TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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A continuous local inverse has derivative reciprocal to a nonzero complex derivative

Statement

Let $U, V \subseteq C$ be open, let $f : U \to V$ be a bijection, and let $g = f^{- 1} : V \to U$ . Fix $a \in U$ and put $b = f (a)$ . If $f$ is complex differentiable at $a$ , $f^{'} (a) \neq 0$ , and $g$ is continuous at $b$ , then $g$ is complex differentiable at $b$ and

$g^{'} (b) = \frac{1}{f^{'} (a)} .$

The existence or continuity of such an inverse is a hypothesis, not a consequence of $f^{'} (a) \neq 0$ here.

Facts & Assumptions

Given: The open sets, bijection, inverse, points, differentiability, nonzero-derivative, and inverse-continuity hypotheses in the Statement.

[F1]

Complex differentiability at $a$ means $(f (a + h) - f (a)) / h \to f^{'} (a)$ through nonzero allowed increments (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions).

[L1]

Complex modulus is definite and multiplicative (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

Proof

technique · direct

1.1

For $w \neq b$ near $b$ , put $z = g (w)$ . Injectivity gives $z \neq a$ , and $\frac{g (w) - g (b)}{w - b} = \frac{z - a}{f (z) - f (a)} = {(\frac{f (z) - f (a)}{z - a})}^{- 1} .$

givenalgebra

2.1

As $w \to b$ , continuity of $g$ gives $z \to a$ . By [F1], the parenthesized quotient in step 1.1 tends to the nonzero number $f^{'} (a)$ .

step 1.1givenF1

3.1

If $q \to q_{0} \neq 0$ , then $q^{- 1} - q_{0}^{- 1} = (q_{0} - q) / (q q_{0}) \to 0$ , with denominators nonzero near the limit by [L1]. Applying this to step 2.1 proves $g^{'} (b) = 1 / f^{'} (a)$ .

step 2.1L1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 36 results over 11 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. Orloff, MIT 18.04 Topic 2, §2.6.1 (standard reference, not scraped)