I tried to compute Bredon cohomology of $\mathbb{S}^\sigma$, where $\sigma$ is a sign representation of $\mathbb{Z}/2$, following first chapter and first construction of cohomology from Bredon's "Equivariant cohomology theories". Could somebody please verify it, at least the result?


Throughout $G$ means $\mathbb{Z}/2$.


So I assume the following: $G$-CW structure is obvious, given by two points with trivial action as 0-cells and two arcs with swapping action as 1-cells. I am using simple coefficient system $\mathcal{L}$ on it, that is my functor from "cellular category" to abelian groups factors through some coefficient system $M$. $M$ consists of two groups $M(\*)$ - trivial $G$-module and $M(G)$ - $G$-module, and an equivariant map $\epsilon:M(\*)\rightarrow M(G)$.


$C^0(\mathbb{S}^\sigma;\mathcal{L})$ consists of the functions $f:\{e^0\_1,e^0\_2\}\rightarrow M(\*)$. Since action of G is trivial on 0-cells, induced action on 0-chains is also trivial, therefore $C^0(\mathbb{S}^\sigma;\mathcal{L})=C^0\_G(\mathbb{S}^\sigma;\mathcal{L})=M(\*)^2$.
$C^1(\mathbb{S}^\sigma;\mathcal{L})$ consists of the functions $f:\{e^1\_1,e^1\_2\}\rightarrow M(G)$. Action on 1-cells is non-trivial (even free), so $C^1\_G(\mathbb{S}^\sigma;\mathcal{L})$ consists of equivariant $f$'s. Thus $C^1\_G(\mathbb{S}^\sigma;\mathcal{L})=M(1)$.


The only non-trivial differential is $\delta :C^0\rightarrow C^1$ and is given by $(\delta f)(\tau)=\pm\epsilon(f(e^0\_1))\mp\epsilon(f(e^0\_2))$. Here $\tau$ of course means any of two 1-dimensional cells.


So $H^0\_G(\mathbb{S}^\sigma;\mathcal{L})=M(\*)$ and $H^1\_G(\mathbb{S}^\sigma;\mathcal{L})=M(G)/M(G)^G$ - but for this I have to assume that $\epsilon$ is an iso on $M(G)^G$.


If this is not "mathoverflow" question, I can ask it also on MathStack.



Bredon cohomology of $\mathbb{S}^\sigma$
In general, you're going to get $H\_G^0(\mathbb S^\sigma;\mathcal L) = M(\*)\oplus\ker\epsilon$ and $H\_G^1(\mathbb S^\sigma;\mathcal L) = M(G)/\operatorname{im}\epsilon = \operatorname{coker}\epsilon$. This is probably easier to see if you compute the reduced cohomology, where the cochain complex becomes simply $\epsilon\colon M(\*)\to M(G)$.


