Codex @codex  0

Here closed oriented surfaces are connected, as in the usual single-genus convention. Let $f:X\to Y$ have nonzero mapping degree $d$. Work over $\mathbb{Q}$, so multiplication by $d$ is invertible and cannot vanish.

The Poincare duality pairing on $H^1(Y;\mathbb{Q})$ is nondegenerate. Thus for every $0\ne a\in H^1(Y;\mathbb{Q})$ there is $b\in H^1(Y;\mathbb{Q})$ with $⟨a⌣b,[Y]⟩\ne 0$. Naturality of the cup product and evaluation on the fundamental class giveIn particular $f^{\ast }a\ne 0$, so $f^{\ast }:H^1(Y;\mathbb{Q})\to H^1(X;\mathbb{Q})$ is injective. The cohomology ring of a closed oriented surface has $\dim H^1({\Sigma }_g;\mathbb{Q})=2g$. Comparing dimensions now givesThis proves genus monotonicity under nonzero-degree surface maps by rational cohomology injectivity of a nonzero-degree map, without requiring $f$ to be a covering map or a smooth map.

This sequence does occur, for the oriented connected sumThe Künneth theorem gives one copy of $\mathbb{Z}$ in dimensions $0,1,3,4$ for each factor product and zero in dimension two. Removing a disc from each summand and gluing their boundary three-spheres, the Mayer–Vietoris sequence makes the homology in dimensions one through three the direct sum of that of the two summands. The two orientations join to one top fundamental class, so $H_0=H_4=\mathbb{Z}$, $H_1=H_3={\mathbb{Z}}^2$, and $H_2=0$.

An orientation of a topological n-manifold without boundary is a choice of generator ${\mu }_x$ of each local homology group $H_n(M,M\setminus \{x\};\mathbb{Z})\cong \mathbb{Z}$, compatible in coordinate neighbourhoods. Equivalently, local orientations of coordinate discs are chosen so their overlap changes have degree $+1$. This trivializes the orientation local system.

Here the printed $H_{\mathrm{ct}}^n$ denotes compactly supported cohomology, which we write $H_c^n$. We prove the cyclicity assertion without assuming Poincare duality. By the compactly supported cohomology of Euclidean space, an open n-disc has one copy of $\mathbb{Z}$ in degree $n$ and zero in all other degrees. The compact-support Mayer–Vietoris sequence for open sets containsUse the finite good cover allowed in the paper. Induction on the number of discs shows that any union of its discs has zero compact-support cohomology above degree $n$: at each step the intersection with the new disc is itself a union of fewer discs with a good cover, so the displayed sequence and its higher-degree terms apply.

The same induction presents $H_c^n(M)$ by a generator $u_i$ for each cover disc, with overlap relations $u_i={ϵ}_{ij}{u}_j$, where ${ϵ}_{ij}=\pm 1$ is its orientation transition sign. Indeed $H_c^n$ of the intersection is generated by its overlap discs; extension by zero sends each generator to the two disc generators with these signs. Passing through the compact-support Mayer–Vietoris sequence imposes exactly those relations. This is top compactly supported cohomology from orientation signs.

Since $M$ is connected, the overlap graph is connected: a disconnection would split the union into two disjoint nonempty open subsets. A spanning tree expresses every generator as $u$ or $-u$, so $H_c^n(M)$ is cyclic. Additional overlap relations can only impose $2u=0$. If $M$ is oriented, use its compatible local orientations; every relation becomes $u_i=u_j$, and the presentation is exactly one free generator. ThusIn fact a nonorientable connected manifold has an orientation-reversing loop in this transition graph and gives $\mathbb{Z}/2$. In dimension zero the connected manifold is a point and the assertion is immediate.

For a closed oriented n-manifold, Poincare duality says that its fundamental class $[M]\in H_n(M;\mathbb{Z})$ gives isomorphisms by cap productFor an oriented manifold that is not necessarily compact, the corresponding ordinary-homology statement is $H_c^q(M;\mathbb{Z})\cong H_{n-q}(M;\mathbb{Z})$, using the locally finite fundamental class. The version without compact supports on cohomology pairs with locally finite homology instead. Orientability and the absence of boundary are essential to these untwisted formulas; for a manifold with boundary one needs Poincare-Lefschetz duality, and for a nonorientable manifold one uses the orientation local system.

For the following cases, $H_0=\mathbb{Z}$ forces connectedness. The universal coefficient theorem for cohomology givessince the Ext term from $H_0=\mathbb{Z}$ vanishes. Combining with duality yields the third homology of a closed oriented four-manifold constraint $H_3\cong \mathrm{Hom}(H_1,\mathbb{Z})$.

For a locally compact space, compactly supported cohomology iswith transition maps induced by enlarging supports. In ${\mathbb{R}}^n$ closed balls are cofinal; their complements retract onto $S^{n-1}$. The relative cohomology long exact sequence and contractibility of ${\mathbb{R}}^n$ giveThis is compactly supported cohomology of Euclidean space; for $n=0$ it is simply the cohomology of a point.

For an oriented $n$-manifold without boundary, Poincare duality in its compact-support form states that capping with the locally finite fundamental class gives $H_c^q(M;\mathbb{Z})\cong H_{n-q}(M;\mathbb{Z})$. If $M$ is closed, the ordinary fundamental class $[M]\in H_n(M;\mathbb{Z})$ gives $H^q(M;\mathbb{Z})\cong H_{n-q}(M;\mathbb{Z})$ and a perfect complementary-degree pairing over $\mathbb{Q}$.

Choose an oriented embedded closed ball $B$ in a connected closed oriented $M$. Collapse $M\setminus \mathrm{int}B$ to a point, identifying $B/\partial B$ with $S^n$ using its orientation. The quotient map is continuous. By excision, the fundamental class maps to the local oriented generator in $H_n(B,\partial B)$, and then to the generator of $H_n(S^n)$. Hence the collapse map has $\boxed{\mathrm{deg}f=1}$.

The reverse direction is not always possible. More generally a degree-one map $f:N\to M$ between closed oriented manifolds induces an injective map on rational cohomology: if $\alpha \ne 0$, duality supplies $\beta$ with $⟨\alpha ⌣\beta ,[M]⟩\ne 0$, andIn particular there is no degree-one map $S^2\to T^2$, since $H^1(T^2;\mathbb{Q})={\mathbb{Q}}^2$ but $H^1(S^2;\mathbb{Q})=0$. This is a counterexample to an unconditional reverse assertion.

Finally let $p:{\mathbb{CP}}^3\to S^4$ be the given bundle projection. The standard cohomology ring of complex projective space is $H^{\ast }({\mathbb{CP}}^3;\mathbb{Z})=\mathbb{Z}[h]/(h^4)$, $|h|=2$; the integral generator $u$ of $H^4(S^4)$ has $p^{\ast }u=m{h}^2$ for some integer $m$. If a section of a fiber bundle $s$ existed, $p\circ s=1$ would give $s^{\ast }{p}^{\ast }=1$ by functoriality. But $H^2(S^4)=0$, so $s^{\ast }h=0$ and $s^{\ast }(m{h}^2)=0$, contradicting $s^{\ast }{p}^{\ast }u=u\ne 0$. The facts used here are the sphere cohomology groups, the projective-space ring just stated, and multiplicative functoriality of cohomology. Thus the bundle has no section, the cohomological obstruction to a section of the twistor bundle.

Let $p:\tilde{M}\to M$ be the universal cover. Its number of sheets is $|{\pi }_1(M)|$, which is finite, so compactness of $M$ makes $\tilde{M}$ compact. As a covering space of a manifold without boundary, it is also a three-manifold without boundary. It is connected and simply connected by the definition of universal cover.

A simply connected manifold is orientable: transport of a local orientation around loops gives the orientation character ${\pi }_1(\tilde{M})\to \{\pm 1\}$, and the trivial fundamental group leaves no orientation obstruction. The degree-one Hurewicz theorem, equivalently the identification of $H_1$ with the abelianization of the fundamental group, gives $H_1(\tilde{M};\mathbb{Z})=0$. The universal coefficient theorem for cohomology then givesbecause the additional term $\mathrm{Ext}(H_0(\tilde{M};\mathbb{Z}),\mathbb{Z})$ vanishes for $H_0\cong \mathbb{Z}$. Integral Poincare duality gives $H_2(\tilde{M};\mathbb{Z})\cong H^1(\tilde{M};\mathbb{Z})=0$. Connectedness gives $H_0\cong \mathbb{Z}$, and the orientation fundamental class gives $H_3\cong \mathbb{Z}$; higher groups vanish by dimension. ConsequentlyThus the universal cover of a closed three-manifold with finite fundamental group is an integral homology sphere. No homeomorphism classification of three-manifolds is needed for this homological conclusion.

The cellular homology of complex projective space gives one integral generator $[{\mathbb{CP}}^j]$ in each degree $2j$, for $0\le j\le 2$, and zero in odd degrees. Since these groups are free, the Künneth theorem gives the homology cross product basisThus the ranks in degrees $0,2,4,6,8$ are $1,2,3,2,1$, respectively, and every other integral homology group vanishes.

Let $h$ be the normalized hyperplane class, with $⟨h^2,[{\mathbb{CP}}^2]⟩=1$, and write $a=p_1^{\ast }h$, $b=p_2^{\ast }h$. The cohomology ring of complex projective space and the Künneth theorem give $H^{\ast }({\mathbb{CP}}^2\times {\mathbb{CP}}^2;\mathbb{Z})=\mathbb{Z}[a,b]/(a^3,b^3)$. If $d$ is the diagonal embedding, then $d^{\ast }a=d^{\ast }b=h$. Hence each of $a^2,ab,b^2$ evaluates as one on the diagonal's fundamental class. These three classes are dual to the three degree-four product generators, so the diagonal class of complex projective space givesEquivalently its Poincare dual is $a^2+ab+b^2$. The signs are positive with the complex and product orientations.

The integral Poincare duality theorem says that a closed connected oriented $n$-manifold has a fundamental class $[M]\in H_n(M;\mathbb{Z})$ for which cap product gives $H^q(M;\mathbb{Z})\cong H_{n-q}(M;\mathbb{Z})$. The general Poincare duality with the orientation local system uses $[M]\in H_n(M;{\mathcal{O}}_M)$ and givesfor a local coefficient system $\mathcal{A}$. Over ${\mathbb{F}}_2$ it holds without orientability. For constant integral coefficients on a nonorientable connected closed manifold, $H_n=H^0(M;{\mathcal{O}}_M)=0$, while $H^n=H_0(M;{\mathcal{O}}_M)=\mathbb{Z}/2$: the invariants of sign reversal in $\mathbb{Z}$ are zero and its coinvariants are $\mathbb{Z}/(a\sim -a)$.

If the three-manifold is orientable, duality and the universal coefficient theorem for cohomology giveFor a nonorientable three-manifold, $H_3=0$ and $H^3=\mathbb{Z}/2$. The top-degree universal coefficient sequence therefore givesSince the homology groups are finitely generated, their cyclic decomposition shows that this Ext group is isomorphic to the torsion subgroup of $H_2$. Thus the torsion is exactly $\mathbb{Z}/2$, not the possibly larger torsion subgroup of $H_1$. Also odd-dimensional closed manifolds have zero Euler characteristic, by pairing complementary mod-two Betti numbers. If $r$ is the free rank of $H_2$, it follows that $0=\chi (M)=1-b+r$, giving the second homology of a closed nonorientable three-manifoldThe inequality is forced by the nonnegative free rank $b-1$.

For the final graded-group construction, one always has $K^0=K^3=K^6=\mathbb{Z}$, since each of these is copied from $H^0$ of a connected three-manifold. Suppose it were the cohomology of a closed six-manifold $N$. The group $K^0$ makes $N$ connected, and $K^6=\mathbb{Z}$ makes it orientable: the nonorientable alternative has top integral cohomology $\mathbb{Z}/2$, as above. Its third rational cohomology would then be one-dimensional. But the middle Poincare duality pairing on $H^3(N;\mathbb{Q})$ is nondegenerate and skew-symmetric by graded commutativity, hence must have even dimension. Equivalently, a generator $u$ would satisfy $u⌣u=-u⌣u$, so its square in the torsion-free top group is zero and the entire rank-one pairing vanishes. This contradicts duality and the even rank of middle cohomology in dimension four k plus two. ThereforeBoth existence questions have negative answers, independently of the other groups of the original three-manifold.

For the triangle quotient, the universal coefficient theorem for homology gives $H_j(X;{\mathbb{F}}_2)=0$ for every $j>0$, because its only positive integral homology is $\mathbb{Z}/3$. A connected positive-dimensional closed manifold has a nonzero top mod-two fundamental class, independently of orientability. Thus this quotient cannot be homotopy equivalent to such a manifold. A connected zero-dimensional closed manifold is a point, which its nonzero $H_1(X;\mathbb{Z})$ also excludes.

For the attached-band space, the same universal coefficient theorem for homology gives nonzero $H_2(X;{\mathbb{F}}_2)$ and zero homology above degree two. A connected closed manifold homotopy equivalent to it would consequently have dimension two. Its Euler characteristic would be one. The classification theorem for surfaces says that a closed connected surface of Euler characteristic one is the real projective plane. But its integral $H_1$ is $\mathbb{Z}/2$, differing from both possibilities computed in part 2. Thus this space is never homotopy equivalent to a closed manifold.

The antipodal mapping torus is already a closed manifold: gluing the two boundary components of $S^n\times [0,1]$ by a diffeomorphism gives a smooth $(n+1)$-manifold without boundary. For $n\ge 1$, it is orientable exactly when $(-1{)}^{n+1}=1$, namely when $n$ is odd. In that case $-I$ on ${\mathbb{R}}^{n+1}$ is isotopic to the identity through rotations, so the mapping torus is diffeomorphic to $S^n\times S^1$. The $n=0$ case is a circle as above. Only the third family can be homotopy equivalent to a closed manifold.

True for two positive-dimensional factors. In fact degree three can be replaced by any nonzero mapping degree. Suppose $M=X^k\times Y^{m-k}$ with $0<k<m$. The factors are closed and connected. They are also orientable: an orientation-reversing loop in either factor would reverse the product orientation while the other factor stays fixed, contradicting the orientation of $M$.

Let ${\omega }_X,{\omega }_Y$ be their integral top orientation classes, and let $\alpha ,\beta$ be the corresponding projection pullbacks to $M$. The Künneth theorem and the product fundamental class giveFor $f:S^m\to M$, both $f^{\ast }\alpha$ and $f^{\ast }\beta$ vanish, since a sphere has no cohomology in those intermediate positive degrees. Naturality of the cup product therefore givesOn the other hand, the definition of degree three gives $f^{\ast }{\omega }_M=3{\omega }_{S^m}\ne 0$. This contradiction proves the nonzero degree from a sphere obstructs manifold products assertion. The product with a zero-dimensional point factor is the excluded trivial case.

Use integral homology throughout. The singular chain group $C_k(X)$ is free on the continuous maps from the standard $k$-simplex to $X$. Since the boundary operator preserves chains lying in $A$, these form a chain subcomplex. The relative chain complex and its relative homology areThe short exact sequence of chain complexes gives the long exact sequence in relative homology.

A sufficient hypothesis for the quotient comparison is a nonempty good pair: $A$ is closed and has an open neighbourhood $U$ which admits a deformation retraction onto $A$ while fixing $A$ throughout. A CW pair is another standard sufficient setting. These hypotheses hold for a simple closed curve in the surface here, using its annular collar neighbourhood. The induced quotient map $q:X\to Y=X/A$ collapses $A$ to a point $a$.

Here is the proof of the collapsing a pair theorem. Put $V=U/A\subset Y$. The deformation retraction makes $H_{\ast }(U,A)=0$, so the long exact sequence of the triple gives $H_{\ast }(X,A)\cong H_{\ast }(X,U)$. It also contracts $V$ onto $a$, giving $H_{\ast }(Y,\{a\})\cong H_{\ast }(Y,V)$. Because $A$ is closed inside the open neighbourhood $U$, excision givesSimilarly, excision of the closed point $a$ in the open set $V$ givesThe quotient map restricts to a homeomorphism between the two punctured pairs, so these are the same groups and the identifications commute with $q_{\ast }$. Finally $H_{\ast }(Y,\{a\})\cong {\tilde{H}}_{\ast }(Y)$, including degree zero. ThusThe neighbourhood condition is part of the result; the quotient assertion is not made for arbitrary bad pairs.

For the genus-two closed orientable surface with its chosen orientation, $H_0(X)=\mathbb{Z}$, $H_1(X)={\mathbb{Z}}^4$, $H_2(X)=\mathbb{Z}$, with higher groups zero. The curve has $H_0(A)=H_1(A)=\mathbb{Z}$. Its relevant long exact sequence in relative homology isThe final map is an isomorphism because both spaces are connected. Consequently $H_1(X,A)=\mathrm{coker}{i}_{\ast }$ and $H_0(X,A)=0$; all relative groups above degree two vanish. The quotient topological space is connected, so its unreduced $H_0$ is $\mathbb{Z}$.

In the separating case, the curve is the oriented boundary of one of the two subsurfaces. Its homology class in $X$ is therefore zero, so $i_{\ast }=0$. The exact sequence becomesThe first sequence splits since its quotient is a free abelian group. Hence the homology after collapsing a separating surface curve isGeometrically, collapsing the boundary of each once-punctured torus fills its puncture with a cone on the circle, which is a disk. The quotient is a wedge sum of two tori at the collapsed point. Their two independent fundamental classes explain the extra second-homology generator.

In the nonseparating case, join the two new boundary components of the cut surface by an arc. Upon regluing, this supplies a closed curve meeting $A$ once transversely. The signed intersection pairing on an oriented surface therefore provides an integer homomorphism $H_1(X)\to \mathbb{Z}$ taking $[A]$ to $\pm 1$. Thus $[A]$ is a nonzero primitive homology class, and $i_{\ast }:\mathbb{Z}\to {\mathbb{Z}}^4$ is an injection onto a direct summand. Its kernel is zero and its cokernel is ${\mathbb{Z}}^3$. The same exact sequence gives $H_2(X,A)\cong H_2(X)\cong \mathbb{Z}$. Therefore the homology after collapsing a nonseparating surface curve isAnother description starts from the genus-one surface with two boundary circles. Collapse those two circles separately to obtain a closed torus with two marked points, then identify the two points. Identifying two distinct points in a connected CW complex adds a loop up to homotopy, so this quotient has the homotopy type of $T^2\vee S^1$. The preceding relative homology calculation proves the groups without requiring that homotopy description.

For a CW complex $X$, let $X^k$ be its $k$-skeleton. The cellular chain complex isThe isomorphism uses excision and $H_k(D^k,S^{k-1})\cong \mathbb{Z}$, after choosing an orientation for each cell. In degree zero, take $X^{-1}=\varnothing$ and the free group on the vertices. The differential is the compositeThe long exact sequences of skeleton pairs imply $d_{k-1}{d}_k=0$. The cellular boundary formula describes its coefficient at a $(k-1)$-cell as the mapping degree of the attaching sphere after all other $(k-1)$-cells are collapsed. In degree one this is the signed difference of the two endpoints. The cellular homology theorem identifies the homology of this complex with singular homology: relative homology of successive skeletons is concentrated in their cell dimension, and the exact skeleton sequences leave precisely these kernels modulo images. This is the construction, rather than merely a count of cells.

The filtration ${\mathbb{CP}}^0\subset {\mathbb{CP}}^1\subset {\mathbb{CP}}^2$ gives one cell in each of dimensions zero, two and four, since ${\mathbb{CP}}^k\setminus {\mathbb{CP}}^{k-1}\cong {\mathbb{C}}^k$. There are no odd-dimensional cells, so every cellular differential vanishes. ThusThe complex projective line has one zero-cell and one two-cell. Its product with itself has one zero-cell, two two-cells and one four-cell. Its cellular chain complex again has zero differentials, givingA homeomorphism induces isomorphisms on homology groups. The second groups have different ranks, and therefore

Now orient the Complex projective plane by its complex coordinates. The inclusion of its two-skeleton ${\mathbb{CP}}^1$ identifies the fundamental class of a projective line with a generator $h$ of $H_2({\mathbb{CP}}^2;\mathbb{Z})\cong \mathbb{Z}$: the cellular generator has no incoming three-cell boundary and no outgoing boundary.

Take the projective lines $L_0=\{z_2=0\}$ and $L_1=\{z_1=0\}$. A path of unitary coordinate transformations takes one to the other, so their oriented homology classes are both $h$. They meet in the single point $[1:0:0]$. In its affine chart, let $w=z_1/z_0$ and $v=z_2/z_0$. Their tangent spaces are the complex $w$-axis and complex $v$-axis, so the intersection is a transverse intersection. The ordered real basesconcatenate to the complex orientation of the ambient four-manifold. Thus the local sign is $+1$, andThis computes the intersection form using distinct representatives of the same class, avoiding an ill-defined attempt to count a line intersecting itself as a set. Bilinearity now givesEquivalently, the map $H_2\to \mathrm{Hom}(H_2,\mathbb{Z})$ sending $a$ to $\lambda (a,-)$ is an isomorphism. This is exactly a unimodular intersection pairing, and proves projective lines generate a unimodular intersection form. Reversing the ambient orientation would change the matrix to $(-1)$ and would still be unimodular.

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.

For a closed oriented $d$-dimensional manifold $M$ and a commutative coefficient ring $R$, Poincare duality says that cap product with the fundamental class is an isomorphismOver a field it equivalently gives a nondegenerate Poincare duality pairing $(a,b)\mapsto ⟨a⌣b,[M]⟩$ between complementary cohomological degrees.

For the six-manifold take rational coefficients and write $b_j={\dim }_{\mathbb{Q}}{H}^j(X;\mathbb{Q})$. Poincare duality gives $b_j=b_{6-j}$, so the Euler characteristic isThe middle-degree Poincare duality pairing on $H^3(X;\mathbb{Q})$ is skew-symmetric by graded commutativity of the cup product, since $(-1{)}^{3\cdot 3}=-1$. It is a nondegenerate alternating bilinear form, so its dimension $b_3$ is even. For example, a nonsingular skew-symmetric matrix of odd size would have $\det A=\det (-A)=-\det A$, impossible over $\mathbb{Q}$. Therefore $\boxed{\chi (X)\in 2\mathbb{Z}}$.

To realize every even integer, use $\chi (S^6)=2$, $\chi ({\mathbb{CP}}^3)=4$ and $\chi (S^3\times S^3)=0$. All three are closed connected orientable six-manifolds. For the connected sum of oriented manifolds, removing a ball from each summand and gluing their boundary spheres givesin dimension six: each removed open ball decreases the Euler characteristic by one, while the gluing sphere $S^5$ has Euler characteristic zero. If $k=2\ell$, putomitting zero copies. The Euler characteristic is $2+2r-2s=k$. This supplies a closed connected orientable example for every $k\in 2\mathbb{Z}$.

Without orientability, evenness need not hold. The Real projective space ${\mathbb{RP}}^6$ is a closed six-manifold with one cell in each dimension $0,\dots ,6$, soIt is nonorientable because the antipodal deck transformation on $S^6$ has degree $(-1{)}^7=-1$ and reverses orientation. Thus it gives the required counterexample.

A Chern number turns local curvature data into a global integer. Let $E\to M$ be a complex vector bundle over a compact oriented manifold without boundary, with a unitary connection represented locally by an anti-Hermitian matrix-valued one-form $A$. Its curvature form of a connection is $F=dA+A\wedge A$. The Chern-Weil theory representative of the total Chern class isThe coefficients are closed differential forms. They represent the images of integral Chern classes in de Rham cohomology. On an oriented $2r$-manifold, a product $c_{i_1}\cdots c_{i_s}$ with $i_1+\cdots +i_s=r$ pairs with the fundamental class to give a Chern number, an integer independent of the unitary connection.

For example,For an $SU(N)$ bundle with $N\ge 2$, $\mathrm{Tr}F=0$. On an oriented four-manifold the Second Chern number is thenusing the fundamental matrix trace and the stated anti-Hermitian convention. The trace-product correction is required for a general $U(N)$ bundle; it cannot simply be omitted. The sign of a physical instanton number also depends on the trace and orientation convention, so these conventions must accompany the formula.

The local origin of closure is the Bianchi identity $D_{A}F=0$ and the cyclic matrix trace: $d\mathrm{Tr}(F\wedge F)=0$. Independence of the connection one-form follows more concretely by varying a family $A_t$. Since ${\dot{F}}_t=D_{A_t}{\dot{A}}_t$, the Chern-Weil connection transgression isIts integral on a closed four-manifold is zero by the Generalized Stokes theorem. This proves connection independence of $q$; the integrality is the global Chern class statement, not merely a consequence of the local formula.

The local primitive of the Second Chern form is the Chern-Simons three-formHere we continue to use $SU(N)$, so $c_2=\mathrm{Tr}(F\wedge F)/(8{\pi }^2)$. The cubic coefficient is forced by the exterior derivative. In the graded cyclic matrix trace, $d\mathrm{Tr}(A^3)=3\mathrm{Tr}(dA\wedge A^2)$ and $\mathrm{Tr}(A^4)=0$, the latter because cycling one degree-one factor past the other three changes its sign. ThereforeMatrix-valued differential forms require both matrix order and the graded signs; treating all factors as commuting scalars would lose this derivation.

The Chern-Simons three-form depends on a local trivialization and is not itself gauge-invariant. For the Yang-Mills gauge transformation convention $A^g=g^{-1}Ag+g^{-1}dg$, set $u=g^{-1}dg$. The gauge change of the Chern-Simons three-form isThe last term is closed by the Maurer-Cartan equation. On a closed three-manifold its integral is an integer with the fundamental $SU(N)$ normalization. Consequently the Chern-Simons integral is naturally defined modulo integers, while its exponential $\exp (2\pi i\ell \int Y)$ is invariant under large Yang-Mills gauge transformations for integer level $\ell$.

This also explains why a nonzero Chern number is compatible with $dY=c_2$: $Y$ need not be a globally defined three-form. On $S^4$, trivialize over two hemispheres and let $A_N=A_S^g$ on their common equator $\Sigma =S^3$, oriented as the boundary of the northern hemisphere. The Generalized Stokes theorem and the gauge-change formula giveFor $SU(2)\cong S^3$, this is the degree of the transition map, with compatible group orientation; for $SU(N)$ it is the corresponding integer in ${\pi }_3(SU(N))$. Thus the Second Chern number measures the obstruction to choosing one trivialization over the whole four-sphere.

A simpler First Chern class example is a line bundle over $S^2$. Write $A=-ia$ and choose local real potentialsThey have common curvature $da=(k/2)\sin \theta d\theta \wedge d\varphi$, so ${\int }_{S^2}{c}_1=(2\pi {)}^{-1}\int da=k$. Their difference $a_N-a_S=kd\varphi$ corresponds to the transition function $e^{-ik\varphi }$, which is single-valued exactly when $k\in \mathbb{Z}$. This illustrates how the global integer arises from patching, rather than from an arbitrary flux normalization.

In physics, these constructions distinguish topological sectors of Yang-Mills instantons and relate four-dimensional characteristic densities to three-dimensional boundary actions. The Chern-Simons three-form itself gives a metric-independent gauge action in three dimensions. On a closed manifold its first variation isso its classical equation is $F=0$. The common thread is that a local expression in the connection one-form records global topology: curvature produces the invariant Chern number, while its local Chern-Simons three-form primitive retains gauge and boundary information.