From: Alexander A Borisov <borisov@math.psu.edu>
Subject: Re: not invertible function
Date: Fri, 17 Mar 2000 12:52:26 -0500
Newsgroups: sci.math.research
Summary: [missing]

On Thu, 16 Mar 2000, manuel wrote:

>     I was trying to find a function f  such that:
> 
> 0) f : IR^2 -> IR^2 ( define from the plane to the plane)
> 1) im f = IR^2 ( f is onto)
> 2) f is diferentiable and the jacobian is positive.
> 3) if is not one to one.
> 
> Can you find f ? Can the 2 components of your f be polynomials?
> 
> manuel
> 
> 
> 
> 
> 
> 
Hi,

This is an addition to my previous post, this time regarding
the polynomial case.

As David Wright kindly informed me, there is a clever
example by Sergey Pinchuk of a polynomial mapping from
$R^2$ to $R^2$ which has positive determinant and is not
1-to-1. It is not clear to me if this map is onto, but probably
it is. This is the reference:

Sergey Pinchuk,
A counterexample to the strong real Jacobian conjecture,
Mathematische Zeitschrift, 217 (1994), 1-4.

The construction is explicit, but very sophisticated.
The resulting polynomials have degrees 10 and 40.
You can try to see for yourself if the map is onto or not.

Alexander Borisov
http://www.math.wustl.edu/~borisov
==============================================================================

From: Alexander A Borisov <borisov@math.psu.edu>
Subject: Re: not invertible function
Date: Fri, 17 Mar 2000 10:11:22 -0500
Newsgroups: sci.math.research

On Thu, 16 Mar 2000, manuel wrote:

>     I was trying to find a function f  such that:
> 
> 0) f : IR^2 -> IR^2 ( define from the plane to the plane)
> 1) im f = IR^2 ( f is onto)
> 2) f is diferentiable and the jacobian is positive.
> 3) if is not one to one.
> 
> Can you find f ? Can the 2 components of your f be polynomials?
> 
> manuel
> 
> 
> 
> 
> 
> 


Hi,

I don't know the answer in the case of polynomials, and could not
estimate the difficulty of that case. It could be relatively easy or it
could be as hard as the Keller's Jacobian Conjecture.

If the functions are not required to be polynomials, then such map is not
hard to construct. Moreover, a map exists such that the preimage of every
point is finite.

Here is the sketch of the construction. The map will be a composition of
two maps. First of all, we map $R^2$ 1-to-1 to the upper half plane
sending $(x,y)$ to $(x,e^y)$.

Then we construct a map from the upper half plane onto $R^2$.
More precisely, we construct a map from the closed upper half
plane. It is done as follows. The map on the boundary will be a
parametrization of a smoothening of the following curve. It starts by a
ray where $y=-1/2$, and $x$ goes from $-\infty to \sqrt{3}/2$ Then it goes
CLOCKWISE around the unit circle until it reaches $(\sqrt{3}/2, 1/2). Then
it goes back to $-\infty$ along $y=1/2$. At the point $(x,y)$, $y>0$, the
map is defined as follows. First we look where the point $(x,0)$ goes.
Then we take a normal to the image of the real line, that points to the
left, and go $y$ in that direction.

It is quite clear that you can make this map infinitely smooth and that it
will have a positive Jacobian and will not be 1-to-1.

Alexander Borisov
http://www.math.wustl.edu/~borisov