The two-form of a potential presentation is closed and nondegenerate, and each generalized-coordinate momentum chart carries the same symplectic structure: ${\mathrm{\Phi }}_X^{\ast }\lambda =\sum P_{a}d{q}^a$ and ${\mathrm{\Phi }}_X^{\ast }\omega =\sum d{q}^a\wedge d{P}_a$

The set $W=Q\times {\mathbb{R}}^{3N}$ is a nonempty open subset of ${\mathbb{R}}^{2n}$ with $n=3N\ge 1$, hence a phase domain, and by [F2] its two-form is $\omega ={\sum }_{a}d{x}^a\wedge d{p}_a$, exactly the standard two-form of $W$ in the sense of [F3]; therefore [F3] gives $d\omega =0$, nondegeneracy of $\omega$ at every point of $W$, and the symplectic-manifold property of $(W,\omega )$: clause 1 holds.

For $(q,P)\in \mathrm{\Omega }\times {\mathbb{R}}^r$ and a tangent vector $(\delta q,\delta P)\in {\mathbb{R}}^r\times {\mathbb{R}}^r$, the derivative of ${\mathrm{\Phi }}_X$ computed by the chain rule [F5] is $(DX(q)\delta q,{\mathrm{\partial }}_q(\mathsf{M}DX{G}^{-1})(q)P\delta q+\mathsf{M}DX(q)G(q{)}^{-1}\delta P)$, so the $\delta x$-component of the image is $DX(q)\delta q$; since $\lambda$ acts on tangent vectors of $W$ by ${\lambda }_{(x,p)}(\delta x,\delta p)=p\cdot \delta x$, the pullback [F4] gives $({\mathrm{\Phi }}_X^{\ast }\lambda {)}_{(q,P)}(\delta q,\delta P)=(\mathsf{M}DX(q)G(q{)}^{-1}P)\cdot (DX(q)\delta q)=P^{\mathsf{T}}G(q{)}^{-1}DX(q{)}^{\mathsf{T}}\mathsf{M}DX(q)\delta q=P^{\mathsf{T}}\delta q$, where the last equality substitutes $G=D{X}^{\mathsf{T}}\mathsf{M}DX$; since $(\delta q,\delta P)$ is arbitrary this is the one-form ${\sum }_a{P}_{a}d{q}^a$, the first identity of clause 2.

The map ${\mathrm{\Phi }}_X$ is $C^2$ by [F1]. To justify the derivative of a pullback at this regularity, write a smooth one-form on the Cartesian chart as $\lambda ={\sum }_{\alpha }{a}_{\alpha }(y)d{y}^{\alpha }$, with $y=(x,p)$. For any $C^2$ map $\mathrm{\Phi }$ with components ${\mathrm{\Phi }}^{\alpha }$, the coefficients of ${\mathrm{\Phi }}^{\ast }\lambda$ are $b_i=(a_{\alpha }\circ \mathrm{\Phi }){\mathrm{\partial }}_i{\mathrm{\Phi }}^{\alpha }$. In ${\mathrm{\partial }}_j{b}_i-{\mathrm{\partial }}_i{b}_j$, the terms $(a_{\alpha }\circ \mathrm{\Phi })({\mathrm{\partial }}_{ji}{\mathrm{\Phi }}^{\alpha }-{\mathrm{\partial }}_{ij}{\mathrm{\Phi }}^{\alpha })$ cancel by equality of mixed second derivatives of $\mathrm{\Phi }$; the remaining terms are exactly the coordinate coefficients of ${\mathrm{\Phi }}^{\ast }(d\lambda )$. Thus $d({\mathrm{\Phi }}^{\ast }\lambda )={\mathrm{\Phi }}^{\ast }(d\lambda )$ for this $C^2$ map by a direct coordinate calculation. Applying this calculation to ${\mathrm{\Phi }}_X$ and using step 1.2 gives ${\omega }_X={\mathrm{\Phi }}_X^{\ast }\omega =-d({\mathrm{\Phi }}_X^{\ast }\lambda )=-d({\sum }_a{P}_{a}d{q}^a)={\sum }_{a}d{q}^a\wedge d{P}_a$, proving the second identity of clause 2.

The image $\mathrm{\Omega }\times {\mathbb{R}}^r$ of the coordinates $(q,P)$ is a phase domain: $\mathrm{\Omega }$ is nonempty and open by [F1] and $r\ge 1$, so $\mathrm{\Omega }\times {\mathbb{R}}^r$ is a nonempty open subset of ${\mathbb{R}}^{2r}$. By step 2.1 the form ${\omega }_X$ equals its standard two-form, which by [F3] is closed and nondegenerate; therefore ${\mathrm{\Phi }}_X^{\ast }\omega ={\omega }_X$ is closed and nondegenerate, proving clause 2. If $r=3N$ and $DX$ is invertible, then $D{\mathrm{\Phi }}_X$ is invertible: its block-triangular derivative has diagonal blocks $DX$ and $\mathsf{M}DX{G}^{-1}=D{X}^{-\mathsf{T}}$. The inverse function theorem The Euclidean inverse function theorem supplies a $C^1$ local inverse $g$; differentiating ${\mathrm{\Phi }}_X\circ g=\mathrm{i}\mathrm{d}$ gives $Dg=(D{\mathrm{\Phi }}_X\circ g{)}^{-1}$. Since $D{\mathrm{\Phi }}_X$ and $g$ are $C^1$, the right side is $C^1$ (matrix inversion is smooth on invertible matrices by the cofactor-over-determinant formula), hence $g$ is $C^2$. Thus these restrictions are genuine $C^2$ phase charts.

Let $X_1,X_2$ and their full-dimensional chart neighborhoods be as in clause 3. By step 3.1 each chart map and its local inverse are $C^2$, so $\mathrm{\Theta }={\mathrm{\Phi }}_{X_2}^{-1}\circ {\mathrm{\Phi }}_{X_1}$ is $C^2$. Pointwise functoriality of pullback [F4] and step 1.2 give ${\mathrm{\Theta }}^{\ast }{\omega }_{X_2}={\omega }_{X_1}$. Applying the same coordinate calculation as in step 2.1 to the smooth one-form ${\lambda }_{X_2}$ and the $C^2$ map $\mathrm{\Theta }$ yields $d({\mathrm{\Theta }}^{\ast }{\lambda }_{X_2})={\mathrm{\Theta }}^{\ast }(d{\lambda }_{X_2})$. Since ${\omega }_{X_j}=-d{\lambda }_{X_j}$, it follows that $d({\lambda }_{X_1}-{\mathrm{\Theta }}^{\ast }{\lambda }_{X_2})=0$. This proves the stated closed-one-form conclusion. For these particular cotangent chart maps, pointwise functoriality also gives ${\mathrm{\Theta }}^{\ast }{\lambda }_{X_2}=({\mathrm{\Phi }}_{X_2}\circ \mathrm{\Theta }{)}^{\ast }\lambda ={\mathrm{\Phi }}_{X_1}^{\ast }\lambda ={\lambda }_{X_1}$, so the difference is actually zero. This stronger conclusion uses their common Cartesian one-form, rather than inferring exactness from closedness.