Q: How to show uniqueness of the non-degenerate solution of $g(x)=0$? Let $g(x): \mathbb{R}^n \rightarrow \mathbb{R}^n$. We call a root $x_*$ of $g(x)=0$ non-degenerate if $Jg(x_*)$ is invertible, where $Jg(x_*)$ is the Jacobian at $x_*$.
How can we show if $x_*$ is non-degenerate, then $g(x) = 0$ has a locally unique solution.
My prof. said to me it can be shown using the Inverse function theorem but I have no idea.

A: We know by the inverse function theorem that if $Jg(x_*)$ is invertible, then $g$ is locally bijective (for some neighborhood $U$ containing $x_*$), meaning that is injective there as well. Therefore, if there exits another $x$ such that $g(x)=0$ in $U$, then $g(x)=g(x_*)$, and so $g$ would not be injective in $U$, giving us a contradiction. Therefore, $x_*$ must be locally unique (in the neighborhood $U$ given to us by the inverse function theorem).
