# Abelian category equivalent to a non-abelian category [closed]

I was told that if we have an equivalence of categories $$F : \mathcal{A} \rightarrow \mathcal{B}$$ with $$\mathcal{A}$$ abelian, then it is not necessarily true that $$\mathcal{B}$$ is also abelian.

I would like to know if there are nice examples of an abelian category $$\mathcal{A}$$ which is equivalent to a non-abelian category $$\mathcal{B}$$.

Furthermore, are there any conditions over $$F$$ or $$\mathcal{B}$$ so that we have "$$F : \mathcal{A} \rightarrow \mathcal{B}$$ is an equivalence and $$\mathcal{A}$$ is abelian implies $$\mathcal{B}$$ is abelian"?

• This is a surprise for me. Never heard this before.. May be I don’t know what exactly is an equivalence of categories :D Have you tried to write down some examples of categories, equivalence of categories and see if you can cook up something... Sorry for the not so useful comment... – Praphulla Koushik Mar 26 at 14:02
• 6 up votes for such a comment is making me nervous :D.. I read the comment now, it sounds like a satirical comment.. This is not intended to be a satirical comment.. Please do not read it as a satirical comment... – Praphulla Koushik Mar 27 at 14:57
• Has the coronavirus infected the central nervous systems of so many of the honorable members of math overflow? This question is uninteresting. If your definition of abelian category is a category satisfying some caterogical property, then the answerr is trivially yes, exactly as "a group isomorphic to an abelian group is abelian". If your definition of an abelian category is a category with additional structure satisfying some property, then the question does not make any sense, because $\mathcal B$ is not given any structure. Bugs Bunny seems the only non-crazy person around. – Joël Mar 29 at 0:31
• @Joël Note that Bugs explicitly claims the answer could be yes, inconsistent with your options, which I think everyone else agrees about. – Kevin Carlson Mar 29 at 1:01

What you were told is wrong, for we have the following:

Proposition. If two categories are equivalent and one of them is abelian, then so is the other.

A proof (and some related results) can be found in Satz 16.2.4 in H. Schubert, Kategorien II, Springer, 1970 (likewise in the English version https://www.amazon.com/Categories-Horst-Schubert/dp/3642653669, under the same numbering).

• Just to make very clear: the proof is not difficult — it could be a slightly boring homework question in any course that introduces Abelian categories. All the components of the definition of Abelian category are pretty straightforwardly invariant under equivalence of categories for all of the definitions of Abelian category I know. If the OP has read some definition of Abelian category for which this isn’t straightforward, I’d be interested to hear what that definition is. – Peter LeFanu Lumsdaine Mar 28 at 16:54
• @PeterLeFanuLumsdaine :the first time I read the definition of an abelian category, it was as an $\mathbf{Ab}$-enriched category with such-and-such properties, so the $\mathbf{Ab}$-enrichment was taken as additional structure in that definition (that was essentially because the documents I was reading wanted to prove some general stuff about $\mathbf{Ab}$-enriched categories before going on to abelian ones). With this in mind, if you've never seen how the $\mathbf{Ab}$-enrichment actually follows from the properties of your abelian category, it's not obviously clear that it should be invariant – Maxime Ramzi Mar 29 at 11:23
• @Maxime: That’s the main definition I think of too; but even without realising that the enrichment is determined by the other properties (which I agree is not at all obvious), I think it’s pretty straightforward to see that Ab-enrichment transfers along équivalences of categories. The only subtle aspect of this is having the confidence to read “invariant under equivalence” as something you can assert about a structure, not just a property. – Peter LeFanu Lumsdaine Apr 1 at 3:08

Here is a manifestly invariant definition of an abelian category $$\mathcal{C}$$. It is a category with finite limits and colimits such that:

1. (It is pointed) the map from the initial to the final object is an isomorphism; we denote by 0 any object which is both initial and final.
2. (It is semiadditive) the map $$X \amalg Y \to X\times Y$$, given on $$X$$ by components $$(\mathrm{id}_X, X \to 0 \to Y)$$ and on $$Y$$ by components $$(Y \to 0\to X, \mathrm{id}_Y)$$, is an equivalence. We denote by $$X\oplus Y$$ the coproduct or product, identified as above. This equivalence produces an abelian monoid structure on all hom sets, where addition arises from $$X \to X\oplus X \stackrel{f\times g}{\to} Y\oplus Y \to Y$$.
3. (It is additive) the shearing map $$X\oplus X \to X\oplus X$$, given by adding the identity map to the projection onto the first component followed by inclusion, is an equivalence. Equivalently, each hom-monoid has the property that it is group-like.
4. (first isomorphism theorem) if $$f: A \to B$$ is arbitrary, then the map $$A/\mathrm{ker}(f) \to \mathrm{ker}(B \to B/A)$$ is an isomorphism.

Being an abelian category is a property not structure.

• Off course, but this is not the definition here: ncatlab.org/nlab/show/abelian+category – Bugs Bunny Mar 26 at 14:36
• And this is the only meaning I can possible give to the statement in question... – Bugs Bunny Mar 26 at 14:37
• Just for reference: this is the definition in Freyd's Abelian categories (so hardly an obscure one :)) – Denis Nardin Mar 26 at 14:48
• Note that 2. and 3. are actually consequences of 1. and 4. – Arnaud D. Mar 27 at 9:22

As a discerning dissenting voice, let me say that it might be true, depending on your definitions.

Take abelian $$A$$. Let $$B:=A$$ with forgotten enrichment in abelian groups. Then the identity functor is equivalence, but $$B$$ is not abelian because it is not even additive.

There are two definitions in play. The first is abelian category. The standard (ncatlab or Wikipedia) definition of abelian category asks for the category to be pre-additive, which requires enrichment in abelian groups that can be forgotten. An alternative definition, mentioned in other answers, is from Freyd's book: there abelian is an inherent property rather than an additional structure. This definition would make the statement false.

The second definition is non-abelian category. There is no accepted definition. One possibility is to think of this as just a category. Together with the first definition of abelian category, this makes the statement true. An alternative is to think of a category that cannot be abelian (essentially the negative of the second definition). This choice would make the statement false.

• "Dissenting voice", I think. – Robert Furber Mar 26 at 14:46
• What is an equivalence between a category and a category enriched in abelian groups? – Lennart Meier Mar 26 at 14:53
• 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. – Kevin Carlson Mar 26 at 15:07
• 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. – Peter LeFanu Lumsdaine Mar 28 at 17:02
• "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. – Todd Trimble Mar 28 at 18:45