In the paper *On the Castelnuovo-Mumford regularity of the cohomology ring of a group*, Symonds describes the following space.


Let $G = (\mathbb{Z}/2\mathbb{Z})^2 = \{1,a,b,ab\}$ be an elementary abelian $2$-group, and let $Z = \{ z\_1, \dots, z\_6\}$ be a discrete space with six elements. Let $G$ act on $Z$ such that each rank one subgroup of $G$ is a stabilizer of some element of $Z$. If I'm not mistaken, this means that the action is for example given as follows:
$$a \cdot z\_1 = b \cdot z\_1 = z\_2;\\
a \cdot z\_3 = z\_4; \; b \cdot z\_3 = z\_3; \; b \cdot z\_4 = z\_4;\\
b \cdot z\_5 = z\_6; \; a \cdot z\_5 = z\_5; \; a \cdot z\_6 = z\_6.$$


Let $SZ$ be the suspension of $Z$, with the induced $G$-action.


**Question.** How to compute the equivariant cohomology $H^\*\_G(SZ)$ as an $H^\*\_G$-module (with coefficients $\mathbb{F}\_2$)?


I believe I could compute the equivariant cohomology $H^\*\_G(Z)$, using the Mayer–Vietoris sequence for equivariant cohomology (it split as a disjoint union of three $G$-spaces), the Künneth theorem, and the known results about the equivariant cohomology of spaces with a free (resp. trivial) action. But unfortunately there's no suspension isomorphism for equivariant cohomology, as far as I can tell...


Of course I know the definition of equivariant cohomology; the classifying space $BG$ is $(\mathbb{RP}^\infty)^2$, and $H^\*\_G = \mathbb{F}\_2[x,y]$ is a polynomial algebra on two variables of degree $1$. The total space of the universal bundle is $(S^\infty)^2$ with $a$ (resp. $b$) acting by the antipodal action on the first (resp. second) factor, and $H^\*\_G(SZ) = H^\*(EG \times\_G SZ)$. But this isn't exactly helpful, or at least I don't see how to compute the equivariant cohomology just from that.



Computing the equivariant cohomology of a specific $(\mathbb{Z}/2\mathbb{Z})^2$-space
The standard way to compute equivariant cohomology of a $G$-space $X$ is to use the spectral sequence of the fibration
$$X\to EG\times\_G X\to BG,$$
where the projection is induced by $X\to \ast$. With $\mathbb{F}\_2$ coefficients this takes the form
$$
E\_2^{p,q} = H^p(BG; H^q(X;\mathbb{F}\_2))\Rightarrow H^\*\_G(X;\mathbb{F}\_2).
$$
In your case, the space $X=SZ$ is connected and one-dimensional, so there are only two non-trivial rows: $q=0$ (which just gives $H^\*(BG;\mathbb{F}\_2) = H^\*\_G$) and $q=1$ (the cohomology of $G=\mathbb{Z}/2\times\mathbb{Z}/2$ with coefficients in the module $H^1(SZ;\mathbb{F}\_2)$). Hence the $E\_2$-term is computable.


There is only one possible non-trivial differential $$d\_2:H^p(BG;H^1(SZ;\mathbb{F}\_2))\to H^{p+2}(BG;H^0(SZ;\mathbb{F}\_2))=H^{p+2}(BG;\mathbb{F}\_2).$$


However, the fact that $SZ$ has $G$-fixed points (the suspension points) implies that the above fibration admits a section, and so the induced map $H^\*(BG;\mathbb{F}\_2)\to H^\*\_G(SZ;\mathbb{F}\_2)$ is injective. This agrees with the edge homomorphism of the spectral sequence, implying that $d\_2$ is in fact trivial. Hence $E\_2=E\_\infty$. There are no extension problems, since we are using field coefficients, so this gives the equivairiant cohomology as a $\mathbb{F}\_2$-module.


I believe it also gives that $H^\*\_G(SZ;\mathbb{F}\_2)$ is free as an $H^\*\_G$-module.


