4
$\begingroup$

I'm fixing some type of structure $\Sigma$ (possibly multi-sorted, with functions symbols and relation symbols, though assuming it single sorted with only relation symbols wouldn't change anything). Let $A$ and $B$ be two $\Sigma$-structure. Is there a connection between the following to notion :

  1. The locale $[A,B]_\Sigma$ classyfing morphisms of $\Sigma$-structure form $A$ to $B$.

  2. Morphism of $\Sigma$-structure from $A$ to an ultrapower of $B$.

Intuitively both are way to make formal the idea that "their ought to be a morphism from $A$ to $B$, up to potential (infinite) cardinality obstruction"

As a first concrete step, the question would be:

Is it true that $[A,B]_\Sigma \neq \varnothing$ if and only if there exists a morphism of $\Sigma$-structure from $A$ to an ultrapower of $B$ ?

But Ideally (and if the above is indeed true) I would like something more concrete that explain how to go back and forth between a morphism to an ultrapower and some sort of witness that the locale is non-trivial (like maybe points of a compactification or something like this).

If you are a model theorist, I guess you can replace $[A,B]_\Sigma \neq \varnothing$ by the existence of a forcing extention of the base set theory in which there is a map from $A$ to $B$ and that should give something that can be translated back to my question.

I'm giving an explicit example in order to fix the idea :

Let $\Sigma$ be single sorted with a single binary relation $R$, and take $A$ and $B$ two $\Sigma$ structure where $R$ is interpreted as the relation $\neq$. Then $[A,B]_{\Sigma}$ is the classifying locale of injection from $A$ to $B$ wich is non trivial as soon as "$B$ is infinite or $|A| \leqslant |B| < \infty$ ". And this is also the condition under which (I'm assuming choice here) you will be able to get an injection from $A$ to some ultrapower of $B$.

Improve this question
$\endgroup$
15
  • $\begingroup$ My hunch is that there may be an answer to this question to be found by considering Makkai's theory of ultracategories. I'd suggest looking at Lurie's treatment of the subject for inspiration. $\endgroup$
    – Tim Campion
    Commented Mar 15, 2022 at 18:16
  • $\begingroup$ @TimCampion well, funny that you mention this : that question did arise while thinking about Ultracategories and Makkai conceptual completness. But I'm not entierely sure there is a concrete link. At lest I couldn't find one. $\endgroup$
    – Simon Henry
    Commented Mar 15, 2022 at 18:21
  • 1
    $\begingroup$ @SimonHenry, the existence of a homomorphism of $A$ into a reduced power of $B$ (quotient by a filter not necessarily ultrafilter) is equivalent to the existence of a pure homomorphism $B \to A$. If one takes elementary embeddings instead of homomorphisms, then an ultrapower is needed. $\endgroup$
    – godelian
    Commented Mar 15, 2022 at 18:26
  • $\begingroup$ @godelian , Hi ! I don't know what "pure homomorphism" mean in this context and google doesn't give a clear answer. Can you clarify/give a pointer ? $\endgroup$
    – Simon Henry
    Commented Mar 15, 2022 at 18:31
  • 2
    $\begingroup$ @SimonHenry A homomorphism $B \to A$ is pure (also called $\omega$-pure) if it reflects the validity of positive existential sentences with parameters from $B$, i.e., if $A \models \exists \overline{x}\phi(\overline{x}, \overline{c})$ implies $B \models \exists \overline{x}\phi(\overline{x}, \overline{c})$ for $\phi$ a conjunction of atomic formulas. $\endgroup$
    – godelian
    Commented Mar 15, 2022 at 18:36

1 Answer 1

Reset to default
2
$\begingroup$

I believe there are situations where the relevant locale is trivial, but there are many homomorphisms post-ultrapower.

For example, take $\mathcal{B}=(\mathbb{N};<)$ and let $\mathcal{A}$ be a nontrivial ultrapower of $\mathcal{B}$. There are no genuine homomorphisms from $\mathcal{A}$ to $\mathcal{B}$, and moreover this remains true in any forcing extension (since no homomorphism of strict linear orders can "move" infinitely-far-apart points to within finite distance of each other); per the comments, I think this means that the relevant locale is trivial too, but I'm not very familiar with locales. On the other hand, we trivially have lots of homomorphisms from $\mathcal{A}$ into ultrapowers of $\mathcal{B}$.

EDIT: Well after the fact, I've run into a result which while not directly related to this question may still be of similar interest, so I've decided to mention it here. John Bell's paper Isomorphism of structures in $S$-toposes. Bell shows that for structures $\mathcal{A},\mathcal{B}$ in the same language, the following are equivalent:

  • $\mathcal{A}\cong\mathcal{B}$ in some forcing extension (that is, $\mathcal{A}\cong_{\infty\omega}\mathcal{B}$ by earlier results).

  • $\mathcal{A}$ and $\mathcal{B}$ "become isomorphic" in some topos over $\bf{Set}$ (which Bell calls "$S$").

I think this result adds some nuance to the observations above, although I'm a bit out of my element here.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Adding $+$ to the structure, I'm completely convinced this is a counter-example. I don't quite know without $+$, but maybe it is as well. $\endgroup$
    – Simon Henry
    Commented Mar 16, 2022 at 16:01
  • $\begingroup$ @SimonHenry I think even without $+$ this should work if the forcing/locale connection holds. The point is that if $f$ is a homomorphism of strict linear orders, then $f$ cannot map infinitely far apart points to finitely far apart points (since $f$ must be injective). But $\mathcal{B}$ is "locally finite" (= any two points are finitely far apart; I think I'm actually misusing the term?) while $\mathcal{A}$ is not. $\endgroup$
    – Noah Schweber
    Commented Mar 16, 2022 at 16:09
  • $\begingroup$ Ah yes, that seems right. Thanks you ! $\endgroup$
    – Simon Henry
    Commented Mar 16, 2022 at 16:11
  • $\begingroup$ @SimonHenry I've folded that into the answer. $\endgroup$
    – Noah Schweber
    Commented Mar 16, 2022 at 16:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .