Local homeomorphism from the inverse function theorem

Statement

Let $U \subseteq \mathbb{R}^n$ be open, and let $f \colon U \to \mathbb{R}^n$ be of class $C^1$ with $Df(x)$ invertible for each $x \in U$. Then each $a \in U$ has an open neighbourhood $V \subseteq U$ such that $f(V)$ is open and $f$ restricted to $V$ is a homeomorphism onto $f(V)$. In particular, $f$ is an open map: $f(O)$ is open for each open set $O \subseteq U$.

Remarks

This follows from the inverse function theorem, since a $C^1$ inverse is continuous. For each $a$, the hypothesis is needed only at $a$: if $Df(a)$ is invertible, $f$ is a homeomorphism from a neighbourhood of $a$ onto an open set. A local homeomorphism need not be injective. The map $(x, y) \mapsto (e^x \cos y, e^x \sin y)$ is a local homeomorphism from $\mathbb{R}^2$ onto $\mathbb{R}^2 \setminus \{0\}$, but it is not injective.

Explanations

L1 human

Where the derivative is invertible, the map neither folds nor crushes space in a small region. So small open sets go to open sets, and the map can be undone continuously on each small region.