I'm trying to understand the proof of the Barr-Diaconescu theorem about Boolean covers for Grothendieck sites. Precisely, the versions you can find in Jardine's book "Local Homotopy Theory" or in Mac Lane - Moerdijk "Sheaves in Geometry and Logic", which are essentially the same. That is,

Theorem. For every Grothendieck topos $\mathcal{E}$ there exists a complete Boolean algebra $\mathcal{B}$ and a topos morphism $p: \mathbf{Sh}(\mathcal{B}) \longrightarrow \mathcal{E}$ such that the inverse image functor $p^* : \mathcal{E} \longrightarrow \mathbf{Sh}(\mathcal{B}) $ is faithful.

Particularly, I'm having troubles with something that must be definitively elementary. In both books the first step goes like this: you take a Grothendieck site $\mathcal{X}$ such that $\mathcal{E} = \mathbf{Sh}(\mathcal{X})$ and you consider the poset of strings of arrows $\mathbf{String}({\mathcal{X}})$ of $\mathcal{X}$, where a string is a sequence of composable arrows in $\mathcal{X}$,

$$ s = \left( U_n \stackrel{\alpha_n}{\longrightarrow} U_{n-1} \stackrel{\alpha_{n-1}}{\longrightarrow} \dots \stackrel{\alpha_1}{\longrightarrow} U_0 \right) $$

These strings carry a natural partial order, both books say (and here, perhaps, begins my problem):

For two strings $s$ and $t$ we write $t \leq s$ if $t$ prolongs $s$ to the left.

That is, for a string $s$ like the one above, $t$ must be of the form

$$ t = \left( U_{n+m} \longrightarrow \dots \longrightarrow U_{n+1} \longrightarrow U_n \stackrel{\alpha_n}{\longrightarrow} \dots \longrightarrow U_0 \right) \ . $$

So far so good, but, if I understand everything correctly, now this poset $\mathbf{String}({\mathcal{X}})$ must also be a frame. Because: (1) Mac Lane-Moerdijk says so in his theorem 1; (2) also because the second step of the Diaconescu-Barr theorem (theorem 3.39 in Jardine's book, lemma 3 in Mac Lane-Moerdijk's one) says:

Proposition. For every frame $\mathcal{S}$ there exists a complete Boolean algebra $\mathcal{B}$ and a topos morphism $i: \mathbf{Sh}(\mathcal{B}) \longrightarrow \mathbf{Sh}(\mathcal{S})$ such that the inverse image functor $i^*: \mathbf{Sh}(\mathcal{S}) \longrightarrow \mathbf{Sh}(\mathcal{B}) $ is faithful.

Hence, you put $\mathcal{S} = \mathbf{String}({\mathcal{X}}) $ in this proposition and you are done: you have the Diaconescu-Barr theorem proved for you. Notice, however, that in the proof of this proposition you do indeed need the frame structure of $\mathcal{S}$.

Ok, so my question is the following:

Question. Which are the meets $\wedge$ and joints $\vee$ of the frame $\mathbf{String}({\mathcal{X}})$?

Neither book says anything about them, so I imagine this must be an easy exercise -that I'm unfortunately unable to solve.

Let me explain my lack of understanding to you.

(1) As for joints, I would say that, given strings

$$ s = \left( U_n \stackrel{\alpha_n}{\longrightarrow} U_{n-1} \stackrel{\alpha_{n-1}}{\longrightarrow} \dots \stackrel{\alpha_1}{\longrightarrow} U_0 \right) $$


$$ t = \left( V_m \stackrel{\beta_m}{\longrightarrow} V_{m-1} \stackrel{\beta_{n-1}}{\longrightarrow} \dots \stackrel{\beta_1}{\longrightarrow} V_0 \right) $$

their meet could be something like the $0$-string

$$ s \vee t = U_0 \sqcup V_0 \ , $$

where $\sqcup$ is the coproduct in the category $\mathcal{X}$: by definition, we have morphisms $U_0 \longrightarrow U_0 \sqcup V_0 \longleftarrow V_0$, so both $s$ and $t$ prolong $s\vee t$ to the left. Right.

But: the existence of coproducts is not included in the definition of the Grothendieck site $\mathcal{X}$, is it?

(2) As for meets, my lack of understanding is even worse: if $s \wedge t$ should prolong both $s$ and $t$ to the left, then I cannot imagine how every pair of strings $s$ and $t$ could have a meet.

Because, if there is a string $s\wedge t$ that prolongs both $s$ and $t$ to the left, then shouldn't $s$ and $t$ be essentially the same string (except for having, perhaps, some arrows deleted in one of them with respect to the other)?

I guess the answers to my questions are pretty obvious, but at this moment I'm stuck with them. What am I missing? Thank you for your help, even if you embarras me a lot.

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    $\begingroup$ There is no claim that $\text{String}(\mathcal{X})$ is a frame, at least not in Mac Lane-Moerdijk. And indeed it's not (for example, there is no bottom element). $\endgroup$ – Todd Trimble Sep 6 '16 at 12:31
  • $\begingroup$ @Trimble. Yep, you're right. Although, on the one hand, the statement in Mac Lane-Moerdijk says: "there exists a locale...". But, on the other hand, as godelian points out, it's true: $\mathbf{String}(\mathcal{X})$ needs not be a locale (a frame) but just a poset, and then the topos $\mathbf{Sh}\mathbf{String}(\mathcal{X}) $ is localic because of their theorem 5.1. Maybe the authors should have mention theorem 5.1 in this proof. Anyway, thank you both for helping me in this issue. $\endgroup$ – d.t. Sep 6 '16 at 12:53
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    $\begingroup$ But they did mention theorem 5.1! At least in the 2nd edition they did. $\endgroup$ – Todd Trimble Sep 6 '16 at 13:05
  • $\begingroup$ @Trimble. Again, you're right. Page 514, third paragraph. My bad: read the sentence and the whole proof several times, and instead the conclusion eluded me. :-( $\endgroup$ – d.t. Sep 6 '16 at 14:44

I think that the missing step is to build the locale $X$ from your poset of strings. But you should not look for a frame structure in the poset; rather, you are supposed to apply Theorem 5.1 in Maclane-Moerdijk to get that locale. If the site of the topos is a poset, the proof of that theorem (which uses in turn Giraud's theorem in the appendix of the same book) says that the topos is equivalent to the localic topos $\mathcal{S}h(X)$, where $X$ is the poset of subobjects of $1$, hence equipped with the operations you want.

  • $\begingroup$ Thank you, godelian. I think you're right: I forgot the theorem you mention. $\endgroup$ – d.t. Sep 6 '16 at 12:54

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