Is a smooth transformation of a plane domain onto a plane domain with everywhere nonzero Jacobian determinant necessarily a bijection?

Question

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. 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.

Answer 1

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.

Answer 2

Here is an elementary answer, which shows how the bijectivity of $f$ may be violated, even when $U$ and $V$ are both simply connected domains (and $f\colon U\to V$ is a smooth surjection with the Jacobian determinant nonzero everywhere).

The idea is very simple: First of all, it is easy to get a smooth non-bijective map, say $f_1$, from a simply connected open set $U_1\subseteq\mathbb R^2$ to a simply connected open set $V\subseteq\mathbb R^2$ with the Jacobian determinant of $f_1$ nonzero everywhere and one point in $V$ missing from the image $f_1(U_1)$. The idea then is just to extend $f_1$ to a map $f\colon U\to V$ in order to smear over the missing point -- keeping, for $f$, the smoothness and nonzero Jacobian determinant properties. Below, the smear is represented by the function $f_2$, which extends the function $f_1$ to $f$ (while the image set of $f_1$ misses a point in $V$).

Let $U:=(0,1)\times(0,4\pi+\pi/2)$ and $V:=D$, the open unit disk. For $(r,t)\in U$, let $$f(r,t):= \begin{cases} f_1(r,t) & \text{ if}\quad 0<t\le2\pi, \\ f_2(r,t) & \text{ if}\quad 2\pi<t<4\pi+\pi/2, \end{cases} $$ where $$f_1(r,t):=r\,(\cos t,\sin t),$$ $$f_2(r,t):=(-2a(t),0)+(a(t) + r (1 - 4 a(t))) \,(\cos t,\sin t),$$ and $$a(t):=\frac14\, \exp\Big(-\frac\pi{t - 2\pi}\Big).$$

Shown below are the (properly shaded) "parametric plots", that is, the sets $S_1:=\{f_1(r,t)\colon 0<r<1,0<t\le2\pi\}$ (left), $S_2:=\{f_2(r,t)\colon 0<r<1,2\pi<t\le4\pi+\pi/2\}$ (right), as well as both of these two sets superimposed together, forming the set $S:=f(U)=S_1\cup S_2=\{f(r,t)\colon 0<r<1,0<t<4\pi+\pi/2\}$.

It is not hard to see that the map $f$ is smooth and the Jacobian determinant of $f$ is nonzero (actually, $>0$) everywhere.

It is also easy to see that $S\subseteq D=V$, $S_1=D\setminus\{(0,0)\}$, whereas $f(r_*,4\pi)=f_2(r_*,4\pi)=(0,0)$, where $r_*:=1/(4 (\sqrt e-1))=0.385\dots$. So, $f$ is surjective.

However, as seen from the pictures, $f$ is not bijective. Formally, for instance, $f(r-(r+1/4)/\sqrt e,2\pi)=f(r,4\pi)$ for all $r\in(r_*,1)$.

Answer 3

As Alexandre Eremenko pointed out, in general the answer is in the negative. However, in a comment the OP asked a modified question:

What if we assume that both $U$ and $V$ are simply connected?

The following result is in a positive direction:

If, $V$ is simply connected and $U$ is connected, and $f$ is proper (i.e. preimages of compact sets are compact), then then $f$ is a bijection.

Under the given assumptions $f$ is a covering map, see https://math.stackexchange.com/q/860351/798404

A covering space is a universal covering space if it is simply connected. Assuming that $V$ is simply connected, it is its universal cover and $\operatorname{id}:V\to V$ is a covering map. Now we use the fact:

If the mapping $p: D \to X$ is a universal cover of the space $X$ and the mapping $f : C \to X$ is any cover of the space $X$ where the covering space $C$ is connected, then there exists a covering map $g : D \to C$ such that $f ∘ g = p$.

In our situation $D=X=V$, $p=\operatorname{id}$, $C=U$ and $f$ is $f$. Therefore there is $g:D\to C$ i.e. $g:V\to U$ such that $f\circ g=p$ i.e., $f\circ g=\operatorname{id}$. That proves that $f$ is a bijection (and hence a diffeomorphism).

You can find basic statements about universal cover and covering maps in https://en.wikipedia.org/wiki/Covering_space#Lifting_properties

If you want to learn more, take almost any book in topology. For example:

M. A. Armstrong, Basic Topology.

Answer 4

Another trivial counterexample.

Consider the rational parametrization of the real nodal plane cubic curve $y^2=x^2(x+1)$:

$$\gamma:t\mapsto (t^2-1,t(t^2-1))\;.$$

We have $\dot{\gamma}(t)=(2t,3t^2-1)$. Consider a normal $\nu(t)=(1-3t^2,2t)$.

(image from Wikipedia)

A "thickening"

$$F:(t,s)\mapsto\gamma(t)+s\nu(t)$$

will do, for example restricting to $U=(-3/2,3/2)\times(-\epsilon,\epsilon)$, $V=F(U)\subseteq\mathbb{R}^2$, for $\epsilon<<1$.