The set of Locales is a subset of the set of Toposes.

Question

Given an ordered set $(Z, \leq)$, we can consider the category $C_Z$ with $Ob(C)=Z$ and for each $r,e\in Z$, $\hom_{C_Z}(r,e)$ is a singleton if $r\leq e$ and is $\emptyset$ if $\neg(r\leq e)$. The idea is that ordered sets are special kinds of categories (with no more than one morphism between any two objects).

An ordered set is called a frame if it admits arbitrary joins and binary meets and binary meets distribute over joins. A morphisms between frames is a monotone function preserving arbitrary joins and binary meets. A Grothendieck topos is a category satisfying the so-called Giraud axioms, which are very similar to the axioms of a frame (binary limits; arbitary small colimits, which are stable under pullbacks).

1. Is $Z\mapsto C_Z$ a fully faithful functor from the category of frames to the category of Grothendieck toposes (with geometric morphisms)?

2. If yes, why is it more common to embed frames into Grothendieck toposes via $Z\mapsto Sh(Z)=$ topos of sheaves on $Z$? Is this embedding somehow related to the functor $Z\mapsto C_Z$?

3. Does 1. imply that the category of frames is equivalent to the category of Grothendieck toposes of the form $C_Z$ for some ordered set?

4. Why are frames required to be ordered sets, i.e., such that $\leq$ is antisymmetric: if $r\leq e\leq r$, then $r=e$ and not merely preordered sets? Because for Grothendieck toposes we don't require them that isomorphic objects are equal (which is the categorical analogue).

Answer 1

No, for a frame $Z$, the category $C_Z$ is never a Grothendieck topos (unless it has only one object).

In terms of Giraud's axioms, $C_Z$ fails the axiom that coproducts should be disjoint. In a frame, $1+1=1$, while in a topos, $1+1=1$ implies $0=1$, which implies that the topos is trivial.