All questions with tag [math: abelian-categories]


What are exact sequences, metaphysically speaking?

Why is it all-natural or valuable to arrange things (of some ideal group) right into specific series? Specific series are common - and also I've experienced them sufficient to recognize that they can give a really valuable and also reliable structure to function within. Nonetheless, I have no suggestion what this structure absolutely is, or why ...
2022-07-24 09:17:29

Is there an abelian category of topological groups?

There are great deals of reasons that the group of topological abelian teams (i.e. inner abelian teams in $\bf Top$ ) is not an abelian group. So I'm wondering: Is there a "suitably well behaved" subcategory of $\bf Top$ , claim $\bf T$ , such that $\bf Ab(T)$ is an abelian group? My first hunch was to seek well acted topological ...
2022-07-16 17:10:17

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

Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories. Suppose $\mathcal{A}$ has enough injectives, and consider a universal (cohomological) $\delta$-functor $T^\bullet$ from $\mathcal{A}$ to $\mathcal{B}$. By the theory of derived functors, we know that that $T^n (A) = 0$ for all injective objects $A$ and all $n \ge 1$ and that $T^n$ is eff...
2022-07-16 16:57:14

Rows have the same kernel in a pullback square.

Suppose for simpleness that we remain in an abelian group. Intend the adhering to square is a pullback: p P ----> A | | m| |n v v B ----> C q Then m and also n have the very same bit, therefore does p and also q, I've been informed (I could have something in reverse). Just how does one confirm this?...
2022-07-10 07:53:45

Mod-$R$, Mod-$S$ and Mod-$R \otimes S$

Let $R,S,T$ be commutative rings and also think that $R,S$ are $T$ - algebras. In a response to this inquiry, Pierre - Yves Gaillard offers an instance of an $R \otimes_T S$ - component that can not be created as the tensor item of an $R$ - component and also an $S$ - component (there, $T=k$, $R=S=k^2$ where $k$ is an area). I'm interested in ...
2022-07-06 03:34:29

$\operatorname{Func}(J,Ab)$ has enough injectives.

I am trying to show that the functor category $\operatorname{Func}(J,Ab)$ has enough injectives (meaning that for each $F\in \operatorname{Func}(J,Ab)$ there is an injective object $I\in \operatorname{Func}(J,Ab)$ and a monomorphism $F\to I$ in $\operatorname{Func}(J,Ab)$), where $Ab$ is the category of abelian groups and $J$ is the category $$...
2022-07-06 00:58:52

Morita equivalence of acyclic categories

(Crossposted to MathOverflow.) Call a group acyclic so the identification morphisms are invertible and also the endomorphism monoid of every object is unimportant. Allow $C, D$ be 2 limited acyclic groups. Intend that they are Morita matching in the feeling that the abelian categories $\text{Fun}(C, \text{Vect})$ and also $\text{Fun}(D, \tex...
2022-07-04 17:24:07

Image of projective objects.

Let $A$ and also $B$ be 2 abelian categories. Think that there exist a functor $F$ in between them which is specific, complete and also essencially surjective. If $x$ is a projective object in $A$, after that $F(x)$ is a projective object in $B$?
2022-07-04 17:22:06

Equivalent module categories

Let $A$ and also $B$ be rings and also allow $A\text{-mod}$ and also $B\text{-mod}$ be their abelian component groups. Allow $F:A\text{-mod}\to M\text{-mod}$ and also $F':B\text{-mod}\to A\text{-mod}$ be functors which manage an equivalence in between both component groups (i.e. such that $F\circ F'\simeq \operatorname{id}_{A\tex...
2022-07-03 04:45:06

Every chain complex is quasi-isomorphic to a $\mathcal J$-complex

I located this in "Algebra & Topology" by Schapira, yet I'm unable to confirm it: Suppose $\mathcal J$ is a cogenerating family members in an abelian group $\mathbf A$. After that for any kind of favorable facility $X^\bullet\in \text{Ch}(\mathbf A)$ the exists a favorable facility made by $\mathcal J$ - things, and also a seemin...
2022-06-29 01:01:08

Proving the snake lemma without a diagram chase

Suppose we have 2 brief specific series in an abelian group $$0 \to A \mathrel{\overset{f}{\to}} B \mathrel{\overset{g}{\to}} C \to 0 $$ $$0 \to A' \mathrel{\overset{f'}{\to}} B' \mathrel{\overset{g'}{\to}} C' \to 0 $$ and also morphisms $a : A \to A', b : B \to B', c : C \to C'$ ma...
2022-06-25 13:06:28

Derived functors of torsion functor

Let $A$ be a domain. For every $A$-module $M$ consider its torsion submodule $M^{tor}$ made up of elements of $M$ which are annihilated by a non zero-element of $A$. If $f \colon M \to N$ is a homomorphism, then $f(M^{tor}) \subseteq N^{tor}$, so call $f^{tor} \colon M^{tor} \to N^{tor}$ the induced map. We have a covariant functor $^{tor}$ from...
2022-06-09 19:34:56

The smallest subobject $\sum{A_i}$ containing a family of subobjects $A_i$

In an Abelian group $\mathcal{A}$, allow $A_i$ be a family members of subobjects of an object $A$. Just how to show that if $\mathcal{A}$ is cocomplete (i.e. the coproduct constantly exists in $\mathcal{A}$), after that there is a tiniest subobject $\sum{A_i}$ of $A$ having every one of $A_i$? Undoubtedly this $\sum{A_i}$ can not be the coprod...
2022-06-09 16:48:24

Abelian categories and axiom (AB5)

Let $\mathcal{A}$ be an abelian group. We claim that $\mathcal{A}$ satisfies (AB5) if $\mathcal{A}$ is cocomplete and also filtered colimits are specific. In Weibel is Introduction to homological algebra, he mentions (without evidence) that $\mathcal{A}$ pleases axiom (AB5) iff $\mathcal{A}$ is cocomplete and also for all latticeworks $\{ A_i ...
2022-06-09 14:32:13

Is $\mathbf{R}F(Z^\bullet)$ equal to $\mathbf{K}F(Z^\bullet)$ when $Z^\bullet$ consists of $F$-acyclic objects?

I'm not exactly sure just how I can show the following: If $F \colon \mathcal{A} \to \mathcal{B}$ is a left specific functor from an abelian group $\mathcal{A}$ to an abelian group $\mathcal{B}$ , whose acquired functor $\mathbf{R}F$ in the feeling of acquired groups exists, after that the adhering to holds: If $Z^\bullet$ is a facility c...
2022-06-09 14:11:50

Arbitrary products of quasi-coherent sheaves?

I have a brief inquiry: Does the group of seemingly - systematic sheaves on a system have approximate items? I recognize that it does if the system is affine and also I recognize that they will certainly not be isomorphic to the item as $\mathcal{O}_{X}$ - components, yet I have not had the ability to locate a counterexample. Any kind of aid i...
2022-06-09 01:05:04

kernel of cokernel is cokernel of kernel

Possible Duplicate: ¢ Equivalent conditions for a preabelian category to be abelian Let $\mathcal{C}$ be an abelian group, and also take into consideration an arrowhead $f:A\rightarrow B$. In a variety of resources (Vakil is notes, the appendix of Weibel is publication, Wikipedia, individual discussion), I've seen it kept in mind that $\oper...
2022-06-08 17:07:03

Equivalent conditions for a preabelian category to be abelian

Let is deal with some terms first. A group $\mathcal{C}$ is preabelian if: 1) $Hom_{\mathcal{C}}(A,B)$ is an abelian team for every single $A,B$ such that make-up is biadditive, 2) $\mathcal{C}$ has an absolutely no object, 3) $\mathcal{C}$ has binary items, 4) $\mathcal{C}$ has bits and also cokernels. A group $\mathcal{C}$ is abelian if...
2022-06-08 05:49:39

Meaning of "efface" in "effaceable functor" and "injective effacement"

I'm reviewing Grothendieck is Tōhoku paper, and also I wondered concerning the thinking behind the terms "effaceable functor" and also "injective effacement". I recognize that in English, to obliterate something suggests to eliminate it, yet I'm not exactly sure if there is an additional definition in either conversational or...
2022-06-08 05:32:38

Uniqueness of Kernels in Abelian Categories

Refering to Serge Lang is "Algebra" pp. 133 - 134 on Abelian Categories, the adhering to is vague to me. Allow $Q$ be an additive group and also $F\stackrel{f}\rightarrow E$ a morphism. Allow $A \stackrel{\alpha}\rightarrow F$ and also $B \stackrel{\beta}\rightarrow F$ be 2 group - logical bits of $f$ (as specified in the message). It ...
2022-06-08 05:07:26