 Solution

The cotangent fiber translation $f(q,p)=(q,p+{\alpha }_q)$ is a smooth diffeomorphism with inverse $(q,p)\mapsto (q,p-{\alpha }_q)$, and $\pi \circ f=\pi$. For a tangent vector $v$ at $(q,p)$, the intrinsic definition of the canonical one-form on a cotangent bundle givesThus $f^{\ast }\theta =\theta +{\pi }^{\ast }\alpha$. Taking its exterior derivative with the fixed sign convention yieldsIf $\alpha$ is a closed differential one-form, the last term vanishes, proving that $f$ is a symplectomorphism. In fact the converse holds too: the pullback by the zero section shows that ${\pi }^{\ast }(d\alpha )=0$ implies $d\alpha =0$.