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Nov
16 
awarded  Commentator 
Nov
16 
comment 
Transgression in terms of kinvariant for chain complexes
I'm also confused, especially by what "multiplication by" means, as this is a map from H^n(X) to H^(n+1) (BG). But I'm very interested if there is a general description of this transgression map. I suggest that M is meant to be the top cohomology group of X. Also perhaps multiplication means composition product, but then shouldn't it be Ext(M,k) rather than (k,M)? 
Sep
17 
comment 
Generation of cohomology of graded algebras
I may just be confused with the gradings/notation, but doesn't it seem like this should follow directly from the cobar complex being of finite type (since A is)? EDIT: never mind, I was indeed just confused. 
Sep
15 
comment 
Computations in modular cohomology of finite groups
You're welcome! By the way, I just posted the thesis to the Arxiv (arxiv.org/abs/1509.03910) to make it less of a pain to track down. 
Sep
14 
awarded  Nice Answer 
Sep
14 
awarded  Yearling 
Sep
13 
answered  Computations in modular cohomology of finite groups 
Sep
24 
awarded  Autobiographer 
Jan
13 
awarded  Critic 
Mar
3 
awarded  Editor 
Mar
3 
revised 
On the cohomology of a finite covering map
added 13 characters in body 
Mar
3 
comment 
On the cohomology of a finite covering map
Yes, that's a much quicker way to see it than the chainlevel argument I had in mind. Of course spectral sequence for $X\to X/G\to BG$ will not collapse without some assumptions... if you want a totally general answer (i.e. with integer coefficients) there will be no way to avoid doing a spectral sequence. I pointed out in my answer (because I thought the questioner was thinking of it) that the spectral sequence for $G\to X\to X/G$ does collapse, but without giving any useful information  sorry that wasn't clear. 
Mar
2 
comment 
On the cohomology of a finite covering map
Chris: the confusion here is that some people are interpreting $H^*(G)$ to be the singular cohomology of $G$, and others the group cohomology. I don't know which one the questioner intended (but group cohomology would make more sense).

Mar
2 
comment 
On the cohomology of a finite covering map
Right, perfect! Of course, in this case (free action) the Borel construction is homotopyequivalent to the orbit space $X/G$. A small quibble: it's not a principle bundle (the fiber is $X$, not $G$) but merely a fiber bundle with structure group $G$. Of course, it's still a fibration so we get a Serre spectral sequence as desired. 
Mar
2 
comment 
On the cohomology of a finite covering map
(added) The thing that goes wrong in that example is that 2, the order of $G$, is not a unit in the coefficient ring. 
Mar
2 
comment 
On the cohomology of a finite covering map
Careful, hypotheses are needed here (to show that $H^*(X/G)$ is the $G$ invariants of $H^*(X)$ ). Example: take the covering space $S^2\to\mathbb{R}P^2$ with integer coefficients. Then $H^2(X)=0$ but $H^2(X/G)\neq0$.

Mar
2 
comment 
On the cohomology of a finite covering map
Also: I have a vague memory that there's another sequence you can use, in case you want the more general case (I haven't checked this!): If memory serves, there is a different fibration, $X\to X/G\to BG$. The latter map is the classifying map of the principle bundle $X\to X/G$. But studying this would give a relationship involving the group cohomology of $G$ (maybe that's what you meant in the question though....) 
Mar
2 
answered  On the cohomology of a finite covering map 
Jan
10 
answered  Theorems that are 'obvious' but hard to prove 
May
23 
awarded  Supporter 