It is continued from the previous question of me. To show it holds for $\mathbb R^2$, I want to show the following theorem holds:
Let $U\subseteq \mathbb R^2$ be open and let $f:U\rightarrow \mathbb R^2$ be nonconstant and smooth. Then $f$ is an open mapping on $U$, that is, $f^-1$ is continuous on $f(U)$.
Let $U$ be the neighborhood of $x_0$ where $f$ is bijective. Let $w_0\in f(U)$. Then by the surjectivity of $f$, there is a $z_0\in U$ such that $f(z_0)=w_0$. Since $U$ is open, we can take $\delta>0$ so that the closed ball $B=D(z_0,d)$ is contained in $U$. Let $g(z)=f(z)-w_0$ be defined. Then $g$ is differentiable and non-constant on $U$.
Now a problem arises:
In complex case, the identity theorem holds. So, if I use it, I could show that $g(z)$ has an isolated root. However, in $\mathbb R^2$, the identity theorem seems to be false. How can I show it? Is there any other way to prove it?
