Let $X$ be a manifold and $T^\ast X$ the cotangent bundle. Let $\alpha$ denote the tautological $1$-form on $T^\ast X$ so that $(T^\ast X, \omega=-d\alpha)$ is a symplectic manifold. I want to know when a symplectomorphism $g:T^\ast X\to T^\ast X$ is the lift of a function $f:X\to X$ (here the lift of $f$ is $f^\sharp:T^\ast X\to T^\ast X \ ,\ (p,\xi_p)\mapsto (f(p),(f^\ast)^{-1}(\xi_p))$. An exercise in Da Silva's "Lectures on Symplectic Geometry" says this happens if and only if $g^\ast\alpha=\alpha$. I can show that the lift of a diffeomorphism preserves $\alpha$, but I am struggling with the converse. Given $g:T^\ast X\to T^\ast X$ such that $g^\ast\alpha=\alpha$, I so far have shown the following:
If $V$ is the symplectic dual of $\alpha$ (i.e. $\omega(V,\cdot)=\alpha$) then $g$ commutes with the flow of $V$ (i.e. $g\circ\theta_t((p,\xi_p))=\theta_t\circ g((p,\xi_p)))$. I can show that the flow of $V$ is $\theta^{(p,\xi_p)}(t)=(p,e^t\xi_p)$ implying for any $\lambda>0$ if $g(p,\sigma_p)=(q,\tau_q)$ then $g(p,\lambda\sigma_p)=(q,\lambda\tau_q)$ and the continuity of $g$ gives this also holds for $\lambda=0$. The hint in Da Silva's book says that this holds for all $\lambda\in\mathbb{R}$, but I don't understand how this follows. After showing this, the hint says to show the existence of a map $f:X\to X$ such that $\pi\circ g=f\circ\pi$ where $\pi:T^\ast X\to X$ is the projection. I thought the obvious map would be $f(p):=\pi\circ g(p,\sigma_p)$ where $\sigma_p\in T^\ast_p X$ is arbitrary. However, even if the above holds for all $\lambda\in\mathbb{R}$ I don't see why this map is well defined.
With such an $f$, I can show that $f^\sharp=g$. Any help would be greatly appreciated.
