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Suppose a sheaf $F$ on a site $(\mathsf C,J)$ has the property that for each $X$, $FX$ is a subframe of the subobject poset of $X$.

I think $F$ is a subsheaf of $\Omega$ in the sheaf topos $\mathcal E=\mathsf{Sh}(\mathsf C,J)$, but what more can be said about it?

I think it may not be possible to know $F$ is a subframe of $\Omega $ because the internal language sees unions only locally. Is it however true $F$ is a sub-inf-lattice of $\Omega$? How do I find out?

Edit. I changed the title because I had accidentally given the same title as as this MSE question.

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    $\begingroup$ A first remark: You also need to add this assumption that for every morphism $f :X \rightarrow Y$ the induced morphism $F(Y) \rightarrow F(X)$ corresponds to the pullback when you identify element of $F(X)$ and $F(Y)$ with subobjects of $X$ and $Y$ to deduce that $F$ is a subobject of $\Omega$. For the rest of the question, this is more subtle: roughly being a frame internally imply being a frame on each object, but being a frame internally is considerably stronger (see the famous "An Extension of the Galois Theory of Grothendieck") I'll think more about it tomorrow $\endgroup$ Oct 18, 2016 at 23:48
  • $\begingroup$ @SimonHenry What would be a counterexample? That is, a sheaf $F$ with $F(X)\subseteq\operatorname{Sub}(X)$ for all $X$ but induced maps on $F$ not agreeing with those of $\operatorname{Sub}$? And can it happen that $F$ is not even isomorphic to one with agreeing induced maps? $\endgroup$ Oct 19, 2016 at 12:28
  • $\begingroup$ Well you can start with a $F$ that is indeed a subobject of $\Omega$ and then mix every $F(X)$ by some chosen automorphism of $\Omega(X)$ for each $X$. You end up with $F(X)$ being identified with a subset of $\Omega(X)$ for all $X$ but not in a way compatible to functoriality.I'm not sure how to construct an example where it is not possible to make the map agree by changing the maps $F(X) \rightarrow \Omega(X)$ but I don't see any reasons for this not being possible $\endgroup$ Oct 19, 2016 at 12:43
  • $\begingroup$ @SimonHenry Right, I now realized there indeed are such examples. Probably the simplest one is in $\mathbb Z/2\mathbb Z$-sets. A two element set with a nontrivial involution has underlying set isomorphic to that of $\Omega$ but the latter has a trivial action. Presumably there also exist simple localic examples. $\endgroup$ Oct 19, 2016 at 13:39

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It is a slighty tricky question and there is a lot to say, so let's go point by point:

1) As I said in the comment, if you want $F$ to be a subobject of $\Omega$ you need $F(X)$ to identify to a set of subobject of $X$ for each $X$ but you need this functorially in $X$, i.e. if $f : X \rightarrow Y$ is an arrow and $v \in F(Y)$ then you want $f^* v$ to corresponds to the pullback of the subobject corresponding to $v$. This is exactly what you need in order to have a map from $F$ to $\Omega$ (ans $F$ will be a subobject of $\Omega$ if and only if you have such a map and its a monomorphism and monomorphism just objectwise one-one.)

2) The following conditions are equivalent:

-$F$ is internally stable under finite intersection/infimum

-$F(X)$ is stable under finite intersection/infimum for all $X$.

Indeed stability under finite intersection can be rephrased as "contains the top element and is stable under binary intersection" and as binary intersection corresponds to a nice map $\Omega \times \Omega \rightarrow \Omega$ the stability by this map can be just tested on the $F(X)$.

The same statement with finite intersection/infimum replaced with finite union/supremum is true for the exact same reasons, but as we will see below the situation become trickier when one talks about arbitrary union.

3) Let $A$ be an object of your topos which is internally a sup-lattice (for example, $\Omega$ or a sub frame of $\Omega$), and by sup-lattice I mean a poset with arbitrary supremum, not just finite one. Then for each object $X$, $A(X)$ (i.e. $Hom(X,A)$) is itself a supp-lattice, indeed:

If $(a_{i})_{i \in I}$ is a familly of element of $A(X)$, then you have a map of set $a:I \rightarrow Hom(X,A)$ which can be promoted to a map of sheaves $p^* I \rightarrow [X,A]$ where $[X,A]$ is the internal hom object comming from the fact that a topos is cartesian closed ($p_*[X,A] = Hom(A,B)$) where $p$ denotes the canonical geometric morphism from your topos to sets.

Internally you hence have a $p^*I$ indexed familly of maps from $X$ to $A$, as $A$ is internally a sup-lattice, you can internally compute the supremum (indexed by $p^* I$) of this familly to get a single map from $X$ to $A$, as the supremum is uniquely defined, this corresponds externally to a well defined map from $X$ to $A$ denoted $a$. It is then not very hard to see that $a$ is the supremum of the $a_i$ : it is obviously greater (or equal) than each $a_i$ by construction if you have $b$ greater than each $a_i$ then by reasoning internally you will see that it is greater than $a$ as well.

Moreover, if $f :A \rightarrow B$ is (internally) a morphism of sup-lattice then for each $X$, $f_X :A(X) \rightarrow B(X)$ will preserve supremum (it follows easily from the description above of how supremum in $A(X)$ and $B(X)$ are computed.)

Hence, if $F$ is a sub-sup-lattice of $\Omega$, then for each $X$, $F(X)$ is a sub-sup-latice of $\Omega(X)$.

Here again replacing supremum by infimum everywhere won't change anything.

In particular, there is no problem with the fact that (as you said) "the internal language sees unions only locally". Those local unions are enough to compute global union because as supremums are unique the local supremum patch up nicely into globale supremum, the problem is in fact in the other direction:

4) The converse is not true ! In the proof above we have only used a very specific type of internal supremum, those indexed by constant object ($p^*I$) internal frame have a lot of different sort of supremum, and $F$ will be a sub-sup-lattice of $\Omega$ only if it is also stable by those internal supremum. here is the typical example of such a construction:

Let $A$ be a sublattice in a topos,, let $f:X \rightarrow Y$ be a morphism in the topos and let $v : X \rightarrow A$ (or $v \in A(X)$ is you prefer). Then one can define internally a map $f_!v : Y \rightarrow A$ as follow:

$$f_! v := y \mapsto \bigvee_{x \in f^{-1}(y)} v(x) $$

For example, for $A = \Omega$ this corresponds to the fact that if you have a map $f:X \rightarrow Y$ you can either use it to pullback sub-objects of $Y$ into sub-objects of $X$ (which is the ordinary functoriality of $\Omega$ corresponding to $f^*$) or to take the direct the direct image of a sub-object of $X$ in $Y$ (which corresponds to this $f_!$ construction).

Also, it is not very hard to see that $f_!$ is a left adjoint to $f^*$.

Moreover because this construction of $f_!$ is done internally it is automatically "stable under pullback" and this gives a kind of "Beck-Chevalley" condition relating the $f^*$ with the $f_!$.

In particular, if $F$ is a sub-sup-lattice of $\Omega$ it has to be stable by "direct image", i.e. if $S \in F(X)$ and $f:X \rightarrow Y$ then $f_! S \in F(Y)$.

I claim that sub-sup-lattice of $\Omega$ are exactly the sub-object of $ \Omega$ that are objectwise sub-sup-lattice and are stable under direct image like this but I don't know if there is a simple direct proof of this (see my remark at the end).

5) More generally when you have an "internal frame" $A$, you need to have that for each $X$ $A(x)$ is a frame, but you also need to write down a compatiblity between finite intersection in $A$ and those $f_!$ functors which are a special kind of supremum. This conditions is basically the Forbenius condition. Fortunately, when you already know that $\Omega$ is a frame, saying that $F$ is a sub-frame just mean that $F$ is a sub-sup-lattice and is stable under finite intersection, so you don't need to write the associativity condition for $F$ itself which simplify things and the remarks above already solve the problem (you just that $F(X)$ is a frame for all $X$ and that is is stable under direct image)!

6)To conclude, if you are interested in internal sup-lattice and internal frame I strongly recommend you to have a look to the paper by Andre Joyal and Miles Tierney " An Extension of the Galois Theory of Grothendieck" (It is not very easy to find but it is in my opinion the most important paper in topos theory), most specifically in section VI.2 and VI.3 it gives explicit description of what are internal sup-lattice and internal frame in terms of sheaf in the special case of the topos of sheaves over a locale and the topos of pre-sheaf over a small category with finite limits. The second case will show you precisely what are those Beck-Chevalley condition and Frobenius condition I mentioned earlier. It will also trivially imply the claim I made in $(4)$ at least for those two examples of toposes.

It happen that one can extend the description given by Joyal and Tierney of a the categoy of internal frame and internal sup-lattice to an arbitrary site, but this has unfortunately (as far as I know) never appears in the literature. It has been on my "To Do list" for a few years to write this down properly but I never found the time to do so. (which is why I know that the claim I made in $(4)$ is true without knowing if there is a simple proof of it or not)

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  • $\begingroup$ Thank you for the very complete and interesting answer! Forgive me but much of it is beyond me for now (though I definitely intend to study this material and the entirety of your answer soon!); for now I would like to make sure I understand correctly - with the identification you mention in your first point, it is true that a subobject of $\Omega$ which is object-wise a sub-inf-lattice is internally a sub-inf-lattice basically because the internal logic of a topos does not see conjunctions locally? $\endgroup$
    – Arrow
    Oct 20, 2016 at 18:54
  • $\begingroup$ @Arrow : if by "inf lattice" you mean with finite conjunctions the yes. For infinite conjunctions then No, but the converse is true. $\endgroup$ Oct 23, 2016 at 14:22
  • $\begingroup$ I do mean finite conjunctions. I think I'll start calling these things sub-posets with meets. Thanks again for the great answer. $\endgroup$
    – Arrow
    Oct 23, 2016 at 14:25
  • $\begingroup$ Sorry for bugging you again. If you have time, could you take a look at this question? $\endgroup$
    – Arrow
    Oct 23, 2016 at 14:43

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