# Brouwer’s invariance of domain

I am recording here an elementary* result used in the proof of the Teichmueller Existence Theorem.

Theorem (Brouwer). Any injective continuous map from $\mathbb{R}^n$ to itself is open.

*the statement itself sounds elementary enough, but the proof (all known proofs) seem to require some amount of machinery from algebraic topology. Note that one corollary of the theorem is that Euclidean spaces of different dimensions are not homeomorphic, another statement which appears elementary (and intuitive), but is difficult to prove without additional machinery.

It seems that the category of continuous maps contains enough curious things—Peano and other space-filling curves and the like—to quite thoroughly upset our intuitions, although going to the rather more restrictive category of smooth things does restore much of our intuition.

Proof of Theorem: It suffices to show that any such map sends open balls to open balls, or, more narrowly, that for any injective continuous map $f: B^n \to \mathbb{R}^n$$f(0)$ lies in the interior of $f(B^n)$ (where $B^n$ denotes the closed unit ball.)

Let f be as in the hypothesis of the Theorem. $f: B^n \to f(B^n)$ is a continuous bijection between compact Hausdorff spaces and hence a homeomorphism. $f^{-1}: f(B^n) \to B^n$ is continuous; by the Tietze extension theorem, $f^{-1}$ may be extended to a continuous map $G: \mathbb{R}^n \to \mathbb{R}^n$.

G has a zero at $f(0)$, and moreover we have the following

Lemma: If $\tilde{G}: f(B^n) \to \mathbb{R}^n$ is a continuous map with $\|G - \tilde{G}\|_\infty \leq 1$ (i.e. is a small perturbation of G), then $\tilde{G}$ has a zero in $f(B^n)$.

Proof of Lemma: Applying Brouwer’s fixed-point theorem to the function $x \mapsto x - \tilde{G}(f(x)) = (G - \tilde{G})(f(x))$ (from the closed unit ball to itself) yields a $x \in B^n$ s.t. $x = x - \tilde{G}(f(x))$, i.e. $\tilde{G}(f(x)) = 0$.

If $f(0)$ were not an interior point of $f(B^n)$, we may construct a small perturbation of G that no longer has a zero on $f(B^n)$, contradicting the above lemma.

At this point I refer the interested reader to Terry Tao’s blogpost (where the gist of this proof was also taken from; he was interested in Hilbert’s fifth problem, whose solution also makes use of invariance of domain) for the details of the construction.

Corollary Any proper injective continuous map $f: \mathbb{R}^n \to \mathbb{R}^n$ is a homeomorphism.

Proof of Corollary: By invariance of domain, f is open. Now it suffices to show that f is surjective (and hence bijective) for it to be a homeomorphism; but surjectivity follows from f being a proper map into a metrizable space, and hence closed—given $E \subset X$ closed and a sequence of points $f(x_n) \in f(E)$ accumulating to $y \in Y$ and $\epsilon > 0$, $f(x_n) \in B(y,\epsilon)$ (the closed ball) for all $n \geq N = N(\epsilon) \gg 0$. By properness, we have, for all $n \geq N$, $x_n \in K \subset E$ with K compact. Then, up to subsequence, $x_n \to x \in K$. By continuity, $f(x_n) \to f(x) = y$, so $y = f(x) \in f(E)$.