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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:

For connected oriented closed manifolds of dimension $n$, the degree of a map between oriented manifolds is the integer determined bywhere the brackets are the chosen fundamental classes. Functoriality gives $\mathrm{deg}(g\circ f)=\mathrm{deg}g\mathrm{deg}f$.

Give $S^n$ its boundary orientation as the boundary of the unit ball in ${\mathbb{R}}^{n+1}$. The antipodal map is the restriction of $-I$. Its ambient determinant is $(-1{)}^{n+1}$, and it takes the outward normal at $x$ to the outward normal at $-x$. Thus its effect on the boundary orientation has this same sign, givingThe original PDF has $A:S^n\to S^n$, correcting the extra prime on the target in the TeX transcription.

For even dimension $n=2k\ge 2$, cellular homology of real projective space gives $H_n({\mathbb{RP}}^n;\mathbb{Z})=0$. Thus the composite map on top homology factors through zero, andZero is attained by constant maps.

For odd dimension $n=2k+1\ge 3$, Real projective space is orientable: the antipodal deck transformation has degree $+1$. Orient it so that the double covering map $q:S^n\to {\mathbb{RP}}^n$ has degree $2$. Since $S^n$ is simply connected, the lifting criterion for a covering space gives $f=q\circ \tilde{f}$ for a map $\tilde{f}:S^n\to S^n$. Hencewhich is even. Every even integer occurs: collapse the complement of an oriented embedded disk in ${\mathbb{RP}}^n$ to obtain a degree-one map $g:{\mathbb{RP}}^n\to S^n$, and choose $\tilde{f}$ of any prescribed integer degree $m$. Maps of all integer degrees on spheres are obtained, for example, by suspending the circle maps $z\mapsto z^m$. Setting $f=q\circ \tilde{f}$ gives degree $2m$. ThusThese are the degrees of maps factoring through real projective space. If $k=0$ is permitted in the odd-dimensional clause, there is an exception: ${\mathbb{RP}}^1\cong S^1$, so every integer degree occurs. The lifting argument requires dimension at least two. The initial connected-manifold degree definition excludes $S^0$; with the usual reduced-homology definition for self-maps of $S^0$, a factorization through the one-point ${\mathbb{RP}}^0$ has degree zero.

Finally suppose $\mathrm{deg}f=p$ is prime. Fix any prime $q\ne p$ and work over $F={\mathbb{F}}_q$. We claim that $f^{\ast }:H^r(N;F)\to H^r(S^n;F)$ is injective. If $a\ne 0$, Poincare duality supplies $b\in H^{n-r}(N;F)$ with $⟨a⌣b,[N{]}_F⟩\ne 0$. Naturality of the cup product and evaluation givesbecause $p$ is invertible in $F$. Hence $f^{\ast }a\ne 0$, proving cohomological injectivity of a map of invertible degree.

The intermediate cohomology of the sphere is zero, so $H^r(N;F)=0$ for $0<r<n$. Over a field, the universal coefficient theorem for cohomology identifies this with the dual of $H_r(N;F)$, so these homology groups vanish as well. The universal coefficient theorem for homology then givesEach integral group is a finitely generated abelian group, by the permitted finite CW complex model. Its decomposition can have no free summand, since that would survive modulo $q$, and no torsion summand divisible by any prime $q\ne p$. Thus it is a finite $p$-primary group. There are only finitely many intermediate degrees, so the exponents of their finite cyclic summands have a common bound $p^a$. Taking $a\ge 1$ also covers the case in which all the groups vanish. ThereforeThis proves that a prime-degree sphere map forces primary torsion, including one uniform exponent for all the degrees.

An $R$-orientation of a rank-$d$ real vector bundle $E\to B$ is a locally coherent choice of generator of $H^d(E_b,E_b\setminus \{0\};R)$ in every fibre. Equivalently, it is represented by a Thom class $u_E\in H^d(D(E),S(E);R)$ restricting to the chosen generator on each fibre pair. If $e(E)\in H^d(B;R)$ is the Euler class, the Gysin sequence of its unit sphere bundle contains

The diagonal quotient defining $L(p)$ is the three-dimensional lens space as a circle bundle $S^1\to L(p)\to S^2$ with Euler class $p$ times a generator. With integral coefficients, the only nontrivial Euler-class map is multiplication by $p:H^0(S^2;\mathbb{Z})\to H^2(S^2;\mathbb{Z})$. Exactness givesModulo $p$, the Euler class vanishes, so the same Gysin sequence giveswith all other groups zero.

For the coefficient sequence $0\to \mathbb{Z}\stackrel{p}{\to }\mathbb{Z}\to {\mathbb{F}}_p\to 0$, the long exact sequence from a coefficient sequence containsThe last map is zero and both middle groups have order $p$, so $\delta$ is an isomorphism. Reduction $H^2(L(p);\mathbb{Z})\to H^2(L(p);{\mathbb{F}}_p)$ is likewise an isomorphism. Their composite is the Bockstein isomorphism for a three-dimensional lens space

If $a^{\prime }=na$ is another generator, linearity of the Bockstein homomorphism and bilinearity of the cup product giveMoreover $t(a)\ne 0$: $\beta (a)$ is nonzero, and the Poincare duality pairing $H^1\times H^2\to H^3\cong {\mathbb{F}}_p$ is nondegenerate. If $h:L(p)\to L(p)$ is an orientation-reversing homotopy equivalence and $h^{\ast }a=na$, naturality givesCancelling the nonzero $t(a)$ yields $n^2\equiv -1(\mathrm{mod}p)$. Thus $-1$ must be a quadratic residue modulo $p$.

The short exact coefficient sequenceinduces the integral Bockstein homomorphismIf $\rho :H^{i+1}(X;\mathbb{Z})\to H^{i+1}(X;\mathbb{Z}/m)$ is reduction modulo $m$, thenEquivalently, if an integral cochain $a$ lifts a modulo-$m$ cocycle and $\delta a=mb$, then $\tilde{\beta }[a]=[b]$ and $\beta [a]=[b\mathrm{mod}m]$.

For integral lifts $a,c$ of classes $x,y$, the cup product coboundary formula isDividing by $m$ and reducing modulo $m$ proves the Bockstein derivation rule

Now assume that the stated closed five-manifold $M$ exists. Its top integral cohomology makes it connected and orientable. The long exact sequence from a coefficient sequence for multiplication by $p$ shows thatand that reduction $H^3(M;\mathbb{Z})\to H^3(M;\mathbb{Z}/p)$ is an isomorphism. HenceThe same coefficient sequence gives $H^4(M;\mathbb{Z}/p)=0$.

Choose $0\ne x\in H^2(M;\mathbb{Z}/p)$ and put $y=\beta x\ne 0$. By Poincare duality over $\mathbb{Z}/p$, the pairingis nondegenerate. Both factors are one-dimensional, so $x⌣y\ne 0$. But $x⌣x=0$ because $H^4(M;\mathbb{Z}/p)=0$, while the derivation rule and graded commutativity of the cup product giveThis contradicts $p>2$. Therefore no such manifold exists.

Part 1 makes $q:S^d\to N$ a degree-one map of closed oriented $d$-manifolds. The cohomological injectivity of a degree-one map embeds $H^{\ast }(N;\mathbb{Z})$ into $H^{\ast }(S^d;\mathbb{Z})$. Hence $N$ has no cohomology in degrees $0<i<d$; Poincare duality and the fundamental class giveThus $N$ is an integral homology sphere.

The long exact sequence of the pair $(N,N\setminus \{p\})$ has local relative homology $\mathbb{Z}$ only in degree $d$, and the map $H_d(N)\to H_d(N,N\setminus \{p\})$ is an isomorphism because it sends the fundamental class to the local orientation. Exactness now givesfor every $i$. Since $N\setminus \{p\}\cong S^d\setminus X$, the latter has the integral homology of a point.

A local orientation of a manifold at $x\in M$ is a generator ofAn $R$-orientation is a locally coherent choice of such generators. An R-fundamental class is a class $[M{]}_R\in H_d(M;R)$ whose image in every one of these local homology groups is a generator. Those images vary coherently under the restriction maps between small coordinate balls, so an $R$-fundamental class determines an $R$-orientation.

For a closed $R$-oriented $d$-manifold, Poincare duality says that cap product with its fundamental class is an isomorphismfor every $k$.

Choose a generator $[S^d]\in H_d(S^d;\mathbb{Z})\cong \mathbb{Z}$ from the homology of a sphere. For every $x\in S^d$, the long exact sequence of $(S^d,S^d\setminus \{x\})$, together with the contractibility of $S^d\setminus \{x\}\cong {\mathbb{R}}^d$, shows thatThe chosen generator is therefore a local generator at every point and is a $\mathbb{Z}$-fundamental class.

The standard CW complex structure on ${\mathbb{CP}}^n$ has one cell in every even dimension from $0$ to $2n$, and ${\mathbb{CP}}^k$ is its $2k$-skeleton. Collapsing that subcomplex leaves one zero-cell and one cell in each dimension $2i$ for $k+1\le i\le n$. All cellular boundaries vanish, so

Let $x\in H^2({\mathbb{CP}}^n;\mathbb{Z})$ be the usual generator. For $i>k$, choose $u_i$ whose pullback under the quotient map is $x^i$. Naturality of the cup product and the cohomology ring of complex projective space giveTogether with the unit, this determines the ring; equivalently, its reduced part is the ideal $(x^{k+1})/(x^{n+1})$ with the inherited multiplication. This is the cohomology ring of a collapsed projective subspace.

If $n=k+1$, the quotient has just a zero-cell and a $2n$-cell, so it is $S^{2n}$ and is a compact manifold. Conversely, suppose $n\ge k+2$ and the quotient is homotopy equivalent to a compact manifold. Its top cohomology is $\mathbb{Z}$, so that manifold must be closed, orientable, and $2n$-dimensional. But $H^{2n-2}\cong \mathbb{Z}$ while $H^2=0$, contradicting Poincare duality. Therefore

The cap product is the chain operationobtained by evaluating the cochain on the front $k$-face of a singular simplex and retaining its back $(p-k)$-face, with the standard sign convention. It descends to homology and cohomology.

A fundamental class $[M]\in H_d(M;\mathbb{Z})$ restricts at every $x\in M$ to the local generator of $H_d(M,M\setminus \{x\};\mathbb{Z})$ selected by the orientation. Poincare duality states thatis an isomorphism for every $k$ and coefficient ring $R$.

Let $0<k<d$ and $\alpha \in H^k(M;\mathbb{Z})$. Its restrictions to the contractible sets $U$ and $V$ vanish. The cap-product support lemma for a two-set cover, proved by representing $[M]$ with small simplices and replacing the restricted cocycles by coboundaries, therefore givesPoincare duality makes cap product with $[M]$ injective, so $H^k(M;\mathbb{Z})=0$ for every intermediate degree. Applying duality again gives $H_i(M;\mathbb{Z})=0$ for $0<i<d$. Finally the connected oriented closed manifold has

No. Its top nonzero group would force dimension three. Poincare duality requires $H_2\cong H^1$, and the universal coefficient theorem gives $H^1\cong \mathrm{Hom}(H_1,\mathbb{Z})=\mathbb{Z}$, since $H_0$ is free. This contradicts the proposed vanishing of $H_2$. Equivalently its listed Betti numbers give Euler characteristic $-1$, impossible for a closed oriented odd-dimensional manifold.

No. The dimension would again be three. The universal coefficient theorem gives $H^1=0$, since both $\mathrm{Hom}(H_1,\mathbb{Z})$ and $\mathrm{Ext}(H_0,\mathbb{Z})$ vanish. Poincare duality then forces $H_2=0$, contradicting the proposed torsion group. In general the second integral homology of a connected closed oriented three-manifold is free abelian, being dual to its first homology.

With the usual second-countability convention, an $n$-manifold is a Hausdorff second-countable space in which every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^n$. A local orientation of a manifold at $x$ is a choice of generator ofExcision and a coordinate ball identify this relative group with $H_n({\mathbb{R}}^n,{\mathbb{R}}^n\setminus \{0\})$. An orientation of a topological manifold is a coherent choice of such generators: near each point the choices are induced by one relative class on a neighbourhood, equivalently by an atlas with orientation-preserving transition maps.

For a connected closed oriented manifold, its fundamental class $[M]\in H_n(M;\mathbb{Z})$ is the unique class mapping to every chosen local generator. Here closed means compact without boundary. Poincare duality states that cap product with this class is an isomorphismOver rational coefficients it gives $b_k=b_{n-k}$. ThereforeWhen $n$ is odd this says $\chi =-\chi$, so $\chi (M)=0$. Torsion does not contribute to the Betti numbers or this Euler characteristic.

Write $M$ for either connected closed oriented manifold and $D$ for its embedded closed ball. A precise relative form of Poincare duality for a compact locally contractible subset isThis is Poincaré-Alexander duality. It applies to the printed closed-ball complement without replacing $D$ by its interior. Since $D$ is contractible, the relative group is $\mathbb{Z}$ at $j=n$ and zero otherwise. Naturality identifies $H_n(M)\to H_n(M,M\setminus D)$ with restriction $H^0(M)\to H^0(D)$, which is an isomorphism because both spaces are connected. Equivalently, the fundamental class restricts to the relative orientation generator on the ball.

In the long exact sequence of a pair, the only potentially exceptional part is thereforeExactness gives $H_n(M\setminus D)=0$ and the desired isomorphism in degree $n-1$. In every lower degree both adjacent relative terms are zero, so the same exact sequence gives the remaining isomorphisms. ThusFor the pieces $M_i\setminus \mathrm{int}{D}_i$ used in the gluing, the same conclusion follows from excision and the pair $(D^n,S^{n-1})$, or from the radial deformation of a punctured ball to its boundary. These pieces also have zero top homology, and the image of their boundary sphere's fundamental class is zero: it is the boundary of their relative oriented fundamental chain.

Cap product with the locally finite fundamental class of an $n$-manifold without boundary gives the displayed duality, including a possibly nontrivial orientation local system ${\mathcal{O}}_M$. It is the noncompact version of Poincare duality; for a compact manifold Borel-Moore homology is ordinary homology. Locally it is the relative fundamental-class calculation in an oriented coordinate ball; compatibility on overlaps and the Mayer–Vietoris sequence give the global statement. Since the chain complex of locally finite homology is zero in negative degrees, this duality implies $H^k(M;\mathbb{Z})=0$ for $k>n$.

For a closed oriented $d$-dimensional manifold over a field $F$, the cup product pairingis a perfect pairing between complementary degrees. The cap product with the fundamental class is an isomorphism by Poincare duality, and the universal coefficient theorem for cohomology over $F$ identifies the complementary cohomology with the full dual space of the resulting homology. Evaluation is therefore a perfect pairing. Taking the top-degree part of the same formula defines a nondegenerate bilinear form on the full graded cohomology. A boundary requires relative groups and Poincare-Lefschetz duality instead.

If $S^n\to N$ has prime degree $p$ and $N$ is a closed connected oriented $n$-manifold, every integral homology group of $N$ in degrees strictly between zero and $n$ is a finite $p$-primary abelian group. Consequently one power of $p$ annihilates all these groups. The proof combines Poincare duality over each field ${\mathbb{F}}_q$, $q\ne p$, with the universal coefficient theorem for homology and finite generation.

For a simply connected closed four-manifold with second Betti number one, Poincare duality and the universal coefficient theorem for cohomology make $H_2$ torsion-free of rank one. The integral intersection form is unimodular, so its one-by-one matrix is $[1]$ or $[-1]$. Hence the square of a generator $u\in H^2$ generates $H^4$, giving the displayed cohomology ring. This is a ring isomorphism; it need not preserve a preassigned orientation class.

For a continuous map $f:M\to N$ of nonzero mapping degree between closed connected oriented manifolds of the same dimension, $f^{\ast }:H^q(N;\mathbb{Q})\to H^q(M;\mathbb{Q})$ is injective in every degree. Pair a nonzero class $a$ with a complementary class $b$ using Poincare duality. Naturality gives $⟨f^{\ast }a⌣f^{\ast }b,[M]⟩=(\mathrm{deg}f)⟨a⌣b,[N]⟩\ne 0$, so $f^{\ast }a\ne 0$. Over coefficients of positive characteristic the argument requires that the degree remain nonzero in that field.

A simply connected closed three-dimensional manifold is orientable, since its orientation monodromy factors through its trivial fundamental group. Its first integral homology is the abelianization of that fundamental group and is zero. The universal coefficient theorem for cohomology gives $H^1(M;\mathbb{Z})=0$, and Poincare duality gives $H_2=0$, $H_3=\mathbb{Z}$. Connectedness gives $H_0=\mathbb{Z}$, and duality gives zero homology above dimension three. Thus it is an integral homology sphere, without needing any homeomorphism classification.

For positive $p,q$, quotient $S^p\times S^q$ by $(x,y)\mapsto (-x,-y)$. This is a free double covering space action. Its closed manifold quotient is orientable exactly when $p+q$ is even, because the product mapping degree is $(-1{)}^{p+q+2}$. If $p,q\ge 2$, the product is simply connected and the quotient has fundamental group $\mathbb{Z}/2$. For $p=q=2$, the quotient has integral cohomology groups $\mathbb{Z}$ in degrees $0,4$, $\mathbb{Z}/2$ in degrees $2,3$, and zero elsewhere. Every positive-degree cup product is zero. These conclusions follow from Poincare duality, the Euler characteristic under a finite covering, and the universal coefficient theorem for cohomology.