2
votes
1answer
48 views

If direct limits of matrices are isomorphic, is the direct limit of the transpose matrices also isomorphic?

On the one hand, the following conjecture seems reasonable, but on the other hand it doesn't seem natural because some objects are being dualised while others are not. I would appreciate if anyone ...
4
votes
0answers
42 views

Commutativity of direct and inverse limits

In exercise 5.34(iv) of Homological Algebra book by Rotman one is asked to prove that direct limits and inverse limits do not necessarily commute. I have two questions : 1.) Is it true that ...
5
votes
1answer
67 views

The splitting lemma and uniqueness

For the sake of concreteness, let's restrict discussion to the category of abelian groups. Throughout, $$ 0 \to A \overset{q}{\to} B \overset{r}{\to}C \to 0$$ is a short exact sequence. One part of ...
1
vote
1answer
61 views

Do the circle groups have any interesting stand-alone descriptions?

By the circle groups, I mean firstly the circle group $\mathbb{T} \subseteq \mathbb{C}$ of all complex numbers having modulus $1$, and secondly the commutative group $\mathbb{S} = \mathbb{T} \cap ...
4
votes
1answer
33 views

Can this be proved purely on base of UMP?

Let $A,B$ be abelian groups and let $P$ serve as a product with projections $p_{A}:P\rightarrow A$ and $p_{B}:P\rightarrow B$. Let $C$ be an abelian group and let $f:C\rightarrow A$ and ...
6
votes
2answers
133 views

Computing easy direct limit of groups

How do I start computing easy direct limit of groups: 1) $\mathbb{Z} \overset{1}\longrightarrow \mathbb{Z} \overset{2}\longrightarrow \mathbb{Z} \overset{3}\longrightarrow \mathbb{Z} ...
5
votes
0answers
101 views

When is $K_0(i)$ an injection?

Suppose that $\mathcal A$ and $\mathcal B$ are two abelian categories such that $\mathcal A$ is a full subcategory of $\mathcal B$. If $i: \mathcal A\rightarrow\mathcal B$ is the inclusion functor, ...
3
votes
0answers
78 views

Directed Colimits exact in the category of abelian groups

Starting right from the defintions, what would be the shortest way to prove, that the category of abelian groups, $\mathcal{Ab}$, has exact directed limits (This means for every directed set $I$ is ...
3
votes
1answer
144 views

Cocartesian squares in the category of abelian groups.

Recently, I've been doing a recap of (basic) category theory and found an old exercise I seem to be unable to solve. The question is as follows. Let $A, B$ be abelian groups, $A'<A$ and $B'<B$ ...
7
votes
1answer
472 views

Equivalences and isomorphisms of short exact sequences

In case it's necessary, I'm working in the category $\mathbf{Ab}$ of abelian groups. My question concerns what I find to be a strange way of viewing the elements of the Ext group $\mbox{Ext}(A,B)$ of ...
5
votes
1answer
122 views

Additive category and zero map

Let $A$ be an additive category. Namely $A$ has a zero object, $A$ has finite products and coproducts, and Every Hom-set is an Abelian group such that composition of morphisms is bilinear. ...
3
votes
1answer
135 views

The free abelian group monad

The forgetful functor $U : \mathsf{Ab} \to \mathsf{Set}$ is monadic, this follows from Beck's monadicity theorem or some other general result. Anyway, I would like to prove this directly, thereby ...
2
votes
0answers
66 views

Inductive vs projective limit of sequence of split surjections II

This question is a follow-up of this earlier question I asked. Let $$ A_1\twoheadrightarrow A_2\twoheadrightarrow A_3\twoheadrightarrow A_4\twoheadrightarrow \cdots $$ be an inductive sequence of ...
3
votes
1answer
88 views

Inductive vs projective limit of sequence of split surjections

Let $$ A_1\twoheadrightarrow A_2\twoheadrightarrow A_3\twoheadrightarrow A_4\twoheadrightarrow \cdots $$ be an inductive sequence of abelian groups, the connecting homomorphisms of which are ...
9
votes
4answers
388 views

Coproducts in $\text{Ab}$

I am currently trying to understand why finite products and coproducts in the category $\text{Ab}$ coincide. In fact, I'm not even sure I can show it. My question is the following: Is there an ...
5
votes
1answer
254 views

Adjoint of forgetful functor between category of vector spaces and category of abelian groups

I've just found out about the forgetful functor between the category of vector spaces and the category of abelian groups. It maps a vector space to it's additive abelian group. My question is, is ...
4
votes
1answer
227 views

Is this the free abelian group functor?

Let $\mathbb{Z}(.) : \mathbf{Set} \to \mathbf{Ab}$ be the functor that assigns to any set $S$ the set of maps $\mathbb{Z}(S) := \{ z: S \to \mathbb{Z} \; | \; z(s)=0 \mbox{ for almost all } s \in S ...
3
votes
1answer
153 views

Simplify the category of finite abelian groups

Consider the category $\mathsf{FinAb}$ of finite abelian groups. The structure theorem tells us that we can write down a skeleton for this category (a set of representatives for the isomorphism ...
0
votes
1answer
291 views

Show that fiber products exist in the category of abelian groups.

Show that fiber products exist in the category of abelian groups. In fact, If $X, Y$ are abelian groups with homomorphisms $f: X \to Z$ and $g: Y \to Z$ show that $X \times_z Y$ is the set of all ...
3
votes
1answer
341 views

Calculating Hom(A,B)

I have been studying modules and homological algebra as of late but somehow I have missed how to calculate Hom(A,B) for abelian groups, modules and Hom(A,_)/Hom(_,B) for exact sequences. I have no ...
0
votes
1answer
72 views

$\mbox{Hom}(A,\mathbb{Z}) \ne 0$ if and only if $A$ has an infinite cyclic direct summand

If $A$ is an abelian group, then $\mbox{Hom}(A,\mathbb{Z}) \ne 0$ if and only if $A$ has an infinite cyclic direct summand. The hint is to use If $F$ is a free abelian group and $g:B \to F$ is a ...
13
votes
2answers
943 views

Why is every abelian group the direct limit of its finitely generated subgroups?

I'm taking classes in homological algebra now, and the book (together with the lecturer) seem to assume more category theory than I already know. A "fact" that is used freely in the book ...