# Inverse Function Theorem/ Polynomial

Let $F: \mathbb{C}^n \to \mathbb{C}^n$ be of the form $F(x_1, \dots, x_n)= (F_1(x_1, \dots, x_n), \dots, F_n( \dots , x_1, \dots, x_n) )$ for some $F_1,\dots , F_n \in \mathbb{C}[X_1, \dots, X_n ]$, which I will call polynomial, with constant Jacobian determinant $\det DF$.
Then by Implicit Function Theorem $F$ has a local inverse $F'$ around $0$. Since $F$ is analytical $F'$ is also analytical. We also know that $DF' = (DF)^{-1}= \frac{1}{\det DF} \text{adj}(DF)$ but the adjugate is computed in terms of the minors of $DF$, $\det DF$ is constant and thus $DF'$ is given by polynomial functions in its components, in particular $D^k F =0$ for some $k\in \mathbb{N}$. Thus $F'$ is polynomial because of its Taylor expansion. Since $F'$ is polynomial it is defined on the whole space $\mathbb{C}^n$.
Now we have $F' \circ F\mid_U= \text{id}$ for some open neighborhood $U$ of $0$, but this implies $F' \circ F= \text{id}$ since $F' \circ F$ is a polynomial.