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I am glad to see that a general question like Is there a relationship between model theory and category theory? receives quite a lot attention and no down-votes for being too general and unspecific. So I feel encouraged to pose some follow-up questions (firstly restricted to first-order model theory).


I hope the following statements are sufficiently sensible, precise and correct.

  • Each first-order theory $T$ with signature $\sigma$ unambigously defines a class of (ZF-)models.

  • This class of models of $T$ together with the $\sigma$-homomorphisms form a category (the category of models of $T$).

  • Two first-order theories with two arbitrary signatures may define equivalent categories of models.

Definition: A first-order theory $T$ provides a model of a category $C$ if the category of models of $T$ is equivalent to $C$.

  • Each category $C$ defines a (possibly empty) set of first-order theories: the set of all $T$ which provide a model of $C$.


[Remark: I had to work this question over, since it seemed to be ill-posed.]

Old version: Can infinite concrete categories be specified other than as categories of (ZF-)models of some (possibly higher-order) theory? Examples?

New version (explicitly restricted to first-order theories):

Given an infinite category of models of a first-order theory $T$. Can this category - or one equivalent to it - be specified/represented/given independently of any first-order theory $T$ and its (ZF-)models?

Remark: $T$ of course can be specified/represented/given independently of its models: as a set of formulas.

Why is the notion of models of a (concrete) category so uncommon? (Maybe because the answer to the first question is "No"?)

Is there a genuine model-theoretic notion of two equivalent theories if these have two arbitrary signatures?

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I don't understand the conditions of your question. How would you axiomatize, say, the category of compact Hausdorff spaces and proper maps between them? Please be a bit more specific about the language of your theories. What does "higher-order theory" refer to precisely? –  Andrej Bauer Jan 18 '10 at 10:29
I agree. The question you're asking is not the question that you want to know the answer to. Your question is ill-defined. However, as asked, both Charles and I gave examples, but after Charles's answer, you changed your question, which is really one of those things that's against the rules. –  Harry Gindi Jan 18 '10 at 10:44
What you should really do is accept Charles's trivial but correct answer to your question. Then you should look up some terminology to make sure you're being precise, because even now, my answer is still the correct answer to your question. –  Harry Gindi Jan 18 '10 at 10:46
Hans, it's not right to keep changing your question after it's already been answered. Either accept Charles's answer, my answer, or I'm going to flag this post for abuse. –  Harry Gindi Jan 18 '10 at 13:34
The policy on MO is if somebody comes up with a counterexample because you forgot to put something in your assumptions, then you are somewhat obliged to accept that answer. It's not like we could have assumed you meant infinite. It's not like somebody jumped out and said, "ah, yes, the 0 ring, of course". –  Harry Gindi Jan 18 '10 at 14:09
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3 Answers

up vote 3 down vote accepted

Sure, by direct construction. Rings, preorders, the category of paths of a given graph, etc. But that's not what you wanted to know, is it?

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I forgot to specify that I'm interested only in infinite categories. Those direct constructions must rely on some underlying set then, don't they? (How to define an infinite preorder without some "concrete" underlying set?) –  Hans Stricker Jan 18 '10 at 9:55
+1 for the correct answer before the op changed the question. –  Harry Gindi Jan 18 '10 at 10:48
+1 for the correct answer before I changed the question. (Can there be direct constructions of infinite categories?) –  Hans Stricker Jan 18 '10 at 10:59
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As for the OP's last question, "Is there a genuine model-theoretic notion of two equivalent theories if these have two arbitrary signatures:"

Sure, it's called bi-interpretability. See pp. 1378-1379 of this article for a definition of what this means for structures. For theories, we say that T_2 is interpretable in T_1 if there is a family of interpretation formulas (as in the linked definition) such that for any model M of T_1, these formulas define an interpretation in M of some model of T_2. Similarly, as above, we can define what it means for two theories to be bi-interpretable.

If you make the class of models of T into a category by declaring the morphisms to be the elementary embeddings (which seems very natural to me), then it follows directly from the definition that any two theories that are bi-interpretable (without parameters) have equivalent categories of models (via the natural "interpretation functors" which translate back and forth between the two languages). .

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Thanks, but I'm sorry, the link doesn't work. Could you please give it another try? –  Hans Stricker Jan 28 '10 at 9:37
@John: I really would appreciate the working link, since I can find "bi-interpretability" only for models/structures and classes of them, but not for theories. –  Hans Stricker Jan 28 '10 at 13:11
@Hans: OK, tried linking to another page. –  John Goodrick Jan 28 '10 at 17:47
Thanks a lot!.. –  Hans Stricker Jan 28 '10 at 18:57
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Yes. The category of topological spaces over the category of sets is not algebraic and therefore not a category of models over the category of sets. It is a category fibered over the category of sets. The adjunction beteween the forgetful functor and its left adjoint is not monadic because the left adjoint produces discrete spaces. That is, the category of models over the monad induced by the adjunction is not equivalent to Top.

Tom Leinster's book Higher Operads, Higher Categories has a description of this phenomenon on page 8. He also gives a reference to Mac Lane's categories for the working mathematician, but you can find that yourself.

Edit: To answer your third second question, the reason that you don't hear people talk about models in the way you're asking is because they're called algebras over a monad or more generally algebras over an operad.

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I think he's looking at general first-order theories, but your argument that topological spaces are not an algebraic category only shows that topological spaces cannot be axiomatized as an algebraic theory (i.e., operations + equational axioms). They could still be aximatized in a richer language, except I don't understand what richer language Hans has in mind. –  Andrej Bauer Jan 18 '10 at 10:31
He asked about first order models with a signature, which are exactly varieties of algebras, all of which can be described by a monadic adjunction between the base category and the category we're looking at. –  Harry Gindi Jan 18 '10 at 10:39
Ah yes. But also note the parenthetical remark about higher-order theories. That threw me off. –  Andrej Bauer Jan 18 '10 at 11:39
@HS: your question does not seem well-defined to me. The usual way to "specify" the category of groups is to say that its objects are groups and its morphisms are homomorphisms of groups. I would say that this does not "rely" on any model-theoretic notions whatsoever: one does not need to know any model theory in order to understand or make this definition (and probably the majority of people who work with this definition do not know or care about model theory). It happens that you can recover this category as the category of models of a certain theory: so what? –  Pete L. Clark Jan 18 '10 at 13:05
Sorry, but "How can an infinite graph be given?" is a conversation-ender for me: I cannot interpret that question in a meaningful but nontrivial way within the context of mainstream mathematics. If you are taking some non-standard approach, then you need to be ten times clearer and more explicit about what your assumptions are. Also I find the style of your discourse to be puzzling: "by accident" is not a paraphrase of "it happens", and I never said "given an infinite graph" (I said similar things, but that's not what quotation marks mean.) Maybe someone else can help you out... –  Pete L. Clark Jan 18 '10 at 16:13
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