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View $S^1$ as the unit circle in the complex plane and let $S^0$ act by complex conjugation. What is the Borel equivariant cohomology $H^*_{S^0}(S^1;{\mathbb{Z}})$ of this action?

I ask this question as an analog of the following well-known equivariant cohomology of $S^1$ on $S^2$ by rotation about the $z$-axis. We have $H^*_{S^1}(S^2;{\mathbb{Z}})\cong {\mathbb{Z}})[x]\oplus {\mathbb{Z}})[y]$ where $x$ and $y$ are in $H^2$. One proof goes by decomposing $S^2$ equivariantly into ${D^+} \cup_{S^1} D^-$ where $D^+$ and $D^-$ are the upper and lower hemispheres respectively. Here the intersection ${D^+} \cap D^- = S^1$ is path-connected so the Mayer-Vietoris sequence applies.

Trying to apply the same method to my above problem leads to $S^1=I\cup_{S^0} I$ where $S^0$ is disconnected. This is reminiscent of a classical problem related to Brown's work on groupoids and the van Kampen theorem. . .

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Mayer-Vietoris sequence doesn't need connected intersections. In your case, the $S^0$-equivariant cohomology of $S^0$ is the ordinary cohomology of a point and you get $H_{S^0}^*(S^1;\mathbb{Z})=H^*(\mathbb{R}P^\infty)\oplus H^*(\mathbb{R}P^\infty)$ (for $*>0$; for $*=0$ you get of course $\mathbb{Z}$).

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