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This is based on a few previous questions.

Can one characterize ultrapowers in the category of L-structures (modeling a fixed complete theory, say) and elementary embeddings?

Previous posts showed that you can do this for ultraproducts in the category of structures and homomorphisms, but not generally for ultraproducts in the other category.

Martin Sleziak
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Pteromys
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Ultraproducts cannot be characterized in the category of models and homomorphisms (see R. Börger, Remarks on categorical notions of ultraproducts, Abstracts, Sussex Category Meeting 1982). His example adds to the theory of fields a 0-ary relation symbol $R$ which is interpreted as true or false. Models are either $A^+$ or $A^-$ where $A$ is a field and $R$ is interpreted in $A^+$ as true and in $A^-$ as false. Let $F$ be the functor on models sending $A^+$ to $A^+$ and $A^-$ to $A^-$ if $A$ has the characteristic $\neq 0$ and $A^+$ to $A^-$ and $A^-$ to $A^+$ otherwise. Then $F$ is an isomorphism on the category of models. Let $\mathcal U$ be a non-principal ultrafilter on the set of primes and $A=\prod\limits_\mathcal U A_p$ where $A_p$ is a field of characteristic $p$. Then $\prod\limits_\mathcal U A_p^+=A^+$, $F(A^+)=A^-$ and $\prod\limits_\mathcal U F(A_p^+)=A^+$. Thus $F$ does not preserve ultraproducts, hence ultraproduts are not definable in the category of models and homomorphisms.

This example also shows that ultraproducts cannot be characterized on the category of models and elementary embeddings. It suffices to take algebraically closed fields instead of fields where homomorphisms coincide with elementary embeddings.

Jiří Rosický
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  • The question is specifically about ultrapowers, not ultraproducts. The corresponding question about ultraproducts is here: https://mathoverflow.net/questions/450222/ultraproducts-in-the-category-of-structures-and-elementary-embeddings . – Emil Jeřábek Jul 17 '23 at 07:04
  • I agree that this works for the category of structures and embeddings or elementary embeddings, but I don't see it for homomorphisms: there is a homomorphism from $A^-$ to $A^+$ but not in the other direction - what does $F$ do to this homomorphism when $A$ has characteristic $0$? – Alex Kruckman Jul 28 '23 at 13:48
  • Also, regarding the last sentence: for algebraically closed fields, elementary embeddings are the same as embeddings, and embeddings are the same as homomorphisms. When we add the $0$-ary relation, elementary embeddings still agree with embeddings, but they no longer agree with homomorphisms (since homomorphisms must preserve, but not necessarily reflect, the $0$-ary relation). – Alex Kruckman Jul 28 '23 at 13:51
  • Maybe Börger was working with the "wrong" definition of homomorphism, as given in Enderton's book, for example, by which homomorphism must reflect truth of relation symbols, in addition to preserving them. Unlike the "correct" category of L-structures and homomorphisms, the category with these strong homomorphisms does not have products, so the obvious categorical characterization of the ultraproduct (as a directed colimit of products indexed by sets in the ultrafilter) doesn't work. – Alex Kruckman Aug 15 '23 at 13:39