Bredon cohomology of a permutation action on $S^3$ I've seen a couple of similar questions asking to verify computations of Bredon cohomology here and here, so I will ask one such question myself.
Let $\mathbb{Z}/2$ act on $S^3\subset \mathbb{C}^2$ by restriction of a permutation action on $\mathbb{C}^2.$ I wanted to compute Bredon cohomology $\mathcal{H}^*_{\mathbb{Z}/2}(S^3;\underline{\mathbb{Z}}).$
I have a cell decomposition based on a decomposition of complex $1$-dimensional disk into $3$ cells: $\mathbb{D}=D\sqcup T\sqcup *.$ Here $T\sqcup *=S^1=\partial \mathbb{D}$ and $D$ is the interior of $\mathbb{D}.$ Then we have a decomposition of $S^3=\mathbb{D}\times S^1 \cup S^1\times \mathbb{D}$ into cells compatible with the $\mathbb{Z}/2$ action.
The fixed point set of an action is a circle given by $\{z_1=z_2\}\cap S^3\subset \mathbb{C}^2.$ Since the orbit category of $\mathbb{Z}/2$ consists of $*$ and $\mathbb{Z}/2$ there are the following equivariant chains:
\begin{array}{|c|c|c|c|}
\hline
\operatorname{dim} &*& \mathbb{Z}/2 & \operatorname{cells corresponding to} \underline{C}_n(S^3)(\mathbb{Z}/2)\\
\hline
0 & \mathbb{Z} & \mathbb{Z} & * \times *\\
1 & 0 & \mathbb{Z}\oplus\mathbb{Z},\quad \begin{pmatrix} 1 \\ 0\end{pmatrix}\xrightarrow{\overline{1}} \begin{pmatrix} 0 \\ 1\end{pmatrix} & T\times *,*\times T\\
2 & 0 & \mathbb{Z}\oplus \mathbb{Z} \oplus \mathbb{Z},\quad \begin{pmatrix} 1 \\ 0\\0\end{pmatrix}\xrightarrow{\overline{1}}\begin{pmatrix} 0 \\ 1\\0\end{pmatrix};\;\begin{pmatrix} 0 \\ 0\\1\end{pmatrix}\xrightarrow{\overline{1}}\begin{pmatrix} 0 \\ 0\\-1\end{pmatrix} & D\times *, *\times D, T\times T\\
3 & 0 & \mathbb{Z}\oplus \mathbb{Z},\quad \begin{pmatrix} 1 \\ 0\end{pmatrix}\xrightarrow{\overline{1}} \begin{pmatrix} 0 \\ 1\end{pmatrix} & D\times T, T\times D\\
\hline
\end{array}
So it seems that the cochains valued in $\underline{\mathbb{Z}}$ are:
\begin{array}{|c|c|}
\hline
\operatorname{dim} & \\
\hline
0 & \mathbb{Z}\\
1 & \mathbb{Z}\\
2 & \mathbb{Z}\\
3 & \mathbb{Z}\\
\hline
\end{array}
Since $(T\times T)^*=0$ in cochains, we have $\mathcal{H}^3_{\mathbb{Z}/2}(S^3;\underline{\mathbb{Z}})=\mathbb{Z}.$ Differential $d_1$ is an isomorphism since $\partial(D\times *)=T\times *.$ It seems that $\mathcal{H}^*_{\mathbb{Z}/2}(S^3;\underline{\mathbb{Z}})=H^*(S^3;\mathbb{Z}).$
It is a bit odd to me that the quotient is a homological sphere. Sure, the group $\mathcal{H}^3_{\mathbb{Z}/2}(S^3;\underline{\mathbb{Z}})=\mathbb{Z}$ since orientation is preserved, but maybe I've missed some $2$-torsion in lower degrees?
 A: Your final answer is correct, but the cell structure you're using isn't a $G$-CW structure: $T\times T$ can't be used as a cell in this way.
I would approach it like this: The action of $G = {\mathbb Z}/2$ on $\mathbb{C}\times\mathbb{C}$ can be written as the representation $\mathbb{C}\oplus\mathbb{C}^\sigma$, where $G$ acts trivially on $\mathbb{C}$ and by negation on $\mathbb{C}^\sigma$. The sphere $S(\mathbb{C}\oplus\mathbb{C}^\sigma)$ is also the one-point compactification $S^{1+2\lambda}$, where $\lambda$ denotes the real line with $G$ acting by negation. This has a $G$-CW structure with


*one $G$-fixed 0-cell,

*one $G$-fixed 1-cell,

*one $G$-free 2-cell, and

*one $G$-free 3-cell,

so that the skeleta are $*$, $S^1$, $S^{1+\lambda}$, and $S^{1+2\lambda}$. From here you can work out that the $\underline{\mathbb{Z}}$-cochain complex is
$$
\mathbb{Z} \xrightarrow{0} \mathbb{Z} \xrightarrow{1} \mathbb{Z} \xrightarrow{0} \mathbb{Z}.
$$
A way to check that the answer is correct is to write
$$
H_G^n(S^{1+2\lambda}) \cong \tilde H_G^n(S^0) \oplus \tilde H_G^n(S^{1+2\lambda})
\cong \tilde H_G^n(S^0)\oplus \tilde H_G^{n-1-2\lambda}(S^0)
$$
and then use the known calculation of the $RO(G)$-graded cohomology of a point (originally due to Stong (unpublished), since published in various places).
