Is there a full-rank map with connected graph and simply connected image that is not injective? 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.
 A: In fact there is a simple smooth non injective immersion of an open rectangle onto an open disk of the plane. 
Imagine we want to paint a large full circle on a wall by a paint roller. So we are applying a  rectangle on a plane disk. Can we do it by a non injective immersion? Yes: make a regular spiral starting from the boundary till a small hole is left, then wander inside the disk, make a curve and go back to the hole to fill it. (I hope  this explanation is sufficient, otherwise I will have to provide an analytic representation).
A: Actually, taking a look at the paper you linked to: the "continuation property" is not as simple as you think it is (and is what I was thinking of in terms of a "completeness" condition). Pietro's map is explicitly one that satisfies all the conditions you alluded to, even the bound on the inverse Jacobian, but fails the "continuation property". For domain being exactly $X = \mathbb{R}^n$ the continuation property holds essentially because paths cannot run off out of the domain $X$ from the side and  the condition on the bound of the inverse Jacobian. 

The following is a modification of Pietro's construction. The blue line is the path traveled by the center of the paint roller. Assume that the radius of curvature of the "turns" is smaller than the half-width of the paint roller, but larger than half the distance between successive horizontal lines. (This condition is available by "jumping at least two lines per turn.) This will then give a mapping of an open, long and skinny rectangle to a rectangle with "rounded corners".

(Personally I prefer to think of this as lawn-mowing patterns rather than wall-painting ones; connectivity and small curvature are more important there.)
