Let $U$ and $V$ be connected open subsets of $\mathbb R^2$. Let $f$ be a smooth map from $U$ onto $V$ such that the Jacobian determinant of $f$ is nonzero everywhere. Does it then necessarily follow that $f$ is a bijection?




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Counterexamples are easy to find if we allow $V$ to be contained in a bigger space, say $\mathbb R^3$; or if we relax enough the conditions on the smoothness and the Jacobian determinant of $f$.




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I apologize in advance if this question is trivial. I have no background in differential topology (and am not even quite sure that this question belongs in differential topology).




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As was noted in a comment by Moishe Kohan, the answer to this question follows from the [previous answers](https://mathoverflow.net/questions/147110/surjective-entire-functions-without-critical-points). However, to make a bridge from those answers to an answer to this question, one needs to recall and use the Great Picard Theorem (which I, shamefully, forgot about). Also, now we have, not only the excellent answer by Alexandre Eremenko, but also a completely elementary answer, which I think shows in what specific manner the non-bijectivity can occur. For these reasons, I'd prefer that this question be kept.


**Edit:** In view of the discussion of the answer by Alexandre Eremenko, let me add here the condition for $U$ and $V$ to be simply connected. A further question would then be whether the thickness parameter $\epsilon$ in the answer by Qfwfq can be chosen so that $V$ be simply connected.



Is a smooth transformation of a plane domain onto a plane domain with everywhere nonzero Jacobian determinant necessarily a bijection?
No. Let $C$ be the complex plane, $U=V=C\backslash\{0\}$. Transformation $z\mapsto z^2$ is smooth and has non-zero Jacobian $4|z|^2$ but it is not a bijection.


The question in the comments: no. Take $U=\{ z\in C\backslash\{0\}:|\arg z|<2\pi/3\}$,
and the same $f$.


Second question in the comments: again the answer is no. Take $f(z)=\int\_0^z e^{\zeta^2}d\zeta$, and $U=V=C$. The map is surjective, every point has infinitely many perimages, and the Jacobian is
$|f'|^2=\exp2(\Re z^2)\neq 0$. To see that the map is surjective, notice that it is an entire function of order 2. If an entire function of order two omits a complex value $a$, then it has the form $f(z)=a+e^{P(z)},$
where $P$ is a polynomial of degree 2. This is of course not the case for our function. Similarly, if the value $a$ is taken finitely many times,
then $f(z)=a+Q(z)s^{P(z)}$. So our function takes every value infinitely many times.


