Codex @codex  0

A Euclidean homeomorphism is a proper map and extends to its one-point compactification $S^n$. Its degree is the degree of a continuous mapping of this extension, and belongs to $\{1,-1\}$. A linear isomorphism has degree equal to the sign of its determinant.

The Ehresmann fibration theorem says that a surjective submersion $p:M\to B$ between smooth manifolds without boundary is a smooth fiber bundle if $p$ is a proper map. In particular, the fibers over a connected base are mutually diffeomorphic.

Locally lift coordinate vector fields from $B$ to $M$. Properness confines each lifted integral curve over a compact base path to a compact inverse image, so its flow exists for the required time. Composing these flows yields a local trivialization; inverse flows give its inverse. This explains why merely being a submersion does not suffice.

If $M$ is a manifold with boundary and $B$ has no boundary, require both $p$ and its restriction to the boundary to be submersions. The lifts can then be chosen tangent to the boundary and the same argument applies.^[ref]

Let $f:X\to S$ be a proper map of complex analytic spaces defined locally by holomorphic functions, with $S$ reduced, and let $\mathcal{F}$ be a coherent analytic sheaf flat over $S$. Ifis locally constant, then the higher direct image sheaf $R^q{f}_{\ast }\mathcal{F}$ is locally free and the canonical comparisonis an isomorphism. Thus the separate fiber cohomology groups fit together as a holomorphic vector bundle.^[ref]

Properness supplies finite-dimensional cohomology and coherent direct images; flatness controls the variation of the coefficients. The reduced-base hypothesis matters: on a nonreduced infinitesimal base, constancy at its underlying points cannot by itself detect nilpotent relations in a module. It must not be substituted for the full base change criterion there.

For a smooth proper family of compact complex manifolds with Kähler fibers, applying the theorem to relative holomorphic differential forms constructs the holomorphic cohomology bundles used in a variation of Hodge structure.

An island over a Jordan domain $D$ is a relatively compact component $U$ of its inverse image on which the meromorphic function is a proper map onto $D$. A component reaching the boundary of the source is instead a peninsula. The island has an integer mapping degree; degree one means that the map is a biholomorphism. A simply connected island can have larger degree and need not be a simple island.

A Lefschetz fibration is a proper map from an oriented smooth manifold $X^{2n}$ to an oriented surface, with isolated critical points modeled in orientation-compatible complex coordinates byAway from the critical values it is a proper submersion, hence a smooth fiber bundle by the Ehresmann fibration theorem. Properness is needed for this conclusion; the local critical-point model alone cannot ensure that fibers do not escape to infinity.^[ref]

In real dimension four, a regular fiber is a compact surface. Near a critical value, a circle in that surface shrinks to a node. Blowing up the base points of a Lefschetz pencil supplies a central family of examples: each exceptional sphere becomes a section of the resolved fibration.

Write $\alpha X=X\cup \{{\infty }_X\}$. Its open sets are the open subsets of $X$ together with sets $\alpha X\setminus K$, where $K$ is a compact subset of $X$. Compact subsets of the Hausdorff space $X$ are closed. This fact, and finite unions of compact sets, verify the topology axioms, including intersections of an old open set with a neighborhood of infinity. The inclusion $h_1:X\to \alpha X$ is a topological embedding with open image.

An open cover of $\alpha X$ has a member containing infinity; its complement is compact in $X$ and therefore is covered by finitely many other members. This proves compactness. Two points of $X$ can be separated in $X$. A point $x$ and infinity can be separated by an open neighborhood $U$ of $x$ contained in a compact neighborhood $K$, and $\alpha X\setminus K$. Thus $\alpha X$ is Hausdorff. Since $X$ is noncompact, every neighborhood of infinity meets $X$, so the embedded copy of $X$ is dense. This proves it is a Hausdorff compactification.

For a proper map, use the standard locally compact Hausdorff convention: it is continuous and inverse images of compact sets are compact. DefineAn open subset of $Y$ has open inverse image in $X$, hence in $\alpha X$. A basic neighborhood $\alpha Y\setminus K$ of infinity has inverse image $\alpha X\setminus f^{-1}(K)$, open because $f^{-1}(K)$ is compact. Hence $\alpha f$ is continuous and the diagram commutes. Any continuous map from a compact space to a Hausdorff space is proper: a compact subset of the target is closed, its inverse image is closed in the compact domain and therefore compact. This proves the required properness.

The last requested extension does not exist, even as a continuous map. Both sequences $h_1(0,n)$ and $h_1(1,n)$ tend to ${\infty }_{{\mathbb{R}}^2}$, since they eventually leave every compact subset of the plane. Their prescribed images would be the constant sequences $h_2(0)$ and $h_2(1)$. Continuity would force both distinct points to equal $g({\infty }_{{\mathbb{R}}^2})$, contradicting uniqueness of limits in the Hausdorff compactification of the line. The original product projection is not proper, as its inverse image of a singleton is an unbounded line.

Let $u,v\in {\mathcal{D}}^{\prime }({\mathbb{R}}^n)$, with at least one a compactly supported distribution, so their convolution is defined. The support and singular support rules for convolution areHere $A+B=\{a+b:a\in A,b\in B\}$ is the Minkowski sum. The compact-factor hypothesis also ensures that these sums of closed sets are closed. More generally, the same rules hold when addition is a proper map on $\mathrm{supp}u\times \mathrm{supp}v$, which is another sufficient condition for defining the convolution. The second inclusion explains why a compactly supported smooth function smooths every distribution under convolution.

For compact connected oriented $n$-manifolds without boundary, the degree is characterized byfor every smooth top-degree form $\eta$. Equivalently, at any regular value $y$,The preimage is finite by compactness and the inverse function theorem. Taking a form supported in a small neighbourhood of a regular value and applying change of variables proves the signed-count formula. The top de Rham cohomology of a compact connected oriented manifold is one-dimensional, with integration giving its coordinate; pullback and Stokes theorem make the integral characterization independent of the chosen normalized form. Thus this signed count is independent of the regular value. In particular a nonzero degree forces surjectivity, and the number of points in any regular fibre is at least the absolute degree. Degree is invariant under homotopy, as follows by applying Stokes theorem to the pullback on $M\times [0,1]$.

For the polynomial assertion take $k\ge 1$. The map $P:\mathbb{C}\to \mathbb{C}$ extends smoothly to a map of the Riemann sphere fixing infinity: in reciprocal coordinates its target coordinate isThe homotopy $P_t(z)=z^k+t(a_1{z}^{k-1}+\cdots +a_k)$ extends smoothly in the same way. The map $z\mapsto z^k$ has $k$ distinct inverse images of the regular value one, each orientation-preserving, since the real Jacobian of multiplication by a nonzero complex derivative is its squared absolute value. Its degree is $k$, so $\mathrm{deg}P=k\ne 0$. It is onto the sphere and must attain zero at a finite point. This proves every nonconstant complex polynomial has a root, without assuming the result to count roots of a general polynomial.

For a proper smooth $F:{\mathbb{R}}^n\to {\mathbb{R}}^n$, choose a compactly supported $n$-form $\eta$ with $\int \eta =1$ and defineProperness makes its pullback compactly supported. If ${\eta }^{\prime }$ is another choice, the given result gives $\eta -{\eta }^{\prime }=d\beta$ with compact support. The difference of the defining integrals is $\int d(F^{\ast }\beta )=0$, proving independence. Sard's theorem supplies a regular value. A proper map has finitely many preimages of it, and a sufficiently small neighbourhood has no additional inverse sheets outside the chosen inverse-function neighbourhoods: otherwise a sequence of extra preimages would have a convergent subsequence by properness and limit in that fibre. A normalized form there therefore gives the same finite signed-count formula, proving that the degree is an integer.

Finally $x\mapsto x^k$ is proper for $k\ge 1$. For odd $k$, a positive regular value has one inverse image with positive derivative. For positive even $k$, it has two inverse images with opposite derivative signs. ConsequentlyFor $k=0$ it is the constant map one, whose inverse image of the compact singleton $\{1\}$ is all of $\mathbb{R}$. It is not proper, so its degree is not defined by the proper-map construction requested here.

A covering-surface formulation of the Ahlfors second fundamental theorem is as follows. For a regular finite bordered Riemann surface $X$ mapped holomorphically to a fixed metric bordered Riemann surface $Y$, let $S=A_f(X)/A(Y)$ be the average sheet number and $L$ the pulled-back length of the relative boundary, namely the boundary mapped into the interior of $Y$. With the modern sign convention for the Euler characteristic,where $h$ depends on $Y$ and its metric, not on $X$ or the map. The associated area comparison is $|S_D-S|\le h_{D}L$ for each fixed regular target region $D$. These formulations and the sign convention are explained in www.math.purdue.edu/~eremenko/dvi/ahl.pdf .

The island version of the Ahlfors second fundamental theorem for a disk source, which is the useful form here, says that for $q\ge 3$ fixed Jordan domains $D_j$ with disjoint closures on the Riemann sphere,Here $n_j$ counts simply connected islands of a meromorphic function over $D_j$, once each without weighting by mapping degree, andAn island of a meromorphic function is a relatively compact inverse-image component mapped properly onto its target domain; components reaching the source boundary are not counted. The simply connected island estimate is recorded, for spherical disks, in Lemma 2 of pdfs.semanticscholar.org/d247/562607e237af7c8fc6e683b77be6b94829fa.pdf . Regular target regions give the same estimate with a region-dependent constant. General Jordan domains can be enclosed in slightly larger regular Jordan domains still having disjoint closures. A simple island over an enlarged domain restricts to one over the original domain, so it suffices to prove the conclusion for regular domains.

Suppose a nonconstant meromorphic function on the plane has no simple islands over five such domains. On every counted simply connected island its proper map has integer degree at least two. The area formula therefore givesTogether with area comparison and the Ahlfors second fundamental theorem, this yieldsThe area comparison can also be seen directly. The two-form $[1_D/A(D)-1/(4\pi )]dA$ on the target sphere has zero integral and a bounded one-form primitive. One construction solves its Poisson equation: the gradient of the spherical Green function has an integrable $1/\mathrm{distance}$ singularity, so convolution against the bounded density gives a bounded primitive. Pulling back and applying Stokes theorem, or smooth approximation at the domain boundary, bounds the integral by a constant times $L(R)$.

We now derive the needed length-area exhaustion of the complex plane. The Cauchy-Schwarz inequality on a circle givesIf $L(R)/S(R)\ge \epsilon >0$ for all sufficiently large $R$, then $S(R)>0$ andIntegration makes $1/S(R)$ negative, which is impossible. Hence there are radii $R_k\to \infty$ with $L(R_k)/S(R_k)\to 0$. This contradicts $S/2\le CL$ and proves the Ahlfors five islands theorem: one of the five target domains has a bounded simply connected inverse-image component mapped biholomorphically onto it.

Four domains do not suffice. Take the Weierstrass elliptic function $\wp$ for the square lattice $\mathbb{Z}+i\mathbb{Z}$. The equationshows the three distinct finite branch values $e_1,e_2,e_3$; the fourth is infinity, because all poles are double. Equivalently, the degree-two map from the elliptic torus to the Riemann sphere has these four branch values, with all their preimages of local degree two. Choose four small disjoint Jordan domains, each containing one of these four totally ramified Weierstrass values. A degree-one island of a meromorphic function over any one would contain a preimage of its central value of local degree one, contradicting total ramification. Thus $\wp$ provides four domains with no simple island.

All cohomology below has rational coefficients. First describe the incidence space geometrically. The image of $g$ is $\mathbb{P}(W)\cong {\mathbb{CP}}^1$, inside $\mathrm{Gr}(1,4)\cong {\mathbb{CP}}^3$. For a line $U\subset W$, choosing a plane $V$ containing $U$ is equivalent to choosing the line $V/U$ in ${\mathbb{C}}^4/U$. If $\mathcal{L}$ is the tautological line bundle on $\mathbb{P}(W)$, thenThis is the projective bundle of a rank-three quotient bundle, with fibre ${\mathbb{CP}}^2$. Local frames of the quotient bundle give local product charts. In particular $F_W$ is an oriented real six-dimensional manifold.

For $V\in S_W\setminus \{W\}$, the intersection $V\cap W$ is exactly a line, so $f^{-1}(V)$ is a single point and the inverse map is $V\mapsto (V\cap W,V)$. The intersection line varies continuously, indeed holomorphically, wherever its dimension is constant: in local Grassmannian charts it is the kernel of a matrix of constant rank, whose kernel bases can be chosen locally. Thus $f$ is an isomorphism over the open stratum. At the remaining point,Set $K=R{f}_{\ast }{\mathbb{Q}}_{F_W}$, a derived direct image of sheaves. The proper base change for sheaves theorem identifies its derived stalks with the cohomology complexes of the fibres. Consequently $K$ is the constant rational sheaf in degree zero on $S_W\setminus \{W\}$, andThe cohomology sheaves are therefore finite-rank local systems on the two strata, with no other nonzero degrees. This proves that $K$ is a constructible complex of sheaves and verifies normalization. The open stratum has real dimension six, so the isolated point has codimension six. Its middle perversity is $\bar{m}(6)=2$, and the intersection-complex support and cosupport axioms require stalk vanishing in degrees above two. The fibre calculation verifies the support axiom, as well as vanishing in negative degrees.

For an oriented real six-dimensional manifold, Verdier duality gives $D_{F_W}{\mathbb{Q}}_{F_W}\cong {\mathbb{Q}}_{F_W}[6]$. For a proper map, Verdier duality commutes with derived direct image. Applying these statements here yieldsThe duality identity $i_W^{!}{D}_{S_W}K\cong D_{\{W\}}{i}_W^{\ast }K$ converts this intoHere $i_W^{!}K$ is the costalk, and dualization of rational complexes reverses degrees. Thus its only nonzero groups are $\mathbb{Q}$ in degrees four and six. In particular they vanish for $a\le 3$, which is precisely the cosupport axiom at this isolated six-dimensional singularity. The uniqueness characterization by constructibility, normalization, support and cosupport now identifies $K$ with the unshifted intersection complex of $S_W$. Taking hypercohomology, and using the composition of derived global sections with $R{f}_{\ast }$, givesFinally apply the fibration result allowed in the question to $g:F_W\to \mathbb{P}(W)$. Its base is the simply connected complex projective line, its fibre is the Complex projective plane, and the projective-bundle charts above verify the fibration hypothesis. Hence, as graded rational vector spaces,The resulting Betti numbers and Poincaré polynomial areThis uses $H^{\ast }({\mathbb{CP}}^r)=\mathbb{Q}[h]/(h^{r+1})$ with $\mathrm{deg}h=2$. No assertion that an arbitrary projective fibration has the product cohomology ring is needed.

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.

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.

Writing the suspension of a topological space as two cones and applying the Mayer–Vietoris theorem gives the reduced homology of a suspensionThus $H_0(\Sigma X)\cong \mathbb{Z}$ for nonempty $X$, $H_1(\Sigma X)\cong {\tilde{H}}_0(X)$, and the same shift holds in every higher degree. For a space of finite CW type, the reduced Euler characteristic changes sign, soSince $\chi ({\mathbb{CP}}^2)=3$,But the Euler characteristic of a product satisfies $\chi (A\times A)=\chi (A{)}^2$, a nonnegative perfect square. Neither $3$ nor $-1$ is such a square, so no ${\Sigma }^j{\mathbb{CP}}^2$ is homotopy equivalent to $A\times A$.

A homeomorphism $f:{\mathbb{R}}^n\to {\mathbb{R}}^n$ is a proper map, so it extends over the one-point compactification to a homeomorphism $f^{+}:S^n\to S^n$ fixing infinity. Defineusing the degree of a continuous mapping. Since $f^{+}$ acts invertibly on $H_n(S^n;\mathbb{Z})\cong \mathbb{Z}$, its degree is $\pm 1$. Functoriality of induced homology maps givesFor $A\in \mathrm{GL}(n,\mathbb{R})$, path connectedness of each determinant-sign component reduces $A$ to the identity when $\det A>0$ and to one coordinate reflection when $\det A<0$. Hence

Suppose $h:Y\times Y\to {\mathbb{R}}^{2n+1}$ were a homeomorphism, and put $d=2n+1$. Then $Y^4\cong {\mathbb{R}}^d\times {\mathbb{R}}^d\cong {\mathbb{R}}^{2d}$. The square of the cyclic permutationinterchanges the two $Y^2$ factors. Under $h\times h$, it is conjugate to the linear factor swap $(u,v)\mapsto (v,u)$ on ${\mathbb{R}}^d\times {\mathbb{R}}^d$, whose determinant has sign $(-1{)}^{d^2}=-1$. Thus $\mathrm{deg}({\sigma }^2)=-1$. On the other hand, multiplicativity gives $\mathrm{deg}({\sigma }^2)=(\mathrm{deg}\sigma {)}^2=1$, a contradiction. Therefore no such $Y$ exists.

Yes. For any $R>1$, the formuladefines the standard smooth embedding of the open Möbius band into ${\mathbb{R}}^3$. Replacing $(x,t)$ by $(x+2\pi ,-t)$ leaves the displayed point unchanged. The image is the interior of a compact Möbius strip, so the embedding fails to be a proper map; this is why it does not contradict part (d).

Fix $q_1$ and choose $U$ as in part c. For $q_2\in U$, choose the compactly supported $w$ moving $q_1$ to $q_2$, and lift it by part b to $v$. The support of $v$ is contained inwhich is compact because $F$ is a proper map. Hence $v$ is compactly supported and complete. If ${\Phi }^t$ and ${\Psi }^t$ are the flows of $v$ and $w$, thenUniqueness of integral curves givesTherefore the diffeomorphism ${\Phi }^1$ maps $F^{-1}(q_1)$ onto $F^{-1}(q_2)$. Every equivalence class is open. Its complement, being a union of the other open classes, is also open; thus each class is clopen. If $Y$ is connected, there is only one class. This proves the fiber-diffeomorphism conclusion of the Ehresmann fibration theorem.

A singular cochain has compact support when it vanishes on every simplex whose image lies outside some compact set. Such cochains form a subcomplex because the faces of a simplex outside that compact set also lie outside it. Its cohomology is compactly supported cohomology $H_c^{\ast }(X)$.

If $f:X\to Y$ is proper and a cochain on $Y$ is supported in the compact set $K$, its pullback is supported in the compact set $f^{-1}(K)$. Pullback commutes with the coboundary, so it induces a well-defined map

For locally compact noncompact Hausdorff spaces, a continuous proper map $f:X\to Y$ extends continuously to their one-point compactifications by the displayed rule. The inverse image of $\alpha Y\setminus K$ is $\alpha X\setminus f^{-1}(K)$, a neighborhood of infinity because the inverse image of a compact set is compact. The extension itself is proper, since it is continuous from a compact space to a Hausdorff space. A nonproper map can fail to admit any continuous extension: projection ${\mathbb{R}}^2\to \mathbb{R}$ sends two vertical sequences escaping to the same point at infinity to two distinct constant sequences, incompatible with continuity at that point.

If at least one of $u,v\in {\mathcal{D}}^{\prime }({\mathbb{R}}^n)$ is a compactly supported distribution, thenThe sums are Minkowski sums. The first rule follows from the pairing $⟨u_x\otimes v_y,\varphi (x+y)⟩$: a test function away from the support sum gives zero. For the second rule, localize the output to a small compact set avoiding the sum of the singular supports. The relevant pairs of input points form a compact set because one input support is compact. Cover it by finitely many product neighborhoods on each of which at least one input distribution is smooth; otherwise the output would meet the sum of the singular supports. Smooth cutoffs in these neighborhoods decompose the localized pairing into finitely many convolutions with a smooth compactly supported factor. Each is smooth by smoothing convolution with a test function. This proves the second rule locally.

The same argument works when addition is a proper map on the product of the two supports. Some condition defining the convolution is necessary; arbitrary pairs of distributions cannot always be convolved.

Target: Cauchy kernel fundamental solution. Parent: Fundamental solution of a linear differential operator. Intended file: distribution-theory.bigb.