Timeline for Can we recognize when a category is equivalent to the category of models of a first order theory?
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 8, 2021 at 7:01 | answer | added | Camell Kachour | timeline score: 0 | |
| Feb 10, 2010 at 20:32 | answer | added | Sridhar Ramesh | timeline score: 3 | |
| Feb 10, 2010 at 14:11 | comment | added | Joel David Hamkins | Sridhar: Part of the point of my question is that Mod(T) is studied by logicians usually without any category theory at all. Are they missing something? If the only way to recognize Mod(T) as a category is by going back to logic, then they are not missing anything, and should carry on without category theory. But if there is some interesting category theoretic properties or distinguishing invariants, which do not amount to a direct translation of the logic into category theory, then it would suggest that logicians should benefit from analyzing Mod(T) category-theoretically. | |
| Feb 10, 2010 at 7:05 | comment | added | Sridhar Ramesh | Just for clarification/making the question more concrete, what exactly do you mean by "In other words, is being Mod(T) a category-theoretic concept?". That is, what is it about something like "There exists a first-order theory (equivalently, a Boolean logos) T such that C is equivalent to Mod(T)?" that would make it not immediately a category-theoretic concept? [It is, after all, a property of categories which is preserved by categorical equivalence] | |
| Jan 28, 2010 at 18:32 | comment | added | Joel David Hamkins | @John. Yes, I agree that I should mean isomorphism classes in that remark. And I'm interested in both versions of the question, either with homomorphisms or elementary embeddings. The latter is more natural in logic and model theory, but the former includes the most canonical category examples, with groups, rings and sets. | |
| Jan 28, 2010 at 18:27 | answer | added | John Goodrick | timeline score: 4 | |
| Jan 28, 2010 at 17:57 | comment | added | John Goodrick | Also, this is an excellent question, one I've thought a lot about myself (in the case where morphisms are elementary maps, rather than arbitrary L-homomorphisms). I wish I had a snappy answer to it by now, but it seems to be a tricky question! | |
| Jan 28, 2010 at 17:55 | comment | added | John Goodrick | To clarify one point: "If Mod(T) is uncountable, then it must be a proper class:" to be strictly correct, you should say, "If Mod(T) has uncountably many isomorphism classes, then it contains a proper class of isomorphism classes," unless by Mod(T) you mean the "embeddability skeleton" containing one representative from each isomorphism class. | |
| Jan 28, 2010 at 9:58 | comment | added | Hans-Peter Stricker | Thanks for the invitation from over there (mathoverflow.net/questions/12180). | |
| Jan 27, 2010 at 19:10 | answer | added | François G. Dorais | timeline score: 14 | |
| Jan 27, 2010 at 17:55 | history | asked | Joel David Hamkins | CC BY-SA 2.5 |