From: Brandsma
Newsgroups: sci.math
Subject: Re: continuous bijections 123
Date: Wed, 21 Oct 1998 09:59:21 +0200
Nick Halloway wrote:
> Hi, I was wondering whether a continuous bijection from an open set of
> R^n --> R^n is a homeomorphism with the range of the function. I read
> that this is true ... can anyone outline how this is proved? and what
> properties of R^n are needed for the proof?
It is indeed true. The result is called Invariance of Domain, and is due
to the Dutch topologist Brouwer. The statement as I know it:
Let n \in N, and let U be an open subset of R^n. For every continuous
injective f:U \to R^n the following hold:
1. f(U) is open in R^n
2. f: U \to f(U) is a homeomorphism.
To prove this we need a lemma:
Let X,Y be closed subspaces of R^n, then any homeomorhism f from X onto Y
has
f(Bd(X)) = Bd(Y). (Bd is the boundary).
[This follows from the Brouwer fixed point theorem: there can be no
retraction from a closed ball onto its boundary, and some general
dimension theory.]
Knowing this, it's easy:
Let x be a point of U. Find some closed ball inside U containing x. f
restricted to this compact closed ball is a homeomorhism (by compactness)
onto its image. So it maps x to the interior of the image of the closed
ball, hence to the interior of f(U). This shows that f(U) is open. Now 2.
is trivial: f is now an open mapping onto f(U) (easy exercise), hence a
homeomorphism: just apply 1 to an open subset V of U ( relatively open is
absolutely open etc...).
We need of R^n that its neighbourhoods are "closed ball-like", in having
no retractions onto their boundaries, and local compactness as well.
Hope this helped,
Henno Brandsma