# 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.
• @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