- explicit counterexample: \begin{aligned} F = (P, Q): \mat...

Prompt

- explicit counterexample: \begin{aligned} F = (P, Q): \mathbb{C}^2 \to \mathbb{C}^2 \quad : \quad & \begin{cases} P(x, y) = x + A(x, y) + x^{16} y^{16} \\ Q(x, y) = y + B(x, y) + x^{56} y^{56} \end{cases} \\ \wedge\quad & \begin{cases} A(x, y) = \sum_{k=2}^{31} c_k x^{\pi_x(k)} y^{\pi_y(k)} \\ B(x, y) = \sum_{m=2}^{111} d_m x^{\phi_x(m)} y^{\phi_y(m)} \end{cases} \\ \wedge\quad & \mathbf{c}, \mathbf{d} \in \mathbf{V}\left( \left\langle \text{Coeff}\left( \frac{\partial P}{\partial x}\frac{\partial Q}{\partial y} - \frac{\partial P}{\partial y}\frac{\partial Q}{\partial x} - 1, \, x^i y^j \right) \right\rangle_{i,j \ge 0} \right) \subset \mathbb{C}^{6856} \\ \wedge\quad & \frac{\partial P}{\partial x}\frac{\partial Q}{\partial y} - \frac{\partial P}{\partial y}\frac{\partial Q}{\partial x} \equiv 1 \\ \wedge\quad & P(0,0) = Q(0,0) = 0 \;\wedge\; \begin{pmatrix} \frac{\partial P}{\partial x}(0,0) & \frac{\partial P}{\partial y}(0,0) \\ \frac{\partial Q}{\partial x}(0,0) & \frac{\partial Q}{\partial y}(0,0) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\ \wedge\quad & \det \begin{pmatrix} \frac{\partial}{\partial x}\left(\frac{P+Q}{\sqrt{2}}\right) & \frac{\partial}{\partial y}\left(\frac{P+Q}{\sqrt{2}}\right) \\ \frac{\partial}{\partial x}\left(\frac{Q-P}{\sqrt{2}}\right) & \frac{\partial}{\partial y}\left(\frac{Q-P}{\sqrt{2}}\right) \end{pmatrix} \equiv 1 \\ \wedge\quad & S_{P,Q} \supseteq \big\{ (u, v) \in \mathbb{C}^2 \mid v^2 - u^7 = 0 \big\} \ne \emptyset \implies \neg\operatorname{Proper}(F) \\ \wedge\quad & [\mathbb{C}(x,y) : \mathbb{C}(P, Q)] = d \ge 5 \implies \exists z_1 \ne z_2 \in \mathbb{C}^2 : F(z_1) = F(z_2) \implies \neg\operatorname{Inj}(P, Q) \\ \wedge\quad & \neg\operatorname{Inv}(P, Q) \end{aligned} - solution status counterexample found