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).


1 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.

  • $\begingroup$ Thanks! Still, is the embedding $Z\mapsto Sh(Z)$ in some way related to the observation that locales are "0-toposes" as I explained? Also, what about 4. (why should isomorphic objects in a locale be equal, whereas we don't require toposes to satisfy this condition)? $\endgroup$
    – user964610
    Sep 5, 2021 at 11:06
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    $\begingroup$ @user964610 For any category $C$, the category $\mathrm{Psh}(C)$ of presheaves on $C$ is the free cocompletion of $C$. For a site $(C,J)$, the category $\mathrm{Sh}(C)$ is the free cocompletion, modulo sending coverings to colimits. So you can think of passing from $Z$ to $\mathrm{Sh}(Z)$ as taking the category $C_Z$ and replacing lots of the colimits with new "free" ones (that are more like colimits in the category of sets) - in particular, $1+1\neq 1$ in $\mathrm{Sh}(Z)$. I don't know if this helps to answer your first question. Others may have better answers. $\endgroup$ Sep 5, 2021 at 17:16
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    $\begingroup$ Regarding the second question: For simplicity? Remember, a frame is supposed to be an abstraction of the poset of open sets in a topological space, and this is a poset, not a preorder. Also, one of the convenient things about frames is that they're actually (infinitary) algebraic structures, i.e., meets and joins are defined "on-the-nose", not just up to isomorphism. $\endgroup$ Sep 5, 2021 at 17:17
  • $\begingroup$ That definitely helps a lot. Thank you so much! $\endgroup$
    – user964610
    Sep 6, 2021 at 20:23

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