I'm reading [this article upon topological field theory][1] and I'm a bit confused about the way he compute equivariant cohomology of $S^2$ wrt $\mathrm{U}(1)$, i.e. $H^\bullet_{S^1}(S^2)$. You can find this on page 92 - 93. In particular there are two issues:

1. It's clear to me the meaning of eq. (10.24), i.e. I agree for that $k$th-cohomology class for $k\geq 2$. But I do not understand how to find that for $k=1$. My reasoning goes as follow. I consider the exact sequence (to avoid cumbersome notation I will write $H^\bullet$ in stead of $H^\bullet_{S^1}$) $$ 0\rightarrow H^0(S^2) \rightarrow H^0(U_1)\oplus H^0(U_2) \rightarrow H^0(U_1\cap U_2) \rightarrow H^1(S^2) \rightarrow H^1(U_1)\oplus H^1(U_2) \rightarrow 0. $$ Then, as explained at the top of page 93, $H^\bullet(U_i) = \mathbb{C}[\Omega]$, moreover since $U_1\cap U_2$ is a deformation retract of $S^1$ on which $S^1$ acts freely, so that $H^0(U_1\cap U_2) = \mathbb{C}[\Omega]$. Hence I remain with: $$ 0\rightarrow H^0(S^2) \rightarrow \mathbb{C}[\Omega]\oplus\mathbb{C}[\Omega] \rightarrow \mathbb{C}[\Omega] \rightarrow H^1(S^2) \rightarrow \mathbb{C}[\Omega]\oplus \mathbb{C}[\Omega] \rightarrow 0. $$ Now, is this correct? How can I deduce $H^0$ and $H^1$?
2. Why in (10.25) he specifies $f(0) = g(0)$? Where does this condition come from?

Moreover, in what follow immediately he tries also to compute the same cohomology with Cartan Model. Here what is unclar to me is that he says that a cohmology class is $$ h_1(\Omega)(\mathrm{d}\phi\mathrm{d}\theta + \Omega\cos\theta) + h_2(\Omega) $$ but this seem to be only a possible even-form since the degree of $\Omega$ is 2 and there is a two form inside. How does it work?

Thanks. [1]: http://arxiv.org/pdf/hep-th/9411210v2.pdf

I'm reading this article upon topological field theory and I'm a bit confused about the way he compute equivariant cohomology of $S^2$ wrt $\mathrm{U}(1)$, i.e. $H^\bullet_{S^1}(S^2)$. You can find this on page 92 - 93. In particular there are two issues:

1. It's clear to me the meaning of eq. (10.24), i.e. I agree for that $k$th-cohomology class for $k\geq 2$. But I do not understand how to find that for $k=1$. My reasoning goes as follow. I consider the exact sequence (to avoid cumbersome notation I will write $H^\bullet$ in stead of $H^\bullet_{S^1}$) $$ 0\rightarrow H^0(S^2) \rightarrow H^0(U_1)\oplus H^0(U_2) \rightarrow H^0(U_1\cap U_2) \rightarrow H^1(S^2) \rightarrow H^1(U_1)\oplus H^1(U_2) \rightarrow 0. $$ Then, as explained at the top of page 93, $H^\bullet(U_i) = \mathbb{C}[\Omega]$, moreover since $U_1\cap U_2$ is a deformation retract of $S^1$ on which $S^1$ acts freely, so that $H^0(U_1\cap U_2) = \mathbb{C}[\Omega]$. Hence I remain with: $$ 0\rightarrow H^0(S^2) \rightarrow \mathbb{C}[\Omega]\oplus\mathbb{C}[\Omega] \rightarrow \mathbb{C}[\Omega] \rightarrow H^1(S^2) \rightarrow \mathbb{C}[\Omega]\oplus \mathbb{C}[\Omega] \rightarrow 0. $$ Now, is this correct? How can I deduce $H^0$ and $H^1$?
2. Why in (10.25) he specifies $f(0) = g(0)$? Where does this condition come from?

Moreover, in what follow immediately he tries also to compute the same cohomology with Cartan Model. Here what is unclar to me is that he says that a cohmology class is $$ h_1(\Omega)(\mathrm{d}\phi\mathrm{d}\theta + \Omega\cos\theta) + h_2(\Omega) $$ but this seem to be only a possible even-form since the degree of $\Omega$ is 2 and there is a two form inside. How does it work?