I am trying to solve a question as:
Let $f$ be a continuous map from an open set $S$ in a Banach space $X$ in to a Banach space $Y$. Suppose that for some $x_0$ in $S$, $f'(x_0)$ exists and invertible. Prove that f is one-to-one in some neighborhood of $x_0$.
My question is, how should I use the assumption "$f'(x_0)$ is invertible"? What does it imply?
I assume that "one-to-one" means "injective".
This is not true. Take $X = Y = \mathbb{R}$ and $$f(x) = x + x^2 \, \sin(1/x^2).$$
Then, $f$ is differentiable at $x_0 = 0$ with $f'(0) = 1$, hence, $f'(x_0)$ is continuously invertible. However, in every neighborhood of $x_0$ you find points $x$ with $f'(x) < 0$ and this can be used to see that $f$ is not locally injective.
You need an additional assumption, e.g., that $f$ is strictly differentiable in $x_0$, i.e., for all $\varepsilon > 0$, there is $\delta > 0$ with $$\|f(x) - f(y) - f'(x_0) \, (x - y)\| \le \varepsilon \, \|x - y\|$$ for all $x,y \in B_\delta(x_0)$. Note that this follows if $f$ is continuously differentiable in $x_0$.