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Results tagged with homological-algebra
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user 4008
(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.
16
votes
Accepted
Where can I easily look up / calculate (abelian) group cohomology?
This group is best understood in terms of the universal coefficient formula,
i.e., in terms of the homology of the involved group. Hence, if $A$ is any
abelian group we have $H_1(A)=A$ and the additio …
- 22.6k
14
votes
Accepted
Central extensions of group schemes
If we have a central extension of group schemes $1\rightarrow B \rightarrow C\rightarrow
A\rightarrow1$ with $A$ abelian, then we get a commutator mapping
$\Lambda^2A\rightarrow B$ (of sheaves as $\La …
- 22.6k
14
votes
What is a "block" in an abelian category?
It seems clear to me that blocks should have something to do with the
decomposition of the category as a direct product of subcategories. A
decomposition into a product of two factors corresponds exac …
- 22.6k
14
votes
Accepted
Injection of Ext into H^2
You get a description from the universal coefficient theorem which gives a (split) exact sequence
$$
0\to \mathrm{Ext}(H_1(G),A) \to H^2(G,A) \to \mathrm{Hom}(H_2(G),A) \to 0
$$
and the fact that $H_1 …
- 22.6k
12
votes
A ring such that all projectives are stably free but not all projectives are free?
This is an attempt to complete Tyler's argument. We first note that
$KO^0(S^5)=\mathbb Z$ (note this true for all spheres of dimension $\equiv 5,6,7 \bmod 8$). This means that every topological vector …
- 22.6k
9
votes
Splitting of the Universal Coefficients sequence
I would claim that the splitting (and indeed the whole universal coefficient
theorem) is not really a topological theorem. If we take the homological version
one really works with the chain complex $C …
- 22.6k
8
votes
Accepted
Does Ext commute with direct limit?
For the first question you already have had an answer in Is it true that if $\operatorname{Ext}^{1}_{A}(P,A/I)=0 $ for all $ I$ then $P$ is projective? if $\mathrm{Ext}^1_{\mathbb Z}(P,M)=0$, then it …
- 22.6k
7
votes
Accepted
Coinciding induced maps
The answer is no. Consider the complex over the integers $A$ which is $\mathbb Z$ in degree $0$ and $1$ and the only non-trivial differential being multiplication by $2$ and let $B$ be the same comple …
- 22.6k
5
votes
Accepted
Explicit injective resolutions of (Laurent) polynomial rings
$\newcommand{\C}{\mathbb C}
$I think this is OK.
The first step is the inclusion of $\C[X,Y]$ into its fraction field which is
$\C(X,Y)$. For each irreducible polynomial $f$ (normalised so that the to …
- 22.6k
3
votes
Accepted
$M \oplus N$ is of finite type if $M,N$ are of finite type?
At least if we have a Grothendieck category everything seems OK: Suppose
$0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow0$ is exact with $M'$ and
$M''$ of finite type. Assume $\{M_i\}$ is a di …
- 22.6k