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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.
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
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
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
15
votes
Dual of a bimodule
As explained in more detail in this blog post linked by Jakob in the comments, every $(A, B)$-bimodule $M$ has two natural duals:
If $M$ is finitely generated projective as a left $A$-module, it has …
- 118k
15
votes
Is every "nice" abelian category with enough projectives an additive presheaf category?
The category $[C^{op}, \text{Ab}]$ of $\text{Ab}$-valued presheaves on any (small, for simplicity) $\text{Ab}$-enriched category is about as nice as it gets - locally finitely presentable, Grothendiec …
- 118k
30
votes
intuition for hochschild homology
Slogan: Hochschild homology is a (derived) categorification of the trace.
This means the identity at the end of John Pardon's answer is a categorification of the identity $\text{tr}(AB) = \text{ …
- 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
$$\text{Ext}^1(\mathbb{R}/\mathbb{ …
- 118k
2
votes
Are chain complexes over a field always injective?
More abstractly, in an abelian category, saying either that every object is injective or that every object is projective is equivalent to saying that every short exact sequence splits (semisimplicity) …
- 118k
4
votes
Heuristic explanation of why we lose projectives in sheaves.
We can turn the question around to ask: why do we have projectives in module categories? One answer is that we know we have a plentiful supply of projectives because free modules are projective. Abstr …
- 118k
6
votes
Understanding the purely formal part of the sheaf theoretic (cohomological) framework for re...
Let $G$ be an affine group scheme (e.g. a finite group). There is a stack $BG$ which is more or less determined by either of the following two properties:
Maps to $BG$ are the same thing as $G$-tor …
- 118k
37
votes
Does module Hom commute with tensor product in the second variable?
You can think about tensor products as a kind of colimit; you're asking the hom functor $\text{Hom}_A(L, -)$ to commute with this colimit in the second variable, but usually the hom functor only commu …
- 118k
6
votes
Accepted
Does homotopy invariance of homology follow from the structure of the simplex category $\Del...
Chris's comment suggests that very little about the target category $C$ is being used in the standard argument, but I still think there's something interesting to check, namely what exactly is being u …
- 118k
4
votes
Lyndon–Hochschild–Serre spectral sequence for a non-normal subgroup
I don't think so. The LHS spectral sequence can be thought of as the Serre spectral sequence associated to the fiber sequence
$$BN \to BG \to B(G/N)$$
where $G$ is a group and $N$ is a normal subgro …
- 118k
22
votes
is the tensor product of projective modules again projective?
Recall that $P$ is projective iff $\text{Hom}(P, -)$ is exact. We have
$$\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
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