I'm reading Lecture 10 of Lectures on Symplectic Geometry by Ana Cannas da Silva. Here is the online version.
The theorem in question is Theorem 10.5 on page 59. I'll use the notation from the proof. I'm asking because I'm unable to check that the constructed $\rho_t$ indeed gives the answer. To show the existence of $u_t$ such that $\rho_t^*\alpha_t=u_t\cdot\alpha_0$, it is necessary and sufficient to show that $\rho_t^*\alpha_t$ vanishes on $H_0=\ker\alpha_0$. Thus for $v\in H_0$ we consider $\frac{d}{dt}(\rho^*_t\alpha_t)=\rho_t^*(\mathcal{L}_{v_t}\alpha_t+\frac{d\alpha_t}{dt})$, and we wish to show that it is $0$. However, by contruction we only know that $\rho_t^*(\mathcal{L}_{v_t}\alpha_t+\frac{d\alpha_t}{dt})$ vanishes on $\rho_t^*H_t$, hence we would need $\rho_t^*H_t=H_0$ to finish the proof. But this is just what we set out to prove in the beginning, so the argument doesn't work.
What is the correct missing detail in the proof? Thank in advance!
I'm not sure if the reasoning is correct. You don't need to prove that $\frac{d}{dt}(p_t^*\alpha_t)=0$. You aim to prove that $p_t^*\alpha_t=u_t \alpha_0$, which implies that $\frac{d}{dt}(p_t^*\alpha_t)=d(u_t \alpha_0)$. When trying to solve the differential equation, if you impose that your hypothetic solution $X_t$ is lying in $H_t$ the equation reduces to $$ \iota_{X_t}d\alpha_t|_{H_t}=-\frac{d\alpha_t}{dt}|_{H_t}. $$ Sinde $d\alpha_t$ is nondegenerate in $H_t$, the vector field $X_t$ is uniquely determined and you have a solution. By integrating it you obtain $p_t$ and deduce the expression of $u_t$ from the equation $p_t^* \alpha_t= u_t \alpha_0$. By construction you have $p_t^*\alpha_t=u_t\alpha_0$ which implies that $p_t^*H_t=H_0$.