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The vertical Liouville vector field on a cotangent bundle has complete flow $D_s(x,p)=(x,e^{s}p)$. Its pullbacks scale both the Liouville one-form and the canonical symplectic form by $e^s$. As $s\to -\infty$, each orbit tends to its base point on the zero section. This contraction proves rigidity of cotangent Liouville-form preservation.

First observe that the Liouville one-form is zero as an ambient covector exactly on the zero section. Indeed, ${\alpha }_{(x,p)}=p\circ d\pi$, and $d\pi$ is surjective. Since $dg$ is invertible, $g^{\ast }\alpha =\alpha$ therefore implies that $g$ maps the zero section onto itself. It induces a diffeomorphism $f:X\to X$ defined by $g(x,0)=(f(x),0)$.

The map $g$ is a symplectomorphism, since it preserves $-d\alpha$. Let $E={\sum }_i{p}_i{\partial }_{p_i}$ be the vertical Liouville vector field. With the chosen sign convention,Preservation of both $\omega$ and $\alpha$ forces $g_{\ast }E=E$ because $\omega$ is nondegenerate. Its complete cotangent fiber-dilation flow isUniqueness of integral curves gives $g\circ D_s=D_s\circ g$. If $g(x,p)=(y,P)$, let $s\to -\infty$. Continuity and this commutation yieldThus $y=f(x)$ for every $p$: the whole cotangent fiber over $x$ maps into the fiber over $f(x)$. This step proves that $g$ covers $f$; it was not assumed.

Now write $g(x,p)=(f(x),P(x,p))$. For any $u\in T_{x}X$, choose a tangent vector to $T^{\ast }X$ projecting to $u$. Evaluating $g^{\ast }\alpha =\alpha$ on it givesSince $d{f}_x$ is invertible, $P(x,p)=p\circ (d{f}_x{)}^{-1}$. Thereforeand $f$ is unique. This rigidity of cotangent Liouville-form preservation uses smoothness at the zero section; it follows from the canonical form and its dilation dynamics on the entire cotangent bundle.