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LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-24 (gpt-6-sol)
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Sard on the nonflat critical strata

Statement

Let m,n≥1, let U⊆Rm be open, let f:U→Rn be Cr, and for j≥1 define

Cj:={x∈U:Dαf(x)=0 for every multi-index 1≤∣α∣≤j}.

If 1≤j<r, K⊆Cj∖Cj+1 is compact, and the Morse-Sard conclusion is already known for Cr maps from open subsets of Rm−1 to Rn, then f(K) is null.

Facts & Assumptions

Given: An integer n≥1, a Cr map f:U→Rn, an integer 1≤j<r, and a compact set K⊆Cj∖Cj+1.

[L1]

If a Euclidean map has invertible derivative at a point, it becomes a coordinate there after shrinking (The Euclidean inverse function theorem).

[L2]

A Cr map that is j-flat on a set has a Cr rough-composition extension along a Cr−j parametrization of that set (Rough composition on a flat closed set).

Proof

technique · direct
1.1L1givenchoose

Fix x∈K. [L1, given, choose] Because x∉Cj+1, some partial derivative of order j+1 of some component of f is nonzero at x. After reordering coordinates and components, choose a multi-index α with ∣α∣=j and a component fμ such that, for g:=Dαfμ, one has ∂g/∂x1(x)≠0. Since g is Cr−j, [L1] applied to Φx(y):=(g(y),y2,…,ym) gives a neighbourhood Wx of x and a Cr−j diffeomorphism from Wx onto an open set Ix×Ωx⊆R×Rm−1.

2.1L2step 1.1algebra

Every point of K∩Wx lies in Cj, so g vanishes there by definition of Cj. Hence Φx(K∩Wx)⊆{0}×Ωx. If m=1, then Ωx⊆R0 has at most one point. Thus K∩Wx has at most one point, and its image is null in Rn. Assume m≥2. Define the Cr−j graph parametrization bx:Ωx→U,bx(u):=Φx−1(0,u), and the relatively closed set Ax∗:=bx−1(Cj∩Wx) in Ωx. The map f is j-flat on Cj, and bx(Ax∗)⊆Cj. By [L2] there is a Cr map Hx:Ωx→Rn agreeing with f∘bx on Ax∗ and satisfying DHx=0 there. Thus every u∈bx−1(K∩Wx) is a critical point of the Cr map Hx, and f(K∩Wx)=Hx(bx−1(K∩Wx))⊆Hx(Crit⁡Hx). The induction hypothesis, at the unchanged differentiability order r, gives nullity of this image.

3.1step 2.1givenchoose∎

Finitely many neighbourhoods Wx cover the compact set K, so f(K) is a finite union of null sets and therefore null.

Depends on

Used by

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Sources