Invariance of domain

Remarks

For $n = 1$ the claim follows from the intermediate value theorem, since $f$ is then strictly monotone on each interval in $U$; so let $n \ge 2$. It suffices to show that, for every closed ball $B \subseteq U$, the image $f(B \setminus \partial B)$ of its interior is open. The image $f(\partial B)$ is homeomorphic to the sphere $S^{n-1}$, so by the Jordan–Brouwer separation theorem its complement has two components, a bounded one and an unbounded one. A homology argument shows that $f(B \setminus \partial B)$ is connected and disjoint from $f(\partial B)$ and that $\R^n \setminus f(B)$ is connected, which forces $f(B \setminus \partial B)$ to be the bounded component, an open set. The second claim follows from the first, since $f$ then maps open subsets of $U$ to open sets.

Example: it implies invariance of dimension, that is, $\R^m$ and $\R^n$ are not homeomorphic if $m < n$. Indeed, a homeomorphism $\R^n \to \R^m \subset \R^n$ would have an open image inside the hyperplane $\R^m \times \set{0}$. But this hyperplane has empty interior in $\R^n$.

Both hypotheses are needed. Without injectivity, $f(x) = x^2$ on $\R$ has image $[0, \infty)$, which is not open. The target must have the same dimension: the map $x \mapsto (x, 0)$ from $\R$ to $\R^2$ is continuous and injective, but its image is not open in $\R^2$.

Going deeper

For a student

The theorem shows that the dimension of a topological manifold is well defined. It also shows that a continuous bijection between manifolds of the same dimension without boundary is a homeomorphism. It is false in infinite dimensions: on a Hilbert space, the shift operator is a continuous injective linear map whose image is not open. It is usually proved with homology, through the Jordan–Brouwer separation theorem, or with the Brouwer fixed point theorem.