Let $f:\Omega \to \mathrm{R^n}$ be a continuous map, differentiable at $x_0$ and such that $Df_{x_0}$ is invertible. I want to prove there exists $\epsilon>0$: $f$ is surjective in $B_\epsilon(f(x_0))$
WLOG: $x_0=0=f(x_0)$, $Df(0)=Id$ (the last assumption is justified by the fact that the differential map is invertible). I tried to prove the statement by contradiction.
Suppose $\forall \epsilon >0$ there exists $y \in \overline B_\epsilon(0)$ such that $y \neq f(x) \forall x \in \overline B_{\delta}(0)$ where $\delta$ is such that $f(\overline{B_{\delta}(0)}) \subseteq \overline B_\epsilon(0)$. I distinguish two cases:
- y is a limit point in $\overline B_\epsilon(0)$. In that case I consider a sequence of points $f(x_n) \to y$. Since the closed ball is compact there is a subsequence $x_{n_k} \to \tilde{x} $ therefore by continuity of f: $y=\mathrm{lim}_{k \to \infty}f(x_{n_k})=f(\tilde x)$ which is the desired contradiction.
The second case is the one I am struggling with, that is when y is an isolated point in the epsilon ball. Intuitively I know that in this case $||f(x)-y||\geq r(y)$ and $||f(x)-x||\leq \sigma ||x||$ for some $\sigma \in \mathbb{R^+}$ because it is differentiable in zero so I would like to say that x and y get very close but still they may never touch and I can't use sequence-like arguments as before (at least I can't see them).
Any ideas?I also tried to use the Banach fixed point theorem using $g=x-y+f(x)$ as a possible contraction but it doesn't seem to work in this case because I can't show it is a contraction.