Re: extension theorem required
- From: "Antonio" <anton.cali@xxxxxxxx>
- Date: 10 Feb 2007 05:05:22 -0800
On 10 Feb, 05:07, "Zbigniew Karno" <zbigniew.ka...@xxxxx> wrote:
On 9 Lut, 19:17, "Antonio" <anton.c...@xxxxxxxx> wrote:
Hi there,
Consider an open subset U of a finite dimensional manifold M modelled
on an euclidean space of dimension n, say R^n.
Denote the closure of U in M by Cl(U) and suppose there is an assigned
embedding
f from Cl(U) to R^n.
What is the theorem needed to determine an embedding h: M --> R^n
which extend f: Cl(U) -->R^n, i.e., such that for any p in M, h(x) =
f(x) ?
Regards, Antonio
This depends on global structure of
considered manifold M and in general
it is impossible.
For example consider 1-dimensiomal
manifold (the unit circle)
M = S^1 = {(x,y) in R^2 : x^2 + y^2 = 1}
and an open set U = {(x,y) in M : y > 0}
of M. So, Cl(U) = {(x,y) in M : y >= 0}
in M.
However observe that the projection
p : M --- > R^1, given by p(x,y) = x for
(x,y) in M, is an embedding on Cl(U),
but it is not on M.
Please formulate your question more
precisely.
Best regards,
Z. Karno
I'm going to formulate my question more precisely:
Consider an open subset U of a finite dimensional manifold M modelled
on an euclidean space of dimension n, say R^n.
Denote the closure of U in M by Cl(U) and suppose there is an
assigned embedding f from Cl(U) to R^m, m assigned.
What is the theorem needed to determine an embedding h: M --> R^N
(N sufficently large) which extend f, i.e., such that for any p in M,
h(x) = f(x) ?
Best regards, Antonio
.
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