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