TITLE: Cotangent bundle lift theorem
QUESTION [8 upvotes]: Let $M$ be a smooth manifold and $T^\ast M$ be its cotangent bundle. Consider the tautological 1-form $\theta$ on $T^\ast M$ ($\theta=\sum y_i dx^i$ in local canonical coordinate systems).
A diffeomorphism $f:M\to M$ induces a pull-back lift $F=f^\ast:T^\ast M\to T^\ast M$. 
It seems that we always have $F^\ast\theta=\theta$. But I don't know how to verify this.
I just heard the following cotangent bundle lift theorem:
If a diffeomorphism $F:T^\ast M\to T^\ast M$ preserves $\theta$, then $F=f^\ast$ for some diffeomorphism $f:M\to M$.
This looks too strong to me. Do you know how to prove this?
Thank you!

REPLY [4 votes]: Just as a side-note.  The proposition does not imply that every symplectomorphism of $T^*M$ is a cotangent lift of a diffeomorphism on $M$.  As a counter example, let $\sigma$ be a non-zero closed one-form on $M$ and consider the map $F(x,p) = (x,p + \sigma(x))$.  Notes that $F$ is not the cotangent lift of any diffeomorphism on $M$. Also $F^*d \theta = d(F^*\theta) = d( \sigma^H + \theta) = d\sigma^H + d\theta = d\theta$ where $\sigma^H$ is the one-form on $T^*M$ obtained by horizontal lift.  Needless to say, if $\sigma$ is closed, then so is $\sigma^H$.  Thus $F$ preserves the canonical symplectic form $d\theta$.