# Bredon cohomology of $\mathbb{S}(2\sigma)$

My question refers to Tyler Lawson's answer to this question: Computing Bredon Cohomology of Z/2-spheres?

Namely, I have a problem with understanding 0'th degree. From my calculation it seems that $H^0(\mathbb{S}(2\sigma);M)=M(G)^G$. Here $G=\mathbb{Z}/2$ and $\sigma$ is a sign representation. Thus if we substitute $\mathbb{Z}$ with a sign action for $M(G)$ we will get that 0 degree cohomology is $0$ rather than $\mathbb{Z}$. Where is a mistake?

There's no mistake. Tyler's answer of $\mathbb Z$ in degree 0 holds if $M(G) = \mathbb Z$ with trivial action of $G$, whereas you're correctly getting 0 if $G$ has the nontrivial action. More generally, if you take $\mathbb S(n\sigma)$, the cohomology with constant coefficients $\mathbb Z$ will look like the nonequivariant cohomology of $\mathbb{RP}^n$ with $\mathbb Z$ coefficients, while the cohomology with a coefficient system with $G$ acting nontrivially on $\mathbb Z$ will look like the cohomology of $\mathbb{RP}^n$ with twisted $\mathbb Z$ coefficients, which starts with 0 (then continues with alternating $\mathbb Z/2$ and 0).