advanced29 min studyLesson 464 in Mathematics

Sard's theorem and regular values

Distinguish critical points from critical values and use the correct dimension-dependent differentiability threshold.

On this page

What you’ll understand

  • State Sard's theorem for smooth and sufficiently differentiable maps.
  • Locate critical points and their target values in concrete examples.
  • Explain measure-zero and vacuous regular-value statements without claiming every fiber is nonempty.

A complicated singular fiber can make a smooth map look exceptional, but Sard's theorem says that its critical values occupy a measure-zero subset of the target when enough differentiability is available. The distinction between points and values is essential. The theorem controls the image of derivative-rank failures, not the size of the critical set in the source. It is also a finite-dimensional regularity result, with a differentiability threshold depending on source and target dimensions rather than a generic statement about arbitrary continuous maps.

Criticality is tested in the source and reported in the target

For a C¹ map from an m-manifold to an n-manifold, a source point is critical when its derivative is not surjective. A critical value is the image of at least one critical point. A regular value has no critical preimages, including values with no preimages at all. If m<n, every source point is critical, because no linear map from a smaller-dimensional space can be onto the target tangent space. Sard then says that the entire image has target measure zero under its hypotheses.

When m≥n, a critical set can still be large. A constant smooth map to a positive-dimensional target has every source point critical but only one critical value. Thus the theorem does not say that almost every source point has full rank. Nor does it guarantee that a regular value belongs to the image. To use the regular-value theorem for an interesting fiber, first find a regular target value and separately establish that its preimage is nonempty.

Cf={p∈M:rank⁡(dfp)<n},critical values=f(Cf),y regular  ⟺  f−1(y)∩Cf=∅C_f=\{p\in M:\operatorname{rank}(df_p)<n\},\qquad \text{critical values}=f(C_f),\qquad y\text{ regular}\iff f^{-1}(y)\cap C_f=\varnothing
Here m=dim M and n=dim N≥1. Surjectivity, rather than injectivity, is the relevant derivative test for a regular target value.↔ Scroll wide equations horizontally.

State the differentiability and measure conventions

For smooth maps between finite-dimensional smooth manifolds without boundary, Sard's theorem states that the critical values have measure zero in the target. More generally a Cʳ map satisfies this conclusion when the positive integer r is greater than max(m−n,0), equivalently r≥max(1,m−n+1). This is a sufficient sharp general threshold, not permission to use a C¹ theorem in every dimension. Our standing second-countability assumption permits countable chart reductions, while smooth coordinate changes preserve null sets locally.

Measure zero is interpreted in target coordinate charts; no global Euclidean volume form is needed to define the null-set property. Every nonempty target open set has positive coordinate measure, so regular values are dense. Density does not imply that the critical values form a finite or empty set, and without additional conditions it does not by itself make the set of regular values open. Compactness or properness can supply useful stability statements, but those are extra ingredients beyond the measure-zero conclusion.

f∈Cr(Mm,Nn),r>max⁡(m−n,0)⟹f(Cf) has measure zero in Nf\in C^r(M^m,N^n),\quad r>\max(m-n,0)\quad\Longrightarrow\quad f(C_f)\text{ has measure zero in }N
The manifolds are finite dimensional, Hausdorff, second countable, and without boundary for this statement; n≥1 and r is a positive integer. Smooth maps automatically meet the threshold.↔ Scroll wide equations horizontally.

A generic value is not a complete generic-map theorem

Sard's theorem is often used by introducing parameters and applying it to a projection from a solution manifold. That is how regular-value arguments lead to transversality results. One must first prove that the parameterized solution set is a manifold and check the projection's dimensions and differentiability. Declaring a perturbation generic without that setup skips the mechanism that makes the claim valid. The next lesson makes the tangent-space condition explicit.

For a practical example, calculate the derivative, solve its rank-failure equations, map those solutions into the target, and distinguish the resulting critical values from other levels. A critical value does not force every point of its fiber to be critical, and a critical fiber might still happen to be a manifold. The theorem classifies which values enjoy the regularity guarantee; it does not classify every exceptional fiber or turn a plotted near-zero derivative into an exact rank certificate.

A quartic landscape with only two critical values

For h(x,y)=(x2−1)2+y2h(x,y)=(x^2-1)^2+y^2, find the critical points and critical values. Determine whether −1 and 2 are regular values and what the regular-value theorem says about the latter fiber.

  1. The derivative has components 4x(x2−1)4x(x^2-1) and 2y. It vanishes exactly at (−1,0),(0,0),(1,0). Their h-values are 0,1,0 respectively, so the critical values are {0,1}.

  2. Every other target value is regular. Since h≥0, the value −1 has an empty preimage and is regular vacuously. The value 2 has preimages, for example (1,√2), and none can be critical.

  3. The map has source dimension 2 and target dimension 1, so its nonempty regular fibers are one-dimensional embedded submanifolds. It is polynomial and smooth; even the general finite-regularity theorem would be available for C² maps in these dimensions.

The result

The critical points and values are different sets in different spaces. The level h=2 is a smooth curve, while regularity at −1 says only that its empty fiber has no bad preimages.

Explore the model

Interactive lab

Interactive lab · function

Separate a critical source point from its value

Pause and Reset; set Coefficient a=1,Position x=0 and predict derivative zero and output zero. In separate fresh runs Reset and set a=1,x=1, then Reset and set a=1,x=−1: both outputs are 1 but the slopes are 2 and −2. The quadratic illustrates one critical value and two regular preimages of another value; it is not the quartic landscape or a general Sard-theorem proof.

Separate a critical source point from its valueThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA regular value requires surjectivity at every preimage, while Sard controls the target image of the critical set.

Check your understanding

For a map from a 4-manifold to a 2-manifold, what integer differentiability level meets the stated finite-regularity Sard threshold?

Try it yourself

For F(x,y)=(x,y³−3y), find the critical values in the target plane and show that (0,0) is a regular value with three preimages.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The Jacobian is diagonal. Find its singular y-coordinates, then solve y³−3y=0 at target (0,0).

Compare with the solution

The determinant is 3y²−3, so critical points lie on y=1 or y=−1. Their images are the horizontal target lines with second coordinate −2 or 2. These lines have planar measure zero. The preimages of (0,0) are (0,−√3),(0,0),(0,√3). The second diagonal derivatives are 6,−3,6, all nonzero, so every derivative is onto. The fiber is a zero-dimensional embedded manifold of three points.

Take it with you

  • Sard controls critical values in the target, not the entire critical set in the source.
  • The finite Cʳ threshold depends on the dimension difference and uses a strict inequality.
  • Values outside the image are regular vacuously, so nonempty fibers require a separate check.

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