I want to find a continuously differentiable function $F:X\to Y$, where $X\subseteq\mathbb{R}^n$, $Y\subseteq\mathbb{R}^m$ are open ($n\le m$) with

- ${\rm rk}\, \frac{\partial F}{\partial x}(x) = n$ for all $x\in X$,
- ${\rm graph}\, F = \{\,(x,F(x))\, \mid\, x\in X\,\}$ is connected,
- ${\rm im}\, F = \{\,F(x)\, \mid\, x\in X\,\}$ is simply connected,

$\underline{\rm but}$ $F$ is not injective (if this is possible).

I know that if we additionally claim that a left inverse of the Jacobian, for instance $$ L(x) = \left(\left(\frac{\partial F}{\partial x} (x)\right)^\top \frac{\partial F}{\partial x} (x)\right)^{-1} \left(\frac{\partial F}{\partial x} (x)\right)^\top, $$ is bounded in the sense $$ \|L(x)\| \le \omega(\|x\|) $$ for all $x\in X$, where $\omega:[0,\infty)\to (0,\infty)$ is continuous and satisfies $$ \int_0^\infty \frac{dt}{\omega(t)} = \infty, $$ then $F$ must be injective. I just wonder whether we really need this bound on the left inverse of the Jacobian (in some cases it is definitely too strong). So far I have been unable to find an example (satisfying the three conditions abvove) which violates it and is not injective.

As you may recognize, this question is related to a global implicit function theorem.

nottrue without additional assumptions (there should be some completeness or something like that). $\endgroup$ – Willie Wong Jul 26 '16 at 16:133more comments