# 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.

• Can you add a proof or a reference to the fact that the additional claim renders your aim impossible? – Amir Sagiv Jul 26 '16 at 8:13
• The proof essentially relies on showing that the projection onto the second component $\pi: X\times Y\to Y$ restricted to ${\rm graph }\, F$ is a covering map. Then it is a homeomorphism by the third condition. Now: The first condition ensures that it is locally a homeomorphism. In order to show that it is a covering map we may apply Theorem 2.6 in [O. Gutú and J. A. Jaramillo, “Global homeomorphisms and covering projections on metric spaces,” Math. Ann., vol. 338, pp. 75–95, 2007] by first showing that it has the contnuation property for an appropriate set of paths. But this is similar – Thomas Jul 26 '16 at 9:34
• On a side note, since F si continuous, the graph of F is homeomorphic to its domain, so the second point is equivalent to "X is connected" – Pietro Majer Jul 26 '16 at 14:55
• Between the result that you claim and the counterexample Pietro gave below, something does not quite match up. I am pretty sure there exists a path that satisfies Pietro's description that also satisfies your left inverse bound on the Jacobian. // In fact I am pretty sure what you claim to be true is not true without additional assumptions (there should be some completeness or something like that). – Willie Wong Jul 26 '16 at 16:13
• What do you mean by closed? As topologically a closed set in $\mathbb{R}^n$? Then your assumption of both open and closed means $X$ is in fact the whole space. But then as the set $X$ in both Pietro's and my answers are open and convex (being open rectangles), they are diffeomorphic to $\mathbb{R}^n$. So composing with such a diffeomorphism you get an example. (And the function $L$ clearly fails the criterion you listed in your question.) – Willie Wong Aug 2 '16 at 13:09

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.)

• I like a lot your picture, but what do you mean by "continuation property" here? I can't find any reference in the OP nor comments. – Pietro Majer Sep 14 '16 at 7:26
• @PietroMajer: it was mentioned in this comment by the OP and is defined in the paper he linked to. The statement of the property is (I paraphrase) "a map $f:X\to Y$ between metric spaces is said to have the continuation property if for every pair of paths $p:[0,1]\to Y$ and $q:[0,b) \to X$ with $b\in (0,1]$, such that $p|_{[0,b)} = f\circ q$, one can find a sequence $t_n \nearrow b$ such that $q(t_n)$ converges." – Willie Wong Sep 14 '16 at 13:24
• @PietroMajer: I interpret it to mean something to the effect of "if $f(x_n)$ converges in $Y$, then $x_n$ must also converge in $X$" which is why I think of it more as something like "completeness". – Willie Wong Sep 14 '16 at 13:27

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).