I'm self-studying Cheney's book in Functional Analysis, and this is among the exercises (problem 3.4.5). I think I've proven the claim in the problem, but I guess I've made errors along the way or perhaps there is some easier way to show this:
Let $f: \Omega \rightarrow Y $ be a continuously differentiable map, where $\Omega$ is an open set in a Banach space, and $Y$ is a normed linear space. Assume that $f'(x)$ is invertible for each $x\in \Omega $ and prove that $f(\Omega)$ is open.
Now, to remain faithful to the book's notation and pedagogical approach, I'm gonna bring the Inverse Function Theorem as stated there:
Let $f$ be a continuously differentiable map from an open set $\Omega$ in a Banach space into a normed linear space. If $x_0 \in \Omega$ and if $f'(x_0)$ is invertible, then there is a continuously differentiable function g defined on a neighborhood $N$ of $f(x_0)$ such that $f(g(y)) = y, \forall y \in Y.$
So here is my attempt at the proof:
Let $x_0 \in \Omega \Rightarrow \exists N \subset Y$, s.t., $f(x_0) \in N$ and $f(f^{-1} (N)) =N,\, \, N$ open. (I denote $g$ by $f^{-1}.)$
$f$ is continuously differentiable $\Rightarrow $ it is indeed continuous, and as such: $f^{-1}(N)$ is open in $\Omega$.
Denoting $f^{-1}(N)$ by $V_{x_0}$, we get:
Now, because $f(x_0) \in N \Rightarrow x_0 \in f^{-1} (N)$. (Here is where I get into trouble, as I think I need injectivity to claim this. But that is not assumed in the problem description.)
$\forall x_0 \in \Omega, \exists V_{x_0} \subset \Omega , s.t. x_0 \in V_{x_0} \Rightarrow \Omega = \bigcup V_{x_0} \Rightarrow f(\Omega) = \bigcup f(V_{x_0})$ which is open as each $f(V_{x_0}) = N$.
I feel something's wrong in here, but I cannot pinpoint it. I'd appreciate any help correcting probable mistakes. Thanks.