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A branch of algebraic topology concerning the study of cocycles and coboundaries. It is in some sense a dual theory to homology theory. This tag can be further specialized by using it in conjunction with the tags group-cohomology, etale-cohomology, sheaf-cohomology, galois-cohomology, lie-algebra-cohomology, motivic-cohomology, equivariant-cohomology, ...
10
votes
1
answer
280
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Induced map on $H_4$ of Eilenberg–MacLane spaces
A map $f:K(A, 1) \to K(B,2)$ is given by a degree 2 cohomology class. …
19
votes
1
answer
2k
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Integral cohomology ring of K(Z,3)
Computing the cohomology of Eilenberg Maclane spaces is a feasible but difficult problem in algebraic topology. … What is the integral cohomology ring of K(Z,3)?
You can get quite far with spectral sequence calculations, but if this is worked out in detail somewhere, why reinvent the wheel? …
45
votes
Difference between represented and singular cohomology?
The functor $[-, K(\mathbb{Z}, n)]$ always agrees with singular cohomology.
This brings us to the issue of what exactly a cohomology theory is supposed to be? … But it is in good company Cech cohomology and sheaf cohomology also fail this litmus test, so many people outside of algebraic topology feel uncomfortable with this axiom. …
19
votes
Twistings for other cohomology theories
Twisted forms exist for all multiplicative generalized cohomology theories. … If E is a generalized cohomology theory, represented by a spectrum also denoted E, then the E-cohomology of X coincides with the homotopy classes of maps
$$[ \Sigma^{-i} X, E] $$
i.e. the "E-valued …
34
votes
2
answers
5k
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Example Wanted: When Does Čech Cohomology Fail to be the same as Derived Functor Cohomology?
I want to know exactly how derived functor cohomology and Cech cohomology can fail to be the same. … with the derived functor version of sheaf cohomology. …
16
votes
Why do gerbes live in H^2?
I have a couple things to say.
First, believe your definition of gerbe is slightly incorrect. When you say that your stack is locally isomorphic to $U \times B\mathbb{G}_m$, this isomorphism needs to …
5
votes
1
answer
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Classifying Algebra Extensions over a fixed extension?
There are lots of "Ext groups" in homological algebra which measure extensions of various things. I'm sure there must be a homological algebra machine for computing the following, and I'm hoping that …
5
votes
Representablity of Cohomology Ring
Cohomology in degree n is represented by the (pointed) space K(Z, n), as you pointed out. … Then the product R of all the K(Z, n) where n ranges over all non-negative integers is the representing object for the whole cohomology ring. …