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Define the smooth mapAt $(M,v)$ its derivative in the direction $(A,w)$ isThe restriction to variations $(0,w)$ is the surjective linear map $M:{\mathbb{R}}^n\to {\mathbb{R}}^m$, since $M$ has full row rank. Thus $0$ is a regular value of $\Phi$. The regular level set theorem now shows thatis an embedded submanifold of $X_{m,n}\times {\mathbb{R}}^n$, of dimension $mn+n-m$.

A smooth manifold of dimension $k$ is a Hausdorff, second-countable space with an atlas of smoothly compatible charts to open subsets of ${\mathbb{R}}^k$. A value $y\in Y$ is a regular value of a smooth map $f:X\to Y$ when every $x\in f^{-1}(y)$ has surjective derivative $d{f}_x$.

The inverse function theorem says that a smooth map between equal-dimensional manifolds whose derivative is invertible at a point is a diffeomorphism between neighbourhoods of that point and its image. If $y$ is a regular value of a map from an $m$-manifold to an $n$-manifold, choose coordinates corresponding to an invertible $n$ by $n$ minor of $df$. Applying the inverse theorem to $f$ together with the remaining $m-n$ coordinates makes $f$ the coordinate projection onto the first $n$ coordinates. Its level set is therefore locally ${\mathbb{R}}^{m-n}$. This proves the preimage theorem.

The critical-point set is closed because failure of full rank is the simultaneous vanishing of all maximal minors of $df$. If $X$ is compact, this set is compact, so its image under $f$ is compact and hence closed in the manifold $Y$. Its complement, the set of regular values, is open.

For the printed equations putA rank calculation shows that $(a,0)$ is a regular value when $a>0$: if
$\nabla F_2=\lambda \nabla F_1$, thenAt a point of the level set these relations force $a\le 0$. Hence for $a>0$, the preimage theorem makes $X_a$ a two-dimensional manifold.

As printed, however, the assertion for every $a\ne 0$ is false. If $a<0$, the pointsbelong to $X_a$. Near either point the first equation solves smoothly for $w$, while the second equation isIts positive-$z$ and negative-$z$ sheets meet only at the origin, so deleting the meeting point disconnects every sufficiently small neighbourhood. A punctured neighbourhood in a two-manifold is connected. Thus $X_a$ is not a manifold there.

For $a=0$, Cauchy--Schwarz givesThe first equation requires $w^2=x+y-z^2\ge 0$, so equality must hold. ThereforeThis is the image of a smooth embedding with nonzero derivative, and hence
$\boxed{X_0\text{ is a one-dimensional manifold}}$.

A point $y_0\in Y$ is a regular value of the smooth map $f:X\to Y$ if, for every $x\in f^{-1}(y_0)$, the derivativeis surjective. A value outside the image is regular vacuously.

For a nonvacuous example, takeThe value $1$ is regular: its fibre is the unit circle, and at every $(x,y)$ on that circle,is a nonzero linear functional and hence surjective.

Sard theorem states that the set of critical values of a smooth map between smooth manifolds has measure zero in the target. Equivalently, almost every target point is a regular value.

Assume first that $Y$ is connected; otherwise the definition is made separately on each connected component. Choose a regular value $y\in Y$ and define the degree modulo two by

The following course results make this well-defined.

The same transverse-preimage argument applied to a smooth homotopy proves homotopy invariance. Consequently the definition depends only on $f$, rather than on the chosen regular value.

A critical point of a smooth map is a point where its derivative is not surjective; its image is a critical value. A regular value is a target value all of whose preimages have surjective derivative.

For $f(X)=X^{T}AX$ with $A=A^T$,Thus $X$ is critical exactly when $AX=0$, and every critical point has value zero. Hence zero is the only possible critical value; it is a critical value precisely when $A$ is singular. If $A$ is singular, every point of its kernel is critical, so there can be infinitely many critical points; if $A$ is invertible, only the origin is critical.

A level set at a regular value is a smooth submanifold. By the regular level set theorem, its codimension is the dimension of the target. For determinant-one real symmetric matrices, $D(\det {)}_B(C)=\mathrm{tr}(B^{-1}C)$ is nonzero because $C=B$ gives $n$, so the constraint has codimension one.

A Seifert hypersurface for an oriented codimension-two submanifold is an oriented codimension-one submanifold having it as boundary. For a closed oriented surface in $S^4$, an integral meridional map from the complement to the circle, with angular behavior near the surface, supplies a Seifert hypersurface by taking a regular value.

For a smooth map between oriented unit spheres, integrate the pullback of the target area form. Localizing a normalized top-degree form near a regular value, and using Stokes theorem to discard exact-form differences, identifies this integral with the degree as a sum of local degrees.

For a Hamiltonian group action, a regular value level preserved by a free proper group action has a smooth quotient. The restricted symplectic form descends to a symplectic form on that quotient when the kernel is exactly the orbit directions; this is symplectic reduction.