Yet another question "I compute Bredon cohomology of something and I am not sure, whether it is correct".


So I am taking a sign representation $\sigma$ of cyclic group of order 4, $C\_4$. Then I compactify $\sigma$ to get $\mathbb{S}^\sigma$ with two fixed 0-cells and one 1-cell of type $C\_2$. Then I am taking $\underline{\mathbb{Z}}$, constant $\mathbb{Z}$-coefficients. It seems that
$$
C\_{C\_4}^0(\mathbb{S}^\sigma;\underline{\mathbb{Z}})=\mathbb{Z}\oplus\mathbb{Z} \\ C\_{C\_4}^1(\mathbb{S}^\sigma;\underline{\mathbb{Z}})=0,
$$
therefore $H\_{C\_4}^0(\mathbb{S}^\sigma;\underline{\mathbb{Z}})=\mathbb{Z}\oplus\mathbb{Z}$ and $H\_{C\_4}^1(\mathbb{S}^\sigma;\underline{\mathbb{Z}})=0$. 


Degree 1 looks ok for me, but I am worried about degree 0. Here I calculated similar thing for $C\_2$:
[Bredon cohomology of $\mathbb{S}^\sigma$](https://mathoverflow.net/questions/298217/bredon-cohomology-of-mathbbs-sigma)
and degree 0 answer is different. But it should not change, since sign action of $C\_4$ factors through sign action of $C\_2$.


Also, since I am using constant coefficient system, $H\_{C\_4}^\*(\mathbb{S}^\sigma;\underline{\mathbb{Z}})=H^\*(\mathbb{S}^{\sigma}/C\_4;\mathbb{Z})$ (unless it is true only for cyclic groups of prime order) - so degree 0 cohomology should be only one $\mathbb{Z}$, as I am taking non-equivariant cohomology of a contractible space.


So is my answer correct?



Bredon cohomology of a sign representation for a cyclic group of order 4
No, I don't think so - I think there are more Bredon $1$-cochains than that.


The orbit category $\mathcal{O}C\_4$ looks like
$$
C\_4/e \to C\_4/C\_2 \to C\_4/C\_4
$$
where the automorphism groups of the objects are $C\_4$, $C\_2$ and the trivial group, respectively. The Bredon cochains in degree $i$ are natural transformations from the functor $\underline{C}\_i(\mathbb{S}^{\sigma}):\mathcal{O}C\_4\to Ab$ given by $\underline{C}\_i(\mathbb{S}^{\sigma})(C\_4/H) = C\_i((\mathbb{S}^\sigma)^H)$ to the constant functor $\underline{\mathbb{Z}}$. In degree $i=1$ we have $$ \begin{array}{ccccc}\underline{C}\_1(\mathbb{S}^{\sigma})(C\_4/e) &\leftarrow &\underline{C}\_1(\mathbb{S}^{\sigma})(C\_4/C\_2)& \leftarrow &\underline{C}\_1(\mathbb{S}^{\sigma})(C\_4/C\_4)\newline
\parallel & & \parallel & & \parallel \newline
\mathbb{Z}\oplus\mathbb{Z} & \leftarrow & \mathbb{Z}\oplus\mathbb{Z} & \leftarrow & 0 
\end{array}
$$


with the action of $C\_4$ on the bottom left and $C\_2$ on the bottom centre both generated by $(a,b)\mapsto (-b,-a)$. So if I'm not mistaken there is a Bredon cochain given by mapping $(a,b)$ to $a-b$. 


