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On a sphere centred at the origin, an antipodal pair is the unordered pair $\{x,-x\}$. The antipodal map exchanges its two points.

The Borsuk-Ulam theorem says that recording $n$ continuous real-valued measurements on an $n$-dimensional unit sphere must give identical records somewhere on opposite sides. The measurements can vary wildly; continuity alone forces a coincidence at an antipodal pair.

Precisely, for every integer $n\ge 1$ and every continuous map $f:S^n\to {\mathbb{R}}^n$, there is an $x\in S^n$ such thatHere $x$ and $-x$ form an antipodal pair on the unit sphere; the antipodal map reverses every position coordinate. The target is the Euclidean space ${\mathbb{R}}^n$. Both points have the same entire output, so every one of the $n$ measurements agrees at this common pair. See Hatcher, Algebraic Topology, Corollary 2B.7, p. 176.

Circle example. Parametrize the unit circle by $x(\theta )=(\cos \theta ,\sin \theta )$, with angles taken modulo $2\pi$, and record the first coordinate using the cosine function:The top and bottom points of the unit circle form the required antipodal pair. Their first coordinates agree, while their second coordinates are $1$ and $-1$.

Sphere example. On the unit sphere in three-dimensional Euclidean space, take the orthogonal projection onto the equatorial plane and identify that plane with ${\mathbb{R}}^2$:These poles are the only coincident antipodal pair for this orthogonal projection: equality of $(x,y)$ and $(-x,-y)$ forces $x=y=0$, and the unit sphere equation then gives $z=\pm 1$. The Borsuk-Ulam theorem guarantees an antipodal pair for every continuous map into ${\mathbb{R}}^2$, even when no convenient projection formula is available.

Equal output need not mean equal position coordinates. Keeping all three coordinates gives a continuous map $i:S^2\to {\mathbb{R}}^3$ defined by $i(x,y,z)=(x,y,z)$, the restriction of the identity map of ${\mathbb{R}}^3$. Its outputs at an antipodal pair are always distinct, since $u=-u$ would force $u=0$, which is outside the unit sphere. Thus replacing the target ${\mathbb{R}}^2$ by ${\mathbb{R}}^3$ destroys this guarantee.

Why the circle case follows from the intermediate value theorem. For an arbitrary continuous map $f:S^1\to \mathbb{R}$, write $F(\theta )=f(x(\theta ))$ using the unit circle parametrization above. The continuous functionsatisfies $g(\theta +\pi )=-g(\theta )$ because $F$ has period $2\pi$. If $g(0)=0$, the first antipodal pair already works. Otherwise $g(0)$ and $g(\pi )$ have opposite signs, and the intermediate value theorem supplies a zero between them. At that zero, the two antipodal values of $f$ agree. In the cosine function example, $g(\theta )=2\cos \theta$. This short argument handles only $n=1$. See Hatcher, Algebraic Topology, discussion of Theorem 1.10, p. 32.

An ideal weather interpretation. Model the surface of Earth as a sphere. At one fixed instant, suppose temperature $T$ and pressure $P$ are globally defined continuous scalar measurements on that sphere. Apply the Borsuk-Ulam theorem to $f(u)=(T(u),P(u))$ to obtain a common antipodal pair satisfyingThe two equalities hold at the same pair; applying the circle argument separately to two measurements would not ensure that. This is a conclusion about the ideal continuous model, with both measurements taken at the same instant, rather than an assertion that an exactly matching pair has been empirically measured. Matoušek gives this motivating interpretation and explicitly cautions that physical temperature need not be continuous.^[ref]

Equivalent formulations. A continuous map $h:S^n\to {\mathbb{R}}^n$ is odd when $h(-x)=-h(x)$, extending the symmetry of an odd function. The Borsuk-Ulam theorem equivalently says every such map has a zero. For a general $f$, the difference $h(x)=f(x)-f(-x)$ has this symmetry; conversely, equal antipodal outputs of an odd map must be zero. Another equivalent formulation is that no odd map between spheres $S^n\to S^{n-1}$ exists. Normalizing a nonvanishing odd map by its Euclidean norm would produce such an odd map between spheres. These formulations are useful starting points for algebraic topology; the general proof is beyond the circle argument here.^[ref]

For further study, Jiří Matoušek's Using the Borsuk-Ulam Theorem (2003) develops the theorem and its applications to combinatorics and Euclidean geometry. An early publication is Karol Borsuk's 1933 paper Drei Sätze über die n-dimensionale euklidische Sphäre.

The antipodal map on the two-sphere reverses orientation, so its mapping torus is a nonorientable closed three-manifold. The Wang sequence gives $H_0=H_1=\mathbb{Z}$, $H_2=\mathbb{Z}/2$ and $H_3=0$. The universal coefficient theorem for cohomology therefore gives the cohomology ring $\mathbb{Z}[t,u]/(t^2,tu,u^2,2u)$ with $|t|=1$, $|u|=3$. The base-circle class satisfies $t^2=0$ by pullback; the other positive-degree products vanish for dimension reasons.

The displayed one-form on the unit round sphere has zero integral on every closed geodesic. Indeed, the antipodal map pulls it back to its negative, and the second half of an oriented great circle is the antipodal image of the first. Nevertheless $d\theta =2x_{3}d{x}_3\wedge d{x}_1$ is nonzero on an open set, so the one-form is not a closed differential form and cannot be an exact differential form. Thus a curvature-free extension of one-form injectivity fails.

A nowhere-zero tangent vector field on $S^n$ can be normalized to a unit field $w(x)$ perpendicular to $x$. Then $\cos (\pi t)x+\sin (\pi t)w(x)$ is a homotopy from the identity to the antipodal map. Their mapping degrees are $1$ and $(-1{)}^{n+1}$, respectively, contradicting homotopy invariance of mapping degree when $n$ is positive and even. For $S^0$ the tangent spaces are zero, so every tangent field is zero directly.

If a continuous map $f:S^d\to S^d$ has no fixed point, the formula $H_t(x)=((1-t)f(x)-tx)/\Vert (1-t)f(x)-tx\Vert$ is a homotopy from $f$ to the antipodal map. The denominator could vanish only at $t=1/2$ and $f(x)=x$, which is excluded. The homotopy invariance of mapping degree therefore gives $\mathrm{deg}f=(-1{)}^{d+1}$.

For $n\ge 2$, the Wang sequence and the mapping degree $(-1{)}^{n+1}$ of the antipodal map give $H_0=H_1=\mathbb{Z}$. If $n$ is odd, $H_n=H_{n+1}=\mathbb{Z}$; if $n$ is even, $H_n=\mathbb{Z}/2$ and $H_{n+1}=0$. All other groups vanish. At $n=1$ the mapping torus is a torus with $H_1={\mathbb{Z}}^2$. At $n=0$, the two-point swapping mapping torus is a circle. Every example is already a closed manifold; for $n\ge 1$ it is orientable exactly when $n$ is odd.

Modulo two the antipodal map acts trivially on fiber cohomology. The Wang sequence supplies a class $y$ restricting to the generator of $H^2(S^2;{\mathbb{F}}_2)$. The Leray-Hirsch theorem with classes $1,y$ gives a basis $1,t,y,ty$, so $ty\ne 0$. Pullback from the circle gives $t^2=0$, while dimension gives $y^2=0$. This cohomology ring agrees with that of $S^1\times S^2$, although the original manifold is nonorientable.

An odd map between spheres is a continuous map $f:S^n\to S^m$ satisfying $f(-x)=-f(x)$. Thus it is equivariant for the two antipodal maps. It descends to a continuous map of Real projective spaces.

The outward-normal-first volume form on the unit sphere pulls back under the antipodal map by $(-1{)}^{n+1}$. For $n\ge 1$, the sphere is connected, and its orientation descends to the twofold quotient exactly when this deck transformation preserves it. Thus Real projective space is orientable precisely in odd positive dimensions; the zero-dimensional case is one point and is also orientable.

A vector field on the sphere means a tangent vector field. For $S^{2m}\subset {\mathbb{R}}^{2m+1}$ with $m\ge 1$, suppose $V(x)$ were continuous and nowhere zero. Normalize it to $w(x)=V(x)/\Vert V(x)\Vert$. Tangency means $⟨x,w(x)⟩=0$, and both vectors have norm one. Thereforehas squared norm ${\cos }^2(\pi t)+{\sin }^2(\pi t)=1$. It defines a homotopy through self-maps of the sphere from the identity to the antipodal map.

The identity has mapping degree $1$. The antipodal map on $S^n$ has degree $(-1{)}^{n+1}$: it is the boundary restriction of $-I$ on the oriented $(n+1)$-ball, and this linear map changes orientation by $\mathrm{sgn}\det (-I)=(-1{)}^{n+1}$. For $n=2m$ its degree is $-1$. Homotopy invariance of mapping degree would force $1=-1$, a contradiction. Thus every continuous tangent field on an even-dimensional sphere vanishes somewhere. For $S^0$ the tangent spaces are zero, so the conclusion holds directly. This proves the Hairy ball theorem using degree, rather than assuming it.

For a sphere self-map, the Lefschetz number is $L(f)=1+(-1{)}^n\mathrm{deg}f$. At degree zero this is $1$, so the Lefschetz fixed-point theorem supplies $x$ with $f(x)=x$.

Let $A:S^n\to S^n$ be the antipodal map, $A(z)=-z$, of degree $(-1{)}^{n+1}$. The composite $Af$ also has degree zero, hence Lefschetz number one, so it has a fixed point $y$. The equation $A(f(y))=y$ is $f(y)=-y$. ConsequentlyThe two points are distinct, since a unit vector is never its own negative.

The standard simplicial decomposition of a sphere $F^n$ is the boundary of the cross-polytope with vertices $\pm e_1,\dots ,\pm e_{n+1}$, identified radially with $S^n$. A face chooses distinct coordinate indices and one sign for each; it never contains an antipodal pair.

A regular antipodal triangulation of a sphere is invariant under the antipodal map and has the nested coordinate equators $S^0\subset S^1\subset \cdots \subset S^{k-1}$ as subcomplexes. In particular each equator separates its sphere into two triangulated hemispheres. These conditions are preserved by barycentric subdivision.

For a simplicial map $f:F\to F^n$, a positive alternating simplex of dimension $j$ has distinct image labelsThe opposite alternating pattern is negative; all other simplices, including collapsed images, are neutral.

To prove the antipodal alternating-simplex parity lemma, let $p_j$ count positive $j$-simplices in the equatorial $S^j$. There is exactly one positive vertex in $S^0$, so $p_0=1$. A $j$-simplex has an odd number of positive alternating facets precisely when it is positive or negative alternating. Indeed an alternating sequence has exactly one positive alternating deletion; a nonalternating sequence has either zero or two. For a collapsed image, a positive facet can occur only when exactly one label is repeated; deleting either copy gives two such facets.

Count incidences of positive $(j-1)$-faces with $j$-simplices in one closed hemisphere. Interior faces contribute twice and equatorial faces once. The count modulo two is therefore $p_{j-1}$. It is also the number of positive or negative $j$-simplices in that hemisphere modulo two. Because $f$ is antipodal, each alternating pair $\sigma ,-\sigma$ has one positive and one negative member, with one member in each hemisphere. ThusInduction gives odd $p_k$, proving the existence of a positive $k$-simplex. In particular, an antipodal simplicial map $F\to F^{k-1}$ is impossible: a positive $k$-simplex would require $k+1$ distinct coordinate labels but only $k$ are available.

The Borsuk-Ulam theorem states that every continuous map $g:S^k\to {\mathbb{R}}^k$ has some $x$ with $g(x)=g(-x)$. First rule out a continuous antipodal map $h:S^k\to S^{k-1}$ for $k\ge 1$. By uniform continuity, take a sufficiently fine regular barycentric subdivision so that endpoints of each edge have images less than $2/\sqrt{k}$ apart. Label a vertex by $\mathrm{sgn}(h_i)e_i$, where $i$ is the smallest index maximizing $|h_i|$. Since $h$ is unit length, this maximum is at least $1/\sqrt{k}$; two adjacent vertices cannot have opposite labels. The labels therefore define an antipodal simplicial map to $F^{k-1}$, contradicting the preceding result.

If the stated Borsuk-Ulam theorem failed, the mapwould be just such a continuous antipodal map. This proves the required version; $k=0$ is immediate.

For the Kneser conjecture, equivalently Lovász theorem on Kneser graphs, the claim isThe Kneser graph joins disjoint $r$-subsets of $[N]$. Let $d=N-2r+1$. Colour each $r$-set by its least element in $[d]$ when it has one, and otherwise by one final colour. A colour class indexed by $i$ is an intersecting family because all its members contain $i$; the final class lies in a $(2r-1)$-set and is also intersecting. This proves the upper bound $d+1$.

For the lower bound, first deduce the open-cover Lusternik-Schnirelmann-Borsuk theorem. An open cover $U_0,\dots ,U_d$ of $S^d$ admits a continuous partition of unity ${\varphi }_0,\dots ,{\varphi }_d$ subordinate to it. Apply the Borsuk-Ulam theorem to $({\varphi }_1,\dots ,{\varphi }_d)$. Equal values at $x,-x$ and ${\sum }_i{\varphi }_i=1$ make all values equal. At least one common value is positive, so some $U_i$ contains an antipodal pair.

Now suppose the $r$-sets could be coloured with only $d$ colours, each an intersecting family. Choose $N$ points $v_1,\dots ,v_N\in S^d$ such that every $d+1$ of their vectors are linearly independent. For example, normalize $(1,t_i,\dots ,t_i^d)$ for distinct real $t_i$; the Vandermonde determinant proves this property. The general-position hemisphere count says that at most $d$ points lie on any equator. Since $N-d=2r-1$, at least one of the opposite open hemispheres determined by $x,-x$ contains an $r$-set of the labelled points.

Let $U_i$ be the open set of directions whose positive open hemisphere contains an $r$-set of colour $i$. It contains no antipodal pair, as such a pair would give two disjoint sets of that colour. The compact set $K=S^d\setminus {\bigcup }_{i=1}^d{U}_i$ also contains no antipodal pair: otherwise each of the two opposite hemispheres would have at most $r-1$ points, contradicting the count $2r-1$.

If $K$ is nonempty, compactness gives a positive set distance between $K$ and $-K$; an open neighbourhood $U_0$ of radius less than half that set distance still has no antipodal pair. If $K$ is empty, take $U_0=\varnothing$. In either case $U_0,U_1,\dots ,U_d$ is an open cover with no member containing an antipodal pair, contradicting the Lusternik-Schnirelmann-Borsuk theorem. Hence at least $d+1$ colours are required.

Choose real constants $a_0<a_1<\cdots <a_m$ and define the smooth homogeneous quotientIt is unchanged by nonzero rescaling, so it is well defined on Real projective space. On the unit sphere its differential vanishes on the tangent space precisely when $Ax$ is proportional to $x$, where $A=\mathrm{diag}(a_0,\dots ,a_m)$. The entries are distinct, so the only critical projective lines are $p_i=[e_i]$. In the affine chart $x_i=1$, write $u_j=x_j/x_i$ for $j\ne i$. ThenIts Hessian at zero is diagonal with entries $2(a_j-a_i)$. Exactly $i$ are negative, and none vanish. Thus this is a Morse function with one critical point $p_i$ of index $i$ for every $0\le i\le m$.

Choose the quotient of the round metric on the unit sphere. The downward gradient flow has sphere equation ${\dot{x}}_j=-2(a_j-f(x))x_j$ and projective solutionNormalization to the unit sphere supplies the common positive factor. As $t\to +\infty$, the smallest index with nonzero coordinate dominates; as $t\to -\infty$, the largest dominates. ConsequentlyThese are its unstable manifolds and stable manifolds. If $j\ge k$, the tangent directions in $W^u(p_j)$ allow coordinates $0,\dots ,j$, and those in $W^s(p_k)$ allow $k,\dots ,m$, modulo the common projective line. Together they span the full tangent space at every intersection. If $j<k$ the intersection is empty. Thus the flow is a Morse-Smale gradient flow, with no generic perturbation needed.

Orient each unstable manifold using its affine coordinates. The Morse-Smale complex has $C_i=\mathbb{Z}{p}_i$; its differential counts oriented unparametrized connecting trajectories of index difference one. A trajectory from $p_i$ to $p_{i-1}$ has only coordinates $x_{i-1},x_i$ nonzero, and there are exactly two, distinguished by the sign of $x_{i-1}/x_i$. Each sign branch is a single orbit modulo time translation.

To calculate their signs rather than merely count two, use the characteristic closed disk of the unstable cell: the closed upper hemisphere of $S^i$ with $x_i\ge 0$, followed by projection onto ${\mathbb{RP}}^i$. Its interior is $W^u(p_i)$, and its boundary map is the antipodal double cover $S^{i-1}\to {\mathbb{RP}}^{i-1}$. After collapsing ${\mathbb{RP}}^{i-2}$, its degree on the $(i-1)$-cell is the sum of the degrees of the two sheets. Comparing the sheets by the antipodal map on $S^{i-1}$ gives relative orientation $(-1{)}^i$: it is induced by $-I$ on ${\mathbb{R}}^i$, whose determinant is $(-1{)}^i$. Thus the two trajectory signs add to $1+(-1{)}^i$, up to the choice of generator orientations. The projective quadratic Morse complex is thereforeIn particular ${\partial }_1=0$, as required for the connected one-dimensional skeleton. The signed trajectory count agrees with this attaching-map incidence calculation, so it computes integral rather than only mod-two homology.

For $m\ge 1$, taking kernels modulo images givesFor example, an odd interior-degree generator has zero outgoing differential but twice itself is the image from the next degree; an even positive-degree differential is injective. At the top there is no incoming differential, leaving a copy of $\mathbb{Z}$ exactly when $m$ is odd. For $m=0$ the space is one point and only $H_0=\mathbb{Z}$ remains. This recovers the cellular homology of real projective space by explicitly constructing its Morse-Smale complex and signs.

The space is the mapping torus of the antipodal map $A:S^2\to S^2$. Its mapping degree is $(-1{)}^3=-1$, so transporting a fiber orientation once around the base circle reverses it. $X$ is nonorientable.

On integral fiber homology, $A_{\ast }$ is the identity in degree zero and minus the identity in degree two. The Wang sequence, whose relevant map is $1-A_{\ast }$, therefore givesThe universal coefficient theorem for cohomology shifts the torsion contribution into degree three, because ${\mathrm{Ext}}_{\mathbb{Z}}(\mathbb{Z}/2,\mathbb{Z})=\mathbb{Z}/2$. HenceLet $t\in H^1(X;\mathbb{Z})$ be pulled back from the base circle, and $u$ the nonzero degree-three class. The square of the circle class vanishes, so $t^2=0$. All other products of positive-degree classes land above dimension three and vanish. This gives the cohomology ring of the antipodal two-sphere mapping torusModulo two, $A^{\ast }$ is the identity on fiber cohomology. The cohomological Wang sequence shows that restriction $H^2(X;{\mathbb{F}}_2)\to H^2(S^2;{\mathbb{F}}_2)$ is onto; choose $y$ mapping to its generator. The classes $1,y$ restrict to a fiber basis, so the Leray-Hirsch theorem makes the total cohomology free over $H^{\ast }(S^1;{\mathbb{F}}_2)$ on those two classes. Write $\bar{t}$ for the mod-two base class. The basis is $1,\bar{t},y,\bar{t}y$, and in particular $\bar{t}y\ne 0$. Since ${\bar{t}}^2=0$ by pullback and $y^2=0$ by dimension, the mod-two cohomology ring of the antipodal two-sphere mapping torus isThe argument of Question 3 works over any field, so its use here over ${\mathbb{F}}_2$ is justified. The ring agrees with that of $S^1\times S^2$; a cohomology ring alone does not determine orientability.

No: the round sphere supplies a counterexample. On the unit round $S^2\subset {\mathbb{R}}^3$, restrict the smooth one-formThe antipodal map $A(x)=-x$ satisfies $A^{\ast }\theta =-\theta$, since $x_3^2$ is unchanged and $d{x}_1$ changes sign. Every unit-speed closed geodesic is a great circle, with $\gamma (t+\pi )=A(\gamma (t))$. Therefore its two half-circle integrals cancel:The same holds for all repeated traversals and either orientation.

However, its exterior derivative iswhich is not zero on the sphere. For example, at $p=(0,1/\sqrt{2},1/\sqrt{2})$ the tangent vectors $u=(1,0,0)$ and $w=(0,1/\sqrt{2},-1/\sqrt{2})$ give $d{\theta }_p(u,w)=1$. Hence $\theta$ is not a closed differential form, and cannot be an exact differential form, because $d(dh)=0$. This is the counterexample to geodesic one-form injectivity on the round sphere; periodic geodesic integrals alone do not imply exactness without an additional geometric hypothesis.

The space is the mapping torus of the antipodal map $A:S^n\to S^n$. For $n\ge 1$, its mapping degree is $\delta =(-1{)}^{n+1}$: $A$ is the restriction of $-I$ on ${\mathbb{R}}^{n+1}$, whose determinant is $(-1{)}^{n+1}$. The Wang sequence givesThe kernel terms are free abelian, so these short exact sequences split as groups. On $H_0(S^n)$ the map is the identity, and on $H_n(S^n)$ it is multiplication by $\delta$. For $n\ge 2$ this givesAt $n=1$, both contributions to degree one are present, giving $H_0=H_2=\mathbb{Z}$ and $H_1={\mathbb{Z}}^2$, with all other groups zero. At $n=0$, the antipodal map exchanges the two points of $S^0$, and the two cylinder intervals join into a single circle; thus $H_0=H_1=\mathbb{Z}$ and higher groups vanish. These are the homology of the antipodal sphere mapping torus cases.

The antipodal map is the restriction of the smooth linear map $-I$, so is smooth. For $n\ge 1$ orient the sphere by the outward-normal convention. Its degree is $(-1{)}^{n+1}$: the ambient determinant is $(-1{)}^{n+1}$, and the outward normal is carried to the outward normal at the antipodal point.

Choose a regular value of $f$, available by Sard theorem; if it is outside the image, the degree is zero already. Otherwise its finite preimage is partitioned into distinct antipodal pairs. Differentiating $f\circ h=f$ shows that their local degree signs obeyFor even $n$, the signs in each pair cancel, so $\mathrm{deg}f=0$. For odd $n$, each pair contributes $2$ or $-2$, so $\mathrm{deg}f$ is even. In neither case can its degree equal the identity's degree one. Since the degree of a map between oriented manifolds is invariant under smooth homotopy, $f$ is not smoothly homotopic to the identity. For $n=0$, the symmetry makes $f$ constant on the two-point discrete sphere; a continuous homotopy into this discrete space cannot change either point's image, so the conclusion still holds.

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.

Call this simultaneous antipodal quotient of two spheres $M$. The simultaneous antipodal map acts freely on $S^2\times S^2$, so $M$ is a connected closed manifold of dimension four. Each antipodal factor has mapping degree $-1$; their product preserves orientation. Hence $M$ is orientable.

The product $S^2\times S^2$ is simply connected, so this double covering space is the universal cover. Consequentlyusing the abelianization of the fundamental group. The Euler characteristic under a finite covering gives $\chi (M)=4/2=2$. Its rational first Betti number is zero, and Poincare duality gives $b_3=b_1=0$ and $b_0=b_4=1$. Thus $b_2=0$ as well.

The universal coefficient theorem for cohomology now gives $H^1(M;\mathbb{Z})=0$ andHere $\mathrm{Hom}(H_2(M;\mathbb{Z}),\mathbb{Z})=0$ because $b_2=0$. Integral Poincare duality further gives $H^3(M;\mathbb{Z})\cong H_1(M;\mathbb{Z})=\mathbb{Z}/2$, and $H^4(M;\mathbb{Z})=\mathbb{Z}$.

Let $a,b,c$ generate these groups in degrees $2,3,4$ respectively. The only potentially nonzero positive-degree cup product is $a^2$. But $2a^2=(2a)a=0$, while $H^4(M;\mathbb{Z})$ has no nonzero torsion subgroup, so $a^2=0$. Every other positive product vanishes by dimension. Thuswith every product of positive-degree elements equal to zero. The degree-three torsion is essential: it would be lost by computing only rational cohomology.