REPRESENTING HOMOMORPHISMS OF CONGRUENCE LATTICES

REPRESENTING HOMOMORPHISMS OF CONGRUENCE LATTICES AS RESTRICTIONS OF CONGRUENCES OF ISOFORM LATTICES ¨ G. GRATZER AND H. LAKSER

Abstract. Let L1 be a finite lattice with an ideal L2 . Then the restriction map is a {0, 1}-homomorphism from Con L1 into Con L2 . In 1986, the present authors published the converse. If D1 and D2 are finite distributive lattices, and ϕ : D1 → D2 is a {0, 1}-homomorphism, then there are finite lattices L1 and L2 with an embedding η of L2 as an ideal of L1 , and there are isomorphisms ε1 : Con L1 → D1 and ε2 : Con L2 → D2 such that ϕ is represented as the restriction map of congruences from L1 to L2 , up to the two isomorphisms. Let us call a lattice isoform, if for any congruence, all congruence classes are isomorphic lattices. In 2003, G. Gr¨ atzer and E. T. Schmidt proved that every finite distributive lattice can be represented as the congruence lattice of an isoform lattice. In this paper we combine the two results, reproving the 1986 result with isoform lattices.

1. Introduction Let η : K2 → K1 be a homomorphism of lattices. There is then a {1, ∧}preserving mapping res(η) : Con K1 → Con K2 , the restriction map, defined by setting a ≡ b (res(η)Θ) iff ηa ≡ ηb (Θ). The mapping res(η) is 0-preserving exactly when η is an embedding. If res(η) is a bijection, and so is a lattice isomorphism, then we say that η is a congruence-preserving extension . Now, in general, even if η is an embedding, res(η) need not preserve joins. However, if η embeds K2 as a convex sublattice of K1 , in particular, as an ideal of K1 , then res(η) also preserves joins, and so is a {0, 1}-preserving lattice homomorphism. A congruence relation Θ on a lattice L is said to be isoform, if all the congruence classes of Θ are isomorphic sublattices of L. The lattice L is said to be isoform, if all of its congruences are isoform. In this paper we prove the following theorem: Theorem 1. Let K1 and K2 be lattices with finite congruence lattices, Con K1 and Con K2 . Let ϕ : Con K1 → Con K2 be a {0, 1}-preserving lattice homomorphism. Then there are isoform lattices L1 and L2 with an embedding η of L2 as an ideal in L1 , and there are congruence-preserving extensions ε1 : K1 → L1 and ε2 : K2 → L2 such that res(ε2 )res(η)Θ = ϕres(ε1 )Θ, Date: Revised: March 25, 2009. 2000 Mathematics Subject Classification. Primary: 06B10; Secondary: 06B15. Key words and phrases. Congruence lattice, congruence-preserving extension, isoform. The research of the first author was supported by the NSERC of Canada. 1

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¨ G. GRATZER AND H. LAKSER

for all Θ ∈ Con L1 . If K2 is finite, then L2 can be taken to be finite. If K1 is also finite, then L1 can also be taken to be finite. By a classical result, which in [1] we call the Dilworth Theorem, every finite distributive lattice can be represented as the congruence lattice of a finite lattice. Combining the Dilworth Theorem with Theorem 1, we then obtain: Theorem 2. Let D1 and D2 be finite distributive lattices, and let ϕ : D1 → D2 be a {0, 1}-preserving lattice homomorphism. Then there are finite isoform lattices L1 and L2 with an embedding η of L2 as an ideal of L1 , and there are isomorphisms ε1 : Con L1 → D1 and ε2 : Con L2 → D2 such that ε2 res(η)Θ = ϕε1 Θ, for all Θ ∈ Con L1 . In the process of proving Theorem 1, we shall develop all the tools needed to prove Theorem 2 without needing to appeal to the Dilworth Theorem. Some of the techniques we utilize were developed in G. Grätzer and E. T. Schmidt [4] and G. Gr¨ atzer, R. W. Quackenbush, and E. T. Schmidt [5]. An aspect of this work, the solution of Problem 6 of G. Grätzer, R. W. Quackenbush, and E. T. Schmidt [5], was written up as a separate paper: G. Grätzer, H. Lakser, and R. W. Quackenbush [3]. In view of the result in this paper, the following problem arises naturally: Problem. Let K1 and K2 be lattices with finite congruence lattices, let K2 be an ideal of K1 . Do there exist isoform lattices L1 and L2 with L2 an ideal in L1 , such that L1 is a congruence-preserving extension of K1 and L2 is a congruencepreserving extension of K2 ? The problem seems to be hard even for finite—rather than congruence-finite— lattices. For the background of this field, the reader should consult the book [1]. Acknowledgement. We would like to thank R. W. Quackenbush for correcting an error in Section 4 and for presenting this paper in the UALT seminar in Winnipeg. 2. The lattices SP and TP Let P be a finite set with two binary relations, %1 and %2 . For each p ∈ P , let there be • a bounded lattice Sp with a separator v1p , that is, with 0p ≺ v1p ≺ 1p , where 0p and 1p are the respective zero and unit of Sp ; • a bounded lattice Tp , a sublattice of Sp , with |Tp | > 3 and with zero 00p and unit 10p , that does not contain v1p and that contains a separator v2p , with 00p ≺ v2p ≺ 10p , which is doubly irreducible in Sp . Later on, in our proof of of Theorem 1, we shall have 00p = 0p but 10p 6= 1p , / Tp . automatically guaranteeing that v1p ∈ Q Q We set SP = ( Sp | p ∈ P ) and TP = ( Tp | p ∈ P ), a subset of SP . The elements of SP will be bold face lower case letters. An element s ∈ SP is written in the form hsp ip∈P . We write sp for sp . In the papers [5] and [3], a single family of lattices Sp , p ∈ P , was considered, with a single relation % on P . If each Sp is simple, then the direct product of

HOMOMORPHISMS OF CONGRUENCE LATTICES

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the lattices Sp is an isoform lattice whose congruence lattice is boolean. Now, any finite distributive lattice is a sublattice of a finite boolean lattice. In those papers it was shown that, by judicious choice of the relation %, it is possible to define a lattice suborder of the direct product order, whose congruence lattice is a sublattice of the boolean congruence lattice of the direct product, thus yielding an isoform lattice with the correct congruence lattice. In this paper, we extend this this idea. Eventually, we shall take the Tp as ideals of the Sp , all the Sp and Tp simple, and use the two relations %1 and %2 to get a lattice suborder of the direct product of the Sp . The lattice suborder is a congruence preserving isoform extension of K1 , with an ideal, which is the induced lattice suborder of the direct product of the Tp ; this lattice suborder is a congruence preserving isoform extension of K2 . Furthermore, the congruences restrict in the desired manner. We proceed to define the desired order on the direct product SP . We define a binary relation ≤ on SP : Definition 3. Let a, b ∈ SP . Then a ≤ b iff ap ≤ bp , for all p ∈ P , and the following two conditions hold: (P1) If p ∈ P and ap = v1p = bp , then aq = bq , for all q ∈ P with p %1 q. (P2) If p ∈ P and ap = v2p = bp , then aq = bq , for all q ∈ P with p %2 q. Lemma 4. The relation ≤ is an order on SP . Proof. That ≤ is reflexive is immediate, since (P1) and (P2) hold if a = b. That ≤ is antisymmetric is also immediate, since it is a subrelation of an antisymmetric relation. We now establish transitivity. Let a ≤ b ≤ c. Then, for each p ∈ P , (1)

ap ≤ bp ≤ cp ,

and so ap ≤ cp . We need only establish (P1) and (P2) for the pair a, c. Let i ∈ {1, 2}, let ap = vip = cp , and let p %i q, for some q ∈ P . Then, by (1), ap = vip = bp and bp = vip = cp . Thus, by (Pi), for a ≤ b, we get aq = bq , and, similarly for b ≤ c, we get bq = cq Thus aq = cq , establishing (Pi) for a, c, thereby completing the demonstration that a ≤ c. Thus ≤ is also transitive. Since v1p ∈ / Tp , for all p ∈ P , the following result is immediate: Lemma 5. Let a, b ∈ TP . Then a ≤ b iff ap ≤ bp , for all p ∈ P and condition (P2) holds for a, b. We now proceed to show that SP is a lattice under the above defined order ≤. We work out the details for the join operation and appeal to the principle of duality to get the meet operation. We define a binary operation ∨ on SP and then proceed to show that a ∨ b is the least upper bound of {a, b} under ≤. For some p ∈ P , we shall have (a∨b)p = ap ∨bp . However, because of conditions (P1) and (P2), for certain p ∈ P , which we call join-singular with respect to a, b, we are forced to have (a ∨ b)p > ap ∨ bp . We shall give an inductive definition of join-singularity in the following definition. For each integer n ≥ 0, we shall define 1-join-singularn and 2-join-singularn . We shall say that p is join-singularn , if it is either 1-join-singularn or 2-join-singularn . We shall say p is i-join singular if it is


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i-join-singularn , for some integer n ≥ 0, and that it is join-singular, if it is either 1-join-singular or 2-join-singular. We proceed with the definition: Definition 6. Let a, b ∈ SP , let p ∈ P , and let i ∈ {1, 2}. Then p is i-join-singular 0 with respect to a, b, if ap ∨ bp = vip and there is a q ∈ P with p %i q and with either ap ≤ bp and aq bq or with bp ≤ ap and bq aq . Let n > 0. Then p is i-join-singular n with respect to a, b, if ap ∨ bp = vip and there is a q ∈ P with p %i q such that q is join-singularn−1 (that is, either 1-join-singularn−1 or 2-join-singularn−1 ) with respect to a, b. p is i-join-singular with respect to a, b, if it is i-join-singularn with respect to a, b for some integer n ≥ 0, and is join-singular with respect to a, b, if it is either 1-join-singular or 2-join-singular with respect to a, b. If p is neither 1-join-singular nor 2-join-singular with respect to a, b, then p is join-regular with respect to a, b. Note that, since both v1p and v2p are join-irreducible in Sp , if ap ∨ bp = vip , then we have either ap ≤ bp or bp ≤ ap . Note also that, since v1p 6= v2p , 1-join-singularity and 2-join-singularity are mutually exclusive. Furthermore, if p is i-join-singular with respect to a, b, then ap ∨ bp = vip . Again, since v1p ∈ / Tp and Tp is a sublattice of Sp , for each p ∈ P , we have: Lemma 7. Let a, b ∈ TP and let p ∈ P . Then, for each integer n ≥ 0, p is join-singularn with respect to a, b iff p is 2-join-singularn with respect to a, b. In particular, p is join-singular with respect to a, b iff p is 2-join-singular with respect to a, b. We now define the binary operation ∨ on SP . Definition 8. Let a, b ∈ SP .   1p , (a ∨ b)p = 10p ,   ap ∨ bp ,

Then a ∨ b ∈ SP is defined by setting if p is 1-join-singular with respect to a, b; if p is 2-join-singular with respect to a, b; if p is join-regular with respect to a, b,

for each p ∈ P . From Definition 6, we see that ap ∨ bp ≤ (a ∨ b)p , for all p ∈ P . By Lemma 7, we have: Lemma 9. Let a, b ∈ TP . Then, for each p ∈ P , ( 10p , if p is join-singular with respect to a, b; (a ∨ b)p = ap ∨ bp , otherwise. In particular, a ∨ b ∈ TP . We now proceed to show that a ∨ b is the least upper bound in hSP , ≤i of {a, b}. First, a preliminary result: Lemma 10. Let a, b ∈ SP , let i ∈ {1, 2}, and let p, q ∈ P with p %i q. If ap ∨ bp = vip and p is join-regular with respect to a, b, then so is q. Proof. If q were join-singularn for some integer n ≥ 0, then p would be i-joinsingularn+1 .


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Lemma 11. Let a, b ∈ SP . Then a ∨ b is an upper bound of {a, b}. Proof. By symmetry, it suffices to show that a ≤ a ∨ b. Clearly, ap ≤ ap ∨ bp ≤ (a ∨ b)p , for all p ∈ P . To conclude that a ≤ a ∨ b, we need only show that conditions (P1) and (P2) hold for the pair a, a ∨ b. Let i ∈ {1, 2}, let p ∈ P with ap = vip = (a ∨ b)p , and let p %i q. Since vip 6= 1p , 10p , it follows, by Definition 8, that p is join-regular with respect to a, b. Thus, vip = ap = (a ∨ b)p = ap ∨ bp , that is, bp ≤ ap . Since p is join-regular and ap ∨ bp = vip , it follows that bq ≤ aq (p is not i-joinsingular0 —see Definition 6). Thus aq = aq ∨ bq . But, since p is join-regular with respect to a, b, since ap ∨ bp = vip , and since p %i q, it follows from Lemma 10 that q is join-regular. Thus, aq = aq ∨ bq = (a ∨ b)q , establishing conditions (P1) and (P2) for the pair a, a ∨ b.

Before proving that a ∨ b is the least upper bound of {a, b}, we prove a preliminary lemma: Lemma 12. Let a, b, c ∈ SP with a, b ≤ c and let p ∈ P . If p is join-singular with respect to a, b, then ap ∨ bp < cp Proof. We proceed by induction. Let i ∈ {1, 2} and let p be i-join-singular0 with respect to a, b. Without loss of generality, we may assume that bp ≤ ap = vip and that there is a q ∈ P with p %i q and with (2)

bq aq .

Now, ap = ap ∨ bp ≤ cp . Assume that ap ∨ bp = cp . Then we have ap = vip = cp . By condition (Pi) for a ≤ c, we get aq = cq , contradicting (2), since (2) implies that aq < aq ∨ bq ≤ cq . This contradiction shows that ap ∨ bp < cp if p is join-singular0 .


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Now, for the induction step, let n > 0 and let aq ∨ bq < cq , for all q ∈ P that are join-singularn−1 with respect to a, b. Let i ∈ {1, 2}, let p be i-join-singularn with respect to a, b, and let us assume that ap ∨ bp = cp . Now, ap ∨ bp = We then have

vip ,

which is join-irreducible. Thus, we may assume that bp ≤ ap . ap = ap ∨ bp = vip = cp .

(3)

Now, since p is i-join-singularn , there is a q ∈ P with p %i q and with q joinsingularn−1 . Then (4)

aq ∨ bq < cq .

But, by condition (Pi) for a ≤ c and by (3), cq = aq ≤ aq ∨ bq ≤ cq , contradicting (4). This contradiction establishes the inductive step, thereby proving the lemma. Lemma 13. For a, b ∈ SP , the element a ∨ b is the least upper bound of {a, b}. Proof. By Lemma 11, we need only show that if c ∈ SP and a, b ≤ c, then a∨b ≤ c. We first show that (a ∨ b)p ≤ cp , for all p ∈ P . Since ap , bp ≤ cp , we conclude that (a ∨ b)p = ap ∨ bp ≤ cp , for all p join-regular with respect to a, b. If p is join-singular with respect to a, b, then, by Lemma 12, (5)

ap ∨ bp < cp .

If p is 1-join-singular, then ap ∨bp = v1p , whereby (5) implies that cp = 1p = (a∨b)p . If p is 2-join-singular, then ap ∨ bp = v2p , whereby (5) implies that cp ≥ 10p = (a ∨ b)p , since v2p is meet-irreducible. Thus (a ∨ b)p ≤ cp , for all p ∈ P . We now establish conditions (P1) and (P2) for the pair a ∨ b, c. Let i ∈ {1, 2}, let p %i q, and let (a ∨ b)p = vip = cp . Since vip 6= 1p , 10p , we conclude, by Definition 8, that p is join-regular with respect to a, b, and so that ap ∨ bp = (a ∨ b)p = vip = cp . Since vip is join-irreducible, we may assume that ap = ap ∨ bp = vip = cp . Since then bp ≤ ap and since p is join-regular, we conclude that bq ≤ aq and, by Lemma 10, q is join-regular. Thus (a ∨ b)q = aq ∨ bq = aq . Since ap =

vip

= cp , we get aq = cq by condition (Pi) for a ≤ c, and so (a ∨ b)q = aq = cq ,


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establishing condition (Pi) for a ∨ b and c. Thus, if a, b ≤ c, then a ∨ b ≤ c, proving that a ∨ b is the least upper bound of {a, b}. By duality, we have the concepts of meet-singularity and the meet operation ∧ on SP : Definition 14. Let a, b ∈ SP , let p ∈ P , and let i ∈ {1, 2}. Then p is i-meet-singular 0 with respect to a, b, if ap ∧ bp = vip and there is a q ∈ P with p %i q and with either ap ≤ bp and aq bq or with bp ≤ ap and bq aq . Let n > 0. Then p is i-meet-singular n with respect to a, b if ap ∧ bp = vip and there is a q ∈ P with p %i q such that q is meet-singularn−1 (that is, either 1-meet-singularn−1 or 2-meet-singularn−1 ) with respect to a, b. p is i-meet-singular with respect to a, b, if it is i-meet-singularn with respect to a, b for some integer n ≥ 0, and is meet-singular with respect to a, b if it is either 1-meet-singular or 2-meet-singular with respect to a, b. If p is neither 1-meet-singular nor 2-meet-singular with respect to a, b, then p is meet-regular with respect to a, b. Definition 15. Let a, b ∈ SP . Then a ∧ b ∈ SP is defined by setting   if p is 1-meet-singular with respect to a, b; 0p , 0 (a ∧ b)p = 0p , if p is 2-meet-singular with respect to a, b;   ap ∧ bp , if p is meet-regular with respect to a, b, for each p ∈ P . We then have: Lemma 16. The order SP is a lattice with join operation ∨ of Definition 8 and meet operation ∧ of Definition 15. The subset TP is a sublattice of SP . Proof. That SP is a lattice under ∨ and ∧ follows from Lemma 13 and its dual. That TP is a sublattice follows from Lemma 9 and its dual. 3. Some sublattices and congruences Q Let Q ⊆ P . Then, Q restricting %1 and %2 to Q, we have the lattices Q SQ = ( Sp | p ∈ Q ) and TQ = ( Tp | p ∈ Q ), as in Section 2. Let x ∈ SP −Q = ( Sp | p ∈ / Q ). We then have a (set) injection ηx : SQ → SP determined by ( ap , if p ∈ Q; (ηx a)p = xp , if p ∈ / Q. Lemma 17. Let x ∈ SP −Q with xp 6= v1p , v2p , for all p ∈ / Q. Let a, b ∈ SQ , and let p ∈ P . Then, for each i ∈ {1, 2}, p is i-join-singular with respect to ηx a, ηx b iff p ∈ Q and p is i-join-singular with respect to a, b. Proof. Clearly, if p ∈ Q and p is i-join-singular with respect to a, b, then p is i-join singular with respect to ηx a, ηx b. Now, let i ∈ {1, 2} and let p ∈ P be i-join-singular with respect to ηx a, ηx b. Since (ηx a)p ∨ (ηx b)p = xp ∨ xp = xp 6= vip ,

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if p ∈ / Q, it follows that p ∈ Q. It suffices then to prove, the statement, for each integer n ≥ 0: (Sn) If i ∈ {1, 2}, if p ∈ Q, and if p is i-join-singularn with respect to ηx a, ηx b, then p is i-join-singularn with respect to a, b. We proceed by induction. We first prove (S0): Let i ∈ {1, 2} , let p ∈ Q, and let p be i-join-singular0 with respect to ηx a, ηx b. Thus, (ηx a)p ∨ (ηx b)p = vip , that is, ap ∨ bp = vip , and, say, (ηx b)p ≤ (ηx a)p , that is, bp ≤ ap , and there is a q ∈ P with p %i q, and with (ηx b)q (ηx a)q . Then q must lie in Q, since q ∈ / Q implies that (ηx b)q = xq = (ηx a)q . So bq aq . Thus p is i-join-singular0 with respect to a, b, establishing (S0). Now, for the inductive step, let n > 0 and assume (S(n − 1)). Let i ∈ {1, 2} and let p ∈ Q be i-join-singularn with respect to ηx a, ηx b. Then, ap ∨ bp = vip , and there is a q with p %i q such that q is join-singularn−1 with respect to ηx a, ηx b. Then q ∈ Q and, by statement (S(n − 1)), q is join-singularn−1 with respect to a, b. So p is then i-join-singularn with respect to a, b, thereby establishing the inductive step. Thus statement (Sn) holds for all n ≥ 0, concluding the proof. Corollary. If x ∈ SP −Q and xp 6= v1p , v2p , for all p ∈ / Q, then ηx is a lattice embedding of SQ into SP . If, furthermore, x ∈ TP −Q , then ηx embeds TQ as a sublattice of TP . Proof. To show that ηx is a lattice embedding, we need only show that ηx preserves join and meet (since ηx is injective). By duality, it suffices to show that ηx preserves join. Let a, b ∈ SQ , let p ∈ P , and consider (ηx a∨ηx b)p . If p ∈ / Q, then, by Lemma 17, p is join-regular with respect to ηx a, ηx b. So (ηx a ∨ ηx b)p = (ηx a)p ∨ (ηx b)p = xp = (ηx (a ∨ b))p , if p ∈ / Q. Now let p ∈ Q be join-regular with respect to ηx a, ηx b. Then, by Lemma 17, p is join-regular with respect to a, b. Then (ηx a ∨ ηx b)p = (ηx a)p ∨ (ηx b)p = ap ∨ bp = (a ∨ b)p = (ηx (a ∨ b))p , if p ∈ Q is join-regular with respect to ηx a, ηx b. Next, let p ∈ Q be 1-join-singular with respect to ηx a, ηx b. Then p is 1-join singular with respect to a, b. Then (ηx a ∨ ηx b)p = 1p = (a ∨ b)p = (ηx (a ∨ b))p , if p ∈ Q is 1-join-singular with respect to ηx a, ηx b. Finally, let p ∈ Q be 2-join-singular with respect to ηx a, ηx b. Then p is 2-join singular with respect to a, b. Then (ηx a ∨ ηx b)p = 10p = (a ∨ b)p = (ηx (a ∨ b))p , if p ∈ Q is 2-join-singular with respect to ηx a, ηx b. Consequently, for all p ∈ P , (ηx a ∨ ηx b)p = (ηx (a ∨ b))p . That is, ηx a ∨ ηx b = ηx (a ∨ b). So, ηx preserves join and, thus, by duality, is a lattice homomorphism.


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The last statement in the corollary follows because TQ is a sublattice of SQ and TP is a sublattice of SP by Lemma 16. A subset Q ⊆ P is said to be a %i -down-set, if whenever p ∈ Q and q %i p, then q ∈ Q. We say that Q is a {%1 , %2 }-down-set, if it is both a %1 -down-set and a %2 down-set. For Q a {%1 , %2 }-down-set, the analog of Lemma 17 holds for arbitrary x ∈ SP −Q : Lemma 18. Let Q ⊆ P be a {%1 , %2 }-down-set and let x ∈ SP −Q . Let a, b ∈ SQ and let p ∈ P . Then, for each i ∈ {1, 2}, p is i-join-singular with respect to ηx a, ηx b iff p ∈ Q and p is i-join-singular with respect to a, b. Proof. It is immediate that if p ∈ Q and p is i-join-singular with respect to a, b, then p is i-join-singular with respect to ηx a, ηx b. It then suffices to prove the statement, for each integer n ≥ 0: (Sn) If i ∈ {1, 2} and if p ∈ P is i-join-singularn with respect to ηx a, ηx b, then p ∈ Q and p is i-join-singularn with respect to a, b. We proceed by induction. We first prove (S0). Let i ∈ {1, 2}, and let p be i-join-singular0 with respect to ηx a, ηx b. Thus (ηx a)p ∨ (ηx b)p = vip with, say, (ηx b)p ≤ (ηx a)p , and there is a q ∈ P with p %i q and with (ηx b)q (ηx a)q . Then, as in the proof of Lemma 17, q must lie in Q. So, since Q is a %i -down-set, p ∈ Q. Then ap ∨ bp = vip , bp ≤ ap , and bq aq . Thus p is i-join-singular0 with respect to a, b, establishing (S0). Now for the inductive step, let n > 0 and assume (S(n − 1)). Let i ∈ {1, 2} and let p be i-join-singularn with respect to ηx a, ηx b. Then (ηx a)p ∨ (ηx b)p = vip and there is a q ∈ P with p %i q such that q is join-singularn−1 with ηx a, ηx b. Then, by (S(n − 1)), q ∈ Q and q is join-singularn−1 with respect to a, b. Since Q is a %i -down-set, p is then in Q and so ap ∨ bp = vip , p %i q, and q is join-singularn−1 with respect to a, b; that is, p is i-join-singularn with respect to a, b. So the inductive step is established. Thus (Sn) holds, for all n > 0, concluding the proof. Corollary. If Q ⊆ P is a {%1 , %2 }-down-set and x ∈ SP −Q , then ηx is a lattice embedding of SQ into SP . Proof. The proof is a verbatim copy of the proof of the Corollary to Lemma 17, with all discussion of TQ and TP deleted, and with each reference to Lemma 17 replaced with a reference to Lemma 18. Lemma 19. Let Q ⊆ P be a %2 -down-set and let x ∈ TP −Q . Let a, b ∈ TQ and let p ∈ P . Then p is join-singular with respect to ηx a, ηx b iff p ∈ Q and p is join-singular with respect to a, b. Proof. By Lemma 7, join-singularity with respect to elements a, b with ap , bp ∈ Tp , for all p ∈ P , is 2-join-singularity. Our lemma is then almost a consequence of Lemma 18. The only difference is that in Lemma 18 Q is a {%1 , %2 }-down-set, while here Q is only a %2 -down-set. However, because here join-singularity is 2-joinsingularity, the relation %1 never appears in the characterization of join-singularity. Thus, here, it suffices that Q be only a %2 -down-set. Then, exactly as for the corollary to Lemma 18, we have the corollary:

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Corollary. If Q ⊆ P is a %2 -down-set and x ∈ TP −Q , then ηx is a lattice embedding of TQ into TP . We now look at some congruences. A subset Q ⊆ P is said to be a %i -up-set (for i ∈ {1, 2}), if p ∈ Q and p %i q imply that q ∈ Q. We say that Q is a {%1 , %2 }-up-set if Q is both a %1 -up-set and a %2 -up-set. For each Q ⊆ P , we have the projection πQ : SP → SQ and its restriction 0 0 πQ : TP → TQ given by (πQ a)p = ap and (πQ a)p = ap , for p ∈ Q. Then we have the following duals, in a category-theoretic sense, of Lemmas 18 and 19. Lemma 20. Let Q ⊆ P be a {%1 , %2 }-up-set and let a, b ∈ SP . Then, for each p ∈ Q and each i ∈ {1, 2}, p is i-join-singular with respect to a, b iff p is i-joinsingular with respect to πQ a, πQ b. Proof. If p ∈ Q is i-join-singular with respect to πQ a, πQ b, then it follows immediately that p is i-join-singular with respect to a, b. It then suffices to prove the statement, for each integer n ≥ 0: (Sn) If i ∈ {1, 2}, p ∈ Q, and p is i-join-singularn with respect to a, b, then p is i-join-singularn with respect to πQ a, πQ b. We proceed by induction. We first prove (S0). Let i ∈ {1, 2} , let p ∈ Q, and let p be i-join-singular0 with respect to a, b. Thus, ap ∨ bp = vip , that is, (πQ a)p ∨ (πQ b)p = vip , and, say, bp ≤ ap , that is, (πQ b)p ≤ (πQ a)p , and there is a q ∈ P with p %i q, and with bq aq . Then q ∈ Q, since Q is a %i -up-set, and so (πQ b)q (πQ a)q . Thus p is i-join-singular0 with respect to πQ a, πQ b, establishing (S0). Now, for the inductive step, let n > 0 and assume (S(n − 1)). Let i ∈ {1, 2} and let p ∈ Q be i-join-singularn with respect to a, b. Then, (πQ a)p ∨ (πQ b)p = ap ∨ bp = vip , and there is a q with p %i q such that q is join-singularn−1 with respect to a, b. Then q ∈ Q, since Q is a %i -up-set, and, by statement (S(n − 1)), q is join-singularn−1 with respect to πQ a, πQ b. So p is then i-join-singularn with respect to πQ a, πQ b, thereby establishing the inductive step. Thus (Sn) holds for all n ≥ 0, concluding the proof. Corollary. Let Q ⊆ P be a {%1 , %2 }-up-set. Then πQ : SP → SQ is a lattice surjection. Proof. By duality, it suffices to show that πQ preserves join. Let a, b ∈ SP , let p ∈ Q, and consider (πQ a ∨ πQ b)p . If p is join-regular with respect to a, b, then, by Lemma 20, p is join-regular with respect to πQ a, πQ b. Then (πQ a ∨ πQ b)p = (πQ a)p ∨ (πQ b)p = ap ∨ bp = (a ∨ b)p = (πQ (a ∨ b))p , if p ∈ Q is join-regular with respect to a, b. Next, let p ∈ Q be 1-join-singular with respect to a, b. Then p is 1-join-singular with respect to πQ a, πQ b. So (πQ a ∨ πQ b)p = 1p = (a ∨ b)p = (πQ (a ∨ b))p , if p ∈ Q is 1-join-singular with respect to a, b.


11

Finally, let p ∈ Q be 2-join-singular with respect to a, b. Then p is 2-join-singular with respect to πQ a, πQ b. So (πQ a ∨ πQ b)p = 10p = (a ∨ b)p = (πQ (a ∨ b))p , if p ∈ Q is 2-join-singular with respect to a, b. Consequently, for all p ∈ Q, (πQ a ∨ πQ b)p = (πQ (a ∨ b))p . That is, πQ a ∨ πQ b = πQ (a ∨ b). So, πQ preserves join and, thus, by duality, is a lattice homomorphism.

0 For πQ we have similar results.

Lemma 21. Let Q ⊆ P be a %2 -up-set and let a, b ∈ TP . Then, for each p ∈ Q, 0 0 p is join-singular with respect to a, b iff p is join-singular with respect to πQ a, πQ b. Proof. By Lemma 7, join-singularity here is 2-join-singularity. 0 0 b, then it follows immediately a, πQ If p ∈ Q is join-singular with respect to πQ that p is join-singular with respect to a, b. It then suffices to prove the statement, for each integer n ≥ 0: (Sn) If p ∈ Q, and if p is join-singularn with respect to a, b ∈ TP , then p is 0 0 b. a, πQ join-singularn with respect to πQ We proceed by induction. We first prove (S0). Let p ∈ Q, and let p be join-singular0 with respect to a, b. Thus, ap ∨ bp = v2p , 0 0 0 0 that is, (πQ a)p ∨ (πQ b)p = v2p , and, say, bp ≤ ap , that is, (πQ b)p ≤ (πQ a)p , and there is a q ∈ P with p%2 q, and with bq aq . Then q ∈ Q, since Q is a %2 -up0 0 0 0 set, and so (πQ b)q (πQ a)q . Thus p is join-singular0 with respect to πQ a, πQ b, establishing (S0). Now, for the inductive step, let n > 0 and assume that (S(n − 1)). Let p ∈ Q 0 0 be join-singularn with respect to a, b. Then, (πQ a)p ∨ (πQ b)p = ap ∨ bp = v2p , and there is a q with p %2 q such that q is join-singularn−1 with respect to a, b. Then q ∈ Q, since Q is a %2 -up-set, and, by statement (S(n − 1)), q is then join0 0 singularn−1 with respect to πQ a, πQ b. So p is then join-singularn with respect to 0 0 πQ a, πQ b, thereby establishing the inductive step. Thus (Sn) holds for all n ≥ 0, concluding the proof. 0 Corollary. Let Q ⊆ P be a %2 -up-set. Then πQ : TP → TQ is a lattice surjection. 0 Proof. By duality, it suffices to show that πQ preserves join. 0 0 Let a, b ∈ TP and p ∈ Q; consider (πQ a ∨ πQ b)p . If p is join-regular with respect to a, b, then, by Lemma 21, p is join-regular 0 0 with respect to πQ a, πQ b. Then 0 0 0 0 0 (πQ a ∨ πQ b)p = (πQ a)p ∨ (πQ b)p = ap ∨ bp = (a ∨ b)p = (πQ (a ∨ b))p ,

if p ∈ Q is join-regular with respect to a, b. If p ∈ Q is join-singular with respect to a, b, then p is join singular with respect 0 0 to πQ a, πQ b. So 0 0 0 (πQ a ∨ πQ b)p = 10p = (a ∨ b)p = (πQ (a ∨ b))p ,

if p ∈ Q is join-singular with respect to a, b.


12

0 0 0 Consequently, for all p ∈ Q, (πQ a ∨ πQ b)p = (πQ (a ∨ b))p . That is, 0 0 0 πQ a ∨ πQ b = πQ (a ∨ b). 0 So, πQ preserves join and, thus, by duality, is a lattice homomorphism.

For each {%1 , %2 }-down-set Q ⊆ P , the set P − Q is a {%1 , %2 }-up-set, and so we have the congruence ΘQ = Ker πP −Q on SP . That is, a ≡ b (ΘQ ) iff ap = bp , for all p ∈ / Q. Similarly, for each %2 -down-set Q ⊆ P , we have the congruence Θ0Q = Ker πP0 −Q on TP , whereby a ≡ b (Θ0Q ) iff ap = bp , for all p ∈ / Q. Lemma 22. For each {%1 , %2 }-down-set Q ⊆ P , the congruence ΘQ is an isoform congruence on SP . Proof. The congruence classes of ΘQ are precisely the subsets ηx (SQ ), where x ranges over SP −Q , which, by the Corollary to Lemma 18, are lattices isomorphic to the lattice SQ . Similarly, by the Corollary to Lemma 19, we have: Lemma 23. For each %2 -down-set Q ⊆ P , the congruence Θ0Q is an isoform congruence on TP . We now show that if all the lattices Sp are simple, then the congruences of SP are precisely the congruences ΘQ , as Q ranges over the {%1 , %2 }-down-sets of P . Similarly, if all the lattices Tp are simple, then the congruences of TP are precisely the congruences Θ0Q as Q ranges over the %2 -down-sets of P . We denote by 0 the element h0p ip∈P of SP , and by 00 the element h00p ip∈P of TP . The elements 0 and 00 are the zero of the respective lattices SP and TP . For each Q ⊆ P , we define uQ ∈ SP by setting ( 1p , if p ∈ Q; Q (u )p = 0p , if p ∈ / Q. Note that u∅ = 0. If Q is the singleton {q}, then we write uq for u{q} . Similarly, u0Q ∈ TP is defined by setting ( 10p , if p ∈ Q; 0Q (u )p = 00p , if p ∈ / Q, and we write u0q for u0{q} . Observe that, just from looking at the order ≤, _ (6) uQ = ( up | p ∈ Q ), and (7)

u0Q =

_ ( u0p | p ∈ Q ).

Lemma 24. Let Q ⊆ P be a {%1 , %2 }-down-set, and let a ∈ SP . Then, for each p ∈ P, ( (a ∨ uQ )p =

1p , ap ,

if p ∈ Q; if p ∈ / Q.


13

Proof. Now, a ∨ uQ ≥ uQ . So if p ∈ Q, then (a ∨ uQ )p ≥ (uQ )p = 1p , that is, (a ∨ uQ )p = 1p . Since Q is a {%1 , %2 }-down-set, πP −Q (a ∨ uQ ) = πP −Q a ∨ πP −Q uQ = πP −Q a ∨ 0 = πP −Q a, where 0 here denotes the zero of SP −Q . That is, (a ∨ uQ )p = ap , if p ∈ / Q.

Exactly the same way, we have: Lemma 25. Let Q ⊆ P be a %2 -down-set, and let a ∈ TP . Then, for each p ∈ P , ( 10p , if p ∈ Q; 0Q (a ∨ u )p = ap , if p ∈ / Q. We shall also need the following trivial observation: Lemma 26. Let a, b ∈ SP and let p ∈ P . If ap 6= bp , then (a ∨ b)p > (a ∧ b)p . Proof. By Definitions 8 and 15, (a ∨ b)p ≥ ap ∨ bp > ap ∧ bp ≥ (a ∧ b)p .

Recall that the Tp , p ∈ P , were chosen to each contain more than three elements. Lemma 27. Let p ∈ P , and let Sp be a simple lattice. Let Θ be a congruence on SP such that there are a, b ∈ SP with ap 6= bp and with a ≡ b (Θ). Then 0 ≡ up (Θ). Proof. We note that, by the Corollary to Lemma 17, η0 (Sp ) (where by slight abuse of notation, 0 here is the zero of SP −{p} ), is a sublattice of SP with zero 0 and unit up ; hence it is isomorphic to Sp , and so it is simple. By Lemma 26, we may take b < a. There are two cases to consider: the case where ap 6= v1p , v2p and the case where ap = vip , for some i ∈ {1, 2}. Case 1: ap 6= v1p , v2p . Now, p is meet-regular with respect to a, up since ap ∧ (up )p = ap 6= v1p , v2p . So, (b ∧ up )p ≤ bp < ap = (a ∧ up )p . Thus b ∧ up and a ∧ up are distinct, they are congruent modulo Θ, and they lie in the simple sublattice η0 (Sp ) with zero 0 and unit up . Therefore, 0 ≡ up Case 2: ap =

vip ,

(Θ).

for some i ∈ {1, 2}. Then, since bp < ap , certainly bp ≤ 00p .

Since v2p is doubly irreducible in the simple lattice Sp and |Tp | > 3, there is a w ∈ Tp distinct from 00p , 10p , and v2p . Then bp ∨ w = w and ap ∨ w ≥ 10p . Set ( w, if q = p; wq = 0q , otherwise. Then, b ∨ w ≤ a ∨ w, b ∨ w ≡ a ∨ w (Θ), and (b ∨ w)p = wp = w < 10p ≤ (a ∨ w)p . Then, (a ∨ w)p 6= vip for i ∈ {1, 2}. So, replacing a by a ∨ w and b by b ∨ w, we are back in Case 1, concluding the proof.


14

Exactly the same way (except we need not consider v1p ), we get: Lemma 28. Let p ∈ P , and let Tp be a simple lattice. Let Θ be a congruence on TP such that there are a, b ∈ TP with ap 6= bp and with a ≡ b (Θ). Then 00 ≡ u0p (Θ). For each p ∈ P , we define vp ∈ SP by setting ( v1p , if q = p; (vp )q = 0q , otherwise, and we define v0p ∈ TP by setting ( v2p , if q = p; (v0p )q = 00q , otherwise. Lemma 29. Let p, q ∈ P . If q %1 p, then vq ∨ up = u{p,q} . Proof. If p = q, then, since vp ≤ up , we get vq ∨ up = vp ∨ up = up = u{p,q} . So we may now assume that p 6= q. For each p0  q  v1 , (vq )p0 ∨ (up )p0 = 1p ,   0 0p ,

∈ P, if p0 = q; if p0 = p; otherwise.

Thus, (vq ∨ up )p0 = (u{p,q} )p0 if p0 6= q. Now, (vq )q ∨ (up )q = v1q , q %1 p, (up )q ≤ (vq )q , and (up )p (vq )p , that is, q is 1-join-singular with respect to vq , up . So (vq ∨ up )q = 1q = (u{p,q} )q also. Thus, vq ∨ up = u{p,q} . Similarly, we have: Lemma 30. Let p, q ∈ P . If q %2 p, then v0q ∨ u0p = u0{p,q} . Lemma 31. Let Sp be a simple lattice, for each p ∈ P . Then the congruences of SP are precisely the ΘQ , as Q ranges over the {%1 , %2 }-down-sets in P . Proof. If Q is a {%1 , %2 }-down-set, then ΘQ is a congruence. Now, let Θ be a congruence on SP . Set Q(Θ) = { p ∈ P | a ≡ b

(Θ) and ap 6= bp , for some a, b ∈ SP }

We first show that Q(Θ) is a {%1 , %2 }-down-set. Let p ∈ Q(Θ) and let q %1 p. By Lemma 27, 0 ≡ up (Θ) and so, by Lemma 29, q v ≡ u{p,q} (Θ). Since (vq )q = v1p 6= 1q = (u{p,q} )q , we conclude that q ∈ Q(Θ). Similarly, let q %2 p. By Lemma 28, 00 ≡ u0p (Θ) and so, by Lemma 30, 0q v ≡ u0{p,q} (Θ). Since (v0q )q = v2p 6= 10q = (u0{p,q} )q , we conclude that q ∈ Q(Θ). Thus, Q(Θ) is a {%1 , %2 }-down-set, and so we have the congruence ΘQ(Θ) . We now show that Θ = ΘQ(Θ) . If a ≡ b (Θ), then, by the definition of Q(Θ), it follows immediately that ap = bp , for all p ∈ / Q(Θ), that is, that a ≡ b (ΘQ(Θ) ). Thus Θ ≤ ΘQ(Θ) . Now, let a ≡ b (ΘQ(Θ) ). So, ap = bp if p ∈ / Q(Θ).


15

Then, by Lemma 24, a ∨ uQ(Θ) = b ∨ uQ(Θ) . By Lemma 27, 0 ≡ up (Θ), for all p ∈ Q(Θ). Then, by (6), 0 ≡ uQ(Θ) (Θ) (note that Q(Θ) is finite). Thus, a ≡ a ∨ uQ(Θ)

(Θ)

and

b ≡ b ∨ uQ(Θ)

(Θ).

So, a ≡ b (Θ), showing that ΘQ(Θ) ≤ Θ. Thus Θ = ΘQΘ) for the {%1 , %2 }-down-set Q(Θ), concluding the proof.

Similarly, we have: Lemma 32. Let Tp be a simple lattice, for each p ∈ P . Then the congruences of TP are precisely the Θ0Q as Q ranges over the %2 -down-sets in P . 4. The main technical result Now let P1 and P2 be (disjoint) finite sets with relations σ1 and σ2 , respectively. We also assume that there is a mapping ψ : P2 → P1 that preserves the relations, that is, that, given p, q ∈ P2 with p σ2 q, then (ψp) σ1 (ψq). We set P = P1 ∪ P2 , and we define relations %1 and %2 on P by setting (8)

%2 = σ2

and (9)

%1 = σ1 ∪ σ2 ∪ { hp, ψ(p)i, hψ(p), pi | p ∈ P2 }.

Note that %2 ⊆ %1 , so the concept of a {%1 , %2 }-down-set is identical to that of a %1 -down-set. We assume that we have the lattices Sp with their sublattices Tp , for p ∈ P , as above. We alsoQassume that all the lattices Sp and Tp are simple. Q Q We form the lattices SP1 = ( Sp | p ∈ P1 ), SP = ( Sp | p ∈ P ), and TP2 = ( Tp | p ∈ P2 ). Now, congruence relations on SP1 and TP2 are determined by down-sets of P1 under the restriction of %1 and %2 to P1 and of down-sets of P2 under the restriction of %2 to P2 , respectively. We use the convention that if P 0 ⊆ P , then a %i -downset Q of P 0 is a subset Q ⊆ P 0 that is a down-set of P 0 under the restriction of %i to P 0 . Lemma 33. (a) If Q is a %1 -down-set of P , then ψ −1 (Q ∩ P1 ) is a ρ2 -down-set of P2 . (b) If Q1 is a ρ1 -down-set of P1 , then Q1 ∪ ψ −1 (Q1 ) is a %1 -down-set of P . (c) If Q is a %1 -down-set of P , then Q ∩ P2 = ψ −1 (Q ∩ P1 ). Proof. (a) Let p, q ∈ P2 with p ∈ ψ −1 (Q ∩ P1 ) and q ρ2 p; we must show that q ∈ ψ −1 (Q ∩ P1 ). Now, q σ2 p; thus, ψ(q) σ1 ψ(p), that is, ψ(q) ρ1 ψ(p). As Q is a %1 -down-set of P and ψ(p) ∈ Q, it follows that ψ(q) ∈ Q ∩ P1 . So we conclude that q ∈ ψ −1 (Q ∩ P1 ). (b) Let q %1 p with p ∈ Q1 ∪ ψ −1 (Q1 ); we must show that q ∈ Q1 ∪ ψ −1 (Q1 ). First, let p ∈ Q1 . Then either q ∈ P1 , in which case, q ∈ Q1 , since Q1 is a %1 -down-set of P1 . Or q ∈ P2 , so that q %1 p implies that q = ψ(p) and so q ⊆ ψ −1 (Q1 ). Second, let p ∈ ψ −1 (Q1 ), so that ψ(p) ∈ Q1 . If q ∈ P1 , then q %1 p implies that q = ψ(p) ∈ Q1 . If q ∈ P2 , then q σ2 p, so that ψ(q) %1 ψ(p) and, therefore, ψ(q) ∈ Q1 . We conclude that q ∈ ψ −1 (Q1 ).


16

(c) First, let p ∈ Q ∩ P2 . Since ψ(p) %1 p, it follows that ψ(p) ∈ Q ∩ P1 , and so p ∈ ψ −1 (Q ∩ P1 ). Second, let p ∈ ψ −1 (Q ∩ P1 ), so that ψ(p) ∈ Q ∩ P1 . Since p %1 ψ(p), we have that p ∈ Q and so p ∈ Q ∩ P2 . We have lattice embeddings η1 : SP1 → SP , given by η1 = η0 , where 0 here denotes the zero of SP2 , and η2 : TP2 → SP , given by η2 = η00 , where 00 here denotes the zero of TP1 . Theorem 34. Under the conventions of this section, the lattices SP , SP1 , and TP2 are all isoform lattices. The embedding η1 : SP1 → SP is a congruence-preserving extension. For each %1 -down-set Q of P , res(η2 )ΘQ = Θ0ψ−1 (Q∩P1 ) . Proof. The lattices SP , SP1 , and TP2 are all isoform by Lemmas 22, 23, 31, and 32. Next, we show that the embedding η1 is a congruence-preserving extension. If Q is a %1 -down-set of P , then Q ∩ P1 is a %1 -down-set of P1 , and (10)

Q = (Q ∩ P1 ) ∪ ψ −1 (Q ∩ P1 ),

since Q ∩ P2 = ψ −1 (Q ∩ P1 ) by Lemma 33. Furthermore, if Q1 is a %1 -down-set of P1 , then Q1 ∪ ψ −1 (Q1 ) is a %1 -down-set of P . Thus the mapping Q 7→ Q ∩ P1 from the %1 -down-sets of P to the %1 -down-sets of P1 is a bijection. Now, for each %1 -down-set Q of P , we have res(η1 )ΘQ = ΘQ∩P1 . Thus, by Lemma 31, res(η1 ) is a bijection, that is, η1 is a congruence-preserving extension, as claimed. Finally, if Q is a %1 -down-set of P , then res(η2 )ΘQ = Θ0Q∩P2 = Θ0ψ−1 (Q∩P1 ) , since Q ∩ P2 = ψ −1 (Q ∩ P1 ) by Lemma 33.

5. The proof of Theorem 2 We recall the duality between finite distributive lattices and finite ordered sets. Given a finite distributive lattice D, we consider the ordered set Join D of join irreducible elements of D. There is an isomorphism between D and the lattice of ≤-down-sets of Join D, whereby a ∈ D corresponds to { p ∈ D | p ≤ a }, and, W conversely, the ≤-down-set Q corresponds to Q ∈ D. Furthermore, given a {0, 1}preserving homomorphism ϕ : D1 → D2 of finite distributive V lattices, we have the associated isotone map ψ : Join D2 → Join D1 given by ψp = ( x ∈ D1 | ϕx ≥ p ). Then ϕ corresponds to ψ −1 on the ≤-down-sets of Join D1 : for each a ∈ D1 , we W −1 have ϕa = (ψ { p ∈ Join D1 | p ≤ a }). Now, under the hypotheses of Theorem 2, let P1 = Join D1 with relation σ1 the order ≤ on P1 , and let P2 = Join D2 with the relation σ2 the order ≤ on P2 . Then, as above, ϕ : D1 → D2 determines ψ : P2 → P1 satisfying (ψp) σ1 (ψq), for all p, q ∈ P2 with p σ2 q. As in Section 4, set P = P1 ∪ P2 with the relations %2 and %1 defined on P by (8) and (9) there. For each p ∈ P , let Sp be a finite simple lattice with separator v1p and with a simple ideal Tp with separator v2p such that v1p ∈ / Tp . For example, each Tp can be isomorphic to M3 and each Sp can be isomorphic to the lattice depicted in Figure 1.


17

We then have the lattices SP1 , SP , and TP2 of Section 4. Then η2 embeds TP2 as an ideal in SP . Also, by Theorem 34, SP and TP2 (as well as SP1 ) are isoform lattices.

1p

v1p

v2p

Figure 1. The lattice Sp for Theorem 2. Now, we have an isomorphism ε0 : Con SP1 → D1 whereby ε0 : ΘQ 7→ Q a %1 -down-set of P1 , and an isomorphism

W

Q for

ε2 : Con TP2 → D2 , Θ0Q

whereby ε2 : 7→ Q for Q a %2 -down-set of P2 . By Theorem 34, res(η1 ) : Con SP → Con SP1 is an isomorphism. Thus we have an isomorphism ε1 = ε0 ◦ res(η1 ) : Con SP → D1 . Furthermore, for each %1 -down-set Q of P , W

res(η2 )ΘQ = Θ0ψ−1 (Q∩P1 ) . Then ε2 res(η2 )ΘQ = ε2 Θ0ψ−1 (Q∩P1 ) =

_

ψ −1 (Q ∩ P1 ) = ϕ

_ (Q ∩ P1 ) = ϕε0 ΘQ∩P1

= ϕε0 res(η1 )ΘQ = ϕε1 ΘQ . Since any congruence Θ of SP is of the form ΘQ , for some %1 -down-set Q of P , we thereby have ε2 res(η2 )Θ = ϕε1 Θ, for all Θ ∈ Con SP . Setting L1 = SP , L2 = TP2 , and η = η2 , we get Theorem 2. 6. The proof of Theorem 1 For each i ∈ {1, 2}, set Pi = Join(Con Ki ), the ordered set of join-irreducible congruences of Ki . Then the {0, 1}-preserving homomorphism ϕ : Con K1 → Con K2 determines, as in Section 5, an isotone map ψ : P2 → P1 with _ ϕΘ = ψ −1 { p ∈ P1 | p ≤ Θ }, for each Θ ∈ Con K1 . For each p ∈ Pi , set (11)

Ψ(p) =

_ ( Θ ∈ Con Ki | p Θ ).

Clearly, p Ψ(p). Lemma 35. For each p ∈ Pi , the congruence Ψ(p) ∈ Con Ki is meet-irreducible.

18


Proof. Let Θ1 , Θ2 ∈ Con Ki with Ψ(p) < Θ1 , Θ2 . Then there are p1 , p2 ∈ Pi with p1 ≤ Θ1 , p2 ≤ Θ2 , and with p1 , p2 Ψ(p). But, then, p ≤ p1 , p2 , that is, p ≤ p1 ∧ p2 ≤ Θ1 ∧ Θ2 . But p Ψ(p). Thus Ψ(p) 6= Θ1 ∧ Θ2 . Lemma 36. For each congruence Θ of Ki , ^ Θ = ( Ψ(p) | p ∈ Pi , p Θ ). Proof. By (11), Θ ≤ Ψ(p) for all p Θ, that is, ^ Θ ≤ ( Ψ(p) | p ∈ Pi , p Θ ). V Conversely, let q ∈ Pi , and let q ≤ ( Ψ(p) | p ∈ Pi , p Θ ), that is, q ≤ Ψ(p), for all p Θ. Thus, for each p Θ and all Ψ ∈ Con Ki with p Ψ, we have q ≤ Ψ. Then, taking Ψ = Θ, we get q ≤ Θ. Thus, ^ ( Ψ(p) | p ∈ Pi , p Θ ) ≤ Θ as well.

Lemmas 35 and 36 are standard results in the duality theory of finite distributive lattices, and the mapping Ψ is the standard isomorphism between the ordered set of join-irreducible elements of a finite distributive lattice and the ordered set of its meet-irreducible elements. By Lemma 36, Lemma 37. If Θ ∈ Con Ki and a, b ∈ Ki , then a ≡ b (Θ) iff a ≡ b (Ψ(p)), for all p ∈ Pi with p Θ. By Lemma 35, for each i ∈ {1, 2} and each p ∈ Pi , the congruence Ψ(p) is meetirreducible and the quotient lattice Kip = Ki /Ψ(p) is thus subdirectly irreducible. It is then easy to embed Kip in a bounded simple lattice Tp with a separator v2p , as follows. We add a zero 00P and a unit 10p to Kip . Since Kip is subdirectly irreducible, it contains elements b < a such that any nontrivial congruence collapses a and b. We add a new element c with 00p ≺ c ≺ a, and add two separators d, v2p with 00p ≺ d, v2p ≺ 10p (see Figure 2.). Then Tp is a simple extension of Kip with a separator v2p . We next extend Tp to get Sp by adding a new unit 1p and separators e, v1p with 00p ≺ e, v1p ≺ 1p . Then Sp is a simple lattice with zero 0p = 00p and unit 1p > 10p (see Figure 3.) that contains Tp as an ideal. As in Section 5, set σ1 on P1 to be the order ≤ on P1 , and set σ2 on P2 to be the order ≤ on P2 . We set P = P1 ∪ P2 and define the relations %2 and %1 on P by (8) and (9) of Section 4. We then have the lattices SP1 , SP , and TP2 , and lattice embeddings η1 : SP1 → SP and η2 : TP2 → SP , where η2 embeds TP2 as an ideal of SP . By Theorem 34, SP and TP2 (as well as SP1 ) are isoform lattices, η1 is a congruence-preserving extension, and (12)

res(η2 )ΘQ = Θ0ψ−1 (Q∩P1 ) ,

for each %1 -down-set Q of P . Now, we have the mappings Diag1 : K1 → SP1 , defined by setting (Diag1 a)p = a/Ψ(p) ∈ K1p ⊆ Sp , for each p ∈ P1 , and Diag2 : K2 → TP2 ,


19

Figure 2. The lattice Tp for Theorem 1.

Figure 3. The lattice Sp for Theorem 1.

defined by setting (Diag2 a)p = a/Ψ(p) ∈ K2p ⊆ Tp , for each p ∈ P2 . Lemma 38. The mapping Diag1 : K1 → SP1 is a congruence-preserving extension whereby _ (13) res(Diag1 )ΘQ = Q ∈ Con K1 , for each %1 -down-set Q of P1 = Join(Con K1 ). Proof. First, Diag1 is a homomorphism. Since (Diag1 a)p ∈ K1p , for all p ∈ P1 , and v1p , v2p ∈ / K1p , each p ∈ P1 is join-regular and meet-regular with respect to Diag1 a,


20

Diag1 b, for any a, b ∈ K1 . Thus, for all p ∈ P1 , (Diag1 a ∨ Diag1 b)p = (Diag1 a)p ∨ (Diag1 b)p = a/Ψ(p) ∨ b/Ψ(p) = (a ∨ b)/Ψ(p) = (Diag1 (a ∨ b))p . That is, Diag1 preserves ∨. Thus, by duality, Diag1 is a homomorphism. Next, each congruence of SP1 is of the form ΘQ for a uniquely determined %1 down-set Q of P1 , and, for a, b ∈ K1 , Diag1 a ≡ Diag1 b (ΘQ ) iff a ≡ b (Ψ(p)),

for all p ∈ P1 with p ∈ /Q

a ≡ b (Ψ(p)),

for all p ∈ P1 with p

iff _

Q

iff _ a ≡ b ( Q) by Lemma 37, thereby establishing (13). Now, res(Diag1 ) is then surjective since W Θ = ( p ∈ P1 | p ≤ Θ ), for each Θ ∈ Con KW 1 , and res(Diag1 ) is injective since there is only one %1 -down-set Q of P1 with Θ = Q, the set Q = { p ∈ P1 | p ≤ Θ }. Thus Diag1 is a congruence-preserving extension, concluding the proof. By virtually the same proof, we have: Lemma 39. The mapping Diag2 : K2 → TP2 is a congruence-preserving extension whereby _ (14) res(Diag2 )Θ0Q = Q ∈ Con K2 , for each %2 -down-set Q of P2 = Join(Con K2 ). Now, set L1 = SP , and set L2 = TP2 . Then, as noted above, L1 and L2 are isoform lattices. Define ε1 : K1 → L1 , by setting ε1 = η1 ◦ Diag1 , define ε2 : K2 → L2 , by setting ε2 = Diag2 , and define η : L2 → L1 , by setting η = η2 . Then ε1 is a congruencepreserving extension, since η1 and Diag1 both are, and ε2 (= Diag2 ) is a congruencepreserving extension. Furthermore, η embeds L2 as an ideal in L1 . Now, let Θ ∈ Con L1 . Then Θ = ΘQ , for a uniquely determined %1 -down-set Q of P . Then res(ε2 )res(η)Θ = res(ε2 )res(η2 )ΘQ = res(ε2 )Θ0ψ−1 (Q∩P1 ) by (12), and res(ε2 )Θ0ψ−1 (Q∩P1 ) = res(Diag2 )Θ0ψ−1 (Q∩P1 ) =

_

ψ −1 (Q ∩ P1 )


21

by (14) of Lemma 39. Now, _ _ ψ −1 (Q ∩ P1 ) = ϕ (Q ∩ P1 ) = ϕres(Diag1 )ΘQ∩P1 by (13) of Lemma 38. Calculating further, res(Diag1 )ΘQ∩P1 = res(Diag1 )res(η1 )ΘQ = res(η1 ◦ Diag1 )ΘQ = res(ε1 )ΘQ = res(ε1 )Θ. Thus,

res(ε2 )res(η)Θ = ϕres(ε1 )Θ, for all Θ ∈ Con L1 , concluding the proof of Theorem 1. References [1] G. Gr¨ atzer, The Congruences of a Finite Lattice, A Proof-by-Picture Approach. Birkh¨ auser Boston, 2006. xxiii+281 pp. ISBN: 0-8176-3224-7. [2] G. Gr¨ atzer and H. Lakser, Homomorphisms of distributive lattices as restrictions of congruences, Canad. J. Math. 38 (1986), 1122–1134. [3] G. Gr¨ atzer, H. Lakser, and R. W. Quackenbush, Congruence-preserving extensions of congruence-finite lattices to isoform lattices Acta Sci. Math. (Szeged), to appear. [4] G. Gr¨ atzer and E. T. Schmidt, Finite lattices with isoform congruences, Tatra Mt. Math. Publ. 27 (2003), 111–124. [5] G. Gr¨ atzer, R. W. Quackenbush, and E. T. Schmidt, Congruence-preserving extensions of finite lattices to isoform lattices, Acta Sci. Math. (Szeged) 70 (2004), 473–494. Department of Mathematics, University of Manitoba, Winnipeg, MB R3T 2N2, Canada E-mail address, G. Gr¨ atzer: [email protected] URL, G. Gr¨ atzer: http://server.math.umanitoba.ca/homepages/gratzer/ E-mail address, H. Lakser: [email protected]