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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 $n=2$.

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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.

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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.

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  • $\begingroup$ An intuitive proof isn’t too hard, but what about rigorous existence of that lift? The Real line is disconnected if we remove a point, but the Plane is a different world. $\endgroup$
    – Hulkster
    Nov 14 '19 at 21:17
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    $\begingroup$ You mean, the lift of continuous $S^1 \rightarrow S^1$ functions? That’s probably the example in every algebraic topology book to explain fundamental groups. And the proof really is elementary. I can’t see where else I’m being unrigorous. $\endgroup$
    – Mindlack
    Nov 14 '19 at 21:25
  • $\begingroup$ Thanks for the proof. Just to be clear: when you assume $c \not\in f(D)$ in the second line, this is in order to derive a contradiction, correct? Also, how exactly do we know that U is open? Since $\partial D$ is compact, $\mathbb{R}^2 \setminus f(\partial D)$ is open, but how do we know its path-components are open? (this is probably very simple but I'm drawing a blank). $\endgroup$ Nov 14 '19 at 22:16
  • $\begingroup$ Also $\mathbb{D}$ is the closed disk, right? So I'm assuming the $\overline{D}$ on the fourth line is a typo. And also, is the proof that odd maps $S^2 \mapsto S^2$ aren't null-homotopic much more difficult than for maps $S^1 \mapsto S^1$? $\endgroup$ Nov 14 '19 at 22:19
  • $\begingroup$ Yes: $c \notin f(D)$ is only assumed to derive the contradiction. For the other thing, it’s a “general” topology thing: let $U$ be an open subset in a normed space, then the path-connected components of $U$ are open. The proof is simple: if $x \in U$, $U$ contains a small ball around $x$, which is path-connected. No, $D$ is the open unit disk and the $\overline{D}$ isn’t a typo, just a minor technical detail. $\endgroup$
    – Mindlack
    Nov 14 '19 at 22:20

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