Re: Looking for a surjection or R^2

In article
<1149925147.272877.32240@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"denis feldmann" <denis_feldmann@xxxxxxxx> wrote:

The World Wide Wade wrote:
In article
<waderameyxiii-45674F.23201008062006@xxxxxxxxxxxxxxxxxxxxxxxx>,
The World Wide Wade <waderameyxiii@xxxxxxxxxxxxxxxxxxxx> wrote:

In article <4488be04$0$7763$7a628cd7@xxxxxxxxxxxxxxxxxxxxx>,
Denis Feldmann <denis.feldmann.asupprimer@xxxxxxxxxxxxxxxx>
wrote:

I am looking for a smooth surjection of R^2 onto itself, not bijective,
with jacobian everywhere not zero. My best example so far is an
holomorphic function (with the obvious isomorphism between C and R^2)
like the antiderivative of exp(z^2) (which is surjective by Picard
theorem), but I would like something simpler, more explicit, and where
the proof of surjectivity uses only elementary calculations...

Let f be any smooth function on R such that f(y) = 0 if y <= 0,
f' <= 0 on [0,Pi], f' >= 0 on [Pi,2Pi], f(2Pi) < 0, and f(y) =
f(2Pi) for y >= 2Pi. Writing z = x + iy, define F(z) = e^z +
f(y). Then the Jacobian of F at x + iy is e^(2x) -
e^x*sin(y)*f'(y), which = 0 iff e^x = sin(y)*f'(y). The right
hand side is always <= 0, so the equation cannot be satisfied.
Therefore the Jacobian of F is never 0. F is surjective because
below the x-axis it maps onto C \ {0},

Here I should have said that below the x_axis, F(z) = e^z, where
it maps many to one onto C \ {0}.

and on the line y = 2Pi it
is the function e^x + f(2Pi); because f(2Pi) < 0, this function
takes on the value 0.

And finally, F is not a bijection because of its behavior below
the x-axis.

Thanks a lot. Are you the author of this construction

Yes.
.

Relevant Pages

  • Re: Looking for a surjection or R^2
    ... the proof of surjectivity uses only elementary calculations... ... Therefore the Jacobian of F is never 0. ... below the x-axis it maps onto C \, ... it maps many to one onto C \. ...
    (sci.math)
  • Re: Looking for a surjection or R^2
    ... The World Wide Wade wrote: ... with jacobian everywhere not zero. ... the proof of surjectivity uses only elementary calculations... ... it maps many to one onto C \. ...
    (sci.math)
  • Re: Looking for a surjection or R^2
    ... holomorphic function (with the obvious isomorphism between C and R^2) ... the proof of surjectivity uses only elementary calculations... ... Therefore the Jacobian of F is never 0. ...
    (sci.math)
  • Re: Looking for a surjection or R^2
    ... The World Wide Wade wrote: ... with jacobian everywhere not zero. ... the proof of surjectivity uses only elementary calculations... ... it maps many to one onto C \. ...
    (sci.math)