TITLE: Is a surjective mapping of R2 to itself with full rank derivative everywhere necessarily injective?
QUESTION [6 upvotes]: If $f:\mathbb R^2\rightarrow\mathbb R^2$ has rank 2 derivative everywhere, then by the inverse function theorem it is locally injective. If it is surjective, is it then necessarily globally injective as well?
What if we consider the same case, but for a map of $\mathbb R^{2+}$ to itself (i.e. ($x\geq 0,y\geq 0$)?

REPLY [7 votes]: No, $f$ needn't be globally injective. A counterexample is $$f:\mathbb C\to \mathbb C:z\mapsto \int_0^ze^{t^2}dt$$
Why is that entire  map $f$ surjective?
 Because by Picard's theorem it could at most skip one value $b\in \mathbb C$ i.e. $f(\mathbb C)=\mathbb C\setminus \{b\}$.
  Of course that potential $b$ is nonzero since  $f(0)=0$.
But since $f(-z)=-f(z)$, if it skipped $b$ it would also skip $-b$ so that actually $b$ does not exist and $f$ is indeed surjective.
Why is that entire  map $f$ not injective ?
Because it would then be bijective and the only holomorphic bijections $u:\mathbb C\to \mathbb C$ are of the form $u(z)=az+b$ and have constant derivative $u'(z)=a$.
However $f'(z)=e^{z^2}$ is not constant.
Why does $f$ have maximal real rank  everywhere?
Because $f'(z)=e^{z^2}$ never vanishes.  
Nota Bene
I have used that $\mathbb C$ is just $\mathbb R^2$, endowed with its well-known supplementary complex structure.