Proof of invariance of domain in two dimensions

Question

Invariance of domain states that for a given Euclidean space $\mathbb{R}^n$, any continuous injective map $f$ from an open subset $U \subset \mathbb{R}^n$ into $\mathbb{R}^n$ is an open map, hence a homeomorphism onto its image (i.e. an embedding). Terry Tao has a blog entry about it here:

https://terrytao.wordpress.com/2011/06/13/brouwers-fixed-point-and-invariance-of-domain-theorems-and-hilberts-fifth-problem/

He notes that invariance of domain (and the related invariance of dimension) "can be proven by simple ad hoc means" in low dimensions. E.g. $\mathbb{R}$ is not homeomorphic to $\mathbb{R}^m$ for any $m>1$ since removing a single point of $\mathbb{R}$ disconnects it, but this is not true of $\mathbb{R}^m$.

A rough sketch of the ad hoc proof for invariance of domain in the case $n=1$ would be as follows: The open subsets of $\mathbb{R}$ are precisely the countable disjoint unions of open intervals. Any continuous injection maps such a disjoint union to another such disjoint union, and hence is an open map.

My question is whether there is a simple ad hoc proof of invariance of domain for the case $n=2$. The general proof given in Tao's blog post - which he likes because it avoids any homology theory - relies on Brouwer's fixed point theorem, the Tietze extension theorem, and Weierstrass approximation theorem (as well as using perturbation methods), which is an awful lot of machinery if one only cares about the planar case. I am aware that invariance of domain is a very strong result, and so any general proof will necessarily require a lot of machinery, but I am wondering whether this is also true of the special case $n=2$.

Answer 1

Let $D$ be the unit disk, and $f: rD \rightarrow \mathbb{R}^2$ be continuous injective for some $r > 1$.

Let $c \in \mathbb{R}^2$ be in the path-connected component $U$ of $f(0)$ in $\mathbb{R}^2 \backslash f(\partial D)$ (this is an open neighborhood of $f(0)$). Assume $c \notin f(D)$ for the sake of contradiction.

Consider the application $\alpha:\,s \in \partial D \rightarrow \frac{f(0)-c}{\|f(0)-c\|}\partial D$.

$\alpha$ is constant and homotopic (since $c \notin f(\overline{D})$) to $\beta:\, s \in \partial D \longmapsto \frac{f(s)-c}{\|f(s)-c\|}$.

$\beta$ itself is homotopic (because of the hypothesis on $c$) to $\gamma:\, s \in \partial D \rightarrow \frac{f(s)-f(0)}{\|f(s)-f(0)\|}$.

$\gamma$ is homotopic (because $f$ is injective) to the odd function of $\partial D$, $\delta:\,s \longmapsto \frac{f(s)-f(-s)}{\|f(s)-f(-s)\|}$.

Now, it just remains to prove the elementary theorem: if $f: S^1 \rightarrow S^1$ is odd, $f$ isn’t homotopic to a constant.

The proof isn’t too hard: $f$ lifts to a continuous function $\hat{f}: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(e^{it})=e^{i\hat{f}(t)}$.

Then $\hat{f}(t)-\hat{f}(t+\pi)$ must be in $2\pi \mathbb{Z}+\pi$ for all $t$, so is a nonzero constant $c$. So $\hat{f}(2\pi)-\hat{f}(0)=2c \neq 0$, therefore $f$ has a nonzero degree, so $f$ can’t be homotopic to a constant.

$\,$

$\,$

In other words, for the original $f$, $f(D)=U$ is open. By suitable transformations, the conclusion follows.

This proof generalizes to higher dimensions, but it requires the following version of Borsuk-Ulam’s theorem in dimension $n$: an odd continuous function $S^{n-1} \rightarrow S^{n-1}$ isn’t homotopic to a constant.

Answer 2

An elementary, but long, proof of Brouwer's Invariance of Domain Theorem for $\mathbb{R}^2$, requiring only basic topology and the Jordan Curve Theorem, can be found in the proof of Theorem 6.1 on pp. 94-98 of A. I. Markushevich's Theory of Functions of a Complex Variable, Three Volumes in One, 2nd Edition, Chelsea Publishing Company, 1977 (see here).

However, I suspect there's a flaw in the proof, specifically the statement "$O$ can contain no interior points of $D_1$" in the last paragraph on p. 97. This statement holds if $O$ happens to be connected, but what if $O$ is not connected? If anyone can explain this to me, or confirm it is indeed an error in the proof, I'll be very grateful.