Timeline for Abelian category equivalent to a non-abelian category
Current License: CC BY-SA 4.0
29 events
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| Mar 30, 2020 at 5:02 | comment | added | Kevin Carlson | @BugsBunny Well, no. If you will kindly look a bit more closely, I specified that homomorphisms must preserve the unit. Under the analogy, you'll recall we're discussing equivalences between abelian and "non-abelian" categories, which automatically preserve the biproducts that we're analogizing here to the unit in a ring. | |
| Mar 30, 2020 at 1:56 | comment | added | Bugs Bunny | @Kevi Carlson The map $A\rightarrow A \oplus B$ is a homomorphism in one and not in another. | |
| Mar 29, 2020 at 19:11 | comment | added | Kevin Carlson | @Patriot I have tried to understand Bugs’ explanations and I still cannot see how one can consistently claim the question just discussed is anything other than meaningless or negative. If a unit is part of the signature, then it’s meaningless to ask whether $B$ “is unital.” If not, then the answer is that $B$ must be unital. Bugs, do you really think that the model theory of rings which happen to admit a unit and homomorphisms which preserve the unit is different from that of rings with a unit in their theory’s signature? How? The categories of models are isomorphic! | |
| Mar 29, 2020 at 14:36 | comment | added | Patriot | Even if @BugsBunny's point of view presented is non-standard, it is by no means ridiculous, and certainly not deserving of all these downvotes. If anything, I applaud BugsBunny's openness of mind to step outside the definitions imposed by the majority for the sake of evaluating what may have led the OP to his question. | |
| Mar 29, 2020 at 13:47 | comment | added | Bugs Bunny | @Dan Peterson Excellent example! If "unital" means a ring with 1, then yes. If "unital" means a ring with 1 in signature, then no. | |
| Mar 29, 2020 at 13:37 | comment | added | Dan Petersen | @BugsBunny Then how would you answer the following question? "Suppose $A$ is a unital ring, $B$ a not necessarily unital ring, $A \cong B$ an isomorphism of not necessarily unital rings. Must $B$ be unital?" | |
| Mar 29, 2020 at 10:42 | comment | added | Denis Nardin | @BugsBunny The model theory is independent for all definitions I know: a functor of semiadditive categories has an enriched structure iff it preserves (co)products, and such a structure is unique | |
| Mar 29, 2020 at 8:16 | comment | added | Bugs Bunny | @Dan Petersen I understand your point completely. Now try to understand mine. Think of a unital associative ring. Translated to this case, your point is that unital is inherent property. My point would be that taking "unital" out of the signature changes many things, for instance, its representation theory or model theory. Going back to the original question, model theory of abelian categories will probably depend on whether enrichment is part of the signature. | |
| Mar 28, 2020 at 23:12 | comment | added | Todd Trimble | @DanPetersen +1. Yes, that is precisely the point (and essentially the point made by Dylan, of course). | |
| Mar 28, 2020 at 22:28 | comment | added | Dan Petersen | The point is that the Ab-enrichment is uniquely determined by the fact, which is implies by the axioms of an abelian category, that the sum of two arrows $f$ and $g$ from $X$ to $Y$ is the composite $X \to X \oplus X \to Y \oplus Y \to Y$ where the middle arrow is $f \oplus g$, and the other two arrows come from $\oplus$ being a categorical biproduct. | |
| Mar 28, 2020 at 22:24 | comment | added | Dan Petersen | @BugsBunny The definition from e.g. Dylan's answer is equivalent to the one you're using! The point is that although many books define an abelian category to be a pre-additive category satisfying various extra properties, which makes it seem as an Ab-enrichment has to be specified as part of the data of an abelian category, this is in fact not the case. | |
| Mar 28, 2020 at 22:16 | history | edited | Bugs Bunny | CC BY-SA 4.0 |
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| Mar 28, 2020 at 22:04 | comment | added | Bugs Bunny | Well, that is a completely different objection to my answer. This deserves clarification in the answer. | |
| Mar 28, 2020 at 21:54 | comment | added | Todd Trimble | Actually, I don't find "non-abelian category" unclear. I can't think of a realistic situation where I would ever use that phrase, except maybe in response to a beginner question, but a category either is abelian or it isn't, and there's only one way (up to canonical iso's, etc.) that a category can be an abelian category. It's not in the same league as equipping a set with a group structure. | |
| Mar 28, 2020 at 20:16 | comment | added | Peter LeFanu Lumsdaine | [cont’d from prev] …or “Can a group be bijective with a set that is not a group?”. I think the typical mathematician’s response to this latter question would be to first note that “a set that’s not a group” is slightly bad writing, but to take it as meaning “…a set that can’t be given a group structure”, and as such answer with “No: given a bijection from (the underlying set of) a group $G$ to some other set $X$, we can transfer the structure from $G$ to $X$.” I think your interpretation of the question is not what OP meant, and not how most mathematicians would understand it. | |
| Mar 28, 2020 at 20:12 | comment | added | Peter LeFanu Lumsdaine | @BugsBunny: My objection to this answer is nothing to do with the specific definition of Abelian category used by Wikipedia/nlab/Peter Freyd; nor is it anything to do with Abelian categories being property-like structure. My objection is that you interpret the OP’s “…a non-Abelian category” (which is, admittedly, a slightly unclear phrase) as being “a category not currently considered as an Abelian category”. I think this is clearly not the standard mathematical practice; compare analogous questions like “Can a non-even number be equal to an even one”, [cont’d] | |
| Mar 28, 2020 at 19:08 | comment | added | Todd Trimble | On further reflection: what Bugs might mean is that, for example, there is more than one way to endow an object $Z$ with the structure of being a product of objects $X, Y$ (using different choices of projection maps), so that Person A's notion of how $Z$ is a product doesn't match Person B's notion. But that doesn't matter: we say that $F$ preserves products if the canonical map $F(X \times Y) \to FX \times FY$ is an isomorphism. With this understanding of what preservation of (property-like) categorical structure means, all equivalences preserve zero objects, biproducts, and so on. | |
| Mar 28, 2020 at 19:05 | comment | added | Kevin Carlson | @BugsBunny Another try: if you require that an abelian category $A$ be given along with its canonical enrichment in abelian groups, then there is no such thing as a functor from $A$ to a category $B$ whose hom-objects are mere sets. You've tried to make a definition as "a functor out of the underlying ordinary category of $A$", but by your own definition the underlying ordinary category of $A$ is not abelian. Thus the question becomes meaningless, rather than having a positive answer. | |
| Mar 28, 2020 at 18:45 | comment | added | Todd Trimble | "Being abelian" is property-like structure (see ncatlab.org/nlab/show/…) that is preserved by any equivalence. One could go through this step by step. Is having a zero object preserved by equivalence? Check. Is having biproducts preserved by equivalence? Check. And so on. Now it's quite true that such property-like structure is not preserved by arbitrary functors. But functors that are equivalences? Certainly. | |
| Mar 28, 2020 at 17:44 | comment | added | Bugs Bunny | I politely disagree and stand by my answer. I suggest that some of those, making snippy comments and downvoting my answer, go to Wikipedia and ncatlab and change the definition there to Freyd's definition. If this gets accepted by the Wikipedia cabal, I will withdraw my answer. | |
| Mar 28, 2020 at 17:02 | comment | added | Peter LeFanu Lumsdaine | The absurdity of this answer is clear if we say “Can an even number be equal to a non-even number? Take even $a$. Let $b := a$, but we forget that it is even.” Just because we forget that a number is even doesn’t make it not even. Just because we forget that a category is abelian doesn’t make it not abelian. When you forget the Abelianness, what you haven’t isn’t a non-Abelian category, it’s a category that hasn’t yet been equipped with Abelian structure, but could be. | |
| Mar 26, 2020 at 15:08 | history | edited | Bugs Bunny | CC BY-SA 4.0 |
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| Mar 26, 2020 at 15:07 | comment | added | Kevin Carlson | Well, if you make a wrong definition then you can prove all kinds of things...As we’re doing category theory here it shouldn’t be too controversial that there is no such thing as a bijection between a set and an abelian group, rather than with the underlying set of an abelian group. So you call $A$ abelian after forgetting the enrichment, but refuse to call $B$ abelian even though it admits a canonical enrichment-these choices aren’t consistent with each other. | |
| Mar 26, 2020 at 15:02 | history | edited | Bugs Bunny | CC BY-SA 4.0 |
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| Mar 26, 2020 at 15:01 | comment | added | Bugs Bunny | @Robert Furber Both | |
| Mar 26, 2020 at 15:01 | comment | added | Bugs Bunny | @Lennart Meier It is just a category equivalence. $Hom(X,Y)\rightarrow Hom(FX,XY)$ is a bijection between a set and an abelian group. | |
| Mar 26, 2020 at 14:53 | comment | added | Lennart Meier | What is an equivalence between a category and a category enriched in abelian groups? | |
| Mar 26, 2020 at 14:46 | comment | added | Robert Furber | "Dissenting voice", I think. | |
| Mar 26, 2020 at 14:27 | history | answered | Bugs Bunny | CC BY-SA 4.0 |