The drift, kick and velocity-Verlet maps are canonical: $(DC{)}^{\mathsf{T}}JDC=J$ and $C^{\ast }\omega =\omega$

Clause 1, derivative and flow. The map $D_h(q,p)=(q+h{M}^{-1}p,p)$ is affine, hence $C^{\mathrm{\infty }}$, with constant Jacobian $\left(\begin{array}{cc}{I}_d& h{M}^{-1}\\ 0_d& I_d\end{array}\right)$ read off from the partial derivatives by [F1]. The curve $t\mapsto (q+t{M}^{-1}p,p)$ has derivative $(M^{-1}p,0)=X_A$ at its own value, and at $t=0$ it equals $(q,p)$; if its full segment between $0$ and $h$ stays in $U$, uniqueness in [F4] identifies its endpoint with the time-$h$ local flow value of $X_A$. Writing $C:=D{D}_h(q,p)$, block multiplication using [F1] and the fact that the blocks $I_d$, $h{M}^{-1}$ commute with each other gives $C^{\mathsf{T}}JC=\left(\begin{array}{cc}{I}_d& 0\\ h{M}^{-1}& I_d\end{array}\right)\left(\begin{array}{cc}0& I_d\\ -I_d& -h{M}^{-1}\end{array}\right)=\left(\begin{array}{cc}0& I_d\\ -I_d& 0\end{array}\right)=J,$ the middle factor being $JC$.

Clause 2, derivative and flow. The first block row of $D{K}_h$ is $(I_d,0_d)$ because $K_h$ leaves $q$ unchanged; the second block row is $(-h{H}_V(q),I_d)$ because $p-h\mathrm{\nabla }V(q)$ has $q$-gradient $-h{H}_V(q)$ and $p$-gradient $I_d$ by [F1]; $V\in C^3$ makes $\mathrm{\nabla }V$ of class $C^2$ (and the Hessian entries $C^1$), so the displayed formula for $K_h$ is $C^2$. The curve $t\mapsto (q,p-t\mathrm{\nabla }V(q))$ has derivative $(0,-\mathrm{\nabla }V(q))=X_B$ at its own value and equals $(q,p)$ at $t=0$, so, when its full segment up to $h$ stays in $U$, [F4] identifies its endpoint with the time-$h$ local flow value of $X_B$. With $C:=D{K}_h(q,p)$ and $S:=H_V(q)$, block multiplication using $S^{\mathsf{T}}=S$ from [F2] gives $C^{\mathsf{T}}JC=\left(\begin{array}{cc}{I}_d& -hS\\ 0& I_d\end{array}\right)\left(\begin{array}{cc}-hS& I_d\\ -I_d& 0\end{array}\right)=\left(\begin{array}{cc}-hS+hS& I_d\\ -I_d& 0\end{array}\right)=J,$ the middle factor being $JC$.

Clause 3, composition. Since $V$ is $C^3$, its gradient is $C^2$, so each kick is $C^2$ and the drift is smooth. The admissible composition domain is an intersection of finitely many preimages of the open set $U$ under these continuous maps, hence open; their composition $V_h=K_{h\mathrm{/}2}\circ D_h\circ K_{h\mathrm{/}2}$ is $C^2$ on it. At each point there, the chain rule of [F1] gives $D{V}_h=C_1{C}_2{C}_3$ with $C_1=D{K}_{h\mathrm{/}2}$ evaluated at $D_h(K_{h\mathrm{/}2}(q,p))$, $C_2=D{D}_h$ evaluated at $K_{h\mathrm{/}2}(q,p)$, and $C_3=D{K}_{h\mathrm{/}2}$ evaluated at $(q,p)$. Each factor satisfies $C_i^{\mathsf{T}}J{C}_i=J$ by steps 1.1-1.2, so $D{V}_h^{\mathsf{T}}JD{V}_h=C_3^{\mathsf{T}}(C_2^{\mathsf{T}}(C_1^{\mathsf{T}}J{C}_1)C_2)C_3=C_3^{\mathsf{T}}(C_2^{\mathsf{T}}J{C}_2)C_3=C_3^{\mathsf{T}}J{C}_3=J\mathrm{.}$

Clause 3, form preservation. If the composition domain is empty, the pointwise matrix and form assertions are vacuous. Otherwise it is a phase domain, and the identity of step 2.1 is exactly the matrix condition of [F3] for the $C^2$ map $V_h$, so $V_h^{\ast }{\omega }_U={\omega }_U$ on its admissible open domain. The explicit inverse factors are $D_{-h}$ and $K_{-h}$; reversing the palindrome gives $V_{-h}$ as the inverse on the image wherever the compositions exist. Restriction to these open domains makes $V_h$ a $C^2$ diffeomorphism onto its open image, with $C^2$ inverse $V_{-h}$, preserving the form. Thus it is a local $C^2$ canonical transformation in Finite-regularity differential forms, Hamiltonian fields and canonical maps in phase charts, without assuming a smooth potential.