From: lrudolph@panix.com (Lee Rudolph)
Subject: Re: find a continuous real function
Date: 4 Jul 2000 08:55:08 -0400
Newsgroups: sci.math
Summary: [missing]

ags@seaman.cc.purdue.edu (Dave Seaman) writes:

>In article <8jq1qi$ejb$1@nnrp1.deja.com>,  <amspeter@my-deja.com> wrote:
>>Let R be the field of real numbers.
>
>>I need a continuous bijection f:RxR -> RxR such that f^-1 is continuous
>>and f(a,0)=(a^2, (a-1)^2 ), for all real number a in R.
>
>	f(x,y) = (x^2+y,(x-1)^2+y)

To be more general: Suppose K is an algebraically closed field
of characteristic 0 (for instance, C), and (p,q): K -> K^2 is
a "polynomial embedding of the line in the plane" (for instance,
(p,q)(t)=(t^2,(t-1)^2)), that is, a pair of polynomials over K
such that 
   (1) for all s, t in K, if (p,q)(t)=(p,q)(s), then t=s
(so (p,q) is an injection of sets), and 
   (2) for all t in K, (p'(t),q'(t)) is not (0,0) (so (p,q) is 
a "smooth immersion").  Then by a theorem of Abhyankar & Moh and 
Suzuki, there are "polynomial automorphisms of the line and the 
plane" A:K -> K and B:K^2 -> K^2 (for instance, A(t)=t and 
B(x,y)=(x^2+y,(x-1)^2+y)), that is, polynomial bijections whose 
inverses are also polynomials, such that B(t,0)=(p(A(t)),q(A(t)).
In other words, up to automorphism there is only one embedding
of the line in the plane.  (It's also true, and a theorem of
Zaidenberg & Lin, that up to automorphism every polynomial
*injection* of K in K^2, which is *not* an embedding, is
of the form (p,q)(t)=(t^m,t^n) with m,n>1, GCD(m,n)=1.)

To apply the A&M-S theorem over a non-algebraically-closed field
(like R), you simply have to check that the given polynomial
embedding stays an embedding when the scalars are extended to
the algebraic closure.

As to *finding* B, well, that can be automated.  A theorem of
Jung (which is a corollary of A&M's proof of the theorem;
maybe even simply of the truth of the theorem, it's been 
some time since I've checked) tells you that the polynomial
automorphisms of K^2 are generated by the linear automorphisms
GL(2,K) and the "polynomial shears" B(x,y)=(x,y+r(x)), where
r is an arbitrary polyomial.  So if you want to find B for
(p,q)(t)=(t^2,t^2-2t+1), first find a linear map B_1 which
makes the degree of one component strictly smaller than 
that of the other (say, B_1(x,y)=(x,y-x)), then find a 
(conjugate of a) polyomial shear with r a suitable monomial 
to reduce the larger of the two degrees (say, B_2(x,y)=(x-(y^2/4),y)),
and iterate as necessary.  (The existence of r is entailed by
the hypothesis that (p,q) is an embedding, via the A&M proof.)
Here, B_1(t^2,(t-1)^2) = (t^2,-2t+1), B_2(t^2,-2t+1)=(t-(1/4),-2t+1),
so we can let B_3(x,y) be any linear automorphism which takes
(t-(1/4),-2t+1) to (t,1), and B_4 be the shear B_4(x,y)=(x,y-1).
Then B = B_4 o B_3 o B_2 o B_1 works (and for a suitable
choice of B_3, you will get Dave's f). 

Lee Rudolph