Chern classes of reduced space for Hamiltonian circle action I have a question about Chern class of symplectic reduction.
Let $(M,\omega)$ be a smooth compact symplectic manifold with a Hamiltonian circle action.
Let $H : M \rightarrow \mathbb{R}$ be the corresponding moment map.
When $0$ is a regular value of $H$, how can I prove that a Kirwan map $\kappa : H^*_{S^1}(M) \rightarrow H^*(H^{-1}(0)/S^1)$ maps the equivariant first Chern class of $M$ to the first Chern class of $H^{-1}(0)/S^1$? 
I saw this argument in the paper "Floer homology on the extended moduli space" by Manolescu and Woodward (Lemma 4.4), but the precise proof is not involved.
Perhaps (I guess) the proof of the statement seems to be an exercise, but I don't know what I missed. 
Thank you in advance.  
 A: I am not offering an answer, but I suspect that the answer ultimately comes from the behaviour of Chern classes under pullbacks. More precisely, if $L\rightarrow M$ is an $S^1$-equivariant complex line bundle with equivariant first Chern class $c_1^{S^1}(L)\in H_{S^1}^2(M)$, then the $S^1$-equivariant first Chern class of the restricted bundle $L\vert_{\mu^{-1}(0)}\rightarrow\mu^{-1}(0)$ is precisely $i^*(c_1^{S^1}(L))\in H_{S^1}^*(\mu^{-1}(0))$, where $i:\mu^{-1}(0)\rightarrow M$ is the inclusion. The image of $c_1^{S^1}(L)$ under the Kirwan map is then the image of $i^*(c_1^{S^1}(L))$ under the isomorphism $H_{S^1}^*(\mu^{-1}(0))\rightarrow H^*(\frac{\mu^{-1}(0)}{S^1})$. 
A: To complete Peter's proof, we have to compare $A=TM|_{\mu^{-1}(0)}$ with 
$i^*T(\mu^{-1}(0)/S^1)$.  The latter is a sub quotient of the former: we pass to the subbundle $B$ tangent to $\mu^{-1}(0)$, and kill the vectors $C$ tangent to the $S^1$-orbits.  The key observation is that the symplectic form shows that $A/B$ and $C$ are dual (since $B$ is symplectic orthogonal to $C$).  I'm assuming you have a compatible metric which puts a complex structure on $A/B\oplus C\cong \mathbb{C}C$ making this a complex line bundle with trivial Chern class, since it is self-dual.
