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In model theory, one typically defines only embeddings of structures and isomorphisms, but I haven't seen a definition of general structure homomorphisms. Is there some particular reason behind that? Of course, there are a lot of theories whose typical homomorphisms of their respective models don't provide well behaved categories (for example the category of fields), but at least in some special cases (maybe in universally axiomatized theories), it should be possible to apply category-theoretic constructions to the models, which might benefit the theory.

Dominik
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    Model theorists are not category theorists, and the questions they care about are not the questions category theorists care about. But of course you can always define the category of models of a theory, and people who think about categorical logic (http://en.wikipedia.org/wiki/Categorical_logic) do stuff like this. – Qiaochu Yuan Jun 17 '13 at 19:47
  • Sometimes there is no obvious notion of homomorphism. For instance, what's the appropriate notion of morphism for Banach spaces? – Zhen Lin Jun 17 '13 at 19:50
  • @Zhen: a weak contraction! – Qiaochu Yuan Jun 17 '13 at 19:52
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    I don't get the premise of the question. Of course homomorphisms are studied in model theory. – Martin Brandenburg Jun 17 '13 at 21:27
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    To define a structure-homomorphism $f \colon \mathcal M \to \mathcal N$, relax the definition on relations $\mathcal M \models \phi(\bar x) \iff \mathcal N \models \phi(f(\bar x))$ into $\mathcal M \models \phi(\bar x) \implies \mathcal N \models \phi(f(\bar x))$. The model-theorists are interested principally in embeddings because it is the correct notion to make extensions. The general notion of structure-homomorphim is more about universal algebra. – Pece Jun 18 '13 at 09:24

1 Answers1

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I'm aware of two notions of homomorphism (former I've encountered in my introductory classes to logic and model theory, the latter I've seen in a logic textbook).

The usual one, which I've seen called also strong homomorphism, is a function $\Phi\colon M\to N$ between two structures of the same signature such that:

  1. For any constant symbol $c$ we have $\Phi(c^M)=c^N$.
  2. For any function symbol $f$ we have $f^N(\Phi(x))=\Phi(f^M(x))$ for all tuples $x$ of suitable length.
  3. For any relation symbol $r$ we have $r^N(\Phi(x))\iff r^M(x)$ for all tuples $x$ of suitable length.

Then you can define a monomorphism, epimorphism, and isomorphism in the usual manner, and the range of a homomorphism will always be a substructure, and for monomorphisms, the structure will be isomorphic with the initial one. Elementary embedding is just a monomorphism whose range is an elementary submodel.

The other one (weak homomorphism) is where in the third point you only ask for $r^M(x)\implies r^N(\Phi(x))$ (and not the other way around). For algebraic structures those two coincide, of course.

Of course, arbitrary homomorphisms don't preserve first-order properties in general, so they're rarely considered in general model theory, and are not really the natural notion of morphism from pure model-theoretical viewpoint. On the other hand, a monomorphism from a structure to an (elementarily equivalent) structure with quantifier elimination will automatically be an elementary embedding.

Instead, there's the notion of elementary embedding, and maybe more importantly elementary partial function. You probably could define the notion of partial homomorphism, but that would be a rather complicated, recursive definition and I doubt it would really get us closer to elementary functions in general context.

That said, it makes sense to consider the category of models of a given complete theory with elementary maps as morphisms. See for example this article by Lascar.

tomasz
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  • Wow, so many "of course" and "elementary" (not just in your answer). Only missing "the right definition of morphism in model theory is that of an elementary function", which you wrote in a comment to another question. Do you mean something like "element wise" when you write "elementary" here, or does this refer to the non-logical symbols of the first order language, i.e. analogous to how it is used in "elementarily equivalent structures". – Thomas Klimpel Apr 17 '14 at 14:33
  • @ThomasKlimpel: I count two “of course”, and they both seem pretty elementary to me. ;) A function is elementary if it preserves true formulas, so $\models \varphi(a_1,\ldots,a_n)\iff \models \varphi(f(a_1),\ldots,f(a_n))$, but the domain does not have to be the entire model. You might think of it as a partial elementary embedding. – tomasz Apr 17 '14 at 14:51
  • @ThomasKlimpel: And as for the remark I made over at the other post, this is something I've only come to believe later, after writing this answer. – tomasz Apr 17 '14 at 15:00
  • @ThomasKlimpel: But actually, there's one other notion of morphism which makes sense, albeit in a different category: given a model, you can consider the category of definable sets with definable functions, or some generalizations thereof (like type-definable sets etc.). Alas, definable functions are not, in general, anything like homomorphisms (though you can certainly have a definable homomorphsim). – tomasz Apr 17 '14 at 15:37
  • So the first "of course" should be translated as "because there are no relational symbols for algebraic structures". I got confused, because this weak homomorphism notion is the appropriate notion of morphism for universal Horn structures, and in that case there can be relational symbols. – Thomas Klimpel Apr 17 '14 at 15:50
  • @ThomasKlimpel: Yes, that's right. I do know that different notions of embeddings occur in some contexts, especially if we're concerned with finite structures, like wit Fraisse limits (as elementary functions between finite structures are almost isomorphisms)... I'm not saying that other categories are inconceivable or useless, just that I believe in model theory as I know it, elementary functions provide the notion of morphism. :) – tomasz Apr 17 '14 at 16:35