Skip to main content
14 votes
Accepted

How much data does a category contain?

It indeed looks like you need extra information to answer the question whether a given category $\mathcal{C}$ is an abelian category, but this is actually not true. An abelian category is an additive ...
Daniël Apol's user avatar
  • 5,987
13 votes
Accepted

Why is the category of finitely generated modules over a non-noetherian ring not abelian?

Claim: The inclusion $R\text{-mod}\hookrightarrow R\text{-Mod}$ preserves kernels. Once this is known, it follows that a kernel in $R\text{-mod}$, if it exists, must be isomorphic to the ...
Hanno's user avatar
  • 19.6k
13 votes
Accepted

Coimage and Image in Abelian Categories

Notice firstly that $f(\operatorname{ker}f) = 0$ and $(\operatorname{coker}f)f = 0$. Use this to argue that $f = (\operatorname{im}f)f''(\operatorname{coim} f)$. Once you've done that, we want to ...
Rylee Lyman's user avatar
  • 3,930
12 votes
Accepted

What's wrong with my understanding of the Freyd-Mitchell Embedding Theorem?

You don't show that $\mathcal{L}^{op}$ actually is a category of modules, just that any small subcategory of it fully and exactly embeds in one. Indeed, an abelian category with a projective ...
Eric Wofsey's user avatar
12 votes
Accepted

Commutative ring category is not additive category

The ring $0$ is not the zero object. Indeed a zero object is an object which is both terminal and initial but in rings with unit the initial object is $\mathbb{Z}$ and the terminal one is $0$. ...
Sov's user avatar
  • 1,048
11 votes
Accepted

Convolution in Hopf algebras

You can try "Hopf algebras. An introduction" by Dascalescu, Nastasescu, Raianu. They have a detailed account with full proofs and further examples and exercises. Kassel's book "Quantum groups" and ...
KonKan's user avatar
  • 7,354
11 votes
Accepted

Where have I used commutative (coproduct in $\mathbf{Ab}$ vs. $\mathbf{Grp}$)

The typical proof that your $A \times B$ satisfies the universal property of the coproduct in $\mathbf{Ab}$ is as follows: if $f: A \to C$ and $g: B \to C$ are two homomorphisms of abelian groups, ...
Joshua Mundinger's user avatar
10 votes

Why do universal $\delta$-functors annihilate injectives?

In the 10 years since I asked this question, I learned some other theories of derived functors, and the fact that there has been no answer has been forthcoming leads me to believe that the question ...
Zhen Lin's user avatar
  • 90.6k
10 votes
Accepted

Opposite of a category of modules is not a category of modules

Thinking about Grothendieck categories is making things way more complicated than necessary to prove just that $\bf{R-Mod}^{op}$ is not a module category. You can prove this by simply thinking about ...
Eric Wofsey's user avatar
10 votes

Homological algebra using nonabelian groups

A good setting for generalizing homological algebra is that of homological categories. There is a book on the subject due to Borceux and Bourn references at the linked article, as well as a very ...
Kevin Carlson's user avatar
10 votes
Accepted

Writting objects in abelian category as limit of injective objects

This is false. To my mind it's a little easier to think about the dual question: in an abelian category with enough projectives, is every object a filtered colimit of projective objects? In the ...
Qiaochu Yuan's user avatar
10 votes
Accepted

Definition of a presheaf on a topological space with values in an abelian category

No, that does not follow from the definition as a contravariant functor, and it is incorrect to add it to the definition of a presheaf. I've never seen a source that does this, and it would be ...
Kevin Carlson's user avatar
9 votes

Is there an abelian category of topological groups?

This was alluded to in the comments and may not be what you're looking for, but it surely deserves mention that you can take $\mathbf{T}$ to be the category of compact Hausdorff spaces. The category $...
Eric Wofsey's user avatar
9 votes

A characterization of AB5 categories. Does this result show up in the literature?

We will deduce Proposition 1 from the upcoming two lemmas. Lemma 2. Let $F:\mathcal{A}\to\mathcal{B}$ be an additive functor between preabelian categories. The following are equivalent: $F:\mathcal{...
Elías Guisado Villalgordo's user avatar
9 votes
Accepted

Is taking (co)limits exact in an Abelian category?

In complete generality you only preserve one side of the short exact sequence. Taking colimits is a right exact operation while taking limits is a left exact operation and in general neither is two-...
Geoff's user avatar
  • 1,594
8 votes

Is this categorical definition of homology correct and, furthermore, used to teach homology in some book?

This is exactly how Michael Barr develops homology theory in section 2.4 of his book Acyclic Models.
Vladimir Sotirov's user avatar
8 votes
Accepted

In an abelian category, does every family of subobjects have an intersection?

Here is a counterexample. Let $k$ be your favorite field, and let $C$ be the category of "eventually constant" sequences of vector spaces over $k$. That is, an object of $C$ is a sequence $(V_n)$ of ...
Eric Wofsey's user avatar
8 votes

Long exact sequence of cohomology group "without" Snake lemma

Let's see that the long exact sequence theorem is equivalent to the snake lemma. From the short exact sequence we have an induced diagram $$\require{AMScd}\begin{CD}&&L^n/B^n(G,L)@>>&...
Rafael's user avatar
  • 3,829
8 votes
Accepted

Zero element of Ext$^n$ groups

If $A$ and $B$ are objects in the abelian category $\mathcal{A}$, the (equivalence class) of the exact sequence $$0\to A\to A\oplus B\to B\to 0$$ is the zero object in $\text{Ext}^{1}(B,A)$. For $n&...
Zeek's user avatar
  • 1,867
8 votes
Accepted

The category of monoid objects in an abelian category

The category of monoid objects in an abelian category $A$ with respect to the cartesian product is just $A$ again; this follows from the fact that 1) every object is canonically and uniquely a ...
Qiaochu Yuan's user avatar
8 votes
Accepted

Was Grothendieck familiar with Buchsbaum's exact categories while writing Tohoku?

In a letter dated July 13th,1955 Serre informed Grothendieck that Buchsbaum (a Samuel Eilenberg Ph.D student) had developed a "system very similar to your [Grothendieck's] abelian classes", ...
Georges Elencwajg's user avatar
7 votes

Why are hearts of t-structures of triangulated categories abelian?

I understand your question as being a reference request. This is a theorem first proved in Beilinson, Bernstein and Deligne's paper Faisceaux pervers, published in Astérisque. The result in question ...
Pierre-Guy Plamondon's user avatar
7 votes
Accepted

Colimit of a direct system of monomorphisms

Not necessarily. For instance, let $\mathcal{A}$ be the category of finite abelian groups and take the direct system $$\mathbb{Z}/2\mathbb{Z}\stackrel{2}{\to}\mathbb{Z}/4\mathbb{Z}\stackrel{2}{\to}\...
Eric Wofsey's user avatar
7 votes

A chain complex is contractible iff it is split exact

A couple of housekeeping things: First, the title of your question is false; there exist chain complexes which are exact but not split. For instance, one can check that if $R=\mathbb{Z}$, then the ...
Eric's user avatar
  • 1,632
7 votes

Motivation for spectrum of an Abelian category

The basic motivation for this definition is that if $\mathbf{A}$ is the category of modules over a commutative ring $R$, then the $\sim$-equivalence classes of $\operatorname{Spec}(\mathbf{A})$ are ...
Eric Wofsey's user avatar
7 votes
Accepted

Limits and $Hom(-,Y)$-functor in abelian categories

Yes, it is true in any abelian category. In fact, moreover, it's true in every category full stop that $$\hom(\text{colim}_i x_i,y)\cong \lim_i\hom(x_i,y).$$ (in category theory, a projective limit is ...
ziggurism's user avatar
  • 16.9k
7 votes
Accepted

Extension group and Baer sum

The Baer sum is not that surprising. Here is how I see it: The group structure on Hom-set may be recovered from the biproduct $\oplus$. Indeed, if $f:A\to B$ and $f':A\to B$ are two parallel arrows, ...
Roland's user avatar
  • 12.6k
7 votes
Accepted

Can we use the internal logic of a category to do diagram chases "as in $\mathbf{Ab}$" ?

This MathOverflow question and answer and the referenced slides and the comments both by Ingo Blechschmidt seem directly aimed at your question. See also the nLab page on diagram chasing whose section ...
Derek Elkins left SE's user avatar

Only top scored, non community-wiki answers of a minimum length are eligible