Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [homological-algebra]

Homological algebra studies homology and cohomology groups in a general algebraic setting, that of chains of vector spaces or modules with composable maps which compose to zero. These groups furnish useful invariants of the original chains.

A chain complex is a sequence of abelian groups, vector spaces, or modules, with linear maps connecting them which compose to zero.

Homological algebra is the study of chain complexes and their homology groups.

5114 questions
5
votes
1 answer

How to prove $\mathrm{Im}(\mathrm{Ext}_R^1(g,A'))=\mathrm{Ker}(\mathrm{Ext}_R^1(f,A'))$

I'm reading MacLane's "Homology" and got stuck at the proof of the following fact. Theorem. Let $E:0\xrightarrow{}A\xrightarrow{f}B\xrightarrow{g}C\xrightarrow{}0$ be a short exact sequence of left $R$-modules. Let $A'$ be a left $R$-module, then…
Norbert
  • 56,803
5
votes
1 answer

Is each $F$-acyclic resolution homotopic to a projective resolution?

Here is an excerpt from some notes I stumbled upon online: In fact, the elementary homological algebra proof that right derived functors’ definitions do not depend upon the choice of injective resolution shows that, if $$ 0 \longrightarrow …
user153312
5
votes
1 answer

Quasi-isomorphism and homotopical equivalence

I am currently studying some homological algebra, I have a couple of questions concerning the notion of quasi-isomorphism and homotopical equivalence. For two complexes on an abelian category $X^\bullet$ and $Y^\bullet$ to be homotopically…
Pgatti
  • 932
4
votes
3 answers

Are all cyclic groups $\mathbb{Z}/n\mathbb{Z}$ cotorsion?

Is it correct that all cyclic groups $\mathbb{Z}/n\mathbb{Z}$ are cotorsion, i.e., $Ext^1(\mathbb{Q},\mathbb{Z}/n\mathbb{Z})=0$? I believe yes, since $Ext^1(\mathbb{Q}/\mathbb{Z},\mathbb{Z}/n\mathbb{Z})\simeq\mathbb{Z}/n\mathbb{Z}$, so I suspect…
Jacqueline Pauwels
  • 1,077
4
votes
1 answer

Compute $\mathrm{Tor}_{n}^{\mathbb{Z}_{8}}(\mathbb{Z}_{4},\mathbb{Z}_{4})$

Let $\mathbb{Z}_{4}$ as a $\mathbb{Z}_{8}$-module. How can I prove $\mathrm{Tor}_{n}^{\mathbb{Z}_{8}}(\mathbb{Z}_{4},\mathbb{Z}_{4})=\mathbb{Z}_{4}$? Please see this.
someone
  • 535
4
votes
0 answers

Does this square of $\delta$-functors anti-commute?

Let $\mathcal{A}$, $\mathcal{B}$ be abelian categories, and let $F^\bullet: \mathcal{A} \rightarrow \mathcal{B}$ be a cohomological $\delta$-functor. Recall that a cohomological $\delta$-functor is a series of additive functors $F^{n}: \mathcal{A}…
Theone
  • 333
4
votes
2 answers

How can we motivate homotopy in homological algebra?

I'm writing some notes on homological algebra for some readers which are experienced in algebra (had a course in commutative algebra using categorical notions) but who don't necessarily know algebraic topology. With this in mind, I began to wonder…
Gabriel
  • 4,513
4
votes
2 answers

If $A$ has enough projectives, then so does the category $Ch(A)$ of chain complex over $A$

This is Exercise 2.2.2. from Weibel's An Introduction to Homological Algebra. Suppose $A$ is an abelian category, if $A$ has enough projectives, then so does the category $Ch(A)$ of chain complex over $A$. Weibel's hint is to use the fact: a chain…
Cille
  • 401
4
votes
0 answers

Are homology and cohomology really dual to each other?

I don't remember if I've already seen this question even here or in MO or in my mind. This is partly related to questions arose about differences between homology and cohomology; I'm wondering if some kind of difference can appear before computing…
fosco
  • 11,814
4
votes
2 answers

when to use projective vs. injective resolution

I am a bit confused about when I should use projective versus injective resolutions to calculate derived functors. Am I correct in thinking that for right exact functors, the left derived functor is defined using projective resolutions and for left…
bob
  • 247
4
votes
1 answer

Projective dimenson of tensor product

I've been struggeling for some time with the following problem Let $k$ be a field and $A$ and $B$ two $k$-algebras. We can then view the tensor product $A\otimes_k B$ as a $k$-algebra by $(a_1\otimes b_1)\cdot (a_2\otimes b_2)=(a_1a_2\otimes…
M88
  • 87
  • 3
4
votes
2 answers

Is the mapping cone a split complex?

This question arise from the comment in Corollary 1.5.4 in Weibel's An introduction to Homological Algebra: a cochain map $f\colon B\to C$ is a quasi-isomorphism if and only if the mapping cone complex is exact; hence this reduces questions about…
Pedro
  • 4,994
4
votes
1 answer

Is $\mathbb{Q}/\mathbb{Z}$ decomposable?

I would like to see if this $\mathbb{Q}/\mathbb{Z}$ as a $\mathbb{Z}$ module is decomposable. I'm trying to come up with a decomposition that looks like: $$ \mathbb{Q}/\mathbb{Z} = \bigoplus_{p} \left(\prod_{k} \frac{1}{p^k} \mathbb{Z} \right).…
nekodesu
  • 2,724
4
votes
1 answer

Hom cochain complex of two chain complexes

Does anybody know of a good reference (preferably online) where I can find a good, rigorous description of $Hom_R(C_\bullet,D_\bullet)_\bullet$, which is a cochain complex where the module is the nth degree is given as $\Pi_{i-j=n}Hom_R(C_i,D_j)$,…
Jonathan Beardsley
  • 4,503
4
votes
2 answers

Proving the isomorphism in homology

Let $A$ be a chain complex and let $B \subseteq A$ be a sub complex. Assume there is a chain map $\alpha: A \to A$ such that the following conditions are fulfilled (i) $\alpha$ is chain homotopic to the identity map on $A$ (ii) $\alpha(B) \subseteq…
user502427
  • 161
1 2 3
…
25 26