CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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A $C^{1}$ map with everywhere-invertible derivative is open

Statement

Let $n \geq 1$ , let $U \subseteq R^{n}$ be open, and let $f : U \to R^{n}$ be $C^{1}$ . Suppose $D f (x)$ is invertible for every $x \in U$ . Then $f$ maps every open subset of $U$ to an open subset of $R^{n}$ . Thus $f$ is an open map (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

Facts & Assumptions

Given: The map in the Statement and an open subset $O \subseteq U$ .

[L1]

At every point with invertible derivative, there are open sets $V, W \subseteq R^{n}$ on which the map restricts to a $C^{1}$ diffeomorphism onto $W$ (The Euclidean inverse function theorem).

Proof

technique · direct

1.1

Let $y \in f [O]$ and choose $x \in O$ with $f (x) = y$ . Apply [L1] at $x$ and replace its source neighbourhood by its intersection with $O$ ; the image of that smaller neighbourhood is an open neighbourhood of $y$ contained in $f [O]$ .

L1givenchoose

2.1

Thus every point of $f [O]$ is interior. If $O = \emptyset$ , its image is the open empty set, so $f [O]$ is open in every case.

step 1.1∎

Depends on

Used by

Dependency tree · two levels

18 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Basic Analysis II, Corollary 8.5.2 (standard reference, not scraped)

A C1C1 map with everywhere-invertible derivative is open