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The topological degree measures an oriented net number of sheets of a map. Its simplest integer-valued setting is a map $f:M\to N$ between connected oriented closed manifolds of the same positive dimension $n$. The hypotheses matter: an integer sign count needs orientations; compactness prevents preimages from escaping; and the same dimension makes a regular fiber discrete. A fundamental class gives the definition for continuous maps:Here $H_n(M;\mathbb{Z})$ and $H_n(N;\mathbb{Z})$ are generated by their chosen fundamental classes. Reversing either orientation changes the sign of the degree of a map between oriented manifolds. Disconnected sources contribute a sum over components; disconnected targets require a degree for each target component rather than one universal integer.

For a smooth map, choose a regular value $y$. The Sard theorem guarantees plentiful choices. Each point of $f^{-1}(y)$ has an invertible derivative, so the inverse function theorem makes it an isolated point. The fiber is finite by compactness of $M$. An oriented coordinate chart defines $\mathrm{sgn}\det d{f}_x$, and the degree as a sum of local degrees becomesAn empty fiber contributes zero. The sign is independent of the chosen oriented charts. Thus the mapping degree counts sheets with signs, not simply the number of preimages: orientation-preserving and orientation-reversing sheets cancel.

A differential-form argument both proves that this count is independent of $y$ and relates it to the topological definition. Around a regular value, take a small neighborhood $V$ whose full preimage is a disjoint union of neighborhoods mapped diffeomorphically onto $V$. Compactness rules out additional preimages approaching $V$ from elsewhere. Choose a smooth top-degree form $\eta$ supported in $V$, with ${\int }_N\eta =1$. The change of variables formula givesAlthough $\eta$ need not be everywhere positive, it represents the same normalized top cohomology class as any normalized volume form. The top de Rham cohomology of a compact connected oriented manifold says that two top forms with equal integrals differ by $d\beta$. Their pullback integrals agree, because the Generalized Stokes theorem gives ${\int }_{M}d(f^{\ast }\beta )=0$. Therefore the degree by integration of a pullback volume form isfor every smooth top form $\eta$. Pairing de Rham cohomology with fundamental classes identifies this integer with the homological definition. This proof also shows that every regular value gives the same signed sum.

The homotopy invariance of mapping degree makes it stable under continuous deformation. Homologically this follows from homotopic maps inducing the same map on homology. For a smooth homotopy $F:M\times [0,1]\to N$, a direct proof uses $d\eta =0$ and the Generalized Stokes theorem on the cylinder to getwith the cylinder orientation chosen to give the indicated boundary signs. Smooth approximation extends the geometric description to continuous maps, while the fundamental class definition already applies without differentiability. The multiplicativity of mapping degree follows from functoriality and givesA constant map has degree zero when $n>0$. If the mapping degree is nonzero, the map is onto: a point outside its image is a regular value with empty preimage. An orientation-preserving diffeomorphism has degree $1$, and an orientation-reversing one has degree $-1$. An orientation-preserving $k$-sheeted covering map has degree $k$; the signs of sheets must be included for other orientation conventions.

Examples show both the strength and the limits of the invariant. On the circle, $e^{i\theta }\mapsto e^{ik\theta }$ has degree $k$, including negative integers and zero; it is the winding number. On the sphere $S^n$, an ambient reflection has degree $-1$, and the antipodal map has degree $(-1{)}^{n+1}$, because the determinant of $-I$ on the ambient ${\mathbb{R}}^{n+1}$ is $(-1{)}^{n+1}$. On the torus, an integer matrix $A$ induces a map of degree $\det A$, obtained by pulling back the constant top form. However, degree does not classify general manifold maps: the identity and an integer shear on the two-torus both have degree one but induce different maps on $H_1$, so are not homotopic.

The mapping degree gives useful obstruction and existence arguments. There is no retraction of a closed ball onto its boundary sphere: such a retraction would make the degree-one identity of the boundary null-homotopic, since it would extend across a contractible ball, contradicting homotopy invariance and the degree-zero constant map. This is the degree obstruction underlying the Brouwer fixed-point theorem. Also an everywhere nonzero tangent vector field on an even-dimensional sphere would, after normalization, produce the homotopy $\cos (\pi t)x+\sin (\pi t)v(x)$ from the identity to the antipodal map. Orthogonality makes every value a unit vector, but the endpoint degrees are $1$ and $-1$, impossible. This proves the corresponding Hairy ball theorem obstruction.

There are appropriate extensions rather than an unrestricted integer degree for every pair of manifolds. For a proper map between connected oriented manifolds without boundary of equal dimension, the regular-value count remains finite and is invariant; compactly supported top forms or locally finite homology replace the compact fundamental-class description. Without orientations, degree modulo two counts preimages modulo two and uses mod-two fundamental classes. For manifolds with boundary one uses relative mapping degree for maps of pairs, keeping the boundary away from the chosen target value. Maps of unequal dimensions, or nonproper maps where preimages can escape to infinity, do not in general admit this same signed sheet-count invariant. The central principle is an integer signed count stable under homotopy, with its hypotheses explicitly preserved.

There is a small domain issue in the printed notation. If ${\mathbb{R}}^3/{\mathbb{R}}^2$ means a quotient vector space by the horizontal plane, the displayed formula does not descend to that quotient vector space: $(0,0,0)$ and $(1,0,0)$ represent the same class but give different values. The intended construction is stereographic projection, restricted to the unit sphere with the north pole removed. Interpreting the slash as removal of the plane $z=1$ also supplies a suitable ambient domain. The holomorphic stereographic atlas of the sphere is obtained as follows.

Write $N=(0,0,1)$ and $S=(0,0,-1)$. On $U_N=S^2\setminus \{N\}$ use $\zeta =(x+iy)/(1-z)$. Its inverse, with $\zeta =u+iv$, isThese formulas give a smooth manifold chart from $U_N$ onto $\mathbb{C}$. On $U_S=S^2\setminus \{S\}$ choose the second manifold chartIts inverse is $x=2\mathrm{Re}\eta /(1+|\eta {|}^2)$, $y=-2\mathrm{Im}\eta /(1+|\eta {|}^2)$ and $z=(1-|\eta {|}^2)/(1+|\eta {|}^2)$, so this is also a smooth manifold chart onto $\mathbb{C}$. The conjugation in this second stereographic projection is essential. On the overlap, $x^2+y^2=1-z^2$, soBoth directions of this transition are holomorphic maps on ${\mathbb{C}}^{\times }$, with nonzero derivative. The two manifold charts cover the sphere, hence define a holomorphic atlas, giving precisely the Riemann sphere. If one instead used $x+iy$ in both manifold charts, the transition would be $1/\bar{\zeta }$ and would not be holomorphic.

Orient the sphere by this holomorphic atlas. The given volume form is smooth at infinity: replacing $\zeta$ by $1/\eta$ in its exterior product gives the same expressionMoreover, $id\zeta \wedge d\bar{\zeta }=2du\wedge dv$, so the normalization of this volume form isThus the printed volume form has half the area of the standard round unit sphere; replacing its integral by $4\pi$ would introduce an erroneous factor of two.

For $k\ge 1$, the holomorphic map $f(\zeta )={\zeta }^k$ extends over infinity, since the target reciprocal coordinate is ${\eta }^k$ when the source reciprocal coordinate is $\eta$. Its pullback of a differential form isUsing $s=r^{2k}$ givesThe degree of a map between oriented manifolds is thereforeFor $k=0$, the formula on the finite manifold chart is the constant $1$. Its unique continuous extension is also $1$ at infinity, rather than an undefined expression ${\infty }^0$. Its pullback of a differential form is zero and its degree of a map between oriented manifolds is zero.

For the preimage calculation when $k\ge 1$, choose a regular value $w\in {\mathbb{C}}^{\times }$. There are exactly $k$ distinct roots of ${\zeta }^k=w$. At each root the real Jacobian determinant is $|k{\zeta }^{k-1}{|}^2>0$, so every local contribution to the degree of a map between oriented manifolds is $+1$. Their sum is $k$, agreeing with the integral. The exceptional values $0$ and infinity are avoided because they are branch values when $k>1$. For the constant map, any $w\ne 1$ is a regular value with no preimages, giving the same answer zero. This establishes the degree of a power map of the Riemann sphere for every allowed $k$.

Apply the spherical identity of part (b) with $x=v-v_{\ast }$, $y=\xi$ and $\varphi (s)=e^{-is/2}$. The angular exponential in part (c) can be replaced, after angular integration, by $e^{-i|\xi |(v-v_{\ast })\cdot \sigma /2}$. The full phase is thenThe Fubini theorem factors the two velocity integrals into Fourier transforms. With ${\xi }^{\pm }=(\xi \pm |\xi |\sigma )/2$, the gain integral is exactly ${\int }_{{\mathbb{S}}^2}\hat{f}({\xi }^{+})\hat{f}({\xi }^{-})d\sigma$.

The loss Fourier transform is $\hat{f}(\xi )\hat{f}(0)$. Dividing the gain by the sphere area gives the Bobylev identity for the Maxwell molecule collision operator:The initial Fourier datum is ${\hat{f}}_0$. The surface measure on a sphere here is two-dimensional surface area, not the restriction of ambient three-dimensional Lebesgue measure, which would give the sphere measure zero.

The electric field of this point charge is a translated and scaled version of the radial field in part (ii). If $\mathbf{a}\notin V$ and $\mathbf{a}\notin \partial V$, its divergence vanishes throughout $V$, so the divergence theorem gives zero flux integral.

If $\mathbf{a}$ lies inside $V$, remove a small ball centered at $\mathbf{a}$. On the remaining volume the divergence is zero. Its boundary is $\partial V$ together with the small sphere, whose outward normal for the punctured volume points towards $\mathbf{a}$. That inner sphere contributes $-q/{ϵ}_0$, by the $4\pi$ sphere flux computed in part (ii). Thus the exterior flux must be $q/{ϵ}_0$:This is Gauss's law for a point charge. The assumption that the charge is not on the boundary is essential to these alternatives.

On the outward-oriented sphere of radius $R$, $\mathbf{n}=\mathbf{x}/R$, so $\mathbf{x}\cdot d\mathbf{S}=RdS$. HenceAway from the origin, the divergence and curl of a radial vector field with $f(r)=r^{-3}$ giveIf the enclosed volume excludes the origin and its boundary avoids it, this field is smooth throughout the volume. The divergence theorem then gives $I=0$. The nonzero sphere flux is compatible with zero divergence away from the origin because the enclosed singularity prevents applying that smooth-volume theorem directly to the full ball.

Rotational symmetry of the vector field means $\mathbf{F}(r\mathbf{n})=h(r)\mathbf{n}$: rotations fixing $\mathbf{n}$ force the field to be parallel to $\mathbf{n}$, and rotations between directions make its magnitude depend only on $r$. Its flux integral over the radius-$r$ sphere is $4\pi r^{2}h(r)$. Applying the flux identity from part (i) to the ball, with the spherically symmetric source, givesTherefore the origin-regular spherical Poisson flux law isOnly source values at $s\le |\mathbf{x}|$ occur. Changing the exterior source can change the scalar potential by an interior constant, but cannot change this radial gradient. Regularity on all of ${\mathbb{R}}^3$, inherited from the Poisson equation setting, matters: on a punctured domain alone an additional $C\mathbf{x}/r^3$ would have zero divergence away from the origin and would not be fixed by the local source. It is excluded here by the origin regularity needed for the ball flux identity.

Odd-dimensional spheres cannot carry a symplectic form, by even-dimensionality. For $S^{2m}$ with $m\ge 2$, the second de Rham cohomology group vanishes. Any hypothetical symplectic form would therefore be $\omega =d\alpha$. The symplectic cohomology obstruction givesby the Generalized Stokes theorem. But the symplectic orientation makes ${\omega }^m$ a positive volume form, whose integral on this nonempty compact manifold is positive. This is a contradiction.

The oriented area form on $S^2$ is nondegenerate and automatically closed, since a two-dimensional manifold has no nonzero three-forms. Thus among positive-dimensional spheres, exactly $S^2$ admits a symplectic structure:If zero-dimensional symplectic manifolds are admitted, $S^0$ also qualifies: its zero two-form is nondegenerate on the zero tangent spaces. This is a convention-dependent additional case, not another positive-dimensional example.

Compatible Cauchy data for a wave map are ${\varphi }_0:{\mathbb{R}}^n\to S^2$ and ${\varphi }_1(x)\in T_{{\varphi }_0(x)}{S}^2$: thus $|{\varphi }_0|=1$ and ${\varphi }_0\cdot {\varphi }_1=0$. Smooth localized wave map Cauchy data equal a constant $p\in S^2$ with zero velocity outside a compact set. It is ${\varphi }_0-p$, rather than the sphere-valued map itself, that has compact support. The geometric constraints propagate under the wave map equation.

Smooth compatible wave map Cauchy data give a unique local smooth wave map. Relative to a constant map, Sobolev spaces $H^s\times H^{s-1}$ with $s>n/2+1$ provide a classical local theory. Iteration for the semilinear wave equation, Sobolev algebra and energy estimates give existence, uniqueness and continuous dependence. The smooth continuation criterion for semilinear wave equations extends the solution while these norms stay bounded. The sphere constraint and tangency constraint remain satisfied.

Global regularity for one-dimensional wave maps holds for arbitrary smooth localized compatible data. In retarded and advanced null coordinates $u=t-x$, $v=t+x$, the wave map equation says $D_v{\varphi }_u=D_u{\varphi }_v=0$, with $D$ the target covariant derivative. Hence $|{\varphi }_u{|}^2$ depends only on $u$, and $|{\varphi }_v{|}^2$ only on $v$. Initial derivative bounds persist. Compactness of the sphere and differentiated energy estimates then prevent finite-time loss of smoothness.

The Lusternik-Schnirelmann-Borsuk theorem has the following two equivalent covering formulations. For every integer $d\ge 0$, a cover of the sphere $S^d$ by $d+1$ closed sets has a member containing an antipodal pair. The same assertion holds with open sets in place of closed sets. Thus $d+1$ antipodal-pair-free open sets, or $d+1$ antipodal-pair-free closed sets, cannot cover $S^d$. The dimension-zero case simply says that one set covering the two-point sphere contains both points.

Here is why the two versions agree. A finite open cover of a compact metric space admits a closed shrinking that still covers: sufficiently small closed balls subordinate to the open cover can be grouped according to their containing open member. Applying the closed version to that shrinking proves the open version. Conversely, if a nonempty closed subset $C$ of $S^d$ avoids antipodal pairs, compactness gives positive distance between $C$ and $-C$. A sufficiently small open neighbourhood of $C$ still avoids antipodal pairs. Enlarge each member of a hypothetical closed counterexample in this way; the open version rules it out. Empty members cause no difficulty.

Another common equivalent formulation is the Borsuk-Ulam theorem: every continuous $f:S^d\to {\mathbb{R}}^d$ has $f(x)=f(-x)$ for some $x$, or, equivalently, every continuous odd function $g:S^d\to {\mathbb{R}}^d$ has a zero. For example, if $d+1$ antipodal-pair-free open sets covered $S^d$, a subordinate partition of unity $({\varphi }_1,\dots ,{\varphi }_{d+1})$ would give the odd functionIt cannot vanish, since some ${\varphi }_i(x)>0$, whereas antipodal-pair-freeness forces ${\varphi }_i(-x)=0$. In the other direction, if an odd function $g:S^d\to {\mathbb{R}}^d$ never vanishes, choose $d+1$ vectors $u_i$ forming a regular simplex centred at the origin in ${\mathbb{R}}^d$. For $d\ge 1$, the open sets $\{x:u_i\cdot g(x)>0\}$ cover $S^d$ and none contains an antipodal pair, contradicting the covering theorem.

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.