The divergence-free Jacobian conjecture in dimension two (Q2477820)

The author proves the following particular case of the Jacobian conjecture. Let \(F:\mathbb{C}^{2}\rightarrow \mathbb{C}^{2}\) be a polynomial mapping such that \(F(0)=0,\) \(F^{\prime }(0)=I,\) \(\text{Jac}F\equiv 1.\) If we write \(F(x,y)=\left( x+r(x,y),y+s(x,y)\right) ,\) \(\deg r,\) \(\deg s\geq 2,\) and \[ \begin{aligned} \text{Jac}\left( r,s\right) & =0\text{ (i.e. }\frac{\partial r}{\partial x }\frac{\partial s}{\partial y}-\frac{\partial r}{\partial y}\frac{\partial s }{\partial x}=0), \\ \frac{\partial r}{\partial x}+\frac{\partial s}{\partial y}& =0, \end{aligned} \tag{\(*\)} \] then \(F\) is bijective. The proof is based on the following characterization of condition (\(\ast \)): Equality (\(\ast \)) holds if and only if there exists a linear mapping \(L: \mathbb{C}^{2}\rightarrow \mathbb{C}\) and a polynomial mapping \(Q:\mathbb{C} \rightarrow \mathbb{C}^{2}\) such that \[ F(z)=z+Q(L(z))\text{ \;and \;}L(Q(L(z)))=0,\text{ \;\;}z\in \mathbb{C}^{2}. \]