# A valid proof for the invariance of domain theorem?

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?

• ignoring the proof of the sub-theorem, how do you intend to use it to prove the invariance of domain theorem? Oct 3 '14 at 11:19
• $U$ is locally compact. Oct 3 '14 at 11:33
• Highly related: math.stackexchange.com/questions/284813/… Oct 3 '14 at 14:40
• @PhoemueX why does $\bar{B} \cong f(\bar{B}) \Rightarrow B \cong f(B)$? Oct 7 '14 at 13:01
• Possible duplicate of Is there some elementary proof of invariance of domain? Sep 24 '18 at 11:01

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

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).

• 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. Oct 5 '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. Oct 5 '14 at 17:17
• So here is the link terrytao.wordpress.com/2011/06/13/…. 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." Oct 5 '14 at 17:21
• Is this helpful? maa.org/sites/default/files/pdf/upload_library/22/Allendoerfer/… 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 people.math.sc.edu/howard/Notes/brouwer.pdf. Oct 5 '14 at 21:06