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The invariance of domain theorem states that, given an open subset $U\subseteq \mathbb{R}^n$ and an injective and continuous function $f:U\rightarrow\mathbb{R}^n$ then $f$ is a homeomorphism between $U$ and $f$'s image.

I tried proving it by using another theorem:

if $g:K\rightarrow X$ is injective and continuous, $K$ is compact and $X$ is Hausdorff then $g$ is a homeomorphism between $K$ and $f(K)$.

But I'm not sure on how to prove this (sub)-theorem? or perhaps there exists an easier proof of the invariance of domain theorem?

Eric_
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1 Answers1

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First note that the proof of your "lemma" is easy.

For a bijective continuous map $f:X\to Y$ to be a homeomorphism, it is sufficient for $f$ to be a closed/open map, because then

$$ (f^{-1})^{-1}(A) = f(A) $$

is closed/open for each $A \subset Y$, so that $f^{-1}$ is continuous, whence $f$ is a homeomorphism.

Now note that if $A \subset K$ is closed, where $K$ is compact, then $A$ is compact. Hence, so is $f(A)$. In a Hausdorff space, compact sets are closed, so $f(A)$ is closed, so that $f$ is a closed map.

But this does not proof invariance of domain. To see this, first note that your "proof" would note use the fact that $U \subset \Bbb{R}^n$ and $f : U \to \Bbb{R}^n$ (note that the dimensions match). But without matching dimensions, the theorem is not valid, as the following counterexample (taken from http://en.wikipedia.org/wiki/Invariance_of_domain#Notes) shows:

$$ f : (-1.1\, , \, 1) \to \Bbb{R}^2, x \mapsto (x^2 - 1, x^3 - x). $$

The image of this function (also taken from the same post) is enter image description here

It is an easy exercise to show that $f$ is not a homeomorphism onto its image although it is continuous and injective.

The problem here is that the claim you get is only that each restricted map $f|_K : K \to f(K)$ is a homeomorphism for $K \subset U$ compact. But this only gives you continuity of $f^{-1}|_{f(K)}$. But this does not entail continuity of $f^{-1}$ (as the example shows).

PhoemueX
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  • So you're saying proving it using my "lemma" isn't going to work? or more like, that it doesn't help me so much to prove the main theorem? Because apart from that, I'm a bit at loss in here - not really sure where to start from, and couldn't find any understandable proof for invariance of domain. – Eric_ Oct 05 '14 at 16:04
  • Yes, you will not be able to (easily) prove invariance of domain using the "Lemma". What are your prerequisites? If you know/believe in Brouwer's fixed point theorem, there is a nice proof by Terence Tao of invariance of domain using that (I will look up the link). Apart from this, I only know proofs using the mapping degree or some algebraic topology. – PhoemueX Oct 05 '14 at 17:17
  • So here is the link http://terrytao.wordpress.com/2011/06/13/brouwers-fixed-point-and-invariance-of-domain-theorems-and-hilberts-fifth-problem/. Tao also notes that "Theorem 2 or Corollary 3 [Invariance of domain] can be proven by simple ad hoc means for small values of $n$ [...] but I do not know of any proof of these results in general dimension that does not require algebraic topology machinery that is at least as sophisticated as the Brouwer fixed point theorem." – PhoemueX Oct 05 '14 at 17:21
  • Is this helpful? https://www.maa.org/sites/default/files/pdf/upload_library/22/Allendoerfer/1981/0025570x.di021114.02p00993.pdf Otherwise, I really recommend to look at Brouwer's fixed point theorem (and then at Tao's blog). There is e.g. (A variant of) a proof by Milnor using basically only the change-of-variables formula (and the density of polynomials), e.g. here http://people.math.sc.edu/howard/Notes/brouwer.pdf. – PhoemueX Oct 05 '14 at 21:06