41
votes
Name for abelian category in which every short exact sequence splits
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, ...
38
votes
Accepted
What is a triangle?
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 $...
38
votes
What's there to do in category theory?
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.
...
Community wiki
35
votes
Abelian category equivalent to a non-abelian category
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 ...
26
votes
Accepted
Abelian category equivalent to a non-abelian category
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 ...
24
votes
Accepted
Is the category of left exact functors abelian?
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 ...
23
votes
Accepted
Categorical presentation of direct sums of vector spaces, versus tensor products
One way to think about what the monoidal structure on vector spaces is doing is that it is telling us that vector spaces do not really form a category, or not "just" a category: they form a ...
23
votes
Accepted
Functorial kernel in derived category
Let $\mathcal{C}$ be a stable $\infty$-category. Then $\mathcal{C}$ has a homotopy category $h \mathcal{C}$, which is triangulated. The collection of morphisms $f: X \rightarrow Y$ of $\mathcal{C}$ ...
22
votes
Accepted
What was the error in the proof of Roos' theorem?
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 ...
22
votes
Accepted
Example of an abelian category with enough projectives and injectives which are not dual
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 ...
21
votes
What's there to do in category theory?
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 ...
Community wiki
20
votes
Accepted
Are there (enough) injectives in condensed abelian groups?
Indeed, there are no nonzero injective condensed abelian groups.
Let $I$ be an injective condensed abelian group. We can find some surjection
$$ \bigoplus_{j\in J} \mathbb Z[S_j]\to I$$
for some index ...
18
votes
What's there to do in category theory?
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 ...
Community wiki
18
votes
Accepted
Abelian category with enough injectives but not functorially
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}^{\...
18
votes
Abelian categories that are not monoidal
Let $\mathcal{A}$ be an additive category with a monoidal structure such that the maps
$$ \otimes \colon \mathcal{A}(A,B)\times\mathcal{A}(C,D) \to
\mathcal{A}(A\otimes C,B\otimes D)
$$
are ...
18
votes
Abelian categories that are not monoidal
In the paper
Hovey, Mark, Additive closed symmetric monoidal structures on R-modules, J. Pure Appl. Algebra 215, No. 5, 789-805 (2011). ZBL1223.18005.
Hovey shows the following theorem (Theorem 3.3)
...
17
votes
Accepted
When is the category of finitely presented modules abelian?
Wojowu's idea is right:
Lemma. Let $R$ be a ring, let $\mathbf{Mod}_R$ be the category of (left) $R$-modules, and let $\mathbf{Mod}_R^{\text{fp}}$ be the subcategory of finitely presented modules. ...
16
votes
Accepted
In an abelian category with no nontrivial Serre subcategories, does every short exact sequence split?
The category of finite abelian $p$-groups (where $p$ is your favourite prime) is an abelian category with no proper nonzero Serre subcategories, but not every short exact sequence splits.
15
votes
Accepted
Let $F:\mathscr{A}\to\mathscr{B}$ be an equivalence of Abelian categories. Must $F$ be additive?
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 ...
Community wiki
15
votes
Accepted
Can a category be enriched over abelian groups in more than one way?
You can easily find examples among categories with one element: a category with one element is a (multiplicative) monoid, and $Ab$-enrichment over it is a choice of an addition which turns it into a ...
14
votes
What is a triangle?
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 ...
14
votes
Is every "nice" abelian category with enough projectives an additive presheaf category?
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, ...
14
votes
How exotic can an infinite biproduct in an additive category be?
I think your memory is right on Question 1: an Eilenberg swindle implies that if an infinite bipower $\bigoplus_X A$ of an object $A$ exists in an additive category, then $A=0$. The point is that ...
14
votes
Accepted
Is the category of chain complexes a reflexive or coreflexive subcategory of the category of functors?
Yes, it's reflective and coreflective, under mild assumptions on the codomain category $\mathcal A.$ The adjoints are given, by definition, by Kan extension along the quotient from the abelian group-...
13
votes
Accepted
Concrete examples of Freyd-Mitchell embedding
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\...
13
votes
Accepted
Abelian categories satisfying AB5*
The snarky response would be "the opposite category of any of the categories you could name on the spot". The less-snarky response is to observe that some of these are quite natural. For ...
12
votes
Can $\mathcal O_X$ be recognized abstract-nonsensically?
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 ...
12
votes
Accepted
Is a categorical coproduct of epimorphisms (monomorphisms) always an epimorphism (a monomorphism)?
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}\...
12
votes
Do the isomorphism classes of indecomposable objects in $R{\text{-mod}}$ form a set?
In Conjecture $1_{\infty}$ of
Simson, Daniel, On large indecomposable modules, endo-wild representation type and right pure semisimple rings., Algebra Discrete Math. 2003, No. 2, 93-118 (2003). ...
12
votes
What are abelian categories enriched over themselves?
To make sense of enrichment over a category $V$, you want $V$ to have a monoidal structure. Indeed, you want to be able to compose morphisms so you need a way to go from "something in $\hom(a,b)$ ...
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