Proof. By the corollary above, we need only to prove that
and
are dually equivalent to
. Since our proof is about the restrictions of functors from the proof of Theorem 1 in [
23], we concentrate on objects and morphisms being Heyting. We shall use the notations from that proof.
Step 1: The restriced functor .
Assume is an object of . Then is a basis of the induced topology on and the original topology of is the topology induced from . We are to check the Heyting closure condition for the locally small space . Assume . For , we have , so . Since is a basis of the induced topology of stable under finite intersections, .
Assume
is a morphism of
. Since
we also have
Indeed, for any , we have . Applying this to the Heyting mapping condition, we have .
Take now . We can express its topological closure in (omitting ) as . That is why . Each belongs to . Hence . We get and is a Heyting bounded continuous mapping.
Step 2: The restricted functor .
By the classical Esakia Duality (see Remark 10 in [
21]), the topological space
is Heyting spectral for a Heyting algebra
A. Moreover,
a Heyting spectral map for a morphism
of
.
Step 3: The restricted functor .
For a Heyting locally small space , the intuitionistic implication in the bounded lattice can be introduced by the formula , making it a Heyting algebra by Proposition 11. For a morphism f in , the mapping is a homomorphism of Heyting algebras by Proposition 12.
Step 4: The functor is naturally isomorphic to .
The mapping
is a strict homeomorphism by Proposition 2 of [
23] and an isomorphism of
by Fact 12.
Step 5: The functor is naturally isomorphic to .
Each
is an isomorphism in
as well as an isomorphism of
by Fact 11 in [
21], so an isomorphism of
.
Step 6: The functor is naturally isomorphic to .
Each
is an order isomorphism hence, by ([
32], Exercise IX.4.3), an isomorphism of Heyting algebras. □