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The degree and its local signs. Orient $S^n$ and let $[S^n]$ generate its top reduced homology ${\tilde{H}}_n(S^n;\mathbb{Z})\cong \mathbb{Z}$. The mapping degree is the integer characterized byFor $n\ge 1$ this is the usual top homology definition using the fundamental class; using reduced homology also covers $S^0$.

Suppose $f$ is smooth and $p$ is a regular value. Each $x\in f^{-1}(p)$ has invertible tangent map $D{f}_x$, so the inverse function theorem makes $f^{-1}(p)$ discrete. It is also closed in the compact sphere, hence finite. Choose disjoint small neighborhoods of these points on which $f$ is a local diffeomorphism. The Excision theorem identifies the source local homology with the direct sum of one copy of $\mathbb{Z}$ for each inverse image. The induced local map is multiplication by $+1$ or $-1$ according as $D{f}_x$ preserves or reverses orientation. The map from the global fundamental class to these local orientation classes therefore proves the degree as a sum of local degrees formula:Here the determinant is computed in positively oriented tangent bases. Thus the mapping degree counts inverse images with signs, rather than just their cardinality.

The quotient map. Put $m=n+1$ and write $f=F{|}_{S^{m-1}}$. Give $D^m$ its standard orientation and its boundary the induced orientation. The connecting homomorphismis an isomorphism: the disk has zero positive reduced homology. It sends the relative fundamental class $[D^m,S^{m-1}]$ to $[S^{m-1}]$. If the map on relative homology induced by $F$ multiplies this class by $d$, naturality givesThus $d=\mathrm{deg}f$. Collapsing the boundary gives a sphere $D^m/S^{m-1}\cong S^m$, and the quotient map identifies its top reduced homology with the top relative homology of the disk pair. Give this quotient sphere the orientation determined by that identification. The relation $qF=\tilde{F}q$ now showsThis is the quotient-sphere degree identity.

The graph intersection. Orient $D^m\times D^m$ by the product orientation, orient $N=D^m\times \{0\}$ by its first factor, and orient the graph of a function $\Gamma$ by $x\mapsto (x,F(x))$. An intersection is precisely a zero of $F$, and no such zero lies on the boundary because $F(S^{m-1})\subset S^{m-1}$. At an intersection, transverse intersection means that $D{F}_x$ is surjective, hence invertible. The zeros are consequently isolated and finite.

For the smooth intersection number use the ordered tangent spaces $T_{x}N$ first and $T_{(x,F(x))}\Gamma$ second. Relative to the product basis their concatenated basis has matrixIts determinant is $\det D{F}_x$, so the intersection sign is ${ϵ}_x=\mathrm{sgn}\det D{F}_x$. By the same local Excision theorem argument, now in relative homology at the interior point $0$, the sum of these signs equals the multiplier of $F_{\ast }$ on $H_m(D^m,S^{m-1};\mathbb{Z})$. Therefore the graph intersection formula for mapping degree isIf the tangent spaces are ordered $\Gamma$ first and $N$ second, every sign changes by $(-1{)}^m$; the order above specifies the appropriate convention.

The one-dimensional model $F(x)=2x^3-x$ on $[-1,1]$ illustrates the graph intersection formula for mapping degree: its three zeros have signs $+1,-1,+1$, and its endpoint map has mapping degree $1$ on reduced homology.

The degree of a map between oriented manifolds measures how many times the domain covers the target, with signs recording the local local orientation of a manifold. Let $M,N$ be connected, oriented closed manifolds of the same dimension $n>0$. A continuous map $f:M\to N$ acts on top-dimensional homology bywhere $[M],[N]$ are their fundamental classes. Connectedness and the choices of local orientation of a manifold identify $H_n(N;\mathbb{Z})$ with $\mathbb{Z}$. Reversing the orientation of either manifold changes the sign; reversing both does not.

For a smooth map, Sard theorem supplies a regular value $y$. Its inverse image is discrete and, by compactness, finite. At each $x\in f^{-1}(y)$ the differential is an isomorphism; let its sign be $+1$ or $-1$ according to whether it preserves or reverses the chosen local orientations. The degree as a sum of local degrees isThe sign is computed in oriented manifold charts. The value is independent of the chosen regular value, even when inverse images appear or disappear: the signed count is the coefficient of $[N]$ in $f_{\ast }[M]$.

There is a useful local-density expression for the same topological degree. If $\omega$ is a volume form with ${\int }_N\omega =1$, thenThis degree by integration of a pullback volume form follows first by choosing a smooth top-form supported in a small neighborhood of a regular value, where the inverse branches contribute their orientation signs. Any other normalized top-form differs from it by an exact form: integration identifies $H_{\mathrm{dR}}^n(N)$ with $\mathbb{R}$. The integral of its pullback difference vanishes by Stokes theorem. In particular, for any top-form $\eta$, ${\int }_M{f}^{\ast }\eta =\mathrm{deg}(f){\int }_N\eta$.

A homotopy $H:M\times [0,1]\to N$ preserves this integral, since $d\omega =0$ and Stokes theorem givesThus topological degree is a homotopy invariant. It is multiplicative under composition, because the induced maps on homology compose: $\mathrm{deg}(g\circ f)=\mathrm{deg}(g)\mathrm{deg}(f)$. The identity has degree one, a constant map has degree zero for $n>0$, and an orientation-reversing diffeomorphism has degree minus one. An orientation-preserving finite covering map has degree equal to its number of sheets. Nonzero topological degree forces surjectivity, since an omitted point would be a regular value with an empty inverse image.

For the circle, $e^{i\theta }\mapsto e^{ik\theta }$ has degree $k$, positive or negative. This is its winding number, computable as $(2\pi i{)}^{-1}\int f^{-1}df$. The antipodal map of $S^n$ has degree $(-1{)}^{n+1}$: its extension $-I$ on the ambient $(n+1)$-dimensional vector space has that determinant sign and respects the outward-normal convention. A holomorphic map $z\mapsto z^k$, $k\ge 1$, on the Riemann sphere has degree $k$, whereas its complex conjugate has degree $-k$. These examples show how orientation, rather than simply the number of inverse images, determines the integer.

For maps $S^n\to S^n$, topological degree gives the complete homotopy classification ${\pi }_n(S^n)\cong \mathbb{Z}$. The degree does not classify general manifold maps: the identity of the torus and the map induced by the integer matrix $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ both have degree one, but have different induced maps on $H_1(T^2;\mathbb{Z})$ and so are not homotopic. A nonzero-degree map $S^n\to S^n$ cannot extend continuously to $B^{n+1}$, because such an extension would make the boundary map null-homotopic. In the smooth setting, Stokes theorem gives the same obstruction by applying it to the pulled-back normalized volume form.

The hypotheses can be adjusted, but must be stated. For connected oriented noncompact manifolds, a proper map has a degree defined using compactly supported top-forms, and it is invariant under proper homotopies. For manifolds with boundary one uses relative fundamental classes and maps of pairs, or fixes appropriate boundary conditions. Without an integral orientation one can still count inverse images modulo two, obtaining a mod-two degree. The integer integral formula used below assumes the oriented setting.

In classical field theory, these ideas turn continuous fields into quantized topological charges. Suppose a field on ${\mathbb{R}}^d$ approaches one fixed target value at spatial infinity. The one-point compactification makes it a map $\varphi :S^d\to \mathcal{V}$. When the target $\mathcal{V}$ is an oriented closed $d$-manifold, its topological degree labels topological sectors. More generally the sectors are described by homotopy groups; an integer degree is available only when the domain and target have the appropriate dimensions and orientations. Smooth time evolution preserving the boundary condition is a homotopy, so it cannot change the integer. A change requires a singular field, escape from the allowed target, or a change at the boundary.

A normalized closed target $d$-form gives the pullback-volume representation of a topological current. On spacetime, put $\alpha ={\varphi }^{\ast }\omega$. Since $d\alpha ={\varphi }^{\ast }(d\omega )=0$, its dual current is identically conserved, andis independent of time when there is no flux at infinity. This conservation law follows from geometry without using the field equations; it need not arise from a continuous symmetry through Noether theorem.

A concrete example is the O3 nonlinear sigma model in two spatial dimensions. Its unit-vector field $\mathbf{n}$ approaches a constant at infinity, defining $S^2\to S^2$. The normalized area form of the target gives the degree charge of an O3 sigma-model lump:For the energy normalization $E=\frac{1}{2}\int (|{\partial }_1\mathbf{n}{|}^2+|{\partial }_2\mathbf{n}{|}^2)$, the identities $\mathbf{n}\cdot {\partial }_i\mathbf{n}=0$ giveThis is the Bogomolny degree bound for the O3 sigma model. Choosing the sign appropriate to $Q$ makes the square nonnegative; vanishing of the square gives first-order Bogomolny equations and a sigma-model lump saturating the bound. With the oriented stereographic projectionthe maps $w=z^k$ have $Q=k$ and $E=4\pi k$. Their conjugates have $Q=-k$ with the same energy. Holomorphic rational maps have positive degree equal to their degree as rational maps; taking a reciprocal does not reverse the orientation. Antiholomorphic dependence reverses it.

The Skyrme model supplies a three-dimensional example. A field $U:{\mathbb{R}}^3\to \mathrm{SU}(2)$ with $U\to I$ at infinity is a map $S^3\to \mathrm{SU}(2)\cong S^3$. Take $T_i=-i{\tau }_i$ and $U^{-1}dU={\theta }^i{T}_i$, with ${\theta }^1\wedge {\theta }^2\wedge {\theta }^3$ positive. Since $\mathrm{tr}(T_i{T}_j{T}_k)=-2{ϵ}_{ijk}$, the normalized target volume form isThe integral is one on the unit three-sphere. Consequently the Skyrme baryon number as a mapping degree isThis is the topological baryon number in the Skyrme model; the sign has been fixed by the stated orientation and anti-Hermitian generator convention.

A topological charge alone does not guarantee a stable finite-size solution. The degree and energetic stability of a field configuration concern different properties. For a three-dimensional configuration of size $R$, the two-derivative energy scales as $R$, so it can decrease by shrinking while the topological degree remains fixed for every $R>0$. The limit can be singular. The Skyrme term, with four derivatives, scales as $R^{-1}$ and can balance the shrinking tendency. This is the role of Derrick theorem in distinguishing topological obstruction from energetic stability.

For defects, the relevant boundary map can instead be the sphere surrounding a core. A vacuum manifold equal to $S^1$ gives the integer winding number of a vortex; a vacuum manifold $S^2$ gives the degree of a surrounding $S^2$ for a magnetic monopole. This vacuum-boundary degree as a defect charge obstructs extending the normalized vacuum field through the enclosed ball. A nonzero integer therefore forces the field to leave the vacuum manifold somewhere in the core. This construction does not require the field to take one constant value in every direction at infinity.

Degree also appears in four-dimensional gauge theory through a boundary transition function. For an anti-Hermitian SU(2) gauge connection on ${\mathbb{R}}^4$, write $F=dA+A\wedge A$ and assume finite-action boundary behavior $A\to g^{-1}dg$ on the large bounding three-sphere. In the second-Chern conventionthe identity $d\mathrm{tr}(A\wedge dA+\frac{2}{3}{A}^3)=\mathrm{tr}(F\wedge F)$ and the Maurer-Cartan equation giveThis boundary winding representation of Yang-Mills topological charge relates the Second Chern number to the degree of $g:S^3\to \mathrm{SU}(2)$. The Chern-Simons 3-form turns the bulk integral into the boundary winding integral. Conventions which define the instanton number with the opposite trace sign reverse $k$; the integer quantization is unchanged. A Yang-Mills theta term weights a sector by $e^{i\vartheta k}$, giving periodicity $\vartheta \mapsto \vartheta +2\pi$. Thus the same topological degree that counts oriented inverse images also labels field sectors and expresses their quantized charges as integrals of local densities.

Regard complex matrices as a real vector space of dimension $2n^2$ and Hermitian matrices as a real vector space of dimension $n^2$. Set $F(A)=A{A}^{\ast }$. Its derivative is $D{F}_A(H)=H{A}^{\ast }+A{H}^{\ast }$. At a unitary $A$, any Hermitian $S$ is reached by $H=SA/2$. Hence $I$ is a regular value, and the regular level set theorem makes $\mathrm{U}(n)=F^{-1}(I)$ a smooth real manifold of dimension $\boxed{n^2}$.

The tangent space is the kernel of this derivative. Equivalently,At $I$, therefore, $H$ is skew-Hermitian. The explicit geodesic with initial point $I$ and velocity $H$ is $\gamma (t)=e^{tH}$. The matrix exponential is unitary because $e^{t{H}^{\ast }}{e}^{tH}=I$, and $\dot{\gamma }(0)=H$. Its acceleration is $\gamma H^2$. Since $H^2$ is Hermitian and every tangent vector at $\gamma$ is $\gamma K$ with $K$ skew-Hermitian, the induced real inner product satisfiesIndeed the trace of a Hermitian matrix times a skew-Hermitian matrix is purely imaginary. Its acceleration is therefore always normal to the manifold, which is the geodesic equation for an induced Euclidean metric. Diagonalizing $H=W\mathrm{diag}(i{\lambda }_j)W^{\ast }$ gives the equivalent explicit construction $W\mathrm{diag}(e^{it{\lambda }_j})W^{\ast }$.

Orient both spheres in the standard way and normalize their area forms to total area $4\pi$. The degree of a map between oriented manifolds can be obtained in two ways. For a regular value $y$, use the degree as a sum of local degrees:Each inverse image is isolated by the inverse function theorem, and compactness makes the set finite. The determinant is computed in consistently oriented local coordinates. A second method is spherical degree by area pullback:where the last formula represents $g$ as a unit vector in ${\mathbb{R}}^3$. The pullback of a differential form already contains the signed Jacobian determinant; no extra $\sin \theta$ is to be inserted in that last coordinate expression.

To relate the methods, replace $\omega$ by a smooth top-degree differential form with the same total integral supported in a small neighbourhood of a regular value. Two such top-degree forms with equal integral differ by an exact differential form on $S^2$, by its top-degree de Rham cohomology. Their pullbacks therefore have the same integral by Stokes theorem. Over the chosen neighbourhood, $g$ splits into local inverse branches; the change of variables formula makes the contribution of each branch its orientation sign times $4\pi$. Their sum is precisely the first formula. Thus the area integral is an integer and agrees with the signed inverse-image count.

For a nonconstant rational map, first use common-factor reduction of a rational map so $p$ and $q$ are coprime. Write $k=\max (\mathrm{deg}p,\mathrm{deg}q)$ for these reduced polynomials. A generic finite target value $w$ has inverse images at the roots of $p-wq$: avoiding exceptional values makes its degree $k$ and its roots simple. The fundamental theorem of algebra supplies $k$ roots. A holomorphic map has positive real Jacobian determinant $|R^{\prime }{|}^2$ at a regular point, so every local sign is $+1$. HenceThe source leaves coprimality implicit. In an unreduced representation the answer is $\max (\mathrm{deg}p,\mathrm{deg}q)-\mathrm{deg}\gcd (p,q)$, including degree zero for a constant reduced map. For example $(z^2-1)/(z-1)$ extends to $z+1$ and has degree one, although the unreduced maximum degree is two. Exceptional inverse images at infinity or multiple roots do not change the degree of a rational map of the Riemann sphere.

For the rational map approximation for Skyrmions, use stereographic projection $z=\tan (\theta /2)e^{i\phi }$ and the unit target vectorCombine this rational map with a radial profile to form a special unitary group field:where $\mathbf{\sigma }$ are the Pauli matrices. The endpoint values make $U(0)=-1$ independent of angle and $U(\infty )=1$. Appropriate radial behaviour gives an admissible finite-energy field configuration. With $L_i=U^{†}{\partial }_{i}U$, choose the topological baryon number in the Skyrme model conventionSeparating the radial and angular factors givesThus the degree of a rational map of the Riemann sphere supplies the Skyrmion charge.

In conventional dimensionless massless Skyrme model units, its static energy reduces towith the angular Jacobian of a rational mapThe Cauchy-Schwarz inequality gives $\mathcal{I}\ge k^2$. These formulas follow from the radial strain $|f^{\prime }|$ and the two equal angular strains $\sin f\sqrt{J_R}/r$: the quadratic energy sums their squares and the quartic Skyrme term sums their pairwise products of squares. Minimize the angular integral in the rational map approximation over degree-$k$ maps, then minimize the remaining radial energy with the stated endpoints. This replaces a three-dimensional field minimization by finitely many map coefficients and an ordinary differential equation for $f$.

The method constructs a charge-$k$ variational approximation, with topology built in and with rotational symmetry of a rational map translated into combined spatial and isospin rotations. It is efficient for identifying shapes and providing initial data for unrestricted numerical relaxation. Its restrictions are equally concrete: it uses one radial profile and a holomorphic angular map independent of radius, so it cannot represent arbitrary radial-angular correlations, separated clusters, or all deformations. Apart from the degree-one Skyrmion hedgehog ansatz, it generally does not solve the full field equation exactly. Massive-pion terms can be included in the radial functional but do not remove these restrictions, and multi-shell or unrestricted fields may be needed for larger charges. Approximate energy minima and a final collective-coordinate quantization are distinct steps.

For the sphere homology calculation use singular homology, homotopy invariance of homology, and the reduced Mayer–Vietoris sequence, not cellular homology. For a point, its chain group for singular homology is $\mathbb{Z}$ in every nonnegative degree, and its boundary operator is multiplication by ${\sum }_{i=0}^q(-1{)}^i$, hence alternately zero and one in positive degrees. Its homology is therefore $\mathbb{Z}$ in degree zero and zero above. Singular chains of a disjoint union split into the summand chain complexes. Start with $S^0$, the disjoint union of two points: $H_0(S^0;\mathbb{Z})={\mathbb{Z}}^2$, all positive groups vanish, and ${\tilde{H}}_0(S^0;\mathbb{Z})=\mathbb{Z}$. For $n\ge 1$, remove the north and south poles to obtain an open cover $U,V$ of $S^n$. Each member is contractible, while $U\cap V$ has a deformation retraction to $S^{n-1}$, including the disconnected intersection when $n=1$. The reduced exact sequence therefore suppliesAlso $S^n$ is connected, so its reduced degree-zero group vanishes. Induction givesThe separate $S^0$ calculation avoids conflating its two degree-zero generators.

Choose an orientation of $S^n$, with $n\ge 1$, and let $[S^n]$ be its fundamental class. The mapping degree is the integer determined by $f_{\ast }[S^n]=\mathrm{deg}(f)[S^n]$. To define the local degree at $x_0$, set $y_0=f(x_0)$ and require $x_0$ to be isolated in $f^{-1}(y_0)$. Choose an open neighborhood $B$ containing no other point of that fiber. Then $f$ defines a map of pairsThe degree-$n$ relative homology groups on both sides are $\mathbb{Z}$, by a manifold chart and excision. The local orientation of a manifold chooses their generators. The induced map multiplies these generators by an integer, denoted ${\mathrm{deg}}_{x_0}f$. Excision and naturality show independence of the choice of $B$. Differentiability is unnecessary; if $f$ is smooth and $D{f}_{x_0}$ is invertible, the inverse function theorem gives ${\mathrm{deg}}_{x_0}f=\mathrm{sign}\det D{f}_{x_0}$ relative to the orientations. Without isolation in the fiber this pointwise definition need not apply.

For the degree as a sum of local degrees, assume $y\in S^n$ has finite fiber $F=\{x_1,\dots ,x_s\}$. Disjoint coordinate neighborhoods and excision identifyThe absolute fundamental class maps to the tuple of local orientation generators. Under the relative map induced by $f$, the $j$th generator goes to ${\mathrm{deg}}_{x_j}f$ times the generator at $y$, and the resulting map from the direct sum adds these contributions. On the other hand, naturality says that first applying the absolute map $f_{\ast }$ and then passing to the local group at $y$ gives $\mathrm{deg}(f)$ times that generator. ThusIf the fiber is empty the sum is zero: the map factors through the contractible punctured sphere and has degree zero. If every point of a fiber is isolated, compactness makes that closed fiber finite, so the theorem applies. In particular, a regular value of a smooth map gives the signed count of its inverse images. For $n=0$, define a mapping degree on the reduced generator $[p_{+}]-[p_{-}]$: the identity has mapping degree one, the swap of the two points minus one, and either constant map zero. With local orientation signs positive at $p_{+}$ and negative at $p_{-}$, the same sum formula holds. The connected-sphere convention above is the standard positive-dimensional one.

An embedded smooth $d$-dimensional manifold $X\subset {\mathbb{R}}^N$ is locally carried by a smooth ambient coordinate change to ${\mathbb{R}}^d\times \{0\}$. A map between such manifolds is a smooth map between manifolds if its coordinate representatives are smooth. Its differential $d{f}_x:T_{x}X\to T_{f(x)}Y$ sends the tangent vector represented by a curve $\gamma$ to $(f\circ \gamma {)}^{\prime }(0)$; equivalently it is the derivative in charts.

A regular value $y$ is one for which $d{f}_x$ is surjective at every $x\in f^{-1}(y)$, including vacuous regularity when the fibre is empty. For example zero is not a regular value of $f:\mathbb{R}\to \mathbb{R}$, $f(x)=x^2$, because $d{f}_0=0$.

For an invertible matrix $A$, the derivative of the determinant isAt a determinant-one matrix it is nonzero, since taking $H=A/n$ gives value one. The regular level set theorem therefore makes $S{L}_n(\mathbb{R})={\det }^{-1}(1)$ an embedded manifold of dimension $n^2-1$, also for $n=1$ when it is a single point. Matrix multiplication has polynomial ambient coordinates and restricts to the group, so it is smooth. At the identity,Finally suppose the singular-matrix set were an embedded smooth manifold near zero for $n\ge 2$. Every curve $t\mapsto t{E}_{ij}$ lies in it, since its rank is at most one. Thus every matrix unit belongs to its tangent space at zero. These span ${\mathbb{R}}^{n^2}$, forcing its local dimension to be $n^2$. An embedded manifold of full ambient dimension is locally open, but arbitrarily small scalar multiples of the identity are invertible. This contradiction proves the singular-matrix set is not an embedded smooth manifold. The argument addresses the subset-manifold convention specified at the start.

On $N={\mu }^{-1}(-1/4)$, the diagonal fundamental vector field $X$ is nonzero, since the fixed-point set lies at moment map values $0$ and $-1/2$. The nondegenerate bilinear form ${\omega }_p$ and $d\mu ={\iota }_X\omega$ imply $d\mu \ne 0$, so $-1/4$ is a regular value. By the regular level set theorem, $N$ is a three-dimensional embedded submanifold of the four-dimensional Complex projective plane.

At $p\in N$, set $W=T_{p}N=\ker d{\mu }_p$. Then $\omega (X,v)=0$ for every $v\in W$, so $X\in W^{\omega }$, the symplectic orthogonal complement. This complement is one-dimensional and $d\mu (X)=\omega (X,X)=0$, givingThus the level set is a coisotropic submanifold, neither an isotropic submanifold nor a Lagrangian submanifold. Indeed an isotropic subspace of a symplectic vector space in dimension four has dimension at most two, whereas $\dim N=3$. Its characteristic line field is generated by the diagonal circle action.

An immersed submanifold of $M$ is a manifold $Y$ equipped with an injective immersion $\psi :Y\to M$. It is an embedded submanifold when $\psi$ is also a homeomorphism onto its image with the subspace topology, equivalently when it is a smooth embedding. For irrational $\alpha$, the irrational winding of the torusis an injective immersion $\mathbb{R}\to T^2$ with dense image, and therefore is not an embedding. If $Y$ is compact, however, an injective immersion is a continuous bijection from a compact space to its image in the Hausdorff manifold $M$; its inverse is continuous. Thus the compact injective immersion is an embedding theorem makes $\psi (Y)$ embedded.

Now let $X\subset M$ be embedded, with $\dim M=m$, $\dim X=m-k$, and $p\in X$. Apply the constant rank theorem to its inclusion. After choosing coordinates and reordering them, there is a neighborhood $U$ of $p$ with coordinates $(x^1,\dots ,x^m)$ for whichThis is the slice chart for an embedded submanifold.

It is false that every embedded submanifold is the inverse image of a regular value of a map to a Euclidean space. If a codimension-$k$ submanifold is $f^{-1}(y)$ for a regular value of $f:M\to {\mathbb{R}}^k$, the differentials of the component functions give a global frame of its conormal bundle, so its normal bundle is trivial. The core circle of the Möbius band is embedded but has the nontrivial Möbius normal line bundle. This is the normal-bundle obstruction to being a regular level set.

Finally, an inductive spinning construction gives the requested torus. Place an embedding $F:N^m\hookrightarrow {\mathbb{R}}^{m+1}$ in the half-space $F_{m+1}>0$ and defineThe positive radius makes this map injective, and its differential is injective in both the $N$ and circle directions; compactness then makes it an embedding. Starting with $S^1\hookrightarrow {\mathbb{R}}^2$ and iterating proves the embedding of the n-dimensional torus in codimension one:

Define the smooth functionAt $x\in F^{-1}(1)$ its derivative isThis linear map is surjective because $x\ne 0$; for example, $D_{x}F(x)=2$. Hence one is a regular value, and the regular level set theorem makesan embedded codimension-one submanifold of ${\mathbb{R}}^{n+1}$. Its tangent space is the kernel of the derivative:This is the unit sphere as a regular level set.

The curve is symmetric in both coordinate axes and under interchange of $x$ and $z$. Its shape, including its eight inflection points, is:

Writing $r^2=x^2+y^2$, the surface of revolution has equationLetIts gradient isIf this vanished, then $r^2\in \{0,1\}$ and $z^2\in \{0,1\}$, but at such a point $(r^2-1{)}^2+(z^2-1{)}^2\le 2$, not five. Hence zero is a regular value, and the regular level set theorem shows that $S=F^{-1}(0)$ is a smooth embedded surface.

For a surface of revolution, the two principal curvatures are the curvature of the meridian and the normal curvature of a parallel. Along $r=1$,so the normal is vertical in the meridian plane and the parallel principal curvature vanishes. Therefore the Gaussian curvature, the product of the principal curvatures, is zero there.

These are not the only zero-curvature points. The meridian itself has inflection points, at which its principal curvature vanishes. Setting $u=r^2$ and $v=z^2$, their positive squared coordinates satisfyBesides interchanging $u$ and $v$, the solution is approximatelyEach corresponding meridian inflection point sweeps out another circle on $S$, so $r=1$ does not exhaust the zero set of the Gaussian curvature.

LetAt $x\in F^{-1}(-1)$,This differential is nonzero because $x\ne 0$, so $-1$ is a regular value. The regular level set theorem makes $F^{-1}(-1)$ a three-dimensional embedded submanifold; intersecting it with the open set $\{x_0>0\}$ preserves that property. Therefore $X$ is the three-dimensional hyperboloid model, and

The condition $D{f}_p\ne 0$ says that $0$ is a regular value of $f$. The regular level set theorem therefore makes $f^{-1}(0)$ a smooth submanifold of codimension one in ${\mathbb{R}}^3$, hence a smooth surface.

Forthe gradient isOn $F^{-1}(0)$, $x^2+y^2=\cosh (z^2)\ge 1$, so $x$ and $y$ cannot both vanish. Thus $\nabla F\ne 0$ there, and the given set is a smooth surface.

Not every smooth surface in ${\mathbb{R}}^3$ is a global zero set. Every set $f^{-1}(0)$ is closed because $f$ is continuous, whereas the open unit diskis a smooth surface but is not closed in ${\mathbb{R}}^3$.

On $S^n\times S^n$, consider $q(x,y)=⟨x,y⟩$. At a point of $q^{-1}(0)$, its differential isThe tangent vector $(y,x)$ lies in $T_x{S}^n\oplus T_y{S}^n$ and has $dq(y,x)=2$, so zero is a regular value. The regular level set theorem makes $P=q^{-1}(0)$ a smooth submanifold. Its tangent space is

LetIts gradient isIf this vanishes, the first and third coordinates give $x=y=z$, and the middle coordinate then gives $y=0$. Thus the origin is the only critical point of $F$. It has been removed from $V$, so $0$ is a regular value of the restriction under consideration. The regular level set theorem shows thatis a smooth surface.

A point $y$ is a regular value of $f$ when every derivative $D_{x}f$ for $x$ in $f^{-1}(y)$ is surjective. Sard theorem says that the critical values of a smooth map have measure zero. The stack-of-records theorem says that the inverse image of a regular value is a smooth submanifold of codimension $\dim Y$.

Maps $f$ and $g$ are smoothly homotopic when a smooth $F:X\times [0,1]\to Y$ has endpoint maps $f$ and $g$. Perturb such a homotopy relative to its ends so that it is transverse to $y$. Then $F^{-1}(y)$ is a compact one-dimensional manifold whose boundary is $f^{-1}(y)\bigsqcup g^{-1}(y)$. Every compact one-dimensional manifold has an even number of boundary points, proving the parity formula.

The hypothesis on $U$ says that the degree modulo two of $h$ is one. Suppose no $h(x)$ and $h(-x)$ were antipodal. Thenwould be defined, even, and homotopic to $h$ by normalized straight-line interpolation. For a regular value, the fibres of the even map $k$ occur in antipodal pairs, so its degree modulo two is zero. Homotopy invariance gives a contradiction. Hence some antipodal pair is mapped to an antipodal pair.