$\DeclareMathOperator\Mod{Mod}\DeclareMathOperator\Elem{Elem}$Background/Motivation: Inspired by an interesting question by Joel, I’ve been wondering about the relationship between two very natural ways to define the category of “all models of $T$” where $T$ is a first-order theory.

Let us assume that $T$ is a complete theory with infinite models. Then on the one hand we can define the category $\Mod(T)$ whose objects are all the models of $T$ and whose morphisms are all homomorphisms in the sense of model theory – i.e. functions $\varphi\colon M \rightarrow N$ such that

For any $n$-ary relation $R$ in the language of $T$, if $M \models R^M(a_1, \ldots, a_n)$, then $N \models R^N(\varphi(a_1), \ldots, \varphi(a_n))$;


$\varphi(f^M(a_1, \ldots, a_n)) = f^N(\varphi(a_1), \ldots, \varphi(a_n))$ for any $n$-ary function symbol $f$ in the language of $T$.

Also, one can define another category $\Elem(T)$ whose objects are also all the models of $T$, but whose morphisms are only the elementary embeddings, that is, functions which preserve the truth of all first-order formulas. As a model theorist, I’m more used to thinking about the category $\Elem(T)$, and this latter category arises naturally if one cares about which sets are definable but one does not particularly care about which sets are definable by positive quantifier-free formulas.

Question: What sorts of category-theoretic properties automatically transfer from $\Mod(T)$ to $\Elem(T)$, or from $\Elem(T)$ to $\Mod(T)$?

To be clear, by a “category-theoretic property” I mean something that is preserved by an equivalence of categories.

Another related question is:

Question: Suppose we have a set of category-theoretic properties which we know characterize all the categories $C$ which are equivalent to $\Mod(T)$ for some $T$ [or to $\Elem(T)$ for some $T$]. Can we use this to characterize the categories which are equivalent to $\Elem(T)$ [respectively, $\Mod(T)$] for some $T$?

Here are a couple of basic facts I know, which may or may not be useful here. First of all, every category $\Elem(T)$ is equivalent to a category $\Mod(T')$ for some other theory $T'$ – namely, the “Morleyization” $T'$ of $T$, where we expand the language by adding new predicates for every definable set (and iterating $\omega$ times), thereby forcing $T'$ to have quantifier elimination. However, it is certainly not true that every category $\Mod(T)$ is equivalent to a category of the form $\Elem(T')$ – for instance, $\Mod(T)$ might not have colimits of $\omega$-directed chains, but $\Elem(T)$ always will (by Tarski’s elementary chain theorem).

Addendum: As Joel pointed out, there is a third possible notion of “morphism” for this category: the “strong homomorphisms” $\varphi\colon M \rightarrow N$ which commute with the interpretations of function symbols and have the property that for any $n$-ary relation $R$ in the language of $T$, $$M \models R^M(a_1, \ldots, a_n) \iff N \models R^N(\varphi(a_1), \ldots, \varphi(a_n)).$$

I’d also be interested to learn about any relationships between the category of models with strong homomorphisms and the other two categories above.

  • $\begingroup$ John, Great Question! But isn't there also another category here, corresponding to the version of homomorphism where you only ask for preservation of relations in the forward direction, instead of the if-and-only-if version you have stated? I thought model theorists usually use the term homomorphism only in this weaker sense, where we just ask for R^M(a) implies R^N(phi(a)), instead of iff. $\endgroup$ Feb 7, 2010 at 4:19
  • $\begingroup$ @Joel: Yes, you're right; I've fixed the definition of "homomorphism" in the question. $\endgroup$ Feb 7, 2010 at 9:01
  • 1
    $\begingroup$ I think both kinds of homomorphisms on relational structures are interesting, and could lead to interesting categories. It's like the difference between mapping one graph into another as a subgraph, on the one hand, or insisting that the image is an induced subgraph, on the other. $\endgroup$ Feb 7, 2010 at 13:50
  • $\begingroup$ There's a nice paper "Morphism Axioms"(pdf) by Florian Rabe that suggests that one should specify additional axioms for a first order theory that say what the morphisms between models should be. This means that you don't have to worry about equivalent theories having inequivalent categories of models, because you can just explicitly define the morphisms. $\endgroup$ Sep 19, 2018 at 23:40

1 Answer 1


Well, that's not an answer to your question but perhaps it is still a comment you might useful. I happend to have spent some time thinking about how to define a category of models of a theory, and it appeared to me that a reasonable way is to consider the category of families of models of a theory. There are not much indications that this is a good way, though, but still, you can define something like a model category structure on this category, and, perhaps more importantly, this model category structure does give rise to an interesting set-theoretic invariant (the covering number) in the (model theoretically degenerate) case of the theory of equality (and is an actually model category). I am not sure whether you find this helpful, though, and it is definitely not an answer to your question. Actually, I've written a text about this, sort of a report on whatever little I understand now...

  • $\begingroup$ And so what is this category of families of models? And how should we find about your text, given that you are using the nickname mmm and have no info in your profile? $\endgroup$ Jun 10, 2010 at 23:15
  • $\begingroup$ oh sorry. now i put in a link to my <a href="corrigenda.ru/by:gavrilovich/what:work-in-progress/…> "A construction of a set-theoretic model category". The text is essentially set-theoretic, but i do state a conjecture about AEC as a category of families of models. $\endgroup$
    – mmm
    Jun 11, 2010 at 11:03
  • 2
    $\begingroup$ The link is broken ... $\endgroup$ Jan 9, 2020 at 23:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.