TITLE: If $f:\mathbb{R}^n \to \mathbb{R}^n$ is continuous with convex image, and locally 1-1, must it be globally 1-1?
QUESTION [11 upvotes]: For $f:\mathbb{R}\to \mathbb{R}$ which is continuous, being locally 1-1 implies being globally 1-1, see here.  This is not true for a general mapping $f:\mathbb{R}^n\to \mathbb{R}^n$.  My intuition as to the source of this incongruity is that while continuity preserves connectedness, it does not preserve convexity. 
This illustration (from Buck's Advanced Calculus) of a locally injective, not globally injective function from $\mathbb{R}^2$ to $\mathbb{R}^2$ somewhat supports my intuition...note the image is not convex. If it were convex, it would seem there would need to be some "critical point" (not assuming the existence of the derivative) where $f$ was not locally 1-1.
 
Is there a theorem that if $f: S \to \mathbb{R}^n$ is continuous, for $S\subset \mathbb{R}^n$ convex, with convex image, locally 1-1, then it must be 1-1 on all of $S$?

REPLY [3 votes]: One can modify the pictured example so as to have domain, say, the set of $(x,y) \in \mathbb{R}^2$ with $0 < x < 1$.  Imagine the function being given by taking a line segment of width $1$, mapping the center of the line segment to $f(\frac{1}{2},y)$, and then carrying around the rest of the line segment linearly.  Thus we need a continuous function $g: \mathbb{R} \rightarrow \mathbb{R}^2$ so as to set $f(\frac{1}{2},y) = g(y)$.  If $g$ is smooth with nowhere vanishing derivative, $f$ is locally injective.  It is easy to choose such a $g$ so as to make $f$ surjective -- e.g. make sure that every once in a while $g$ traces out the circle of radius $\frac{n}{2}$ centered at the origin for each $n \in \mathbb{Z}^+$. -- and not injective -- each make sure $g$ does what I said above at least twice for each positive integer $n$.
More precisely: I want the segment to be oriented so that for all $y \in \mathbb{R}$, it is perpendicular to the tangent vector $g'(y)$.  Once I specify its orientation is at $y = 0$ (take it to be horizontal; it doesn't matter) that determines it for all $y$.  
Since the domain of $f$ is diffeomorphic to $\mathbb{R}^2$, we can pull this back to get a (smooth) counterexample $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$.  
Added: The above construction can also be used to give an $f$ such that for all $v \in \mathbb{R}^2$, $f^{-1}(v)$ is countably infinite.  Countably infinite is the largest possible preimage for a locally injective function on $\mathbb{R}^2$ (or any $\sigma$-compact $T_1$ topological space): cover $\mathbb{R}^2$ by countably many compact subsets.  If any preimage were uncountable, its intersection with some compact subset would be uncountable, and thus the preimage of $f^{-1}(v)$ would have an accumulation point, at which $f$ is not locally injective.