LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Local side-preserving extensions of half-space transitions

Statement

Let $n \geq 1$ and $G : U \to V$ be a smooth diffeomorphism between relatively open subsets of $H^{n}$ . At every $p \in U \cap {x_{n} = 0}$ there are Euclidean open neighborhoods $O$ of $p$ and $O^{'}$ of $G (p)$ and a smooth diffeomorphism $\hat{G} : O \to O^{'}$ extending $G$ locally, such that $\hat{G}$ maps the positive, zero, and negative sides of $x_{n} = 0$ onto the corresponding sides in $O^{'}$ .

Facts & Assumptions

[F1]

Smooth invariance of the manifold boundary: A smooth diffeomorphism between relatively open half-space sets carries face points to face points and relative-interior points to relative-interior points; consequently $\partial M$ and $Int M$ are intrinsic.

[F2]

Chain rule for smooth half-space maps: If $f : U \to V$ and $g : V \to W$ are smooth maps between relatively open half-space sets, then $g \circ f$ is smooth and $D (g \circ f)_{p} = D g_{f (p)} \circ D f_{p}$ .

[F3]

Half-space extensions agreeing on a relatively open set have the same derivatives there: If two smooth Euclidean extensions agree on a relatively open subset of $H^{n}$ , then all of their derivatives agree at every point of that subset.

[F4]

The Euclidean inverse function theorem: Let $n \geq 1$ , let $U \subseteq R^{n}$ be open, let $f : U \to R^{n}$ be $C^{1}$ , and let $a \in U$ . If $D f (a)$ is invertible, then there are open sets $V, W \subseteq R^{n}$ with $a \in V \subseteq U$ and $f (a) \in W$ such that $f ∣_{V} : V \to W$ is bijective. Its inverse $g : W \to V$ is $C^{1}$ , and $D g (y) = D f (g (y))^{- 1} (y \in W) .$ Thus $f$ is a local diffeomorphism at $a$ .

[F5]

Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative: Let $a < b$ . Suppose $G : [a, b] \to R$ is continuous on $[a, b]$ and differentiable on $(a, b)$ . If $f : [a, b] \to R$ is Riemann integrable and $f (x) = G^{'} (x) (a < x < b),$ then $\int_{a}^{b} f = G (b) - G (a) .$ No derivative of $G$ at either endpoint is assumed, and the two endpoint values assigned to the integrable extension $f$ do not enter the conclusion.

Proof

Given: The objects and hypotheses in the statement above.

1.1

The face maps into the face and the interior into the interior. The half-space chain rule applied to $G^{- 1} G$ shows that $D G_{p}$ is invertible. Derivatives do not depend on the smooth extensions chosen near $p$ .

F1F2F3

2.1

Write the last component of a local extension as $h (x^{'}, t)$ . On a small face disk $h (x^{'}, 0) = 0$ , so its tangential derivatives vanish. Since $h (x^{'}, t) > 0$ for small $t > 0$ , $\partial_{t} h (p) \geq 0$ . Invertibility and the zero tangential entries in the last row exclude zero, so $\partial_{t} h (p) > 0$ .

step 1.1algebra

3.1

After shrinking to a product neighborhood, continuity makes $\partial_{t} h$ positive there. The one-variable fundamental theorem gives $h (x^{'}, t) = \int_{0}^{t} \partial_{t} h (x^{'}, s) d s = t a (x^{'}, t)$ with $a > 0$ , also for negative $t$ . Hence $h$ has exactly the sign of $t$ .

F5step 2.1

4.1

Apply the Euclidean inverse theorem to the extension and shrink its inverse neighborhoods inside that product neighborhood. The inverse is smooth: its derivative is the inverse derivative matrix composed with the inverse map, and repeated differentiation bootstraps the stated $C^{1}$ inverse to every finite order. The sign identity gives both inclusions of each side equality. For $n = 1$ the tangential row is empty and the same positive derivative argument applies.

F4step 3.1∎

Depends on

Used by

Coordinate independence of chart integrals Theorem

Dependency tree · two levels

22 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

Lee Proposition 16.3, p.404, together with the exact published boundary chain rule and inverse theorem (standard reference, not scraped)