How to prove that a certain action is hamiltonian?

I posted this question here on math.stackexchange.com. I have not had an answer, and I thought it could be more appropriate here.
(Please, If you judge this my opinion is wrong, then I will delete this question)

Reading a paper I had the need to complete a proof, and came up with a certain argument(see below). My question is: at your knowledge, could I reduce it to a special case of some other theorem? I ask this question in order to give a correct reference, instead of my trivial ad hoc argument, in the case the answer is positive.

Given a smooth action $\Psi$ of $\mathbb{T}^k$ on a symplectic manifold $(M,\omega)$, if $\omega$ is exact and there exists a smooth map $\pi:M\to P$ constant on the orbit of $\Psi$ and such that $\zeta_X\lrcorner\omega\in\pi^*(\Omega^1(P)),\forall X\in\textrm{Lie}(\mathbb{T}^k)$,( being $\zeta$ the action of $\textrm{Lie}(\mathbb{T}^k)$ on $M$ induced by $\Psi$), then the $\Psi$ is an hamiltonian action w.r.t. $\omega$.

For completeness, I sketch also the trivial proof:
Let $\eta$ be a primitive of $\omega$, and $\mu_i\in C^\infty(M,\mathbb R)$ be defined by $$\mu_i=\int_0^1 \big(\textrm{Fl}_t^{\zeta_{e_i}}\big)^\ast (\zeta_{e_i}\lrcorner\eta)\,dt.$$ We contend that $d\mu_i=\zeta_{e_i}\lrcorner\omega$, for any $i=1,\ldots,k$.
By H.Cartan's formula and the theorem on Lie derivative we get $$d\mu_i=\int_0^1\frac{d}{dt}\bigg(\big(\textrm{Fl}_t^{\zeta_{e_i}}\big)^\ast\eta\bigg)\,dt-\int_0^1\big(\textrm{Fl}_t^{\zeta_{e_i}}\big)^\ast(\zeta_{e_i}\lrcorner\omega)\,dt.$$ The first integral is identically zero for periodicity.
The second one is $\zeta_{e_i}\lrcorner\omega$, because its integrand is a constant function of $t$, i.e. $\mathcal{L}(\zeta_{e_i}).(\zeta_{e_i}\lrcorner\omega)=0$ by the hypothesis.

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Please, excuse me, even if I exactly copied the text of the quoted question, here I have some difficulties in visualizing the sketched proof. Do you have the same troubles? – Giuseppe May 29 '11 at 19:43
It sometimes helps to include LaTeX in backticks (`). I tried to fix your formulas and did some minor tweaking of the formatting, I hope that's fine with you. By the way: isn't it E. Cartan's formula? – Theo Buehler May 29 '11 at 20:17
@Theo Buehler: Thanks a lot. – Giuseppe May 29 '11 at 20:17
@Theo Buehler: Yes surely it is due to the foundational work of Elié Cartan, but I thought it was named after his son Henri Cartan, for this formula is singled out in his axiomatic presentation of the equivariant cohomology for smooth manifolds acted by a Lie group. Precisely, I found it under this name in Symplectic Geometry and Analytical Mechanics of Libermann and Marle. Thanks again for your help with editing and for your suggestion about LaTeX. – Giuseppe May 29 '11 at 20:37
@Giuseppe: The formula you are calling H. Cartan's formula first appeared, to my knowledge, in É. Cartan's 1922 book "Leçons sur les invariants intégraux", and these are lectures based on a course that he had given a couple of years earlier, so the formula might date from around 1920, when H. Cartan would have been about 16. – Robert Bryant May 29 '11 at 22:25