[proofplan] We first prove the corresponding untwisted statement on $T^*(X\times Y)$ for the map sending $(x,y,\theta)$ to the covector $d_{x,y}\phi$. The nondegeneracy hypothesis makes the critical set $\Sigma_\phi$ a smooth conic manifold of the correct dimension, and the standard nondegenerate phase rank argument shows that the phase map is an immersion. Pulling back the canonical one-form gives $d(\phi|_{\Sigma_\phi})$, because the $\theta$ derivatives vanish on $\Sigma_\phi$; hence the pulled-back symplectic form is zero. Since the immersed image is isotropic of half dimension, it is Lagrangian, and the sign change in the $Y$ covector converts the canonical product symplectic form into $\pi_X^*\omega_X-\pi_Y^*\omega_Y$. [/proofplan]

[step:Pass to the untwisted cotangent bundle of $X\times Y$] Let $M := X\times Y$, and write a point of $M$ as $z=(x,y)$. In a local coordinate chart on $M$, write $z=(z_1,\dots,z_m)$, where $m=\dim X+\dim Y$. Define the untwisted phase map \begin{align*} I_\phi:\Sigma_\phi &\to T^*M \end{align*} \begin{align*} (z,\theta) &\mapsto (z,\partial_z\phi(z,\theta)). \end{align*} Here $\partial_z\phi(z,\theta)\in T_z^*M$ is the covector whose local coordinate expression is \begin{align*} \partial_z\phi(z,\theta)=\sum_{a=1}^{m}\partial_{z_a}\phi(z,\theta)\,dz_a. \end{align*} The assumptions $\partial_x\phi\neq 0$ and $\partial_y\phi\neq 0$ imply that $I_\phi(\Sigma_\phi)\subset T^*M\setminus 0$, and also that the twisted map $F_\phi$ lands in $(T^*X\setminus 0)\times(T^*Y\setminus 0)$. [/step]

[step:Use the regular level set theorem to make $\Sigma_\phi$ a smooth conic manifold] Define the constraint map \begin{align*} G:\Omega &\to \mathbb{R}^N \end{align*} \begin{align*} (z,\theta) &\mapsto \bigl(\partial_{\theta_1}\phi(z,\theta),\dots,\partial_{\theta_N}\phi(z,\theta)\bigr). \end{align*} Then $\Sigma_\phi=G^{-1}(0)$. At every point of $\Sigma_\phi$, the differential \begin{align*} dG_{(z,\theta)}:T_{(z,\theta)}\Omega \to \mathbb{R}^N \end{align*} has rank $N$, because its component covectors are precisely \begin{align*} d(\partial_{\theta_1}\phi),\dots,d(\partial_{\theta_N}\phi). \end{align*} Thus $0$ is a regular value of $G$ along $G^{-1}(0)$. By the regular level set theorem, $\Sigma_\phi$ is a smooth submanifold of $\Omega$ of codimension $N$. Therefore \begin{align*} \dim \Sigma_\phi=\dim X+\dim Y. \end{align*} Since $\Omega$ is conic in $\theta$ and $\phi$ is homogeneous of degree $1$ in $\theta$, each $\partial_{\theta_j}\phi$ is homogeneous of degree $0$ in $\theta$. Hence $G(z,\lambda\theta)=G(z,\theta)$ for every $\lambda>0$ for which $(z,\lambda\theta)\in\Omega$, so $\Sigma_\phi$ is conic. [/step]

[step:Prove that the untwisted phase map is an immersion]Let $(z,\theta)\in\Sigma_\phi$, and let $v\in T_{(z,\theta)}\Sigma_\phi$ satisfy \begin{align*} d(I_\phi)_{(z,\theta)}(v)=0. \end{align*} Since the base component of $I_\phi$ is the projection $(z,\theta)\mapsto z$, this implies \begin{align*} dz_a(v)=0 \end{align*} for every $1\leq a\leq m$. Since the fiber component of $I_\phi$ is $\partial_z\phi$, it also implies \begin{align*} d(\partial_{z_a}\phi)_{(z,\theta)}(v)=0 \end{align*} for every $1\leq a\leq m$. Finally, because $v\in T_{(z,\theta)}\Sigma_\phi$ and $\Sigma_\phi=G^{-1}(0)$, we have \begin{align*} d(\partial_{\theta_j}\phi)_{(z,\theta)}(v)=0 \end{align*} for every $1\leq j\leq N$. The identities $dz_a(v)=0$ show that $v$ is vertical in the $\theta$ direction. Write \begin{align*} v=\sum_{k=1}^{N}v_k\,\partial_{\theta_k} \end{align*} in the chosen local coordinates. The equalities \begin{align*} d(\partial_{z_a}\phi)(v)=0 \end{align*} and \begin{align*} d(\partial_{\theta_j}\phi)(v)=0 \end{align*} together say that the differential of the full gradient $d\phi$ in the vertical direction $v$ is zero. Equivalently, \begin{align*} \sum_{k=1}^{N}v_k\,\partial_{\theta_k}\partial_{z_a}\phi=0 \end{align*} for every $a$, and \begin{align*} \sum_{k=1}^{N}v_k\,\partial_{\theta_k}\partial_{\theta_j}\phi=0 \end{align*} for every $j$. Since $\phi$ is homogeneous of degree $1$ in $\theta$, Euler's homogeneous identity gives \begin{align*} \phi(z,\theta)=\sum_{k=1}^{N}\theta_k\,\partial_{\theta_k}\phi(z,\theta). \end{align*} Differentiating this identity in the vertical direction $v$ and using $\partial_{\theta_k}\phi=0$ on $\Sigma_\phi$ gives no further condition on $\phi$ itself, but differentiating the identities for all first derivatives shows that the covector \begin{align*} \sum_{k=1}^{N}v_k\,d(\partial_{\theta_k}\phi)_{(z,\theta)} \end{align*} vanishes on every coordinate vector $\partial_{z_a}$ and on every coordinate vector $\partial_{\theta_j}$. Hence \begin{align*} \sum_{k=1}^{N}v_k\,d(\partial_{\theta_k}\phi)_{(z,\theta)}=0 \end{align*} as an element of $T^*_{(z,\theta)}\Omega$. By the assumed [linear independence](/page/Linear%20Independence) of these covectors, $v_k=0$ for every $k$. Thus $v=0$, so $d(I_\phi)_{(z,\theta)}$ is injective. Therefore $I_\phi$ is an immersion.[/step]

[guided]The only delicate point is to show that the rank condition on the phase really forces the phase map to be immersive. Fix $(z,\theta)\in\Sigma_\phi$ and suppose that a tangent vector $v\in T_{(z,\theta)}\Sigma_\phi$ is killed by $d(I_\phi)_{(z,\theta)}$. The map $I_\phi$ has two components: the base point $z$ and the covector $\partial_z\phi$. Therefore $d(I_\phi)(v)=0$ gives two families of equations: \begin{align*} dz_a(v)=0 \end{align*} for every base coordinate $z_a$, and \begin{align*} d(\partial_{z_a}\phi)(v)=0 \end{align*} for every $1\leq a\leq m$. The first family means that $v$ has no $z$ component. Thus, in local coordinates, it has the form \begin{align*} v=\sum_{k=1}^{N}v_k\,\partial_{\theta_k}. \end{align*} Since $v$ is tangent to $\Sigma_\phi=G^{-1}(0)$, it also satisfies the linearized constraint equations \begin{align*} d(\partial_{\theta_j}\phi)(v)=0 \end{align*} for every $1\leq j\leq N$. Now combine the two families. The equations $d(\partial_{z_a}\phi)(v)=0$ say that the covector \begin{align*} \sum_{k=1}^{N}v_k\,d(\partial_{\theta_k}\phi) \end{align*} vanishes on the coordinate vector $\partial_{z_a}$, because mixed partial derivatives agree: \begin{align*} \sum_{k=1}^{N}v_k\,d(\partial_{\theta_k}\phi)(\partial_{z_a}) = \sum_{k=1}^{N}v_k\,\partial_{z_a}\partial_{\theta_k}\phi = d(\partial_{z_a}\phi)(v) = 0. \end{align*} The tangent-to-$\Sigma_\phi$ equations say that the same covector vanishes on each coordinate vector $\partial_{\theta_j}$: \begin{align*} \sum_{k=1}^{N}v_k\,d(\partial_{\theta_k}\phi)(\partial_{\theta_j}) = d(\partial_{\theta_j}\phi)(v) = 0. \end{align*} Since the coordinate vectors $\partial_{z_a}$ and $\partial_{\theta_j}$ span $T_{(z,\theta)}\Omega$, we obtain \begin{align*} \sum_{k=1}^{N}v_k\,d(\partial_{\theta_k}\phi)_{(z,\theta)}=0. \end{align*} The nondegeneracy hypothesis is exactly the statement that the covectors $d(\partial_{\theta_1}\phi),\dots,d(\partial_{\theta_N}\phi)$ are linearly independent on $\Sigma_\phi$. Therefore every coefficient $v_k$ is zero. Hence $v=0$, proving that $d(I_\phi)$ is injective at $(z,\theta)$. Since the point was arbitrary, $I_\phi$ is an immersion.[/guided]

[step:Pull back the canonical one-form and prove isotropicity] Let $\alpha_M$ denote the canonical one-form on $T^*M$, with local expression \begin{align*} \alpha_M=\sum_{a=1}^{m}\xi_a\,dz_a \end{align*} in cotangent coordinates $(z,\xi)$. We use the convention \begin{align*} \omega_M=d\alpha_M \end{align*} for the canonical symplectic form. Pulling $\alpha_M$ back along $I_\phi$ gives \begin{align*} I_\phi^*\alpha_M=\sum_{a=1}^{m}\partial_{z_a}\phi\,dz_a. \end{align*} On $\Sigma_\phi$, the defining equations give $\partial_{\theta_j}\phi=0$ for every $j$. Hence the differential of the restricted function \begin{align*} \phi|_{\Sigma_\phi}:\Sigma_\phi\to\mathbb{R} \end{align*} is \begin{align*} d(\phi|_{\Sigma_\phi}) = \sum_{a=1}^{m}\partial_{z_a}\phi\,dz_a + \sum_{j=1}^{N}\partial_{\theta_j}\phi\,d\theta_j = \sum_{a=1}^{m}\partial_{z_a}\phi\,dz_a. \end{align*} Therefore \begin{align*} I_\phi^*\alpha_M=d(\phi|_{\Sigma_\phi}). \end{align*} Taking exterior derivatives and using $d^2=0$, we obtain \begin{align*} I_\phi^*\omega_M = I_\phi^*(d\alpha_M) = d(I_\phi^*\alpha_M) = d^2(\phi|_{\Sigma_\phi}) = 0. \end{align*} Thus $I_\phi(\Sigma_\phi)$ is isotropic with its immersed structure. [/step]

[step:Use the dimension count to conclude the untwisted image is Lagrangian] The symplectic manifold $T^*M$ has dimension \begin{align*} 2\dim M=2(\dim X+\dim Y). \end{align*} The domain $\Sigma_\phi$ has dimension \begin{align*} \dim \Sigma_\phi=\dim X+\dim Y=\frac{1}{2}\dim T^*M. \end{align*} Since $I_\phi$ is an immersion and $I_\phi^*\omega_M=0$, the immersed tangent space \begin{align*} d(I_\phi)_{(z,\theta)}(T_{(z,\theta)}\Sigma_\phi) \end{align*} is an isotropic subspace of $T_{I_\phi(z,\theta)}T^*M$ of maximal possible dimension. Hence it is a Lagrangian subspace. Therefore $I_\phi:\Sigma_\phi\to T^*M$ is a conic Lagrangian immersion, and its immersed image is a conic immersed Lagrangian submanifold of $T^*M\setminus 0$. [/step]

[step:Apply the cotangent sign change on the $Y$ factor] Use the product identification \begin{align*} T^*(X\times Y)\cong T^*X\times T^*Y \end{align*} given by \begin{align*} (x,y;\xi,\eta)\mapsto (x,\xi;y,\eta). \end{align*} Define the fiber sign-change map \begin{align*} S:T^*X\times T^*Y &\to T^*X\times T^*Y \end{align*} \begin{align*} (x,\xi;y,\eta) &\mapsto (x,\xi;y,-\eta). \end{align*} Then \begin{align*} F_\phi=S\circ I_\phi. \end{align*} Let $\alpha_X$ and $\alpha_Y$ denote the canonical one-forms on $T^*X$ and $T^*Y$, and let \begin{align*} \omega_X=d\alpha_X,\qquad \omega_Y=d\alpha_Y. \end{align*} Under the product identification, the canonical one-form on $T^*(X\times Y)$ is \begin{align*} \alpha_M=\pi_X^*\alpha_X+\pi_Y^*\alpha_Y. \end{align*} The map $S$ fixes the $T^*X$ covector and multiplies the $T^*Y$ covector by $-1$, so \begin{align*} S^*(\pi_X^*\alpha_X+\pi_Y^*\alpha_Y) = \pi_X^*\alpha_X-\pi_Y^*\alpha_Y. \end{align*} Taking exterior derivatives gives \begin{align*} S^*\omega_M = \pi_X^*\omega_X-\pi_Y^*\omega_Y. \end{align*} Since $I_\phi$ is Lagrangian for $\omega_M$, the composition $F_\phi=S\circ I_\phi$ is Lagrangian for $\pi_X^*\omega_X-\pi_Y^*\omega_Y$. The conic property follows from the homogeneity of $\phi$: for $\lambda>0$, \begin{align*} \partial_x\phi(x,y,\lambda\theta)=\lambda\,\partial_x\phi(x,y,\theta) \end{align*} and \begin{align*} \partial_y\phi(x,y,\lambda\theta)=\lambda\,\partial_y\phi(x,y,\theta). \end{align*} Thus $F_\phi(x,y,\lambda\theta)$ is obtained from $F_\phi(x,y,\theta)$ by positive fiber dilation in both cotangent factors. Therefore $F_\phi$ is a conic Lagrangian immersion, and $C_\phi=F_\phi(\Sigma_\phi)$ is a conic immersed Lagrangian submanifold with the induced immersed structure. [/step]