Computing Bredon Cohomology of Z/2-spheres? Can anyone suggest me how to calculate explicitly the Bredon Cohomology of the sphere(at least for 2-dimensional case) with antipodal Z/2-action with constant coefficient system associated to integers?
 A: EDIT: per request.
We can be pretty explicit in this case: the $n$-sphere with antipodal action has a $\Bbb Z/2$-equivariant cell structure whose $k$-skeleton is the $k$-sphere.  As a result, we get a cellular chain complex of coefficient systems:
$$
0 \leftarrow C_0 \leftarrow C_2 \leftarrow \cdots \leftarrow C_n
$$
where $C_k$ is the coefficient system associated to $(S^k,S^{k-1})$.  According to Bredon's definitions, we calculate the value of the coefficient system on the trivial orbit $*$ as $H_k((S^k)^{\Bbb Z/2},(S^{k-1})^{\Bbb Z/2})$, which is trivial; we calculate the value on the orbit $\Bbb Z/2$ as $H_k(S^k,S^{k-1}) = \Bbb Z \oplus \Bbb Z$ where the generator $g$ of $\Bbb Z/2$-action swaps the two factors.
Therefore, in each degree the coefficient group is the same coefficient system $C$.  More, one can calculate the cellular chain maps using the same homology calculation (one knows that it has to recover the homology of $S^n$) and find that it is a sequence of maps
$$
0 \leftarrow C \stackrel{1-g}{\leftarrow} C \stackrel{1+g}{\leftarrow} \cdots
$$
Now we apply Hom out to any coefficient system $M$.  You can show $Hom(C,M)$ is $M(\Bbb Z/2)$.  The resulting chain complex is
$$
0 \to M(\Bbb Z/2) \stackrel{1-g}{\to}
M(\Bbb Z/2) \stackrel{1+g}{\to}
M(\Bbb Z/2) \stackrel{1-g}{\to} \cdots
$$
Substituting $M = \Bbb Z$, we get the cochain complex computing the cohomology of $\Bbb{RP}^n$.
This is part of a more general result here when the action is free and properly discontinuous, and the group $\Bbb Z/2$ acts trivially on the value $\Bbb Z$ of the coefficient system on $\Bbb Z/2$.  In these cases, the Bredon cohomology coincides with cohomology of the orbit space.
Therefore, we're getting the cohomology of $(S^n) / \Bbb Z/2 = \Bbb{RP}^n$ with coefficients in $\Bbb Z$ (hence $\Bbb Z$ in degree 0, $\Bbb Z/2$ in even positive degrees, and 0 otherwise).
