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For a generator $a\in H^1(L(p);{\mathbb{F}}_p)$, Poincare duality and the Bockstein isomorphism for a three-dimensional lens space make $t(a)$ nonzero. Replacing $a$ by $na$ multiplies $t(a)$ by $n^2$. An orientation-reversing homotopy equivalence multiplies the evaluation by $-1$, so its existence forces $n^2=-1$ in ${\mathbb{F}}_p$ for some $n$.

A degree-one map $f:M\to N$ between closed connected oriented manifolds of the same dimension induces an injection $f^{\ast }:H^{\ast }(N;R)\to H^{\ast }(M;R)$. The wrong-way map defined using Poincare duality is a left inverse by the projection formula and $f_{\ast }[M]=[N]$.

If $f:M\to N$ is a map between closed connected oriented manifolds of the same dimension and its degree is nonzero in a coefficient field $F$, then $f^{\ast }:H^{\ast }(N;F)\to H^{\ast }(M;F)$ is injective. For a nonzero class $a$, Poincare duality supplies $b$ with $⟨a⌣b,[N]⟩\ne 0$, and naturality gives

For a map between closed oriented manifolds of the same dimension, Poincare duality turns homological pushforward into a degree-preserving map on cohomology. It is adjoint to pullback under the duality pairing: $⟨f^{!}a⌣b,[N]⟩=⟨a⌣f^{\ast }b,[M]⟩$. If $f$ has degree $d$, nondegeneracy gives $f^{!}{f}^{\ast }=d\mathrm{id}$. Maps of different dimensions require the corresponding degree shift.

For $k\ge 1$, put $U={\mathbb{CP}}^{2k}\setminus {\mathbb{RP}}^{2k}$. Its cohomology ring isHere $x$ is the restricted hyperplane class. The cohomological Gysin map of an embedding sends $a^{2j}$ to $h^{k+j}$ and every odd-degree class to zero, by Poincare duality and mod-two restriction from complex to real projective space. Its exact sequence gives one-dimensional groups in even degrees $0,2,\dots ,4k-2$, and zero otherwise. Choose $y$ with $\delta y=a$. The connecting-map module identity gives $\delta (x^{j}y)=a^{2j+1}$ for $0\le j<k$, so these are the upper-half generators. The lower-half generators are $x^j$. The relations $x^k=0$ and $y^2=0$ follow respectively from $i_{!}1=h^k$ and $H^{4k}(U;{\mathbb{F}}_2)=0$. The result is a ring isomorphism with Complex projective space ${\mathbb{CP}}^{k-1}$ times $S^{2k}$; it does not by itself assert a homotopy equivalence of the spaces.

Every closed odd-dimensional manifold has Euler characteristic zero. For an orientable manifold this follows by pairing complementary Betti numbers using Poincare duality; the general case follows from the orientable double cover and Euler characteristic under a finite covering.

For a closed oriented manifold of dimension $4r+2$, Poincare duality pairs all rational Betti numbers except the middle one, making their total contribution to the Euler characteristic even. The middle cohomology degree is odd, so its nondegenerate Poincare duality pairing is an alternating bilinear form and has even dimension. Hence the Euler characteristic is even. Nonorientable manifolds can have odd Euler characteristic, as Real projective space of even dimension demonstrates.

Poincare duality gives the displayed formula for connected closed oriented four-manifolds. Every integer occurs. A connected sum of $r$ copies of the Complex projective plane and $s$ copies of $S^1\times S^3$ has Euler characteristic $2+r-2s$, because the summand characteristics are three and zero. Given any integer $h$, choose $s$ large enough that $r=h-2+2s\ge 0$; when both counts are zero use $S^4$. If simple connectivity is additionally required, the possible values instead begin at two, since $b_1=0$; connected sums of projective planes realize every such value.

For a closed oriented manifold, the rational middle-dimensional cup-product pairing is nondegenerate by Poincare duality. Its degree is odd, so graded commutativity makes it skew-symmetric. A nondegenerate skew-symmetric form has even dimension. In particular a closed oriented six-manifold cannot have a rank-one third cohomology group.

// Target: algebra.bigb

For an oriented proper smooth map $f:M^m\to N^n$ of oriented manifolds, the Gysin homomorphism is a degree-shifting pushforward $f_{!}:H^k(M)\to H^{k+n-m}(N)$, defined using Poincare duality and the appropriate pushforward on homology. For an oriented sphere bundle it is integration along the fiber and participates in the Gysin sequence of a sphere bundle. Oriented embeddings give the corresponding Thom-class construction.

For an oriented compact manifold of dimension $4k$, the intersection form is the bilinear pairing on middle-dimensional homology obtained by representing classes by transverse cycles and counting their signed intersection points. On an oriented four-manifold it is equivalently the cup-product pairing on $H^2$ evaluated on the fundamental class.

The intersection form of a 4-manifold is the four-dimensional case of this middle-dimensional intersection pairing.

The intersection form of a 4-manifold is the middle-dimensional signed-intersection pairing of an oriented compact four-manifold. On a closed manifold it can equivalently be defined on second cohomology by evaluating the cup product on the fundamental class. The general intersection form also applies to higher dimensions.

For a closed $n$-dimensional manifold, Poincare duality holds over ${\mathbb{F}}_2$ without an orientation assumption. The mod-two fundamental class exists because orientation signs disappear. The resulting complementary-degree intersection pairing is a perfect pairing.

Use Poincare duality with coefficients in ${\mathbb{F}}_2$, valid also without orientability. The ranks in complementary degrees agree, and their signs in the alternating sum are opposite. They therefore cancel in pairs. Euler characteristic is independent of the coefficient field, giving the stated integer equality.

Let $M$ be the given closed three-dimensional manifold. It is connected by the meaning of simply connected. An orientation chosen at one point can be transported along paths, and its sign change around loops defines a homomorphism ${\pi }_1(M)\to \{\pm 1\}$. Since the fundamental group is trivial, this monodromy vanishes; thus $M$ is orientable.

The first homology group is the abelianization of the fundamental group, giving $H_1(M;\mathbb{Z})=0$. Also $H_0(M;\mathbb{Z})=\mathbb{Z}$. The universal coefficient theorem for cohomology in degree one givesBoth outer groups vanish: the first because $\mathbb{Z}$ is free and the second because $H_1=0$. Hence $H^1=0$. Integral Poincare duality now gives $H_2(M;\mathbb{Z})\cong H^1(M;\mathbb{Z})=0$ and $H_3(M;\mathbb{Z})\cong H^0(M;\mathbb{Z})=\mathbb{Z}$. Homology in degrees greater than three vanishes by the same duality with negative-degree cohomology. ConsequentlyThese are precisely the homology of a sphere for $S^3$, proving that simply connected closed three-manifolds are homology spheres. No homeomorphism classification of three-manifolds is needed.

Work over $\mathbb{Q}$ and put $b_i=\dim H^i(M;\mathbb{Q})$. For a closed orientable six-dimensional manifold, Poincare duality gives $b_i=b_{6-i}$, soThe remaining Betti number is even. Indeed, the cup product pairingis nondegenerate by duality and skew-symmetric by graded commutativity, since $(-1{)}^{3\cdot 3}=-1$. If its dimension were odd, a matrix $A$ for it would satisfy $\det A=\det (-A^T)=-\det A$, forcing $\det A=0$ in characteristic zero, a contradiction. Thus $b_3$ is even andThis is the Euler characteristic parity in dimensions congruent to two modulo four. If the manifold has several components, apply the argument to each component and add.

The orientability assumption is essential. The Real projective space ${\mathbb{RP}}^6$ has one cell in each dimension from zero to six, givingIt is nonorientable: its double cover $S^6$ has the orientation-reversing antipodal deck transformation, of degree $(-1{)}^7=-1$.

For the construction, ${\mathbb{CP}}^3$ is a closed orientable real six-dimensional manifold with cells in dimensions $0,2,4,6$, hence Euler characteristic four. The product $S^3\times S^3$ is also closed and orientable, with Euler characteristic $\chi (S^3{)}^2=0$. The six-sphere has Euler characteristic two.

In dimension six, removing an open ball subtracts one from the Euler characteristic. To see this, write $M$ as the punctured manifold joined to a closed ball along $S^5$, whose Euler characteristic is zero, and apply the cellular inclusion-exclusion formula. Gluing two punctured manifolds along $S^5$ therefore givesA connected sum of $r$ copies of ${\mathbb{CP}}^3$ and $s$ copies of $S^3\times S^3$ haswith the empty sum interpreted as $S^6$. Given any even integer $n$, takeBoth are nonnegative integers and the displayed Euler characteristic equals $n$. Hence every even integer is the Euler characteristic of a closed oriented six-manifold, including zero and negative integers; the construction is connected.

A simply connected manifold is connected and orientable: its orientation character is a group homomorphism ${\pi }_1(X)\to \{\pm 1\}$, and this group is trivial. Choose an orientation. The Hurewicz theorem in degree one gives $H_1(X;\mathbb{Z})=0$. The universal coefficient theorem for cohomology then gives $H^1(X;\mathbb{Z})=0$ andbecause the possible $\mathrm{Ext}(H_1(X),\mathbb{Z})$ term is zero.

The homology groups are finitely generated for a closed manifold. The displayed dual group is therefore free abelian. Poincare duality identifies $H_2(X;\mathbb{Z})$ with $H^2(X;\mathbb{Z})$, so $H_2$ itself has no torsion. Since the second Betti number is one, both groups are isomorphic to $\mathbb{Z}$. This torsion argument is essential: rank one by itself would not rule out finite summands.

Again by Poincare duality, $H^3(X;\mathbb{Z})\cong H_1(X;\mathbb{Z})=0$, and $H^4(X;\mathbb{Z})\cong H_0(X;\mathbb{Z})=\mathbb{Z}$. Thus the only nonzero cohomology groups are $\mathbb{Z}$ in degrees $0,2,4$.

Let $u$ generate $H^2$. The integral intersection formis unimodular. Indeed, cap product with $[X]$ is the Poincare duality isomorphism to $H_2$, and the preceding universal-coefficient identification makes evaluation the full integral dual. Composing these isomorphisms identifies the pairing with an isomorphism $H^2\to \mathrm{Hom}(H^2,\mathbb{Z})$. In rank one its matrix must therefore be $[1]$ or $[-1]$. Consequently $⟨u^2,[X]⟩=\pm 1$, so $u^2$ generates $H^4$.

All products in degrees above four vanish. Hence the assignment of the degree-two generator giveswhich is the cohomology ring of complex projective space for ${\mathbb{CP}}^2$. If the intersection form is negative, this ring isomorphism reverses the chosen top orientation generator; preserving that additional orientation is not part of the requested ring isomorphism.

This sequence cannot occur. Poincare duality and the universal coefficient theorem for cohomology implyThus a nonzero rank-two third homology group is incompatible with the proposed first homology group.

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 $H_3\cong \mathrm{Hom}(H_1,\mathbb{Z})$.

Use rational coefficients and finite singular chains. For an $N$-dimensional stratified pseudomanifold with filtration $X_j$, the lower middle perversity isA singular simplex $\sigma :{\Delta }^i\to X$ is an allowable singular simplex when, for every $k\ge 2$, ${\sigma }^{-1}(X_{N-k})$ lies in the $(i-k+\bar{m}(k))$-skeleton of ${\Delta }^i$; negative-dimensional skeletons are empty. Let $A_i$ be the span of these simplices in $S_i(X;\mathbb{Q})$. The intersection chains areThe condition on $\partial c$ is imposed after combining equal simplices and cancelling coefficients: the faces of an allowable simplex need not themselves be allowable. If $c\in I{S}_i$, then $\partial c\in A_{i-1}$ and $\partial (\partial c)=0$, so $\partial c\in I{S}_{i-1}$. Thus $I{S}_{\ast }$ is a chain subcomplex of $S_{\ast }$, and $I{H}_i(X)=H_i(I{S}_{\ast })$ defines intersection homology.

Write $v$ for the vertex of $cY$. Its codimension is $2n$, so the vertex condition for an allowable $i$-simplex isIn particular all simplices of degree at most $n$ avoid $v$. For $i<n$, both the degree-$i$ chains and the degree-$(i+1)$ chains that can bound them lie in $cY\setminus \{v\}=Y\times (0,1)$. The two chain complexes therefore have the same cycles and boundaries in degree $i$, proving the low-degree isomorphism.

For $i\ge n$, let $z$ be an allowable $i$-cycle. Cone every simplex of $z$ radially to $v$, using the cone on ${\Delta }^i$ as ${\Delta }^{i+1}$, and write the resulting chain as $cz$. At the vertex, coning an old inverse image in the $(i-n-1)$-skeleton gives a subset of the $(i-n)$-skeleton, and the new cone vertex also lies in this skeleton because $i-n\ge 0$. For every other singular stratum, coning the inverse image increases its permitted skeleton dimension by one, exactly as increasing the simplex dimension does. Thus $cz$ is allowable. The usual cone identity is $\partial (cz)=z-c(\partial z)=z$; since $z$ is allowable, the boundary condition for intersection chains holds as well. Every cycle is consequently a boundary. This proves the intersection homology cone formula here:The assumption $\dim Y=2n-1$ entails $n\ge 1$, so no augmented degree-zero exception arises in this coning argument.

For the suspension of a topological space, the two endpoints are collapsed separately, giving two distinct vertices. Use the following standard computational facts for intersection homology: a product with an interval has the same intersection homology as the original stratified space, via the projection and its stratum-preserving homotopy inverse; and an open cover has a Mayer–Vietoris sequence whose overlap map is the signed pair of inclusion maps. Cover the suspension by its two open cones. Their overlap is $Y\times (-\epsilon ,\epsilon )$. Below degree $n$, the overlap map is, under the interval identifications, $a\mapsto (a,-a)$ in $I{H}_i(Y)\oplus I{H}_i(Y)$, hence is injective with cokernel $I{H}_i(Y)$. In degree $n$ both cone groups vanish and the next overlap map, in degree $n-1$, is still injective. Above degree $n$ both adjacent cone groups vanish. Exactness therefore givesThis differs from ordinary reduced suspension homology: below the cutoff the cycles cannot use either vertex, and in the middle degree coning kills them. If $Y$ is a compact oriented Witt space, its Poincare duality pairs $I{H}_i(Y)$ with $I{H}_{2n-1-i}(Y)$. The displayed groups of the suspension are correspondingly paired in complementary degrees $i$ and $2n-i$, while its middle group is zero. The new vertices have even codimension $2n$ and thus impose no additional Witt space condition; the suspension is again Witt. Its middle intersection form has zero rank and, when its dimension is divisible by four, zero signature.