Equivariant cohomology: $H^∗_{S^1} (S^2)$

Question

[Equivariant cohomology: $H^∗_{S^1} (S^2)$][1]

could you explain with details

in that link [1]: Equivariant cohomology: $H^{*}_{S^{1}}(S^{2})$ why ϕ is surjective, please ?

And even if $V$ is contractible how do we know $H^∗_{S^1}(U)≃H^∗_{S^1}(p−)$

Answer 1

This doesn't really have much to do with equivariant cohomology. Consider (open) pathconnected subspaces $U,V$ such that $U\cap V$ is path-connected. Then $H^{0}(U)\cong H^{0}(V)\cong H^0(U\cap V)\cong \mathbb{R}$. These are generated by constant functions on $U,V$ and $U\cap V$. The map $\phi$ then maps the constant function $f:U\rightarrow \mathbb R$ and the constant function $g:V\rightarrow \mathbb R$ to the constant function $h:U\cap V\rightarrow \mathbb R$, where

$$ h(x)=f(x)-g(x) $$

A constant function $h$ on $U\cap V$ clearly extends to a constant function on $\overline h$ on $U$. Hence $\phi$ maps $(\overline h,0)$ to $a$.