Page 62 - Textos de Matemática Vol. 41

48 Chapter 7. Images and preimages and by 2.2.1 and 1.3.1(H4), f∗(a)→s = s, that is, s ∈ o(f∗(a)).
7.5.3. Proposition. Let f : L → M be an open localic map. For every sublocale T of M,
f−1[T ] = f−1[T ].
Proof. We need to show that
f−1[T ] =↑( f−1[T ])
is the largest sublocale contained in f−1[T] = f−1[↑(T)].
The inclusion f−1[T] ⊆ f−1[T] is obvious because f−1[T] ≥ f∗(T),
that is, f(f−1[T]) ≥ T (since f−1[T] ⊆ f−1[T]).
On the other hand, let S be a sublocale contained in f−1[T]. Then f(s) ≥
T for every s ∈ S, that is, s ≥ f∗(T) = f∗[T] by openness of f. Also by openness, f∗(t) ∈ f−1[T] for every t ∈ T (indeed,
f∗(t) ∈ mf∗[T] ⊆ f−1[T]
since f(x → f∗(t)) = f!(x) → t ∈ T by Proposition 7.3). Hence, for every s ∈ S,
s ≥ f∗[T] ≥ f−1[T]. 7.6. f−1[−] as a coframe homomorphism.

7.6.1. Lemma. Let f : L → M be a localic map. Then for every a,b ∈ M, f−1[c(a) ∨ o(b)] = f−1[c(a)] ∨ f−1[o(b)].
Proof. Since the preimage is a right adjoint and hence preserves meets we have f−1[c(a) ∨ o(b)] ∩ f−1[o(a) ∩ c(b)] = f−1[O] = O.
On the other hand,
f−1[c(a) ∨ o(b)] ∨ f−1[o(a) ∩ c(b)] = f−1[c(a) ∨ o(b)] ∨ (f−1[o(a)] ∩ f−1[c(b)]) = L by monotony and distributivity. Thus,
f−1[c(a) ∨ o(b)] = ¬(f−1[o(a) ∩ c(b)])
= ¬(f−1[o(a)] ∩ f−1[c(b)])
= f−1[c(a)] ∨ f−1[o(b)].