2 Changed "uniquely" to "up to isomorphism" in 3rd para.

For a long time, I've been thinking about this question for the category Mod*(T) of all models of T with elementary maps as morphisms (I'm putting an asterisk to distinguish it from Mod(T) in the original question). I still have more conjectures than answers about these categories, but I can say a few things about them. For example:

Proposition: If Mod*(T) is "linearly ordered" -- i.e. for any two M, N in Mod*(T), either M is embeddable into N, or vice-versa -- then Mod*(T) must have the Schroeder-Bernstein property (any two bi-embeddable models are isomorphic).

[Sketch of proof: First, note T must be complete. By a result of Shelah, T must be superstable -- otherwise, one of his constructions gives that there are many pairs of "incomparable" models neither of which can be embedded into the other. By some other results of Shelah from Classification Theory, we can deduce that T must be unidimensional (it cannot have a pair of orthogonal regular types) and omega-stable. But any omega-stable, unidimensional theory is categorical in aleph_1, and hence categorical in any uncountable cardinal by Morley's Theorem. By the Baldwin-Lachlan analysis of such theories, any model of T is determined uniquely up to isomorphism by a single cardinal-valued "dimension," and bi-embeddable models must have the same dimension, QED.]

I strongly suspect that there is some dichotomy result for Mod*(T) (and probably also for Mod(T)) saying that either it is extremely wild (e.g. as when T is unstable) or relatively "tame" (such as when T is uncountably categorical, and Mod*(T) is just a big tower, modulo isomorphisms). But I'm not sure what's the best way to make this precise.

As an example of the kind of dichotomy that may be true: I conjecture that if Mod*(T) does not have the Schroeder-Bernstein property, then in fact Mod*(T) contains an infinite collection of models which are pairwise bi-embeddable but pairwise nonisomorphic. I can prove this in some special cases (e.g. when T is weakly minimal) but not in general.

1

For a long time, I've been thinking about this question for the category Mod*(T) of all models of T with elementary maps as morphisms (I'm putting an asterisk to distinguish it from Mod(T) in the original question). I still have more conjectures than answers about these categories, but I can say a few things about them. For example:

Proposition: If Mod*(T) is "linearly ordered" -- i.e. for any two M, N in Mod*(T), either M is embeddable into N, or vice-versa -- then Mod*(T) must have the Schroeder-Bernstein property (any two bi-embeddable models are isomorphic).

[Sketch of proof: First, note T must be complete. By a result of Shelah, T must be superstable -- otherwise, one of his constructions gives that there are many pairs of "incomparable" models neither of which can be embedded into the other. By some other results of Shelah from Classification Theory, we can deduce that T must be unidimensional (it cannot have a pair of orthogonal regular types) and omega-stable. But any omega-stable, unidimensional theory is categorical in aleph_1, and hence categorical in any uncountable cardinal by Morley's Theorem. By the Baldwin-Lachlan analysis of such theories, any model of T is determined uniquely by a single cardinal-valued "dimension," and bi-embeddable models must have the same dimension, QED.]

I strongly suspect that there is some dichotomy result for Mod*(T) (and probably also for Mod(T)) saying that either it is extremely wild (e.g. as when T is unstable) or relatively "tame" (such as when T is uncountably categorical, and Mod*(T) is just a big tower, modulo isomorphisms). But I'm not sure what's the best way to make this precise.

As an example of the kind of dichotomy that may be true: I conjecture that if Mod*(T) does not have the Schroeder-Bernstein property, then in fact Mod*(T) contains an infinite collection of models which are pairwise bi-embeddable but pairwise nonisomorphic. I can prove this in some special cases (e.g. when T is weakly minimal) but not in general.