bio  website  

location  Seattle  
age  28  
visits  member for  4 years, 9 months 
seen  Jan 14 '12 at 6:59  
stats  profile views  115 
I'm a grad Student at the University of Washington, studying algebraic topology.
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 
May 18 
comment 
Equality vs. isomorphism vs. specific isomorphism
Also: absolutely anything involving vector bundles. :) 
May 12 
answered  Equality vs. isomorphism vs. specific isomorphism 
May 1 
awarded  Teacher 
May 1 
answered  Is there any geometry where the triangle inquality fails? 