5. Orienta a la organización hacia grandes desafíos La búsqueda de las mejores prácticas que se usan en el entorno lleva a la organización a plantearse
3.1.8 Identificar en qué actividades hacer benchmarking El benchmarking es un instrumento muy versátil Como señala Roland Loesser, un alto ejecutivo de
Lemma 6.4.1(Interpretation of finite SAJandSAMalgebras).
Fix any finite join-semilattice Q∈JSLf and endofunction σ∶Q→Q.
1. (Q, σ)∈SAJf iffσdefines a self-adjoint JSLf-morphism of type Q→Qop, or equivalently:
q2≤Qσ(q1) ⇐⇒ q1≤Qσ(q2) for allq1, q2∈Q.
2. (Q, σ)∈SAMf iffσdefines a self-adjoint JSLf-morphism of typeQop→Q, or equivalently:
σ(q1)≤Qq2 ⇐⇒ σ(q2)≤Qq1 for allq1, q2∈Q. Proof.
1. Given a self-adjointJSLf-morphismσ∶Q→Qopthen it certainly defines a monotone morphism(Q,≤Q)→(Q,≥Q), hence(Revσ)holds. The self-adjoint relationship informs us that:
q2≤Qσ(q1) ⇐⇒ σ(q1)≤Qopq2 ⇐⇒ q1≤Qσ(q2).
Consequently(Exσ2) holds, because for everyx∈ Q we haveσ(x)≤Qσ(x) ⇐⇒ x≤Q σ(σ(x)). Conversely,
suppose we have a functionσ∶Q→Qsatisfying(Revσ)and(Exσ2). Then∀q1, q2∈Qwe have:
q2≤Qσ(q1) Ô⇒ σ(σ(q1))≤Qσ(q2) Ô⇒ q1≤Qσ(q2)
using the order-reversing monotonicity ofσand also q1≤Qσ(σ(q1)). Then by symmetry we have:
σ(q1)≤Qop q2 ⇐⇒ q1≤Qσ(q2) for allq1, q2∈Q.
By Lemma 2.2.7.2 and the fact that Q has all finite joins, we deduce that σdefines a JSLf-morphism of type Q→Qop. Finally, the above equivalence informs us thatσis self-adjoint i.e.σ=σ∗.
2. We have(Q, σ)∈SAMf if and only if(Qop, σ)∈SAJf by Lemma6.2.2, so apply (1).
Note 6.4.2 (Interpretation of infiniteSAJandSAMalgebras).
The above interpretation does not extend to infinite algebras e.g. becauseQop needn’t be a well-defined join-semilattice in the infinite case.
1. Given any join-semilattice Q∈JSL then we have theSAM-algebra(Q, σ)whereσ(q)∶=Q for allq∈Q. Indeed,
(Revσ) holds trivially for this constant map, as does (Cxσ2) because σ(σ(x)) = Q ≤Q x. Thus there are
SAM-algebras whose join-semilattice has no top and/or fails to have binary meets, see Definition 2.2.1.12.d. 2. ConcerningSAJ-algebras(Q, σ), it so happens thatQalways has the top elementσ(Q)because:
Q≤Qσ(q) ⇐⇒ q≤Qσ(Q)
via the adjoint relationship. However,Qneedn’t have binary meets e.g. letQbe the join-semilattice depicted in
Definition2.2.1.12.d and defineσ(q)∶=⊺for allq∈Q. ∎
Lemma 6.4.3(Interpretation of arbitrarySAI-algebras).
1. If(Q, σ)∈SAIthenQis a bounded lattice and σdefines a self-adjointJSL-isomorphismQ→Qop (hence bounded lattice isomorphism) and also its inverse. Furthermore,Qhas the following meet structure:
2. Given anyQ∈JSLand functionσ∶Q→Qthen t.f.a.e. (a) (Q, σ)∈SAI.
(b) σdefines a self-adjoint JSL-isomorphismQ→Qop.
(c) σdefines a self-adjoint JSL-morphism of typeQ→Qop andQop→Q. (d) σis involutive and defines a JSL-morphism of type Q→Qop. Furthermore, in (b) and (d) one may replaceQ→Qop with Qop→Q. 3. EverySAI-morphism defines a bounded lattice morphism.
Proof.
1. Let(Q, σ)be anSAI-algebra. Givenq2≤Qσ(q1)then applyingσyieldsq1=σ(σ(q1))≤Qσ(q2), so that: (⋆) σ(q1)≤Qopq2 ⇐⇒ q2≤Qσ(q1) ⇐⇒ q1≤Qσ(q2) for allq1, q2∈Q.
Since Q has all finite joins, we may apply Lemma 2.2.7.2 to deduce that σ(⋁QX)= ⋀Qσ[X] for every finite subsetX ⊆ω Q i.e. these particular meets exist in Q. But since σ is involutive it is bijective, hence Q has all finite meets. It follows thatQis a bounded lattice andσdefines a join-semilattice morphismσ∶Q→Qop. Now, since σis involutive it is bijective and thus a JSL-isomorphism by universal algebra, and also a bounded lattice isomorphism becauseQ andQop are bounded lattices. Again by involutivenessσ−1(q)=σ(q). Observe
that the join-semilattice isomorphismσ∶Q→Qop between bounded lattices is self-adjoint by(⋆). Concerning Q’s meet structure, sinceσ∶Q→Qop is a bounded lattice isomorphism we have⊺Q=σ(Q), and finally:
σ(σ(q1)∨Qσ(q2))=σ(σ(q1))∧Qσ(σ(q2))=q1∧Qq2.
2. • (a ⇐⇒ b): The implication ⇒ follows by (1). Conversely, (Revσ) follows by taking the underlying
monotone map ofσ∶Q→Qop whereas(Invσ)holds because for everyq∈Qwe have:
σ(q) =σ∗(q) by self-adjointness
=⋁Q{q′∈Q∶σ(q′)≤Qop q} by definition
=⋁Q{q′∈Q∶q≤Qσ(q′)}
=⋁Q{q′∈Q∶q′≤Qσ−1(q)} apply isomorphism =σ−1(q).
• (a ⇐⇒ c): Regarding⇒, this follows becausea⇒b, and the inverse of a self-adjoint isomorphism is itself
self-adjoint. Conversely, taking an underlying monotone map yields(Revσ), and concerning involutiveness we can use the two adjoint relations to deduce that for everyq∈Q:
σ(q)≤Qσ(q) ⇐⇒ q≤Qσ(σ(q)) and σ(q)≤Qσ(q) ⇐⇒ σ(σ(q))≤Qq.
• (a ⇐⇒ d): First of all, ⇒ follows because a⇒b. The converse is immediate by taking the underlying
monotone morphism.
3. Follows because the top and binary meet are definable in terms of , ∨ andσ by (1), hence are preserved by algebra homomorphisms.
Corollary 6.4.4 (SAIis the variety of De Morgan algebras).
By identifying (Q, σ)∈SAIwith the tuple (Q,∨Q,Q,∧Q,⊺Q, σ), the categorySAIis precisely the variety of De Morgan
algebras i.e. bounded lattices L equipped with a unary operationσ∶L→Lsatisfying the equations: σσ(x)≈x σ(x∨y)≈σ(x)∧σ(y) σ(x∧y)≈σ(x)∨σ(y).
Proof. Given (Q, σ)∈SAI then the induced tuple(Q,∨Q,Q,∧Q,⊺Q, σ)is a de morgan algebra by(Invσ)and Lemma
6.4.3.1. Conversely, any de morgan algebra defines an SAI-algebra because (Invσ) by assumption, and moreover
σ(x∨y)≈σ(x)∧σ(y)implies(Revσ):
x≼y ⇐⇒ x∨y≈y Ô⇒ σ(x∨y)≈σ(y) Ô⇒! σ(x)∧σ(y)≈σ(y) ⇐⇒ σ(y)≼σ(x).
The homomorphisms of De Morgan algebras are the bounded lattice morphisms which preserve the unary operation, and using Lemma6.4.3.3 they are precisely theSAI-morphisms.
Corollary 6.4.5 (Order-dual algebras).
1. Given(Q, σ)∈SAJf then (Qop, σ)∈SAJf iff (Q, σ)∈SAIf.
2. Given(Q, σ)∈SAMf then(Qop, σ)∈SAMf iff(Q, σ)∈SAIf.
3. Given(Q, σ)∈SAI thenσdefines an SAI-isomorphism(Q, σ)→(Qop, σ)and also its inverse.
Proof.
1. Given(Q, σ)∈SAIf then(Qop, σ)∈SAIf ⊆SAJf because(Invσ)continues to hold, as does(Revσ)by considering the opposite monotone morphism. Conversely, if(Qop, σ)∈SAJf thenσdefines a self-adjoint morphism of type Q→Qop andQop→Qand hence an SAIf-algebra by Lemma6.4.3.2.
2. Follows because(Q, σ)∈SAMf iff(Qop, σ)∈SAJf.
3. σdefines aJSLf-isomorphismQ→Qop by Lemma6.4.3, and also anSAIf-morphism because σ○σ=σ○σ.
We can also characterise the finiteSAJf andSAMf-algebras in terms of the finite de morgan algebras. Lemma 6.4.6(SAJf andSAMf-algebras as extensions of finite De Morgan algebras).
1. Given any(Q, σ)∈SAJf then we have (σ[Q], σ∣σ[Q]×σ[Q])∈SAIf, in fact:
σ= Qσ∣Q↠×σ[Q]σ[Q]ÐÐÐÐÐÐ→σ∣σ[Q]×σ[Q] (σ[Q])op(σ∣Q×σ[Q])∗ ↣ Qop. 2. Given any(Q, σ)∈SAMf then(σ[Q], σ∣σ[Q]×σ[Q])∈SAIf and moreover:
σ= Qopσ∣Q↠×σ[Q]σ[Qop]ÐÐÐÐÐÐσ∣σ[Q]×σ[Q→] (σ[Qop])op(σ∣Q×σ[Q])∗ ↣ Q. 3. Consequently,
– (Q, σ)∈SAJf iff there exists (R, σ0)∈SAIf and aJSLf-morphism α∶Q→Rsuch thatσ=α∗○σ0○α. – (Q, σ)∈SAMf iff there exists(R, σ0)∈SAIf and aJSLf-morphism β∶R→Qsuch that σ=β○σ0○β∗.
Proof.
1. By Lemma 6.4.1 we know σ defines a join-semilattice morphism σ ∶ Q → Qop. By the (surjection,inclusion) factorisation we have the join-semilattice inclusion-morphism σ[Q]↪ Qop which restricts to a join-semilattice endomorphismσσ[Q]×σ[Q] of typeσ[Q]→σ[Q]. By restriction it satisfies(Revσ), whereas for anyσ(x)∈σ[Q],
σσ[Q]×σ[Q]○σσ[Q]×σ[Q](σ(x))=σσσ(x)=σ(x)
by Lemma6.2.2.3, so that(Invσ)holds. Consequently(σ[Q], σ∣σ[Q]×σ[Q])is a finite de morgan algebra.
Next, the surjective join-semilattice morphism σ∣Q×σ[Q] arises from the other part ofσ’s (surjection,inclusion)
factorisation. Then the respective composite is a well-defined join-semilattice morphism of type Q→Qop. Its action is that ofσbecause:
(σ∣σ[Q]×Q)∗○σσ[Q]×σ[Q]○σ∣σ[Q]×Q(q0)) =(σ∣σ[Q]×Q)∗(σσ(q0)) =⋁Q{q∈Q∶σ∣σ[Q]×Q(q)≤Qopσσ(q0)} =⋁Q{q∈Q∶σσ(q0)≤Qσ(q)} =⋁Q{q∈Q∶q≤Qσσσ(q0)} by adjoint relationship =σσσ(q0) =σ(q0) by Lemma6.2.2.3.
2. Follows because(Q, σ)∈SAMf iff(Qop, σ)∈SAIf.
3. Take any(R, σ0)∈SAIfand anyJSLf-morphismα∶Q→R. Then(Q, α∗○σ0○α)∈SAJfbecauseα∗○σ0○α∶Q→Qop
is a self-adjoint morphism becauseσ0∶R→Rop is. Conversely by (1) everySAJf-algebra arises in this way, in fact we may assumeαis surjective. The second item follows analogously.
It is worth mentioning a related result.
Lemma 6.4.7(Lifting JSLf-quotients and embeddings toSAJf andSAMf).
Fix any finite de morgan algebra (Q, σ).
1. Each surjectiveJSLf-morphism ψ∶R↠Qdefines a SAJ-morphism:
ψ∶(R, ψ∗○σ○ψ)↠(Q, σ)
2. Each injective JSLf-morphism e∶Q↣Rdefines a SAM-morphism:
e∶(Q, σ)↣(R, e○σ○e∗)
Proof.
1. (R, ψ∗○σ○ψ)∈SAJf by Lemma 6.4.6.3. To establish that ψ∶(R, σR)↠(Q, σ)is a well-definedSAJ-morphism we must show thatψ(ψ∗○σ○ψ(r))=σ(ψ(r))for eachr∈R. This follows because for everyq∈Q,
ψ(ψ∗(q)) =ψ(⋁R{r∈R∶ψ(r)≤Qq}) by definition =⋁Q{ψ(r)∶r∈R, ψ(r)≤Qq} by join-preservation
=q by surjectivity.
2. (R, e○σ○e∗) ∈ SAMf by Lemma 6.4.6.3. It remains to establish that e ∶ (Q, σ) ↣ (R, σR) is a well-defined
SAM-morphism. Then for everyq∈Qwe must show thate(σ(q))=e○σ○e∗(e(q)), which follows because:
e∗(e(q)) =⋁Q{q′∈Q∶e(q′)≤Re(q)} by definition
=⋁Q{q′∈Q∶q′≤Qq} injectiveJSL-morphisms are order embeddings =q.
We finish off with an explicit description of the free one-generated algebras. They are finite in each case, whereas the two-generated algebras are already infinite. In fact, a free De Morgan algebra on X amounts to a free bounded lattice onX+X equipped with a natural involution.
Proposition 6.4.8 (Free one-generatedSAJ,SAMandSAI-algebras).
1. The free one-generatedSAI-algebra may be depicted as follows:
σ() x∨σ(x) xoo //σ(x) σ(x∨σ(x)) BC o o ED o o @A / / GF //
More generally, given any set X then the free X-generated SAI-algebra arises as the free bounded lattice on generatorsX+X with inductively defined unary operation:
σX(l(x))∶=r(x) σX(r(x))∶=l(x)
2. The free one-generatedSAJ-algebra (Q, σ) is depicted below: σ() σσ(x∨σ(x)) σ(x)∨σσ(x) σσ(x) x∨σ(x) x∨σ(x∨σ(x)) σ(x) x∨σσ() σ(x∨σ(x)) x σσ() ↦ σ() x ↦ σ(x) σσ() ↦ σ() x∨σσ() ↦ σ(x) σ(x∨σ(x)) ↦ σ(σ(x∨σ(x))) x∨σ(x∨σ(x) ↦ σ(x) σ(x) ↦ σσ(x) σσ(x) ↦ σ(x) x∨σ(x) ↦ σ(x∨σ(x)) σ(x)∨σσ(x) ↦ σ(x∨σ(x)) σσ(x∨σ(x)) ↦ σ(x∨σ(x)) σ() ↦ σσ().
The boxed elements show that the imageσ[Q]⊆Qop is a freeSAI-algebra on the generator σ(x). 3. The free one-generatedSAM-algebra(Q, σ) may be depicted as follows:
x∨σ() ED BC o o σ() x∨σ(x) ED BC o o σσ(x∨σ(x)) x r r ❡❡❡❡❡❡❡❡❡❡ ❡❡❡❡❡❡❡❡❡❡ ❡❡❡ σ(x)oo //σσ(x) σ(x∨σ(x)) @A / / GF// @A / / GF //
The boxed elements show that the imageσ[Q]⊆Q is a freeSAI-algebra on the generator σ(x). Proof.
1. The depicted finite de morgan algebra is a well-defined bounded lattice because:
σ()=⊺ σ(x∨σ(x))=x∧σ(x)
i.e. we have the bounded lattice structure by Lemma6.4.3. Then it is closed under the involutionσand defines a finite de morgan algebra. Since no additional relations were assumed this is a free one-generated algebra. Regarding the more general statement, take any setX and let:
QX ∶=F(X+X) be a free bounded lattice on generatorsX+X i.e. two copies ofX.
We may view its elements as equivalence classes of bounded lattice terms in variablesl(x), r(x)forx∈X. Then
σX is a well-defined involutive bounded lattice isomorphism QX →QopX. This follows by the symmetry of the usual equational presentation of bounded lattices, and the fact that we may bijectively relabel variables, so that
φ≈ψ ⇐⇒ σX(φ)≈σX(ψ). Then given(Q, σ)∈SAIand elementsel∶X →Q, we have a unique bounded lattice morphismα∶QX→Qwhere:
α(l(x))∶=el(x) α(r(x))∶=σ(el(x)),
via the universal property of free bounded lattices sinceQ is a bounded lattice. It remains to establish thatα
preserves the unary operation. First consider the base case:
As for the inductive case, assuming thatα(σX(φ))=σ(α(φ)) holds for allφ∈Φ, then:
α(σX(σX(φ))) =α(φ) σX involutive
=σ(α(σX(φ)) by induction, σinvolutive.
α(σX(⋁QXΦ)) =α(⋀QXσX[Φ]) σX∶QX→Q
op
X a bounded lattice morphism =⋁Qα○σX[Φ]) α∶QX→Qa bounded lattice morphim =⋁Qσ(α[Φ]) by induction
=σ(α(⋁QXΦ) repeating reasoning in reverse.
2. (Sketch) Ignoring the terms we have a well-defined join-semilattice, which is actually distributive. One may verify thatσsatisfies the rules(Revσ)and (Exσ2), so we have a well-defined finiteSAJ-algebra. Now view the
elements as their respective term modulo the equational axioms ofSAJ. The join-structure is compatible using the fact that σ() is the top element by Lemma 6.2.2.4. To see that the unary operation is compatible one
verifies that the action is derivable from the equational laws. It suffices to verify a subset of them.
• σ()≈σ()trivially.
• σ(σσ())≈σ()by Lemma6.2.2.3.
• σ(x∨σσ())≈σ(x). Indeed, sincex≼x∨σσ() we obtainσ(x∨σσ())≼σ(x)via(Revσ). Conversely,
σ(x)≼σ(x∨σσ()) ⇐⇒ x∨σσ()≼σσ(x) via the adjoint relationship, and hence holds using (Exσ2)
and by applying(Revσ)twice.
• σ(x∨σ(x∨σ(x)))≈ σ(x). Firstly since x≼ x∨σ(x∨σ(x)), applying (Revσ) yields half of the desired
equality. Conversely, we need to establish that:
σ(x)≼σ(x∨σ(x∨σ(x))).
Applying the adjoint relationship this is equivalent tox∨σ(x∨σ(x))≼σσ(x). Thenx≼σσ(x)is (Exσ2)
and finallyσ(x∨σ(x))≼σσ(x)follows by applying(Revσ)toσ(x)≼x∨σ(x).
3. (Sketch) Follows by the method used in (2), noting thatσσ()≅in SAMby Lemma 6.2.2.4. That the action ofσ is witnessed by various equational proofs is easier than in (2). One only needs to useσσσ(x)≅σ(x) and
σ(x∨σ())≼σσ()≅via(Revσ).
Corollary 6.4.9. If∣X∣>1 then the freeSAJ,SAMandSAI-algebra onX are infinite.
Proof. By Proposition6.4.8.1, a free SAI-algebra on X is a free bounded lattice onX+X generators equipped with an involution. It is well-known that the free bounded lattice on 3 generators{x, y, z}is infinite e.g. we have the strict <-chain where φ0 ∶=x and φn+1 ∶=x∨(y∧(z∨(x∧(y∨(z∧φn))))) for alln ≥0. Then if ∣X∣ >1 it follows that the freeSAI-algebra on X is infinite. Finally, the freeX-generatedSAJ andSAM-algebra have the free X-generated
SAI-algebra as a quotient, so they are themselves infinite.