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1 vote

Equivariant Coefficient ring action on singular cohomology

$EG$ is also the universal free $G$-space, meaning that, if $X$ is a free $G$-space (let's assume of the $G$-homotopy type of a $G$-CW complex), there is, up to $G$-homotopy, a unique $G$-map $X\to EG …
Steve Costenoble's user avatar
3 votes
Accepted

Bredon cohomology of a permutation action on $S^3$

Your final answer is correct, but the cell structure you're using isn't a $G$-CW structure: $T\times T$ can't be used as a cell in this way. I would approach it like this: The action of $G = {\mathbb …
Steve Costenoble's user avatar
10 votes

Allowing $G$-CW complexes to have more general cells

Let me start with the fact that, in one sense, it's true that Type 1 complexes are all that are "needed." That's true in the sense that complexes built from Type 2 and 3 cells have the $G$-homotopy ty …
Steve Costenoble's user avatar
2 votes

Bredon cohomology of a sign representation for a cyclic group of order 4

To expand on Mark Grant's answer, but looking at it slightly differently: You need to look at the chains as contravariant functors on the orbit category, and they will be projective functors. $\underl …
Steve Costenoble's user avatar
5 votes
Accepted

Calculations of cup products in Bredon cohomology

Frankly, there aren't many calculations out there. Most of the work I know of is on the calculation of the $RO(G)$-graded cohomology of a point, of a projective space, or of $B_GO(n)$. Here are some r …
Steve Costenoble's user avatar