Page 81 - Textos de Matemática Vol. 41

10.6. Another characterization of fitness 67
Conversely, let f∗ be a complete lattice homomorphism, denote by f! the corresponding left adjoint and assume M is subfit. Then f!(a∧f∗(b)) = f!(a)∧b (and by 7.3 f is open). Indeed:
We have always f!(a ∧ f∗(b)) ≤ f!(a) ∧ f!f∗(b) ≤ f!(a) ∧ b. On the other hand, if f!(a)∧b f!(a∧f∗(b)) then, by subfitness, there exists c ∈ L for which
But
(f!(a) ∧ b) ∨ c =
1 ̸= f!(a ∧ f∗(b)) ∨ c.
f∗(f!(a ∧ f∗(b))) ∨ f∗(c) f∗f!(a ∧ f∗(b)) ∨ f∗(c)
(a ∧ f∗(b)) ∨ f∗(c)
(a ∨ f∗(c)) ∧ (f∗(b) ∨ f∗(c)) (a∨f∗(c))∧f∗(b∨c) a∨f∗(c)≥a.
f∗(f!(a ∧ f∗(b)) ∨ c)
= = ≥ = = =
Thus f!(a∧f∗(b))∨c ≥ f!(a) and consequently f!(a∧f∗(b))∨c ≥ f!(a)∨c = 1, a contradiction.
10.5.2. Proposition. Every complemented sublocale of a subfit locale is subfit.
Proof.LetT bethecomplementofasublocaleS⊆L,thatis,S∩T =Oand S ∨ T = L. Let S0 ⊆ S be such that S0 {1} is cofinal in S. Consider the join S0 ∨ T . Let x ∈ L. Then x = s ∧ t for some s ∈ S and t ∈ T . If x < 1 then eithert<1andx=s∧t≤t∈S0∨T ort=1andthens<1andthereisan s0 ∈S0,s0 <1,suchthatx=s≤s0.Thus,S0∨T{1}iscofinalinL{1} and S0 ∨ T = L. Consequently S = S ∩ (S0 ∨ T ) = S ∩ S0 = S0.
10.6. Another characterization of fitness.
Proposition. A locale is fit iff each of its sublocales is subfit.
Proof. The implication ⇒ is trivial.
⇐ : Let each sublocale of L be subfit. We will prove that L satisfies the property (ii) from Proposition 10.3. Let S, T be sublocales of L and let S′ = T ′. Consider the sublocale S ∨ T . By the assumption it is subfit. Consider an element s ∧ t ∈ S ∨ T, s ∧ t < 1. If s < 1 we have s ∧ t ≤ s < 1 with s ∈ S. If s = 1 then t<1andwehaveans′ ∈Ssuchthatt≤s′ <1.Thus,s∧t≤s′ <1with s′ ∈ S again and S {1} is cofinal in S ∨ T {1} and we have, by (ii) in 10.4, S=S∨T.SimilarlyT =S∨T andweconcludethatS=T.
Note. If a space X is T1 then each of its subspaces is T1 and hence subfit. The locale Lc(X) is in such a case generally by far not subfit, though. This is (of course) caused by the sublocales of Lc(X) that are not induced by subspaces.