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Theorem 1.1 in Tibor Bekes Theories of presheaf type (pdf 1, pdf 2) looks like a convenient criterion for whether a given qotient $T^+$ of a geometric theory $T$ of presheaf type is again of presheaf type -- except for the assumption that $T^+$ has enough models in $\mathrm{Set}$ (i.e. $\mathrm{Set}[T^+]$ has enough points). The thing is, I also can't find any mention of this condition in any of the places where the Theorem is applied whithin this paper.

All theories of presheaf type have enough $\mathrm{Set}$-models, so $T$ has enough $\mathrm{Set}$-models, but this doesn't seem to help us with $T^+$. (Recall that $T^+$ might have no $\mathrm{Set}$-models at all while still being consistent.) Coherent theories (no infinitary disjunctions) also have enough $\mathrm{Set}$-models. This justifies the assumption in e.g. Example 3.2 (at least if we start with a coherent $T$), but not in e.g. Examples 3.1, 3.3 and 3.5. (I must say I have additional difficulties understanding Example 3.3. It seems to imply that any propositional (zero sorts) geometric theory is of presheaf type, which is not true.)

Are there other quick reasons why a geometric theory might have enough $\mathrm{Set}$-models? Am I missing something else entirely?

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  • $\begingroup$ Is asking about a specific paper like this allowed on MO? $\endgroup$
    – Matthias Hutzler
    Commented Jun 18, 2021 at 11:22

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It is actually not that simple to contruct geometric theories whose classifying toposes do not have enough points. Of course they exist, as any Grothendieck topos is the classifying topos of something and there are plenty of Grothendieck topos that do not have enough points. But what I mean is that almost all natural example of theories will be classified with a topos having enough point.

There are two fairly similar theorem that guarantee this:

  1. Deligne completness theorem assert that "every Coherent topos has enough points". It implies that any "coherent theory" (that is a geometric theory with not infinite join) is classified by a topos with enough points.

  2. Makkai-Reyes completness theorem assert that a "separable topos", i.e. a topos that admits a countable site with a countable set of basic covering famillies, has enough point. (see theorem 6.2.4 in Makkai and Reyes "First Order categorical logic"). It implies that the classyfing topos of a "countable" geometric theory has enough points. By countable I mean that the theory has a countable signature and a countable set of axioms that uses at most countable joins.

It seems to me that any of these two results already cover all the example mentioned in Beke's paper.

To come back to my initial point, to construct a Geometric theory that do not have enough model, you need to start with some uncountable set of data, for example start from an uncountable set $I$ and look at the theory of surjection $\mathbb{N} \to I$. But pretty much all the theories we want to look at in every day mathematics are countable.

Note: Regarding your comment on example 3.3, the point is that he only considers theories whose axiom are of the form $\forall x, \phi(x)$ where $\phi$ has no quantifier. In the case of propositional theories, it means axiom of the form $\phi$ (with $\phi$ a geometric proposition) instead of the more general geometric sequent $\phi \Rightarrow \psi$ that you can use in a propositional theory.

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    $\begingroup$ Notice that the recent result of Rogers generalizes Deligne's completeness to lfp topoi too. $\endgroup$
    – Ivan Di Liberti
    Commented Jun 18, 2021 at 23:08
  • $\begingroup$ Nice! About Example 3.3: In my understanding, propositional theories can't use quantifiers at all, since a quantifier needs to quantify over one of the sorts of the theory. Am I mixing up terminology? (For example, the language of the theory of a surjection N --> I consists only of proposition symbols "f(n) = y" for all n and y, and the axioms use infinite disjunctions but no quantifiers, right?) $\endgroup$
    – Matthias Hutzler
    Commented Jun 19, 2021 at 9:55
  • $\begingroup$ I guess then, all the fully concrete examples in the paper are covered by these two theorems; and if you want to read the general examples (such as adding "negated positive sentences") as actual corollaries of Theorem 1.1 (as I first did), you need to keep the enough models condition in mind. $\endgroup$
    – Matthias Hutzler
    Commented Jun 19, 2021 at 10:29
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    $\begingroup$ I agree that there is a problem with 3.5 if R is uncountable. I guess there is a way to make it work using that each ring each a directed colimit of countable rings, but I don't think there is a general argument applying. At this point, maybe you need to ask Beke directly. Regarding 3.3, I shouldn't have insisted on 'quantifier-free'. The point is that a general propositional theory has axiom of the form $\phi \Rightarrow \psi$ with $\phi$ and $\psi$ geometric, and here it is justs $\phi$. $\endgroup$
    – Simon Henry
    Commented Jun 19, 2021 at 14:02
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    $\begingroup$ @IvanDiLiberti : I didn't know, thanks ! Though I do not how if it can be used to strengthen my answer: I don't know if there is an easily definable class of theories (other than coherent theories) that has lfp classyfing topos. $\endgroup$
    – Simon Henry
    Commented Jun 19, 2021 at 14:06

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