Equivariant cohomology and Mayer-Vietoris sequence 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:


*

*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 $H^1$?

*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?
 A: You misunderstood the meaning of $H^{\bullet}_{S^1}(U_1) = \mathbb{C}[\Omega]$. This does not mean $H^i_{S^1}(U_1) = \mathbb{C}[\Omega]\forall i$, it means that the cohomology ring (with multiplication given by the cup product) is given by the polynomial algebra in one element that has degree 2. Translated back into the individual degrees $H^i$ this statement is
$$ H^i_{S^1}(U_1) = \begin{cases} \mathbb{C} & i \text{ even} \\  0 & i \text{ odd}\end{cases}$$
The correct exact sequence in the low degrees is
$$ 0\to H^0_{S^1}(S^2)\to \mathbb{C}\oplus\mathbb{C}\to \mathbb{C}\to H^1_{S^1}(S^2) \to 0$$
where now $H^0_{S^1}(S^2)\to \mathbb{C}\oplus\mathbb{C}$ is injective and you need to convince yourself that the map $\mathbb{C}\oplus\mathbb{C}\to\mathbb{C}$ induced by the restrictions $(U_1\to U_1\cap U_2,U_2\to U_1\cap U_2)$ is given by $\mathbb{C}\oplus\mathbb{C}\to\mathbb{C}, (a,b)\mapsto a-b$. This is easiest to understand in the deRham cohomology, since the Mayer-Vietoris map there is always the difference of the forms on the overlap. By exactness and injectivity, the kernel of this map is $H^0_{S^1}(S^2)$, and the kernel is $\{(a,b)\in\mathbb{C}^2\mid a=b\}\cong \mathbb{C}$. Moreover, this map is surjective, so by exactness, $H^1_{S^1}(S^2) = 0$. 
Incidentally, this also answers where the restriction $f(0) = g(0)$ in the expression for $H^\bullet_{S^1}(S^2)$ comes from - $f(0),g(0)$ are the zero degree parts of the cohomology ring elements, and we have just shown that the zero degree is not $H^0_{S^1}(U_1)\oplus H^0_{S^1}(U_2)$, but only the diagonal.
