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(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.
51
votes
8
answers
7k
views
Motivating the category of chain complexes
Let $R$ be a commutative ring. For awhile I have been trying to motivate to myself more fully the definition of and various structures on the category $\text{Ch}(R)$ of chain complexes of $R$-modules …
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 …
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{ …
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 …
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 …
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 …
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{ …
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 …
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 …
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 …
13
votes
2
answers
906
views
Is there an analogue of the Lefschetz fixed point theorem for discrete dynamical systems?
Background/Motivation
Let $(X, f)$ be a discrete dynamical system. For now, $X$ is just a set and $f$ is just a function $f : X \to X$. Suppose that $f^n$ has a finite number of fixed points for ev …
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 …
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 …
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; …
9
votes
3
answers
559
views
For G a Lie group, can I make sense of G/G as a derived manifold in a nice way?
The functor sending a smooth manifold $M$ to its de Rham algebra $\Omega^{\bullet}(M)$ does not send quotients by actions of Lie groups to invariant subalgebras. The example I have in mind is a connec …