33

Here is a manifestly invariant definition of an abelian category $\mathcal{C}$. It is a category with finite limits and colimits such that:
(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.
(It is semiadditive) the map $X \amalg Y \to X\times Y$, given on $X$ by ...

32

To answer your first precise questions:
Yes, every distinguished triangle in $D(A)$ comes from a short exact sequence. For every distinguished triangle $X \to Y \to \mathrm{Cone}(f) \stackrel{+1}\to $ there is a short exact sequence
$$ 0 \to Y \to \mathrm{Cone}(f) \to X[-1] \to 0$$
of complexes in $A$, and our distinguished triangle arises from this one ...

31

There is a majestic paper by Mac Lane
MacLane, Saunders. "Possible programs for categorists." Category Theory, Homology Theory and their Applications I. Springer, Berlin, Heidelberg, 1969. 123-131.
whose opening line is one of the most beautiful I've ever read:
Communication among Mathematicians is governed by a number of unspoken rules. One of these ...

27

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, ...

23

The abelian categories in which all short exact sequences split I would call "split abelian categories", reserving the term "semisimple abelian category" for a more restrictive condition. Roughly, perhaps an abelian category should be called "semisimple" if all objects in it are coproducts of simple objects (a nonzero object is called simple if it has no ...

answered Apr 13 '19 at 0:28

Leonid Positselski

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22

I think I have an example.
Fix a chain of fields $k_\alpha$ indexed by ordinals $\alpha$, where $k_\alpha\subset k_\beta$ is an infinite field extension for all pairs $\alpha<\beta$ of ordinals.
First I'll define an "abelian category" which has large Hom-sets.
An object $V$ will consist of a $k_\alpha$-vector space $V(\alpha)$ for each ordinal $\alpha$,...

answered Oct 15 '14 at 11:31

Jeremy Rickard

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21

Through extended TQFT and the cobordism hypothesis, many questions in topological quantum field theory have been turned into explicit questions about higher category theory.
A TQFT is formalized as a symmetric monoidal functor $Z\colon\mathsf{Bord}_n\to\mathsf C$, where $\mathsf C$ is some symmetric monoidal $(\infty, n)$-category and $\mathsf{Bord}_n$ is ...

20

The category of countable abelian groups is an essentially small abelian category, and has enough projectives and injectives (the countable free abelian groups and the countable divisible groups respectively). However, there is an injective with endomorphism ring $\mathbb{Q}$, but no such projective, so the categories of projectives and injectives can't be ...

18

I apologize in advance for this very long answer. I am pretty sure that many people could write a better version of it. Unfortunately, they are not doing it. So, here we are.
The very beginning of the question asks for some even classical results in category theory that are relevant outside category theory. My knowledge is very limited thus I will list just ...

18

Taking a brief look at the Roos note we see that detailed proofs of the statements aren't provided. There is no argument in the note one could say is wrong. The paper with the corrected statement very carefully points out which parts of the original note do hold and which ones have to be modified and how and gives detailed arguments or precise references for ...

ag.algebraic-geometry at.algebraic-topology ct.category-theory homological-algebra abelian-categories

17

Here is a specific example, though it admits obvious generalizations. Let $R=S=\mathbb{Z}$, and consider the functor from abelian groups to abelian groups defined by
$$T(X) = \mathrm{Ext}^1_{\mathbb{Z}}(\mathbb{Q}_p/\mathbb{Z}_p,X),$$
where $p$ is a prime.
This is right-exact since $\mathrm{Ext}^2_{\mathbb{Z}}\equiv0$. It does not commute with direct ...

17

Since the dual of an abelian category is also an abelian category, the question is equivalent to the same question for projective resolutions.
I will show that the category $\mathbf{Ab}^{\operatorname{f.t.}}$ of finitely generated abelian groups has enough projectives, but no functorial projective cover. The idea is that multiplication by any $n \in \mathbf ...

answered Apr 8 at 22:26

R. van Dobben de Bruyn

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16

The following pair of examples follows the idea of Jeremy Rickard suggested in a comment on Math Stack Exchange under the link. Inverting the arrows, it suffices to construct an example of abelian category $\mathcal A$ such that the category of right exact functors $\mathcal A \to \mathcal Ab$ or $\mathcal A \to k{-}\mathcal{V}ect$ is not abelian (where $k$ ...

answered Apr 29 '18 at 17:59

Leonid Positselski

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I am posting the comment above as an answer.
An equivalence of categories preserves identity morphisms, finite product, and finite coproducts. Thus, it also preserves diagonal morphisms and codiagonal morphisms. In an Abelian category, the finite product equals the finite coproduct. For every $f,g\in \text{Hom}_{\mathcal{A}}(a,a')$, the morphism $f+g$ ...

14

The category $[C^{op}, \text{Ab}]$ of $\text{Ab}$-valued presheaves on any (small, for simplicity) $\text{Ab}$-enriched category is about as nice as it gets - locally finitely presentable, Grothendieck, etc. - and all coproducts of representables are projective (these are the "free" objects), but it won't be a module category in general if $C$ has infinitely ...

ag.algebraic-geometry ct.category-theory homological-algebra abelian-categories locally-presentable-categories

answered Jan 7 '18 at 19:49

Qiaochu Yuan

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13

An interesting class of examples comes from the following construction, also due to Freyd. Let $\mathcal{T}$ be a small triangulated category. Form a new category $\mathcal{A}$, with one object $I(u)$ for each morphism $u$ in $\mathcal{T}$. The morphisms from $I(u:A\to B)$ to $I(v:C\to D)$ are the pairs $(f:A\to C,g:B\to D)$ such that $gu=vf$, modulo ...

answered Apr 20 '13 at 20:02

Neil Strickland

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For each object $c$ in $\mathcal{C}$, let $c^* : [\mathcal{C}, \mathcal{A}] \to \mathcal{A}$ be evaluation at $c$. It is an exact functor, so if a left adjoint $c_! : \mathcal{A} \to [\mathcal{C}, \mathcal{A}]$ exists, $c_!$ will preserve projective objects. Assume $\mathcal{C}$ has $\le \alpha$ morphisms and $\mathcal{A}$ has coproducts for families of $\le ...

11

The derived category of an abelian category has a t-structure, so obviously that's the first thing you want.
To a t-structure corresponds a heart, which is an abelian category, whose derived category might be different than the one you started with.
To further complicate matters, as you noted, you could have a heart whose derived category is actually ...

11

If $A$ is a $k$-algebra, and $M$,$N$ are finite-dimensional $A$-modules, then
$$\operatorname{Ext}^i_A(M,N)\cong\operatorname{Tor}^A_i(M,N^*)^*$$
(where $*$ denotes $k$-dual).
So $\operatorname{Ext}^i_A(M,N)$ must be the dual of a vector space, and so in particular its dimension can't be countably infinite.
For $i=1$ it makes no difference whether we take $...

answered Apr 16 '15 at 12:41

Jeremy Rickard

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11

I think you are a little confused about what your characterization of $R$ does, and this causes problems as you generalize to sheaves.
There is no characterization of $R$ as an element of the category of $R$-modules via abstract nonsense - this is because tensoring by any rank $1$ locally free module is an autoequivalence of this category. Instead, what you ...

11

You really seem to be looking for intuition for the triangulated structures on derived categories of Abelian categories, so here goes:
(Co-)chain complexes are like (Abelianised) pointed homotopy types; homotopy types are typically first encountered in the guise of topological spaces, so it is in the category $\mathbf{Top}_*$ of pointed topological spaces ...

11

Question 1: Yes. The $I$-coproduct-functor $\bigsqcup_I\colon\prod_{i\in I}\mathbf{C}\to\mathbf{C}$ is left-adjoint (its right adjoint is the diagonal functor $\Delta_{\mathbf{C}}^I\colon \mathbf{C}\to\prod_{i\in I}\mathbf{C}$), hence always preserves epimorphisms.
Question 2: No, in general (even if $\mathbf{C}$ is $I$-coproduct-complete). Cocomplete ...

10

In Appendix C (Corollary C.3.3 to be precise) of Neeman's book "Triangulated Categories" an example of an abelian category which is not well-powered is given.
The actual counterexample is given by $A(D(R))$ where $D(R)$ is the unbounded derived category of a discrete valuation ring $R$, and $A(D(R))$ is the category of finitely presented additive functors $...

10

You are correct, the converse also holds. This is a mistake in Weibel, see this answer.

answered Jan 30 '17 at 3:03

Gabriel C. Drummond-Cole

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10

For some abelian categories it is also very easy to describe such a ring quite explicitly if the category you start with is similar enough to a module category.
Let's say you consider $\mathsf{Ch}(A\mathsf{-mod})$ and that $A$ is a $\mathbb{Q}$-algebra. Then $\mathsf{Ch}(A\mathsf{-mod})$ embeds as a full subcategory into $B-\mathsf{mod}$ where
$$B:=A \...

9

There is a construction of Peter Freyd that embeds any triangulated category $\mathcal{T}$ in an abelian category $\mathcal{A}(\mathcal{T})$. Explicitly, we start with the category of arrows in $\mathcal{T}$. Given a morphism $u$ in $\mathcal{T}$, I'll write $I(u)$ for the same thing regarded as an object of the arrow category. Next, we identify two ...

answered Apr 12 '12 at 12:01

Neil Strickland

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9

The category of abelian groups is small-complete, well-powered, and has a cogenerator (e.g., $\mathbb{Q}/\mathbb{Z}$). It follows from the Special Adjoint Functor Theorem that any limit-preserving functor $G: Ab \to Ab$ has a left adjoint $F$. (A proof of the SAFT may be found on this nLab page.) And as Dylan Wilson pointed out in a comment, we then have $G \...

answered Mar 23 '16 at 19:10

8

I don't know what 'explicit' means, but this is all covered in Higher Algebra very well.
If you want a simplicial category at the end of the day, you can either...
Take your favorite $\infty$-category (quasi-category) presentation, and straighten it to a simplicial category. (But no one wants to do this... it doesn't sound like fun.)
Take the category of ...

8

There is a notion of "Bridgeland stability condition" which includes a t-structure. Those have a reasonable moduli space. See papers of Bridgeland for details.

8

Here's an example of a full exact embedding of the module category of a non-Noetherian ring $S$ into that of a Noetherian ring $R$, preserving all direct sums and direct products. So this gives an example of an abelian subcategory of $R\text{-mod}$ that is not injectively closed, but satisfies many additional "niceness" properties.
Let $k$ be a field, and $...

answered May 17 '17 at 10:25

Jeremy Rickard

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