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FALSE: a critical value must have a singular level set
Statement
If is a critical value of a smooth map, then the level set over is singular.
Facts & Assumptions
Given: The smooth map .
The power rule gives the continuous Jacobian row , the continuous-partials theorem identifies it with , and the polynomial map is smooth; a value is critical when its fibre contains a point where the derivative is not surjective (For a natural the function is differentiable everywhere with derivative ; for it is the constant , with derivative ; for a natural the function is differentiable at every with derivative ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative, Euclidean maps are closed under componentwise algebra and composition, Regular and critical points, regular and critical values, and level sets).
A graph of a map is a regular level set for an appropriate defining map (The graph of a Euclidean map is a regular level set).
Refutation
The zero level of is the vertical line . By [L1], vanishes at every point of this fibre, so is a critical value.
The same line is the graph of the zero function over the -axis and hence is smooth by [L2].
Therefore a critical value can have a smooth level set; criticality records a failure of this defining map, not necessarily a singularity of the underlying subset.
Depends on
- Regular and critical points, regular and critical values, and level sets
- The graph of a $C^k$ Euclidean map is a regular level set
- For a natural $n \ge 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $\iota(n)\,x^{\,n-1}$; for $n = 0$ it is the constant $1$, with derivative $0$; for a natural $n \ge 1$ the function $x \mapsto x^{-n}$ is differentiable at every $x \ne 0$ with derivative $-\iota(n)\,x^{-n-1}$; consequently every polynomial function is differentiable at every real, with the derivative computed term by term
- The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case
- If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative
- $C^k$ Euclidean maps are closed under componentwise algebra and composition
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
33 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. M. Lee, Introduction to Smooth Manifolds, critical-value discussion (standard reference, not scraped)