[proofplan] Work in one coordinate chart near $(z_0,\eta_0)$ and shrink the neighbourhoods so that the Hamilton flow and the inverse of $(z,\eta)\mapsto (y(t;z,\eta),\eta)$ are smooth. The Hamilton flow determines a local canonical transformation from initial data $(z,\eta)$ to final data $(y,\xi)$, and the nonsingular projection hypothesis allows this transformation to be written by a type-II [generating function](/page/Generating%20Function) in the independent variables $(t,y,\eta)$. The phase is obtained by subtracting the initial pairing $z\cdot\eta$ from that generating function; its critical equation recovers the inverse relation between $y$ and $z$, while its derivatives recover the covectors and the eikonal equation. [/proofplan]

[step:Shrink to a coordinate neighbourhood where the flow has a smooth inverse]Choose a coordinate chart $(U,\kappa)$ on $Y$ with $z_0\in U$ and identify $U$ with the [open set](/page/Open%20Set) $\kappa(U)\subset\mathbb{R}^n$. Under the cotangent trivialisation induced by $\kappa$, write points of $T^*U\setminus 0$ as $(z,\eta)\in \kappa(U)\times\mathbb{R}^n_0$. After replacing $V$ by a smaller conic neighbourhood of $(z_0,\eta_0)$ and replacing $T_0$ by a smaller positive number, assume that the spatial Hamilton flow is defined on \begin{align*} (-T_0,T_0)\times V. \end{align*} Define \begin{align*} F:(-T_0,T_0)\times V &\to (-T_0,T_0)\times \kappa(U)\times\mathbb{R}^n_0 \end{align*} \begin{align*} (t,z,\eta)&\mapsto (t,y(t;z,\eta),\eta). \end{align*} The Jacobian hypothesis says that the derivative of $F$ in the variables $(z,\eta)$ is invertible at points near $(0,z_0,\eta_0)$, because its block form has upper-left block $\partial y/\partial z$ and lower-right block the identity on the $\eta$-variables. Hence, after shrinking again, there is an open conic set \begin{align*} \Omega_0\subset (-T_0,T_0)\times \kappa(U)\times\mathbb{R}^n_0 \end{align*} and a smooth inverse map \begin{align*} G:\Omega_0 &\to V \end{align*} \begin{align*} (t,y,\eta)&\mapsto (Z(t,y,\eta),\eta) \end{align*} such that \begin{align*} y(t;Z(t,y,\eta),\eta)=y. \end{align*} Define \begin{align*} \Xi:\Omega_0 &\to \mathbb{R}^n_0 \end{align*} \begin{align*} (t,y,\eta)&\mapsto \xi(t;Z(t,y,\eta),\eta). \end{align*}[/step]

[guided]The first task is to make the statement local enough that all coordinates and inverses are honest smooth maps. Choose a chart $(U,\kappa)$ with $z_0\in U$. The chart identifies points near $z_0$ with points of an open subset of $\mathbb{R}^n$, and the induced cotangent trivialisation identifies covectors with elements of $\mathbb{R}^n$. Thus points near $(z_0,\eta_0)$ in $T^*Y\setminus 0$ may be written as $(z,\eta)$ with $z\in\kappa(U)$ and $\eta\in\mathbb{R}^n_0$. The hypothesis gives the map \begin{align*} F:(-T_0,T_0)\times V \to (-T_0,T_0)\times \kappa(U)\times\mathbb{R}^n_0 \end{align*} defined by \begin{align*} F(t,z,\eta)=(t,y(t;z,\eta),\eta). \end{align*} For fixed $t$, the derivative in the variables $(z,\eta)$ is a block [linear map](/page/Linear%20Map) whose upper-left block is $\partial_z y(t;z,\eta)$, whose upper-right block is $\partial_\eta y(t;z,\eta)$, whose lower-left block is the zero map from the $z$-variables to the $\eta$-variables, and whose lower-right block is the identity map on $\mathbb{R}^n$. Therefore its determinant is \begin{align*} \det(\partial_z y(t;z,\eta)). \end{align*} This determinant is nonzero by hypothesis. Therefore the finite-dimensional [inverse function theorem](/theorems/51) applies locally to the variables $(z,\eta)$, uniformly after shrinking the neighbourhood and the time interval. We obtain a smooth inverse of the form \begin{align*} G:\Omega_0 \to V,\qquad G(t,y,\eta)=(Z(t,y,\eta),\eta), \end{align*} where $\Omega_0$ is an open conic neighbourhood of $(0,z_0,\eta_0)$ in the variables $(t,y,\eta)$. The defining relation for this inverse is \begin{align*} y(t;Z(t,y,\eta),\eta)=y. \end{align*} Finally define the transported covector, expressed in the variables $(t,y,\eta)$, by \begin{align*} \Xi:\Omega_0 \to \mathbb{R}^n_0,\qquad \Xi(t,y,\eta)=\xi(t;Z(t,y,\eta),\eta). \end{align*} The point of these definitions is that the final base point $y$ and the initial covector $\eta$ will be the independent variables in the generating function.[/guided]

[step:Write Hamilton equations and record the canonical one form identity]In the coordinates $(t,y,\tau,\xi)$ on $T^*(I\times U)$, the Hamiltonian is \begin{align*} q(t,y,\tau,\xi)=\tau+\lambda(t,y,\xi). \end{align*} Its Hamilton equations, with Hamilton parameter equal to $t$ since $\dot t=\partial_\tau q=1$, are \begin{align*} \frac{d}{dt}y(t;z,\eta)=\partial_\xi\lambda(t,y(t;z,\eta),\xi(t;z,\eta)). \end{align*} \begin{align*} \frac{d}{dt}\xi(t;z,\eta)=-\partial_y\lambda(t,y(t;z,\eta),\xi(t;z,\eta)). \end{align*} Define the lifted time covector as the smooth map $\tau:(-T_0,T_0)\times V\to\mathbb{R}$ given by \begin{align*} \tau(t;z,\eta)=-\lambda(t,y(t;z,\eta),\xi(t;z,\eta)). \end{align*} Let $\Theta$ denote the canonical one-form on $T^*(I\times U)$, so in these coordinates \begin{align*} \Theta=\tau\,dt+\xi\cdot dy. \end{align*} Let $\omega:=d\Theta$ denote the canonical symplectic form in this sign convention. Let $H_q$ denote the Hamilton vector field of $q$ on $T^*(I\times U)$, defined by \begin{align*} \omega(H_q,\cdot)=-dq. \end{align*} Let $\Gamma:(-T_0,T_0)\times V\to T^*(I\times U)$ be the lifted flow graph map defined by \begin{align*} \Gamma(t,z,\eta)=(t,y(t;z,\eta),\tau(t;z,\eta),\xi(t;z,\eta)). \end{align*} Let $\iota:V\to T^*(I\times U)$ be the initial characteristic embedding defined by \begin{align*} \iota(z,\eta)=(0,z,-\lambda(0,z,\eta),\eta). \end{align*} The pullback $\iota^*\omega$ is $d\eta\cdot dz$, because $dt=0$ on $V$ and $dy=dz$ at $t=0$. The map $\Gamma$ is obtained by flowing $\iota(V)$ along $H_q$. More explicitly, tangent vectors to the image of $\Gamma$ are spanned by the flow direction $H_q$ and by vectors obtained by applying the differential of the Hamilton flow to tangent vectors of the initial section $\iota(V)$. The swept-out surface lies in $q^{-1}(0)$: it starts in $q^{-1}(0)$, and $H_q q=\omega(H_q,H_q)=0$, so $q$ is constant along the Hamilton flow. Hence every tangent vector to the image of $\Gamma$ lies in the tangent space of $q^{-1}(0)$, and $dq$ vanishes on those tangent vectors. Therefore, for any tangent vector $W$ to the swept-out surface, \begin{align*} \omega(H_q,W)=-dq(W)=0. \end{align*} Thus all mixed pairings involving the flow direction vanish. On pairs of vectors transported from $T\iota(V)$, preservation of $\omega$ by the Hamilton flow gives the transported form $d\eta\cdot dz$. Combining the transported pairings with the vanishing mixed pairings gives \begin{align*} \Gamma^*\omega=d\eta\cdot dz. \end{align*} Consequently \begin{align*} d(\Gamma^*\Theta-\eta\cdot dz)=\Gamma^*d\Theta-d\eta\cdot dz=0. \end{align*} Thus the smooth one-form $\alpha:=\Gamma^*\Theta-\eta\cdot dz$ on $(-T_0,T_0)\times V$ is closed. After shrinking $V$ and $T_0$ once more, choose $V$ inside a coordinate product of a small ball in the $z$-variables and a contractible conic sector in the $\eta$-variables, so that $(-T_0,T_0)\times V$ is contractible in the chosen coordinates. By the Poincare lemma applied to the closed smooth one-form $\alpha$, there is a smooth real-valued function $A:(-T_0,T_0)\times V\to \mathbb{R}$ such that \begin{align*} d_{t,z,\eta}A=\xi(t;z,\eta)\cdot d_{t,z,\eta}y(t;z,\eta)-\lambda(t,y(t;z,\eta),\xi(t;z,\eta))\,dt-\eta\cdot dz. \end{align*} Here $d_{t,z,\eta}y(t;z,\eta)$ denotes the full differential of the map $(t,z,\eta)\mapsto y(t;z,\eta)$. Normalize $A$ by imposing \begin{align*} A(0,z,\eta)=0. \end{align*}[/step]

[guided]The exactness needed for the generating function comes from the canonical one-form on the extended cotangent bundle, not only from the fixed-time spatial maps. In the coordinates $(t,y,\tau,\xi)$, the canonical one-form is $\Theta=\tau\,dt+\xi\cdot dy$. Along the characteristic lift we define $\tau(t;z,\eta)=-\lambda(t,y(t;z,\eta),\xi(t;z,\eta))$, so the lifted flow map is \begin{align*} \Gamma(t,z,\eta)=(t,y(t;z,\eta),\tau(t;z,\eta),\xi(t;z,\eta)). \end{align*} At $t=0$ the initial embedding is \begin{align*} \iota(z,\eta)=(0,z,-\lambda(0,z,\eta),\eta). \end{align*} On this initial section, $dt=0$ and $dy=dz$, so pulling back $d\Theta$ gives $d\eta\cdot dz$. Now we check why the extra time direction does not spoil closedness. The map $\Gamma$ is the surface swept out by flowing $\iota(V)$ along $H_q$. Its tangent directions consist of the flow direction $H_q$ and transported tangent vectors from the initial section $\iota(V)$. The swept-out surface lies in the characteristic set $q=0$: it starts there, and $H_q q=dq(H_q)=-d\Theta(H_q,H_q)=0$. Therefore $dq$ vanishes on every tangent vector $W$ to the swept-out surface. Since $d\Theta(H_q,\cdot)=-dq$, we get \begin{align*} d\Theta(H_q,W)=0. \end{align*} Thus the pairings involving the time-flow direction vanish. In the directions tangent to $V$, the Hamilton flow preserves $d\Theta$, and the initial pullback is $d\eta\cdot dz$. Hence \begin{align*} \Gamma^*d\Theta=d\eta\cdot dz. \end{align*} Consequently \begin{align*} d(\Gamma^*\Theta-\eta\cdot dz)=\Gamma^*d\Theta-d\eta\cdot dz=0. \end{align*} Thus $\alpha:=\Gamma^*\Theta-\eta\cdot dz$ is a closed smooth one-form on $(-T_0,T_0)\times V$. After shrinking, choose $V$ inside a product of a small coordinate ball in $z$ and a contractible conic sector in $\eta$; then $(-T_0,T_0)\times V$ is contractible in the chosen coordinates. The Poincare lemma therefore applies and gives a smooth function $A:(-T_0,T_0)\times V\to\mathbb{R}$ with $dA=\alpha$. Expanding $\Gamma^*\Theta$ gives \begin{align*} d_{t,z,\eta}A=\xi(t;z,\eta)\cdot d_{t,z,\eta}y(t;z,\eta)-\lambda(t,y(t;z,\eta),\xi(t;z,\eta))\,dt-\eta\cdot dz. \end{align*} The differential $d_{t,z,\eta}y(t;z,\eta)$ is the full differential in all three variables, which is essential because later we substitute $z=Z(t,y,\eta)$ and use $d(y(t;Z(t,y,\eta),\eta))=dy$. We normalize the primitive by adding a constant on each connected component so that $A(0,z,\eta)=0$.[/guided]

[step:Construct the type-II generating function]Define \begin{align*} S:\Omega_0&\to\mathbb{R} \end{align*} \begin{align*} (t,y,\eta)&\mapsto A(t,Z(t,y,\eta),\eta)+Z(t,y,\eta)\cdot\eta. \end{align*} We compute its differential. Since $z=Z(t,y,\eta)$ and $y=y(t;z,\eta)$, the differential identity for $A$ gives \begin{align*} d_{t,y,\eta}S=\Xi(t,y,\eta)\cdot dy+Z(t,y,\eta)\cdot d\eta-\lambda(t,y,\Xi(t,y,\eta))\,dt. \end{align*} Thus \begin{align*} \partial_y S(t,y,\eta)=\Xi(t,y,\eta). \end{align*} \begin{align*} \partial_\eta S(t,y,\eta)=Z(t,y,\eta). \end{align*} \begin{align*} \partial_t S(t,y,\eta)=-\lambda(t,y,\Xi(t,y,\eta)). \end{align*} Since $\partial_yS=\Xi$, this last identity is the Hamilton-Jacobi equation \begin{align*} \partial_t S(t,y,\eta)+\lambda(t,y,\partial_yS(t,y,\eta))=0. \end{align*} At $t=0$, the inverse relation gives $Z(0,y,\eta)=y$, and the normalization of $A$ gives \begin{align*} S(0,y,\eta)=y\cdot\eta. \end{align*}[/step]

[guided]We now build the generating function. The previous step gave a function $A$ satisfying \begin{align*} d_{t,z,\eta}A=\xi(t;z,\eta)\cdot d_{t,z,\eta}y(t;z,\eta)-\lambda(t,y(t;z,\eta),\xi(t;z,\eta))\,dt-\eta\cdot dz. \end{align*} The symbol $d_{t,z,\eta}y(t;z,\eta)$ means the full differential of the map $(t,z,\eta)\mapsto y(t;z,\eta)$, paired with the covector $\xi(t;z,\eta)$. The type-II generating function should use $(y,\eta)$ as independent variables. Since the inverse map gives $z=Z(t,y,\eta)$, define \begin{align*} S:\Omega_0\to\mathbb{R},\qquad S(t,y,\eta)=A(t,Z(t,y,\eta),\eta)+Z(t,y,\eta)\cdot\eta. \end{align*} The added term $Z\cdot\eta$ is chosen so that the unwanted $-\eta\cdot dz$ term in $dA$ cancels when we differentiate. Indeed, differentiate $S$ using the chain rule. The differential of $A(t,Z(t,y,\eta),\eta)$ is obtained by substituting $z=Z(t,y,\eta)$ into the displayed identity for $dA$. Since $y(t;Z(t,y,\eta),\eta)=y$, differentiating this identity gives $d_{t,y,\eta}(y(t;Z(t,y,\eta),\eta))=dy$. Hence the full differential term becomes $\Xi(t,y,\eta)\cdot dy$. The term involving $-\eta\cdot dZ$ also appears. The differential of $Z(t,y,\eta)\cdot\eta$ is \begin{align*} \eta\cdot dZ(t,y,\eta)+Z(t,y,\eta)\cdot d\eta. \end{align*} The two terms $\eta\cdot dZ$ and $-\eta\cdot dZ$ cancel. Since the inverse relation says $y(t;Z(t,y,\eta),\eta)=y$, the remaining differential is \begin{align*} d_{t,y,\eta}S=\Xi(t,y,\eta)\cdot dy+Z(t,y,\eta)\cdot d\eta-\lambda(t,y,\Xi(t,y,\eta))\,dt. \end{align*} Reading off coefficients of the independent differentials $dy$, $d\eta$, and $dt$ gives \begin{align*} \partial_y S(t,y,\eta)=\Xi(t,y,\eta). \end{align*} \begin{align*} \partial_\eta S(t,y,\eta)=Z(t,y,\eta). \end{align*} \begin{align*} \partial_t S(t,y,\eta)=-\lambda(t,y,\Xi(t,y,\eta)). \end{align*} Because $\partial_y S=\Xi$, the last identity becomes \begin{align*} \partial_t S(t,y,\eta)+\lambda(t,y,\partial_yS(t,y,\eta))=0. \end{align*} This is the Hamilton-Jacobi equation. At time $t=0$, the spatial flow is the identity, so $Z(0,y,\eta)=y$. The normalization $A(0,z,\eta)=0$ therefore gives \begin{align*} S(0,y,\eta)=y\cdot\eta. \end{align*}[/guided]

[step:Define the phase and derive its critical set]Define the open conic set \begin{align*} \Omega:=\{(t,y,z,\eta):(t,y,\eta)\in\Omega_0,\ z\in \kappa(U),\ (Z(t,y,\eta),\eta)\in V\}. \end{align*} Define \begin{align*} \phi:\Omega&\to\mathbb{R} \end{align*} \begin{align*} (t,y,z,\eta)&\mapsto S(t,y,\eta)-z\cdot\eta. \end{align*} Because $\lambda$ is homogeneous of degree $1$ in the covector variable, its Hamilton equations are invariant under positive fibre scaling: if $(y(t;z,\eta),\xi(t;z,\eta))$ solves the spatial system with initial data $(z,\eta)$, then $(y(t;z,r\eta),\xi(t;z,r\eta))=(y(t;z,\eta),r\xi(t;z,\eta))$ for every $r>0$ for which both sides are defined. Hence the inverse functions satisfy \begin{align*} Z(t,y,r\eta)=Z(t,y,\eta) \end{align*} and \begin{align*} \Xi(t,y,r\eta)=r\Xi(t,y,\eta). \end{align*} For each fixed $r>0$, define the smooth map \begin{align*} S_r:\Omega_0&\to\mathbb{R} \end{align*} \begin{align*} (t,y,\eta)&\mapsto r^{-1}S(t,y,r\eta). \end{align*} The conic choice of $\Omega_0$ ensures that $(t,y,r\eta)\in\Omega_0$ whenever $(t,y,\eta)\in\Omega_0$. The function $S_r$ has the same differential in $(t,y,\eta)$ as $S(t,y,\eta)$, by the identities for $\partial_tS$, $\partial_yS$, and $\partial_\eta S$. Therefore $S_r-S$ is locally constant. At $t=0$, the normalization $S(0,y,\eta)=y\cdot\eta$ gives $S_r(0,y,\eta)=S(0,y,\eta)$, so this locally constant difference is zero after shrinking to the connected neighbourhood under consideration. Thus $S(t,y,r\eta)=rS(t,y,\eta)$, and $\phi(t,y,z,r\eta)=r\phi(t,y,z,\eta)$. The $\eta$-critical equation is \begin{align*} \partial_\eta\phi(t,y,z,\eta)=\partial_\eta S(t,y,\eta)-z=Z(t,y,\eta)-z. \end{align*} Therefore \begin{align*} \Sigma_\phi=\{(t,y,z,\eta)\in\Omega:z=Z(t,y,\eta)\}. \end{align*} Equivalently, \begin{align*} (t,y,z,\eta)\in\Sigma_\phi \iff y=y(t;z,\eta). \end{align*}[/step]

[guided]The phase is defined by separating the generating function from the initial base variable: \begin{align*} \phi(t,y,z,\eta)=S(t,y,\eta)-z\cdot\eta. \end{align*} Because $\lambda$ is homogeneous of degree $1$ in the covector variable, the spatial Hamilton equations are compatible with positive fibre scaling. If $r>0$, then the same base curve and the rescaled covector curve satisfy the equations with initial covector $r\eta$: \begin{align*} y(t;z,r\eta)=y(t;z,\eta). \end{align*} \begin{align*} \xi(t;z,r\eta)=r\xi(t;z,\eta). \end{align*} Therefore the inverse variables satisfy \begin{align*} Z(t,y,r\eta)=Z(t,y,\eta). \end{align*} \begin{align*} \Xi(t,y,r\eta)=r\Xi(t,y,\eta). \end{align*} For fixed $r>0$, define $S_r:\Omega_0\to\mathbb{R}$ by $S_r(t,y,\eta)=r^{-1}S(t,y,r\eta)$. The conic choice of $\Omega_0$ makes this definition valid whenever $(t,y,\eta)\in\Omega_0$. The identities for the derivatives of $S$ show that $dS_r=dS$. Hence $S_r-S$ is locally constant. At $t=0$, both functions equal $y\cdot\eta$, so after shrinking to the connected neighbourhood under consideration the locally constant difference is zero. Thus $S(t,y,r\eta)=rS(t,y,\eta)$, and consequently \begin{align*} \phi(t,y,z,r\eta)=r\phi(t,y,z,\eta). \end{align*} Now compute the critical equation in the phase variable $\eta$: \begin{align*} \partial_\eta\phi(t,y,z,\eta)=\partial_\eta S(t,y,\eta)-z=Z(t,y,\eta)-z. \end{align*} Thus the critical set is \begin{align*} \Sigma_\phi=\{(t,y,z,\eta)\in\Omega:z=Z(t,y,\eta)\}. \end{align*} Using the defining inverse relation for $Z$, this is equivalent to \begin{align*} (t,y,z,\eta)\in\Sigma_\phi \iff y=y(t;z,\eta). \end{align*}[/guided]

[step:Recover the canonical graph from the phase derivatives]On $\Sigma_\phi$, the identities for $S$ give \begin{align*} \partial_y\phi(t,y,z,\eta)=\partial_yS(t,y,\eta)=\Xi(t,y,\eta)=\xi(t;z,\eta). \end{align*} Also, \begin{align*} -\partial_z\phi(t,y,z,\eta)=\eta. \end{align*} Thus, for fixed $t$ with $|t|<T_0$, \begin{align*} \{(y,\partial_y\phi(t,y,z,\eta);z,-\partial_z\phi(t,y,z,\eta)):(t,y,z,\eta)\in\Sigma_\phi\} \end{align*} is exactly the shrunken local graph \begin{align*} \{(y(t;z,\eta),\xi(t;z,\eta);z,\eta):(z,\eta)\in V\}=C_t. \end{align*} Here $V$ denotes the final conic neighbourhood after the preceding shrinkings, and $\Omega$ was defined from the image of $(-T_0,T_0)\times V$ under $(t,z,\eta)\mapsto(t,y(t;z,\eta),\eta)$, so no points outside this local graph are included. Furthermore, \begin{align*} \partial_t\phi(t,y,z,\eta)=\partial_tS(t,y,\eta), \end{align*} because $z\cdot\eta$ is independent of $t$. Hence, on $\Sigma_\phi$, \begin{align*} \partial_t\phi(t,y,z,\eta)+\lambda(t,y,\partial_y\phi(t,y,z,\eta))=0. \end{align*} This is the required eikonal equation for the full phase with $z$ treated as a passive parameter.[/step]

[guided]On the critical set, the equation $z=Z(t,y,\eta)$ means that $y$ is exactly the final base point obtained by flowing from the initial data $(z,\eta)$: \begin{align*} y=y(t;z,\eta). \end{align*} The $y$-derivative of the phase is the $y$-derivative of $S$, so the derivative identities for $S$ give \begin{align*} \partial_y\phi(t,y,z,\eta)=\partial_yS(t,y,\eta)=\Xi(t,y,\eta)=\xi(t;z,\eta). \end{align*} The $z$-dependence of $\phi$ is only the term $-z\cdot\eta$, hence \begin{align*} -\partial_z\phi(t,y,z,\eta)=\eta. \end{align*} Therefore the set obtained from the phase derivatives is \begin{align*} \{(y,\partial_y\phi(t,y,z,\eta);z,-\partial_z\phi(t,y,z,\eta)):(t,y,z,\eta)\in\Sigma_\phi\}. \end{align*} Substituting the two derivative identities into this set gives precisely \begin{align*} \{(y(t;z,\eta),\xi(t;z,\eta);z,\eta):(z,\eta)\in V\}=C_t. \end{align*} Here $V$ is the final shrunken conic neighbourhood, and $\Omega$ was defined using the image of this same $V$, so the parametrisation neither adds points outside the local graph nor omits points from it. Finally, since $z\cdot\eta$ is independent of $t$, we have \begin{align*} \partial_t\phi(t,y,z,\eta)=\partial_tS(t,y,\eta). \end{align*} On $\Sigma_\phi$, the identity $\partial_y\phi=\partial_yS$ and the Hamilton-Jacobi equation for $S$ give \begin{align*} \partial_t\phi(t,y,z,\eta)+\lambda(t,y,\partial_y\phi(t,y,z,\eta))=0. \end{align*} This is the eikonal equation for the full phase, with $z$ serving as the initial base parameter.[/guided]

[step:Verify nondegeneracy of the phase] It remains to check nondegeneracy. The critical equation is \begin{align*} \partial_\eta\phi(t,y,z,\eta)=Z(t,y,\eta)-z. \end{align*} Let $I_n:\mathbb{R}^n\to\mathbb{R}^n$ denote the identity linear map. The derivative of $\partial_\eta\phi$ with respect to the $z$ variables is the linear map \begin{align*} \partial_z\partial_\eta\phi(t,y,z,\eta)=-I_n. \end{align*} Therefore the differentials of the components of $\partial_\eta\phi$ have rank $n$ along $\Sigma_\phi$. This is precisely the nondegeneracy condition for a phase function with phase variables $\eta\in\mathbb{R}^n_0$. Hence $\phi$ is a nondegenerate homogeneous phase function. Combining this nondegeneracy with the parametrisation in the previous step proves that, after the stated shrinkings, $\phi$ solves the eikonal equation on its critical set and parametrises the canonical graph $C_t$ for every $|t|<T_0$. [/step]