If $f$ has an invertible derivative at some point $x$, then $f$ is 1-1 on some neighborhood of $x$. 
Let $(V,\lvert\cdot\rvert)$ be a normed vector space, $U$ an open subset of $V$, and $f:U\rightarrow V$. Suppose $f$ is differentiable at $\xi$ and $Df(\xi)$ is invertible. Show there exists a neighborhood $N$ of $\xi$ on which $f$ is injective.

Proof$\quad$ Put $T=Df(\xi)$. Since $T$ is invertible, $\lVert T^{-1}\rVert\not=0$, and
$$\lvert x-y\rvert=\lvert T^{-1}(Tx-Ty)\rvert\leq\lVert T^{-1}\rVert\lvert Tx-Ty\rvert\mbox{,}$$
so that
$$\frac{\lvert x-y\rvert}{\lVert T^{-1}\rVert}\leq\lvert Tx-Ty\rvert $$
for any $x,y\in V$.
$\qquad$ Since $f$ is differentiable at $\xi$, we have
$$f(x)=f(\xi)+T(x-\xi)+\lvert x-\xi\rvert R(x-\xi)\mbox{,} $$
where $R(x-\xi)\rightarrow 0$ as $x\rightarrow \xi$. Choose $\delta>0$ so that $\vert x-\xi\rvert<\delta$ implies $\lvert R(x-\xi)\rvert<\frac{1}{2\lVert T^{-1}\rVert}$. It follows then that for $\lvert x-\xi\rvert<\delta$, $\lvert y-\xi\rvert<\delta$,
$$
\begin{aligned}
\lvert f(x)-f(y)\rvert &= \lvert Tx-Ty+\lvert x-\xi\rvert R(x-\xi)+\lvert y-\xi\rvert R(y-\xi)\rvert\\
&\geq \lvert Tx-Ty\rvert-\lvert x-\xi\rvert\lvert R(x-\xi)\rvert-\lvert y-\xi\rvert\lvert R(y-\xi)\rvert \\
&> \frac{\vert x-y\rvert}{\lVert T^{-1}\rVert}-(\lvert x-\xi\rvert+\lvert y-\xi\rvert)\frac{1}{2\lVert T^{-1}\rVert}\\
&> \frac{\lvert x-y\rvert}{2\lVert T^{-1}\rVert}\mbox{.}
\end{aligned}
$$
Is this correct?
 A: This is not true.
Let $f(x) = {x \over 2} + x^2 \sin {1 \over x}$ for $x \neq 0$ and $f(0) = 0$. Note that $f$ is smooth for $x \neq 0$. We see that $f'(0) = {1 \over 2} >0$ and for $x \neq 0$ we have
$f'(x) = {1 \over 2} + 2x \sin {1 \over x} - \cos {1 \over x}$.
We see that there are points $x_+,x_- \neq 0$ arbitrarily close to $0$ such that $f'(x_+)>0$ and $f'(x_-)<0$. Since $f'$ is continuous, this means that $f$ is increasing locally around $x_+$ and decreasing locally around $x_-$.
In particular, $f$ cannot be locally injective at $x=0$.
This is slightly counterintuitive, maybe because we imagine derivatives to be continuous.
A related result is found in
Tao's answer
which is an answer to his question on MO
here.
The difference is (apart from $V=\mathbb{R}^n$) that $f$ is assumed to have an invertible derivative everywhere. (And frankly, that this results in a positive answer is very surprising to me and highlights my lack of understanding things global. Tao's answer is fairly involved.)
