Codex @codex  0

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.

Cells and attaching maps. Regard ${\mathbb{RP}}^k$ as the lines in ${\mathbb{R}}^{k+1}$ and include ${\mathbb{RP}}^{k-1}$ as the lines whose last coordinate is zero. Its complement consists of lines with a unique representative $(x_1,\dots ,x_k,1)$, and is therefore an open $k$-cell. Inductively this makes Real projective space a CW complex with one cell in each dimension from zero to $n$.

A characteristic map is obtained from the northern closed hemisphere of $S^k$: send a unit vector to the line it spans. Its interior maps homeomorphically onto the open cell, while its equator has the antipodal identification. Thus the $k$-cell is attached by the quotient mapwhich is the antipodal two-sheeted covering. This includes the two endpoints of the one-cell attaching to the zero-cell.

The cellular chain complex has $C_k^{\mathrm{cell}}=\mathbb{Z}$ for $0\le k\le n$ and zero otherwise. To compute its differential, follow the attaching map by collapse of the $(k-2)$-skeleton. The resulting map to ${\mathbb{RP}}^{k-1}/{\mathbb{RP}}^{k-2}\cong S^{k-1}$ has two local contributions. They differ by the mapping degree $(-1{)}^k$ of the antipodal map on $S^{k-1}$. With compatible cell orientations,For $k=1$ the two oriented endpoints cancel, giving the same formula. Consecutive differentials compose to zero, as required.

The mod-two cup products. Modulo two every cellular differential vanishes, so cellular cohomology gives a one-dimensional group in each degree $0,\dots ,n$. Let $a\in H^1({\mathbb{RP}}^n;{\mathbb{F}}_2)$ be the Poincare dual of a projective hyperplane. This class is nonzero: a projective line transverse to that hyperplane meets it once. Intersecting $j$ generic projective hyperplanes produces ${\mathbb{RP}}^{n-j}$, and the cup product of their Poincare duals is the Poincare dual of that intersection. In particular, $a^n$ evaluates to one on the mod-two fundamental class. Therefore every $a^j$, $0\le j\le n$, is nonzero, since otherwise multiplying it by $a^{n-j}$ would contradict $a^n\ne 0$. Dimension makes $a^{n+1}=0$. The mod-two cohomology ring of real projective space isFor $n=0$ this simply means the cohomology of a point.

The product with integral coefficients. The final product's coefficients are unstated; take $\mathbb{Z}$ as the default. Dualizing the cellular chain complex above giveswith all unlisted groups zero. The integral Künneth theorem has tensor terms with $i+j=q$ and Tor functor terms with $i+j=q+1$:For these finite free cellular complexes it splits additively, though not canonically. Both the tensor and Tor functor of two $\mathbb{Z}/2$ summands give $\mathbb{Z}/2$, so no order-four summands occur.

For an efficient count, write $F_A(t)$ for the free-rank polynomial and $T_A(t)$ for the number of $\mathbb{Z}/2$ summands in each degree. The integral Künneth torsion polynomial rule isThe last two factors record respectively the tensor contribution in summed degree and the Tor functor contribution one degree lower. The three factors haveThe first two give $F_{23}=1+t^3$ and $T_{23}=2t^2+t^3+t^4+t^5$. Multiplying by the third givesConsequently the integral cohomology of a product of finite real projective spaces in this case isIf the intended coefficients were instead ${\mathbb{F}}_2$, the Künneth theorem over a field gives the dimension polynomialThus the mod-two groups in degrees zero through nine are respectivelyand all other degrees vanish.

Euler number and self-intersection. Orient the rank-two normal bundle ${\nu }_S$ by the orientations of $S$ and $M$: the ordered tangent and normal spaces have the ambient orientation. Choose a smooth normal section $s$ transverse to the zero section. Its zeros are isolated, and their signed number is the evaluation $⟨e({\nu }_S),[S]⟩$ of the Euler class on the fundamental class. Scale the section sufficiently small to lie in a tubular neighborhood. Its graph is a push-off $S^{\prime }$ isotopic to $S$.

The intersections of $S$ with $S^{\prime }$ occur exactly at the zeros of $s$. In a positively oriented local splitting $TS\oplus {\nu }_S$, the basis formed from the tangent spaces to $S$ and to the graph has block matrixIts determinant is $\det Ds$, so the local intersection sign is precisely the local zero index of the section. Summing proves the Euler number equals self-intersection identity

The conic. The given map is the degree-two Veronese map $[u:v]\mapsto [u^2:uv:v^2]$ followed by the projective linear transformationThe Veronese map is injective with nonvanishing differential: on the chart $u\ne 0$ its coordinates are $[1:t:t^2]$, and on $v\ne 0$ they are $[s^2:s:1]$. It is therefore a smooth embedding, and $S$ is a smooth plane conic diffeomorphic to ${\mathbb{CP}}^1\cong S^2$. Its equation in the given coordinates is $z_0^2-z_2^2=4z_1^2$.

Let $L$ be a projective line, with its complex orientation. The cohomology ring of complex projective space gives $[L]\cdot [L]=1$ in ${\mathbb{CP}}^2$. The hyperplane $z_1=0$ meets $S$ at $[u:v]=[1:0]$ and $[0:1]$. The zeros of $uv$ are simple in the respective local coordinates, and complex intersections have positive signs. Hence $[S]\cdot [L]=2$, so $[S]=2[L]$. The self-intersection number and the normal Euler class arewhere $u\in H^2(S^2;\mathbb{Z})$ has $⟨u,[S^2]⟩=1$.

The boundary of the tubular neighborhood. Write $N=\nu (S)$ for the closed disk tubular neighborhood and $E=\partial N$ for its boundary. This distinguishes the disk neighborhood $N$ from the vector normal bundle ${\nu }_S$. The space $E$ is the oriented circle bundle of ${\nu }_S$, with Euler class $4u$. The Gysin sequence containsThe middle map is multiplication by four, yielding $H^1(E;\mathbb{Z})=0$ and $H^2(E;\mathbb{Z})=\mathbb{Z}/4$. The same Gysin sequence gives $H^0(E;\mathbb{Z})=H^3(E;\mathbb{Z})=\mathbb{Z}$. The total space is a closed connected oriented three-manifold, so Poincare duality gives

The exterior. Work first with $W={\mathbb{CP}}^2\setminus \mathrm{int}N$. A collar neighborhood of $\partial W$ shows that its interior, the requested ${\mathbb{CP}}^2\setminus N$, has the same homotopy equivalence type: push the boundary a small positive distance into the collar. The Excision theorem and Thom isomorphism theorem identifyOnly degrees two and four are nonzero, each a copy of $\mathbb{Z}$.

In degree four the map $H_4({\mathbb{CP}}^2;\mathbb{Z})\to H_4(N,E;\mathbb{Z})$ sends the ambient fundamental class to the relative fundamental class of $N$. Under the Thom isomorphism theorem this becomes $[S]$. Thus this map is multiplication by one with compatible orientations. The long exact sequence in homology givesso $H_4(W)=H_3(W)=0$.

In degree two, the Thom isomorphism theorem identifies the map $H_2({\mathbb{CP}}^2)\to H_0(S^2)$ with intersection against $[S]$. A projective line intersects the conic twice, so this map is multiplication by two. The long exact sequence in homology isIt yields $H_2(W)=0$ and $H_1(W)=\mathbb{Z}/2$. In degree zero the relative groups vanish, so $H_0(W)\cong H_0({\mathbb{CP}}^2)\cong \mathbb{Z}$. This computes the homology of the complement of a smooth conic:The normal Euler number $4$, the degree-two intersection map and the degree-four restriction map $1$ play different roles; distinguishing them explains why the boundary has first homology $\mathbb{Z}/4$ while the exterior has first homology $\mathbb{Z}/2$.

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.

Give the closed orientable surface its standard CW complex structure: one zero-cell, $2g$ one-cells, and one two-cell attached by the product of $g$ commutators. The cellular boundary of the two-cell is zero, since every edge occurs once with each orientation in that word; the one-cell boundaries are also zero. HenceHere we use the cellular homology theorem, identifying cellular and singular homology, and the universal coefficient theorem for cohomology: its exact sequence has terms $\mathrm{Ext}(H_{q-1},\mathbb{Z})$ and $\mathrm{Hom}(H_q,\mathbb{Z})$. All the homology groups here are free, so the Ext terms vanish.

The ring structure comes from Poincare duality and algebraic intersection number of curves on an oriented surface. For a closed oriented surface, cap product with its fundamental class identifies degree-one cohomology with degree-one homology; evaluating the cup product of two such classes equals the signed intersection number of their dual one-cycles. Choose the usual $g$ pairs of handle curves, each pair meeting positively once, and distinct pairs disjoint. Their dual classes can accordingly be named $a_1,b_1,\dots ,a_g,b_g$ so that, for the positive orientation class $\omega$,These formulas include squares. More generally, graded commutativity of the cup product kills every degree-one square here because $H^2$ is torsion-free. The unit generates $H^0$, and products involving $\omega$ and any positive-degree class vanish for dimensional reasons. These additive groups and multiplication rules completely describe the cohomology ring of a closed oriented surface, including $g=0$, when there are no degree-one generators. The intersection pairing is integral and unimodular, rather than merely nondegenerate over a field.

Suppose $F:{\Sigma }_g\to {\Sigma }_h$ has degree one. For degree-one classes $u,v$ on the target, naturality of the cup product and the definition of the degree of a map between oriented manifolds giveIf a nonzero $u$ had $F^{\ast }u=0$, nondegeneracy of the Poincare duality pairing would supply $v$ with nonzero right side, a contradiction. Thus $F^{\ast }$ injects the $2h$-dimensional real degree-one cohomology into the $2g$-dimensional source. Hence $g\ge h$. This is the cohomological injectivity of a degree-one map in this setting.

Conversely, for $g\ge h$, express ${\Sigma }_g$ as ${\Sigma }_h\#{\Sigma }_{g-h}$. Collapse the second punctured summand and the joining circle to a point. The quotient of the retained punctured summand by its boundary is homeomorphic to ${\Sigma }_h$, giving a continuous map to that surface. Its restriction to a small oriented disc away from the collapsing region is an orientation-preserving homeomorphism, and a point in this disc has exactly one preimage. The induced map on local top homology, and hence on the fundamental class, has coefficient $+1$. The map therefore has degree one. For $h=0$, the same construction is the familiar collapse of the complement of a disc to obtain $S^2$.

ConsequentlyThis proves both directions of the degree-one maps between closed oriented surfaces criterion. The case $g=h$ also admits the identity map.

Let $m=\dim M$ and $q=m-\dim Y$. Here a closed submanifold means compact and without boundary, which is the condition needed for a compactly supported dual class. The orientations of $M$ and $Y$ orient the normal bundle ${\nu }_Y$. Its Thom class is the unique relative class of degree $q$ restricting to the chosen orientation generator on every normal fiber. A tubular neighborhood identifies this class with one near $Y$ in $M$. Choose a compact normal disc neighborhood and extend the relative class to compactly supported cohomology of $M$. Equivalently, apply Poincare duality $H_c^q(M;\mathbb{Q})\cong H_{m-q}(M;\mathbb{Q})$ to the pushed-forward fundamental class of $Y$:This compactly supported dual of a compact submanifold is independent of the tubular neighborhood and represents intersection with $Y$. Fix the convention $⟨{\epsilon }_Y⌣a,[M]⟩=⟨i^{\ast }a,[Y]⟩$ in complementary degree. Its image in ordinary cohomology restricts to zero on $M\setminus Y$, because it comes from relative cohomology supported near $Y$. If a submanifold is only closed as a subset but is noncompact, its usual dual lies in ordinary cohomology instead; compact support is not asserted in that situation.

Now let $\Delta \subset N\times N$ be the diagonal, with the orientation inherited from $N$, and let $\delta ={\epsilon }_{\Delta }\in H^n(N\times N;\mathbb{Q})$. For each $i$, choose a basis $a_{i,j}$ of $H^i(N;\mathbb{Q})$ and its dual basis $b_{i,j}$ of $H^{n-i}(N;\mathbb{Q})$ satisfying $⟨a_{i,j}⌣b_{i,k},[N]⟩={\delta }_{jk}$. The Poincare duality pairing and the Künneth theorem give the cohomology class of the diagonalThe sign can be checked from the defining identityFor $|u|=i$, moving the degree-$i$ class in the second factor past $u$ contributes $(-1{)}^i$, and interchanging $b_{i,j}$ and $u$ contributes $(-1{)}^{i(n-i)}$; the displayed coefficient cancels their product. Equivalently its exponent is $ni$ modulo two.

The cohomological pushforward between closed oriented manifolds is characterized byfor complementary degrees. This follows immediately from its definition by Poincare duality and naturality of evaluation on pushed-forward homology. In degree $i$, the matrix trace of $f^{!}{g}^{\ast }$ on $H^i(N;\mathbb{Q})$ is consequentlyFor finite-dimensional maps between two possibly different vector spaces, $\mathrm{tr}(AB)=\mathrm{tr}(BA)$. Thus this matrix trace equals that of $g^{\ast }{f}^{!}$ on $H^i(M;\mathbb{Q})$. Pulling back the diagonal formula by $F=(f,g)$ and applying graded commutativity of the cup product now yieldsThis establishes the Lefschetz coincidence number with the requested sign convention.

If $f(m)\ne g(m)$ for every $m$, the map $(f,g)$ factors through $(N\times N)\setminus \Delta$. The diagonal class restricts to zero there, so its pullback is zero and $L(f,g)=0$. Taking the contrapositive proves the Lefschetz coincidence theorem:No transversality or isolated-coincidence assumption is needed. This cohomological argument also applies to continuous maps, which covers the last request.

For a map $f:M\to N$ of degree $d$, the adjoint identity above gives, for all complementary classes $a,b$ on $N$,Nondegeneracy therefore gives $f^{!}{f}^{\ast }=d\mathrm{id}$. Cyclic invariance of matrix trace proves the self-coincidence number of a map of nonzero degree formulaIn particular, ${\mathbb{CP}}^{2k}$ has one even-dimensional cohomology generator in each degree $0,2,\dots ,4k$, so $\chi ({\mathbb{CP}}^{2k})=2k+1$. HenceIf $g$ is homotopic to $f$, homotopy invariance of cohomology gives $g^{\ast }=f^{\ast }$ and thus $L(f,g)=L(f,f)$. The Lefschetz coincidence theorem supplies a point at which $f$ and $g$ agree.

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.

The ordinary Poincare duality statement here is for a compact manifold without boundary, of dimension $d$, oriented over the field $F$. Its fundamental class $[M{]}_F$ induces isomorphismsfor every $q$. For a manifold with boundary the appropriate statement is Poincare-Lefschetz duality with relative groups; the ordinary pairing need not be nondegenerate. Over a field, the universal coefficient theorem for cohomology identifies $H^{d-q}(M;F)$ with the full dual of $H_{d-q}(M;F)$. The cap-cup evaluation identity and duality therefore makea perfect pairing. Explicitly, a nonzero $a$ has a nonzero cap product, and a linear functional $b$ on its homology group takes a nonzero value on that product. The same argument in the other variable proves nonsingularity. On the whole graded cohomology, define $B(a,b)$ by taking the degree-$d$ part of $a⌣b$ before evaluation. For a nonzero component $a_q$ choose $b$ homogeneous of degree $d-q$ to pair nontrivially with it. All other components contribute zero in degree $d$. This proves that the Poincare duality pairing is a nondegenerate bilinear form on $H^{\ast }(M;F)$; it need not be symmetric on all degrees.

Put $r=2n+1$, so $r$ is positive and odd, and orient each $S^r\times S^r$ by its product orientation. The Künneth theorem gives integral cohomology $\mathbb{Z}$ in degrees zero and $2r$, ${\mathbb{Z}}^2$ in degree $r$, and zero elsewhere. Its two degree-$r$ generators $\alpha ,\beta$ have ${\alpha }^2={\beta }^2=0$, $\alpha \beta$ equal to its top orientation class, and $\beta \alpha =-\alpha \beta$.

For a connected sum of oriented manifolds, excision and the long exact sequence for deleting a ball show that deleting a ball removes the top homology class and leaves all lower positive homology groups unchanged. The boundary sphere represents zero in the punctured manifold: it is the boundary of its relative fundamental chain. In the Mayer–Vietoris sequence for the two punctured pieces joined along the separating sphere, the orientation class of the connected sum maps onto that sphere's class. The intermediate positive groups are consequently the direct sums of the groups of the two original manifolds. This also covers $r=1$, where the sphere is a circle and the boundary-class observation is necessary in the middle degree. Iterating and applying the universal coefficient theorem for cohomology givesHere $g\ge 1$; under the conventional extension $W_0=S^{2r}$ the same formula holds with a zero middle group.

Let $\omega$ be the top cohomological orientation class. The connected-sum pinch map to the wedge of the $g$ sphere products gives degree-one projections to each summand. Pull back the two factor classes from summand $i$ to obtain ${\alpha }_i,{\beta }_i$. Their product is $\omega$, since the projection has degree one. Classes from different wedge summands have zero positive-degree cup products, and graded commutativity of the cup product supplies the reversed sign. Thus the full cohomology ring of a connected sum of odd-dimensional sphere products is the graded free abelian group just displayed with multiplicationThe unit is $1$, and $\omega$ times any positive-degree class is zero by dimension. This describes all products, including ${\omega }^2=0$.

The smooth involution is a diffeomorphism, since it is its own inverse. Its fixed set is closed; discreteness and compactness therefore make it finite. The supplied positivity of $\det (I-D{f}_x)$ makes every fixed point nondegenerate with local index $+1$. The Lefschetz-Hopf fixed-point theorem then givesIndeed the degree-zero and top-degree traces are both one, because the manifold is connected and $f$ preserves its orientation; the middle degree is odd.

On $V=H^r(W_g;\mathbb{R})$, the form $\Omega (a,b)=⟨a⌣b,[W_g]⟩$ is a nondegenerate alternating bilinear form by Poincare duality and the oddness of $r$. Thus it is a symplectic vector space of dimension $2g$. Naturality and orientation preservation show that $T$ preserves $\Omega$, and $T^2=I$. For the eigenspaces of a symplectic involution, write $V=V_{+}\oplus V_{-}$: the polynomial $(t-1)(t+1)$ has distinct roots over $\mathbb{R}$. If $v_{+}\in V_{+}$ and $v_{-}\in V_{-}$, then $\Omega (v_{+},v_{-})=\Omega (T{v}_{+},T{v}_{-})=-\Omega (v_{+},v_{-})$, so the two vector subspaces are orthogonal. Each restricted form is nondegenerate, since a vector annihilating its own vector subspace also annihilates the other and hence all of $V$. Their dimensions are therefore even, say $\dim V_{+}=2a$, $\dim V_{-}=2b$, with $a+b=g$. ConsequentlyThis proves the fixed-point congruence for an involution on an odd-sphere connected sum:

All groups in the first calculation have integral coefficients. Let $U=S^n\setminus j(W)$ and $V=S^n\setminus \mathrm{int}j(W)$. A two-sided tubular neighborhood of $j(\partial W)$ identifies $V$ as a compact manifold with boundary, with interior $U$. Pushing its boundary inward in a collar neighborhood gives a homotopy equivalence $U\simeq V$. After thickening across the same collar, excision identifies the relative groupsThe orientation of $S^n$ gives $W$ an orientation. The Poincare-Lefschetz duality isomorphism therefore givesEach component of $W$ has boundary: a closed component would have open image by the local embedding condition and closed image by compactness, hence would occupy the whole connected sphere, leaving no room for the components with boundary. In particular $U$ is nonempty, and $H^n(W)=0$.