)(cid:0) ❅❅❅❅ (a) (b) (c)Fig. 2Every mapping f : A → B induces an equivalence relation Θ on A by defining ( x, y ) ∈ Θif f ( x ) = f ( y ). This equivalence relation is called the kernel of f , usually denoted byker f . The question when for a given poset ( P, ≤ ) and a given mapping F : P → P thequotient set P/ (ker f ) is again a poset is answered in the next theorem.Let ( P, ≤ ) and ( Q, ≤ ) be posets and f : P → Q . Recall that f is called strongly monotone if it is monotone and a, b ∈ P and f ( a ) ≤ f ( b ) imply that there exist a ′ , b ′ ∈ P with f ( a ′ ) = f ( a ), f ( b ′ ) = f ( b ) and a ′ ≤ b ′ . Definition 5.3. Le P = ( P, ≤ ) be a poset. An equivalence relation Θ on P is called an S -equivalence on P if it satisfies the following two conditions for all a, a ′ , b, b ′ ∈ P : If a, b, b ′ , c ∈ P , a ≤ b , b ′ ≤ c and ( b, b ′ ) ∈ Θ then there exist a ′ ∈ [ a ]Θ and c ′ ∈ [ c ]Θ with a ′ ≤ c ′ , (ii) if a, a ′ , b, b ′ ∈ P , a ≤ b , b ′ ≤ a ′ and ( a, a ′ ) , ( b, b ′ ) ∈ Θ then ( a, b ) ∈ Θ . Theorem 5.4.
Let P = ( P, ≤ ) be a poset, f : P → P strongly monotone and Θ an S -equivalence on P and define [ a ]Θ ≤ [ b ]Θ if there exist a ′ ∈ [ a ]Θ and b ′ ∈ [ b ]Θ with a ′ ≤ b ′ .Then (i) ker f is an S -equivalence on P , (ii) ( P/ Θ , ≤ ) is a poset and x [ x ]Θ strongly monotone.Proof. (i) Put Φ := ker f and assume a, b, b ′ , c ∈ P , a ≤ b , b ′ ≤ c and ( b, b ′ ) ∈ Φ. Then f ( a ) ≤ f ( b ) = f ( b ′ ) ≤ f ( c ) . Since f is strongly monotone there exist a ′ , c ′ ∈ P with f ( a ′ ) = f ( a ), f ( c ′ ) = f ( c )and a ′ ≤ c ′ . Hence a ′ ∈ [ a ]Φ, c ′ ∈ [ c ]Φ and a ′ ≤ c ′ proving (i) of Definition 5.3.Next assume a, a ′ , b, b ′ ∈ P , a ≤ b , b ′ ≤ a ′ and ( a, a ′ ) , ( b, b ′ ) ∈ Φ. Then f ( a ) ≤ f ( b ) = f ( b ′ ) ≤ f ( a ′ ) = f ( a )and hence f ( a ) = f ( b ), i.e. ( a, b ) ∈ Φ proving (ii) of Definition 5.3.(ii) We consider the binary relation ≤ on P/ Θ. Obviously, ≤ is reflexive. Assume a, b ∈ P , [ a ]Θ ≤ [ b ]Θ and [ b ]Θ ≤ [ a ]Θ. Then there exist a ′ , a ′′ ∈ [ a ]Θ and b ′ , b ′′ ∈ [ b ]Θ with a ′ ≤ b ′ and b ′′ ≤ a ′′ . Since ( a ′ , a ′′ ) , ( b ′ , b ′′ ) ∈ Θ, we conclude by (ii) of Definition 5.3that ( a ′ , b ′ ) ∈ Θ. This shows [ a ]Θ = [ a ′ ]Θ = [ b ′ ]Θ = [ b ]Θ proving antisymmetryof ≤ . Now assume a, b, c ∈ P , [ a ]Θ ≤ [ b ]Θ and [ b ]Θ ≤ [ c ]Θ. Then there exist a ′ ∈ [ a ]Θ, b ′ , b ′′ ∈ [ b ]Θ and c ′ ∈ [ c ]Θ with a ′ ≤ b ′ and b ′′ ≤ c ′ . Since ( b ′ , b ′′ ) ∈ Θ,we conclude by (i) of Definition 5.3 that there exist a ′′ ∈ [ a ′ ]Θ and c ′′ ∈ [ c ′ ]Θ with a ′′ ≤ c ′′ . Now a ′′ ∈ [ a ]Θ and c ′′ ∈ [ c ]Θ which shows [ a ]Θ ≤ [ c ]Θ proving transitivityof ≤ . Altogether, ( P/ Θ , ≤ ) is a poset. Clearly, x [ x ]Θ is monotone and by thedefinition of ≤ on P/ Θ, this mapping is strongly monotone.
Example 5.5.
Consider the poset P = ( P, ≤ ) visualized in Figure 3: ✉ ✉✉ ✉✉✉ (cid:0)(cid:0)(cid:0)(cid:0)❅❅❅❅ ✁✁✁✁ ❆❆❆❆❆❆❆❆ ✁✁✁✁ a bc d et f : P → P be defined by x a b c d f ( x ) 0 a a c c and put Θ := ker f . Then f is strongly monotone, Θ = { } ∪ { a, b } ∪ { c, d } ∪ { } is an S -equivalence on P and ( P/ Θ , ≤ ) = ( { [0]Θ , [ a ]Θ , [ c ]Θ , [1]Θ } , ≤ ) is again a posetwhere [0]Θ < [ a ]Θ < [ c ]Θ < [1]Θ . References [1] L. R. Berrone, The homomorphism equation on semilattices. Aequationes Math. (2020), 803–816.[2] I. Chajda, Homomorphisms of directed posets. Asian-Eur. J. Math. (2008), 45–51.[3] I. Chajda, M. Goldstern and H. L¨anger, A note on homomorphisms between productsof algebras. Algebra Universalis (2018), Paper No. 25, 7 pp.[4] I. Chajda and ˇS. Hoˇskov´a. A characterization of cone preserving mappings of qua-siordered sets. Miskolc Math. Notes6