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Results tagged with homological-algebra
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user 290
(Co)chain complexes, abelian Categories, (pre)sheaves, (co)homology in various (possibly highly generalized) settings, spectra, derived functors, resolutions, spectral sequences, homotopy categories. Chain complexes in an abelian category form the heart of homological algebra.
6
votes
Accepted
Does the forgetful G-Groups to Groups (nonabelian) have a left adjoint?
Yes. I believe there's a general theorem to the effect that if $T_1 \to T_2$ is any morphism of Lawvere theories, then the corresponding functor $\text{Prod}(T_2, \text{Set}) \to \text{Prod}(T_1, \tex …
- 118k
12
votes
Accepted
Free Objects in Functor Categories
$[C, \text{Ab}]$ admits a forgetful functor to $[\text{Ob}(C), \text{Set}]$ (where $\text{Ob}(C)$ denotes the category with the same objects as $C$ but no non-identity morphisms). This is a direct gen …
- 118k
22
votes
is the tensor product of projective modules again projective?
Recall that $P$ is projective iff $\text{Hom}(P, -)$ is exact. We have
$$\text{Hom}_{A_1 \otimes A_2}(P_1 \otimes P_2, -) \cong \text{Hom}_{A_1}(P_1, \text{Hom}_{A_2}(P_2, -))$$
and a composition of …
- 118k
10
votes
Accepted
A 2-category of chain complexes, chain maps, and chain homotopies?
Part of the problem is that you're not using a convenient definition of chain homotopy. I'll use a differential going down in degree. Let $I$ be the interval object in the category of chain complexes; …
- 118k
15
votes
Compact object and compact generator in a category
Part of the tricky thing about this circle of ideas is that several definitions are not equivalent in full generality but become equivalent with extra hypotheses. For example, a basic result about com …
- 118k
26
votes
Why are injective modules more complicated than projective modules?
Injective modules are of course just projective modules in the opposite category, so it seems to me that the question really is "why is the opposite of a module category more complicated than a module …
2
votes
Are chain complexes over a field always injective?
More abstractly, in an abelian category, saying either that every object is injective or that every object is projective is equivalent to saying that every short exact sequence splits (semisimplicity) …
- 118k
1
vote
Deformations of a complex trivial up to quasi-isomorphism
In general we can consider a differential $d_t$ which depends polynomially on $t$ and whose value at $t = 0$ is our original differential. Simple examples show that we can pick up extra cohomology at …
- 118k
25
votes
Accepted
Any group is a quotient of an acyclic group?
Acyclic groups must in particular have trivial abelianization, so all of their quotients must be perfect.
This is the only obstruction; A.J. Berrick shows in The acyclic group dichotomy (which I just …
- 118k
17
votes
Accepted
The isomorphism class of $\mathrm{Ext}^1_\mathbb{Z}(\mathbb{R}/\mathbb{Z},\mathbb{Z})$
Writing $\mathbb{R}/\mathbb{Z} \cong \mathbb{Q}/\mathbb{Z} \oplus \bigoplus_I \mathbb{Q}$ where $I$ indexes a Hamel basis for $\mathbb{R}$ minus one element, we have
$$\text{Ext}^1(\mathbb{R}/\mathbb{ …
- 118k
7
votes
How to define cohomology of algebraic structures?
There is a tremendous amount of abstract formalism answering this question in various levels of generality depending on what you want to do. I'll pick one in the middle: the machinery of derived funct …
- 118k
6
votes
Accepted
Does homotopy invariance of homology follow from the structure of the simplex category $\Del...
Chris's comment suggests that very little about the target category $C$ is being used in the standard argument, but I still think there's something interesting to check, namely what exactly is being u …
- 118k
12
votes
origin of spectral sequences in algebraic topology
I'm not comfortable enough with spectral sequences to answer this question, but let me answer an easier version of this question with spectral sequences replaced by long exact sequences.
In algebrai …
- 118k
4
votes
Heuristic explanation of why we lose projectives in sheaves.
We can turn the question around to ask: why do we have projectives in module categories? One answer is that we know we have a plentiful supply of projectives because free modules are projective. Abstr …
- 118k
4
votes
Tensor product of linear mappings versus chain complexes
It seems to me that your question is actually about something that is clearer in a more basic context than chain complexes. Fix a field $k$ and let's work instead in the category $C$ of pairs $(V, T)$ …
- 118k