# Bredon cohomology of $\mathbb{S}^\sigma$

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.

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 $\epsilon\colon M(*)\to M(G)$.
• Ok, which agrees with my statement assuming that $\epsilon$ is an iso on $M(G)^G$. Thanks! – Igor Sikora Apr 19 '18 at 12:53