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Put $B=\partial M$ and let $v$ be the cone vertex. Specify the convention away from the Witt case: first take intersection cohomology to mean the hypercohomology of the intersection complex $I{C}_{\bar{m}}$, normalized in degree zero on the regular stratum. The lower middle perversity at the vertex is $\bar{m}(2m+1)=m-1$. The standard local cone calculation for this convention saysand restriction to $U\simeq B$ is the identity below the cutoff. Here this can also be read directly from the construction $I{C}_{\bar{m}}={\tau }_{\le m-1}R{j}_{\ast }\mathbb{Q}$, with $j:\hat{M}\setminus \{v\}\hookrightarrow \hat{M}$: punctured conical neighbourhoods retract onto $B$, and the truncation discards the degrees above $m-1$.

Define relative intersection cohomology as the hypercohomology of the homotopy fibre of restriction, namely $R\Gamma (\hat{M},\hat{U};IC)=\mathrm{Cone}(R\Gamma (\hat{M};IC)\to R\Gamma (\hat{U};IC))[-1]$. Restricting to the pair $(M,U)$ gives a map to the corresponding ordinary relative cohomology, since $IC{|}_M={\mathbb{Q}}_M$. The induced map of exact triangles gives the requested commutative diagram:Both rows are long exact cohomology sequences of a pair. Commutativity includes the connecting homomorphisms, because they arise from the same map of restriction cones, with a common sign convention. The relative vertical maps are isomorphisms: excision removes the common conical neighbourhood, leaving precisely the manifold pair outside it, where $IC$ is constant. Also the collar retracts onto $B$, so $H^{\ast }(M,U)\cong H^{\ast }(M,B)$.

The only possible nonmanifold stratum is $v$, with link $B$ of dimension $2m$. Since intersection homology of a manifold is ordinary homology, the exact necessary and sufficient condition isThe equivalence with the defining $H_m(B;\mathbb{Q})=0$ follows from the universal coefficient theorem for cohomology over $\mathbb{Q}$. In the Witt case the cone restriction is an isomorphism through degree $m$, since both degree-$m$ groups are zero. Comparing the two exact rows, or applying the Five lemma to each five-term segment, gives $I{H}^i(\hat{M})\cong H^i(M)$ for $i\le m$. For $i>m$, both $I{H}^{i-1}(\hat{U})$ and $I{H}^i(\hat{U})$ vanish, so the top row identifies the absolute group with the relative one. HenceWithout the Witt hypothesis, comparison still gives $I{H}_{\bar{m}}^i(\hat{M})\cong H^i(M)$ for $i<m$. At degree $m$ the upper row readsThe lower row identifies the quotient by $\mathrm{im}\delta$ with the image of the relative-to-absolute map. In higher degrees both cone terms vanish. Thus the full lower-middle Deligne answer isFor clarity, if $I{H}^i$ is instead defined as the linear dual of the lower-middle intersection chains of question 1, it corresponds to the complementary upper middle perversity Deligne sheaf. Its cone group retains $H^m(B)$ as well. The same diagram then givesEquivalently the exceptional group is $H^{m+1}(M,B)/\mathrm{im}(H^m(B)\stackrel{\delta }{\to }{H}^{m+1}(M,B))$. These conventions agree under the stated Witt condition, explaining why that condition removes the ambiguity.

Take the solid torus $M=S^1\times D^2$, whose boundary is the two-dimensional torus $B=T^2$. Its ordinary cohomology is $H^0(M)=H^1(M)=\mathbb{Q}$, zero otherwise; Poincare duality for a manifold with boundary gives $H^2(M,B)=H^3(M,B)=\mathbb{Q}$ and all other relative groups zero. The lower-middle Deligne formula givesThus degrees one and two have different dimensions, so there can be no three-dimensional Poincare duality. In the dual-lower-chain convention the dimensions are $(1,1,0,1)$, which likewise fail duality. Both computations exhibit the obstruction $H^1(T^2)={\mathbb{Q}}^2\ne 0$.

An $N$-dimensional stratified pseudomanifold has a closed filtration $\varnothing =X_{-1}\subseteq X_0\subseteq \cdots \subseteq X_N=X$ with each nonempty $X_s\setminus X_{s-1}$ an $s$-dimensional manifold, dense top stratum, and $X_{N-1}=X_{N-2}$. Each point of an $s$-dimensional stratum has a stratum-preserving conical neighbourhood ${\mathbb{R}}^s\times cL$, where $L$ is a compact $(N-s-1)$-dimensional stratified pseudomanifold. The definition is inductive, starting with zero-dimensional spaces. There are no singular codimension-one strata; connectedness of links is not required. One may work in the usual triangulable category used for Witt bordism.

A rational Witt space is such a stratified pseudomanifold for which every stratum of odd codimension $2r+1$ has link $L$ satisfying $I{H}_r^{\bar{m}}(L;\mathbb{Q})=0$. No extra condition is imposed at even codimension. A Witt space with collared boundary has Witt interior and Witt boundary, and a stratum-preserving collar $\partial X\times [0,\epsilon )$ with the product stratification. An orientation means an orientation of the regular stratum, inducing the boundary orientation on its collar.

A three-pronged graph, with its central vertex and open edges as strata, is a stratified space but not a one-dimensional pseudomanifold: the central vertex is a singular codimension-one stratum, and deleting it leaves three local branches instead of the two branches of a one-manifold. For a non-Witt pseudomanifold, take the suspension of $T^2$. Its two vertices have codimension three and link $T^2$, whose middle group is $I{H}_1(T^2;\mathbb{Q})=H_1(T^2;\mathbb{Q})={\mathbb{Q}}^2$. It violates the Witt space condition.

The rational Witt bordism classification in dimensions divisible by four states that, for every $k\ge 1$, the middle intersection form induces an isomorphismHere $W(\mathbb{Q})$ is the Witt group of a field, so a nonsingular symmetric form represents zero precisely when it is metabolic, that is, when it has a half-dimensional totally isotropic subspace. The Complex projective plane gives a nonzero example: its middle group is generated by the degree-two class $h$, with $⟨h^2,[{\mathbb{CP}}^2]⟩=1$. Its form is $⟨1⟩$, of signature one; a metabolic form has signature zero. Thus $[{\mathbb{CP}}^2]\ne 0$ in ${\Omega }_4^{\mathrm{Witt}}$.

For a closed oriented Witt space $X$ of dimension $2k+1$, take the closed coneAway from its vertex it is a stratified product with an interval, so all old Witt space conditions persist. At the new vertex the codimension is $2k+2$, which is even; hence its odd-dimensional link $X$ creates no new Witt condition. The collar $X\times (1-\epsilon ,1]$ is explicit, and the product orientation may be chosen to induce the given orientation on $X$. This is a compact oriented Witt space with collared boundary, proving

Now let $\dim X=2$. Its singular strata are zero-dimensional. Their links are compact one-dimensional pseudomanifolds, which have no singular strata and are therefore nonempty finite disjoint unions of circles. A single circle is allowed, since a chosen stratum need not be intrinsically singular. The Witt condition is automatic in dimension two, because singular strata have even codimension.

Construct the normalization of a two-dimensional pseudomanifold concretely. Remove disjoint small conical neighbourhoods of the finitely many zero-stratum points. The remainder is a compact oriented surface with boundary, with one boundary circle for each local branch. Cap every such circle with a separate disk. The result $\tilde{X}$ is a closed compact oriented topological surface. Map each new disk homeomorphically onto its branch of the original cone, sending all the disk centres belonging to one old singular point to that point. Along the disk boundaries this agrees with the identification with the remainder, so it defines a continuous surjection $\pi :\tilde{X}\to X$. It is a homeomorphism off these finitely many centres, in particular over the prescribed open stratum. Every fibre is a point except over a zero-stratum point, where it is the finite set of its branch centres. Thus every fibre is zero-dimensional, as required.

Each component of $\tilde{X}$ bounds an oriented handlebody: a surface of genus $g$ is the boundary of a three-ball with $g$ one-handles attached. Let $H$ be the disjoint union of these handlebodies, oriented so that $\partial H=\tilde{X}$. Formgluing $x\in \partial H$ to $(\pi (x),0)$. The quotient is compact and Hausdorff, and its only remaining boundary is $X\times \{1\}$, with an explicit product collar. Near regular points of the gluing locus, two half-ball neighbourhoods glue to a three-ball; the induced orientations match after choosing the collar orientation in the usual boundary-gluing convention.

It remains to check the singular links, rather than merely assert that this quotient is a Witt space. If an old point $p$ has $r$ circle branches, the new one-dimensional stratum $\{p\}\times (0,1)$ has link the disjoint union of those $r$ circles. This has codimension two and imposes no Witt condition. At $(p,0)$, each branch supplies a half-three-ball from $H$ and another from the collar. The halves glue to a full three-ball; the different branches share only the positive collar axis. Intersecting with a small link therefore gives $r$ copies of $S^2$ identified at one common point. This is a two-dimensional pseudomanifold with zero middle intersection homology: an allowable one-cycle avoids the common point, so decomposes into cycles in the $r$ punctured spheres, each homeomorphic to ${\mathbb{R}}^2$; each such cycle bounds there. Consequently $I{H}_1({\bigvee }_{a=1}^r{S}^2;\mathbb{Q})=0$. This is exactly the Witt condition at the new codimension-three stratum. The boundary has the required product stratification. Hence $Z$ is a three-dimensional Witt space with collared boundary with $\partial Z\cong X$, proving

For the final product assertion, “product” means a nontrivial product with at least two positive-dimensional factors; a single four-dimensional factor would allow the preceding ${\mathbb{CP}}^2$ counterexample. State the standard rational intersection-homology product fact used here: products of Witt spaces are Witt, their intersection cohomology satisfies the Künneth theorem, and the complementary-degree intersection pairing on elementary tensors is the product of the factor pairings, with the sign $(-1{)}^{\mathrm{deg}\beta \mathrm{deg}{\alpha }^{\prime }}$ for $\alpha \otimes \beta$ and ${\alpha }^{\prime }\otimes {\beta }^{\prime }$. Terms whose degrees in a factor do not add to that factor's dimension pair to zero. Regrouping factors reduces total dimension four to cases $1+3$ or $2+2$.

In the $1+3$ case, write $X=A\times B$. The middle intersection cohomology isEach summand is totally isotropic, because the degrees in $A$ are respectively $0+0$ and $1+1$, neither equal to one. They pair perfectly with one another by factor Poincare duality. Thus either summand has half the total dimension and is a Lagrangian for the middle symmetric form.

For dimensions $2+2$, the middle group is the direct sumThe outside summands are isotropic and pair perfectly with one another. Each factor's degree-one pairing is a nonsingular skew-symmetric form, so choose a Lagrangian subspace $L_A\subset I{H}^1(A)$. Inside the central summand, $L_A\otimes I{H}^1(B)$ is isotropic and has half its dimension. It is orthogonal to both outside summands by their degrees. Thereforeis an isotropic subspace of half the full middle dimension. In both cases the form is metabolic. The Witt bordism classification now proves the desired $[X]=0$ in ${\Omega }_4^{\mathrm{Witt}}$. This also proves the assertion for more factors, using the regrouping just described.

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.

Start with the prime decomposition of a closed orientable three-manifold. Its fundamental group is a free product of the prime-summand groups. A nontrivial abelian group cannot be a free product of two nontrivial groups. The same is true of a nontrivial finitely generated nilpotent group: it has nontrivial center, whereas such a free product has trivial center. Thus all nontrivial fundamental group comes from a single prime summand; simply connected summands do not affect the argument.

A prime but nonirreducible orientable summand is $S^2\times S^1$, giving $\Gamma \cong \mathbb{Z}$. Otherwise the infinite-group prime summand is an irreducible three-manifold. The sphere theorem for three-manifolds gives ${\pi }_2=0$. Its simply connected universal cover is noncompact, so its top ordinary homology vanishes; its first and second homology also vanish by simple connectivity and the Hurewicz theorem. It is therefore acyclic and contractible by the Whitehead theorem. The summand is aspherical, and its closed orientable three-dimensional classifying space gives integral cohomological dimension of a group equal to three: the cellular resolution gives the upper bound and top Poincare duality gives the lower bound.

If $\Gamma$ is torsion-free abelian, finite generation makes it ${\mathbb{Z}}^r$. Its integral cohomological dimension of a group is $r$, as one sees from the standard $r$-torus. The aspherical case therefore has $r=3$. We obtainBoth occur, on $S^2\times S^1$ and the three-torus.

For the infinite torsion-free nilpotent case, the same reduction gives either $\mathbb{Z}$ or an aspherical group of cohomological dimension three. By the Malcev completion of a torsion-free nilpotent group, the latter is a lattice in a simply connected three-dimensional nilpotent Lie group. A nonabelian three-dimensional nilpotent Lie algebra has abelianization of dimension at least two: a nilpotent algebra generated modulo its derived algebra by one element would be generated by that element and thus be abelian. Its derived algebra is therefore one-dimensional, and nilpotence makes that line central. Hence it has a basis with a single nonzero bracket $[Y,T]=X$; the other possibility is an abelian algebra. In the nonabelian case the lattice's center is an isolated infinite cyclic subgroup, generated by $x$, and the quotient by it is ${\mathbb{Z}}^2$. Choose lifts $y,t$ of a basis of that quotient. Their commutator is $x^b$ for a nonzero integer $b$, and normal forms giveThe subgroup $A=⟨x,y⟩\cong {\mathbb{Z}}^2$ is normal and ${\Gamma }_b/A\cong \mathbb{Z}$. In its ordered basis $x,y$, conjugation by $t$ acts byThis is the unipotent torus-bundle group. Indeed, its normal forms multiply according toProjection to the $n,k$ coordinates, followed by inspection of the central $m$ coordinate, proves torsion-freeness. All commutators lie in $⟨x^b⟩$, which is central, proving nilpotence. The mapping torus of the torus automorphism with this matrix realizes every integer $b$ as a closed orientable three-manifold; $b=0$ gives ${\mathbb{Z}}^3$, while $b\ne 0$ gives the Nil cases. Thus the complete list of groups under the nilpotent hypotheses is $\mathbb{Z}$ and the groups ${\Gamma }_b$.

Here is the center argument, with its literal qualification. If a nilpotent group with abelian quotient $\Gamma /A$ were nonabelian, the last nonzero term ${\gamma }_c\Gamma$ of its lower central series would satisfyConsequently $Z(\Gamma )\cap A=1$ forces the group to be abelian. For the specific nontrivial rank-two $A$ in the printed family the hypothesis cannot actually occur: when $b\ne 0$, $x\in Z({\Gamma }_b)\cap A$; when $b=0$, the intersection is all of $A$. The implication is valid, but should not be used to invent an example satisfying this impossible extra condition.

A torsion-free finite extension of a lattice with quotient $C_4$ is realized by the quarter-turn flat three-manifold. On ${\mathbb{R}}^3$ letLet $c$ be unit translation in the third coordinate. Then $sa{s}^{-1}=b$, $sb{s}^{-1}=a^{-1}$, $sc{s}^{-1}=c$, and $s^4=c$. The lattice $\Lambda =⟨a,b,c⟩\cong {\mathbb{Z}}^3$ is normal and the quotient is $C_4$. A representative $a^m{b}^n{c}^k{s}^j$, $0\le j<4$, with $j\ne 0$ has third-coordinate displacement $k+j/4\ne 0$, so has no fixed point; for $j=0$ a nonidentity element is a nonzero translation. This proves torsion-freeness. The action is properly discontinuous, cocompact and orientation preserving, so it gives a closed orientable flat three-manifold with precisely the required extension.

For faithful linear holonomy on the maximal translation lattice, the holonomy groups of closed orientable flat three-manifolds areThe cyclic examples replace the quarter-turn by a planar lattice rotation of order $m=2,3,4,6$ and a translation of $1/m$ along its axis. The square and triangular plane lattices realize these orders; the trace condition $2\cos (2\pi /m)\in \mathbb{Z}$ explains the crystallographic restriction. The remaining case is obtained fromTheir squares translate by one in the first and second directions, and $(\alpha \beta {)}^2$ translates by minus one in the third. Their linear parts generate $C_2\times C_2$. The three nontranslation cosets have half-integral displacement along their respective fixed axes, so every element acts without a fixed point. These constructions realize every member of the orientable three-dimensional Euclidean holonomy classification.

The maximality/faithfulness restriction is essential. Taken literally as a question about arbitrary finite group quotients, the wording has more answers: $\mathbb{Z}$ has every finite cyclic quotient, and ${\mathbb{Z}}^3$ has, for example, quotient $C_5$. An exact criterion for the literal extension question is as follows. Choose an action $Q\to \mathrm{Aut}(A)$ and an extension class $e\in H^2(Q,A)$, where $A=\mathbb{Z}$ or ${\mathbb{Z}}^3$. For every subgroup $⟨q⟩$ of prime order $p$, requireTo prove this, lift $q$ to $g$. Its $p$th power is in $A^q$, and replacing $g$ by $ag$ changes that power by the displayed norm of $a$. Thus a lift of order $p$ exists exactly when this restriction is zero. Any torsion element has a prime-order power, so the condition is both necessary and sufficient. For an oriented flat realization in rank three the linear action must also preserve orientation. This distinguishes a complete extension criterion from the finite faithful-holonomy list.

First correct the group-ring condition in the PDF. For an infinite group, a left-invariant element of $\mathbb{Z}\Gamma$ would have constant coefficients on the whole group; finite support then forces it to be zero. Hence $H^0(\Gamma ,\mathbb{Z}\Gamma )=0$. For a Poincare duality group of positive dimension, the correct condition isIt must be accompanied by the finiteness condition: the trivial module has a finite-length resolution by finitely generated projective modules. The module $D$ has a right group action, possibly an orientation-sign action. The degree-zero exception only applies to the trivial $P{D}^0$ group, not simultaneously to positive-dimensional groups.

Here is how the duality follows. Put $R=\mathbb{Z}\Gamma$ and take a length-$n$ projective resolution $P_{\ast }\to \mathbb{Z}$. Dualize to $P^{\ast }={\mathrm{Hom}}_R(P_{\ast },R)$. Its cohomology is zero except in degree $n$, where it is $D$. Reversing degrees therefore makes $Q_j=P_{n-j}^{\ast }$ a projective right-$R$ resolution of $D$. Finite generation and projectivity give, for every left $R$-module $V$, an isomorphism of cochain complexesTaking cohomology and reversing the grading yieldsSince $D$ has underlying abelian group $\mathbb{Z}$, its action is a sign character $\omega :\Gamma \to \{\pm 1\}$. Twisting the resolution by this character identifies the right side with group homology having coefficient module $D{\otimes }_{\mathbb{Z}}V$, with diagonal left action $\gamma (d\otimes v)=\omega (\gamma )d\otimes \gamma v$. ThusFor an orientable Poincare duality group, $D$ is trivial and this is untwisted Poincare duality. The natural isomorphism can equivalently be realized by cap product with the group fundamental class. The reversed finite projective complex explains why the top group-ring condition produces duality for every coefficient module, rather than just matching integral Betti numbers.

Now let $H$ have finite index in $\Gamma$. There is an isomorphism of left $\Gamma$-modulesOne explicit map sends $r\in \mathbb{Z}\Gamma$ to the $H$-linear function $g\mapsto {\mathrm{pr}}_H(gr)$, where ${\mathrm{pr}}_H$ retains only terms in $H$. Finite index ensures that all such functions arise in this way. Shapiro's lemma gives the isomorphismas abelian groups. If $\Gamma$ is a Poincare duality group, restricting its finite projective resolution to $H$ preserves projectivity and finite generation: the group ring is finite free over $\mathbb{Z}H$. This proves that $H$ is $P{D}^n$.

For the converse one needs the usual ambient torsion-free hypothesis. The standard finite-index resolution lemmas say that type $F{P}_{\infty }$ is preserved under finite-index passage, and that a torsion-free group of finite virtual cohomological dimension has the same integral cohomological dimension as a finite-index subgroup. Thus if $H$ is $P{D}^n$ and $\Gamma$ is torsion-free, $\Gamma$ has a finite resolution by finitely generated projective modules of length $n$. The preceding explicit coinduction and Shapiro calculation transfers exactly the required group-ring cohomology, and proves the finite-index invariance of Poincare duality for torsion-free groups.

Without that hypothesis the claimed converse is false. Take $\Gamma =\mathbb{Z}\times C_2$ and $H=\mathbb{Z}\times 1$. Then $H$ is index two and is $P{D}^1$, while $\Gamma$ contains torsion. Its integral cohomological dimension of a group is infinite: restriction of any putative finite projective resolution to $C_2$ would contradict the nonzero cohomology $H^{2j}(C_2,\mathbb{Z})=\mathbb{Z}/2$ for every $j\ge 1$. Also, orientability need not ascend: the torus subgroup of the Klein-bottle group has trivial orientation action, whereas the full group has a nontrivial one. The invariant theorem concerns the possibly twisted notion of $P{D}^n$.

Finally, a closed aspherical surface has a two-dimensional contractible universal cover and exactly this compact-support duality, so its surface group is $P{D}^2$. Conversely the algebraic condition forces the same local duality dimension and, in the orientable case, a nonsingular skew pairing on first cohomology, the familiar algebra of a closed orientable surface. Infinite-index subgroups behave like noncompact surfaces rather than closed ones. These are strong reasons to expect a surface group; in fact the $P{D}^2$ classification theorem confirms that every integral $P{D}^2$ group is the fundamental group of a closed aspherical surface. This includes the torus and, with orientation twist, the Klein bottle; the sphere and projective plane do not give $P{D}^2$ groups.

For a complex K3 surface, $L=H^2(X,\mathbb{Z})$ is torsion-free and has its integral cup product pairingThe K3 intersection lattice isthe even unimodular integral quadratic lattice of signature $(3,19)$. Here $U$ is the hyperbolic plane with Gram matrixand $E_8(-1)$ is the negative of the positive-definite E8 lattice. Thus the rank is $3\cdot 2+2\cdot 8=22$ and the pairing has determinant of absolute value one.

These invariants can be checked from the topology and Hodge theory. The trivial canonical bundle and $H^1(X,{\mathcal{O}}_X)=0$ give $\chi ({\mathcal{O}}_X)=2$. The Noether formula gives $c_2(X)=24$, which is the topological Euler characteristic. Since $X$ is simply connected, $b_1=b_3=0$, so $24=2+b_2$ and $b_2=22$. Integral Poincare duality gives unimodularity; the universal coefficient sequence and $H_1(X,\mathbb{Z})=0$ give torsion-freeness of $H^2(X,\mathbb{Z})$.

Its second Stiefel–Whitney class is the reduction modulo two of $c_1(X)$, hence zero. The characteristic-class identitymakes the lattice even. Finally the real and imaginary parts of a holomorphic two-form give two positive directions, a Kähler class gives a third, and the primitive real $(1,1)$ classes are negative definite by the Hodge–Riemann bilinear relations. Therefore the signature is $(3,19)$. The classification of indefinite even unimodular lattices then identifies the displayed lattice uniquely. In particular the Picard group is only the integral $(1,1)$ part; it should not be confused with this rank-twenty-two topological lattice.

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.

Write $\dim X=4r+2=2m$, with $m=2r+1$ odd. Rational Poincare duality gives $b_k=b_{2m-k}$. Pairing every term outside the middle yieldsThe Poincare duality pairing on $H^m(X;\mathbb{Q})$ is $B(\alpha ,\gamma )=⟨\alpha ⌣\gamma ,[X]⟩$. It is nondegenerate, and graded commutativity of the cup product gives $B(\alpha ,\gamma )=-B(\gamma ,\alpha )$ because $m$ is odd. In particular $B(\alpha ,\alpha )=0$ over $\mathbb{Q}$. Part (b) therefore makes $b_m$ even, soFor the failure without orientability, ${\mathbb{RP}}^2$ is a closed nonorientable surface with one cell in each dimension zero, one and two; hence $\chi =1-1+1=1$. More generally, ${\mathbb{RP}}^{4r+2}$ is nonorientable and has Euler characteristic one. Thus the orientation hypothesis in the Euler characteristic parity in dimensions congruent to two modulo four is essential.

Choose an orientation for each curve and let $c_i=[C_i]\in H_1(X;\mathbb{Q})$. The algebraic intersection number of curves on an oriented surface gives a nondegenerate alternating bilinear form on the $2g$-dimensional space $H_1(X;\mathbb{Q})$, by Poincare duality. Distinct curves are disjoint, so their intersection numbers are zero; each self-pairing is zero by alternation. Thus the span $W=\mathrm{span}\{c_1,\dots ,c_n\}$ is an isotropic subspace of a symplectic vector space. The assumed linear independence gives $\dim W=n$.

Nondegeneracy identifies $H_1(X;\mathbb{Q})$ with its dual. Restricting functionals to $W$ is surjective, so the symplectic orthogonal complement has dimension $\dim W^{\perp }=2g-n$. Isotropy says $W\subseteq W^{\perp }$, giving $n\le 2g-n$. HenceThe bound is sharp: one meridian on each handle of a genus-$g$ surface gives $g$ pairwise disjoint curves with independent homology classes. This is an algebraic bound on independent classes, not on the number of disjoint separating or mutually homologous curves.

For singular cochains $a\in C^p(X;R)$ and $b\in C^q(X;R)$, define the cup product on a singular $(p+q)$-simplex byThe identity $\delta (a⌣b)=\delta a⌣b+(-1{)}^{p}a⌣\delta b$ makes this descend to cohomology. The induced product is associative, unital and graded-commutative: $ab=(-1{)}^{pq}ba$ in degrees $p,q$.

For finite CW complexes, the integral cohomological Künneth theorem for finite cell complexes is the split short exact sequenceThe splitting need not be natural. If the factor cohomology is torsion-free, the cohomological cross product is an isomorphism, with multiplication $(a\times b)(c\times d)=(-1{)}^{|b||c|}ac\times bd$.

A genus-$g$ closed oriented surface has cellular groups $H^0=\mathbb{Z}$, $H^1={\mathbb{Z}}^{2g}$, $H^2=\mathbb{Z}$, and no higher cohomology. Choose the positive orientation class $\omega$ and a symplectic basis $a_i,b_i$ of $H^1$. The cohomology ring of a closed oriented surface is specified bywith unit $1\in H^0$. These products follow by representing their Poincare duality classes by oriented handle curves: the $i$th meridian meets the $i$th longitude once positively, curves on distinct handles have zero intersection, and swapping two oriented curves reverses the intersection sign. In general, for closed oriented $n$-manifolds and transverse oriented submanifolds of codimensions $p,q$, the cup product of their Poincare-dual classes is the dual class of their oriented intersection. Complementary-dimensional intersections give the signed point count $⟨\alpha \beta ,[M]⟩$.

For ${\Sigma }_2$, the four curves $A_1,B_1,A_2,B_2$ give $a_1{b}_1=a_2{b}_2=\omega$ and all other pairings zero apart from their negatives in reversed order. Thus the degree-one intersection pairing on an oriented surface is two symplectic blocks, and the point class is dual to $\omega$.

For ${\Sigma }_2\times {\Sigma }_2$, let $a_i,b_i,{\omega }_1$ come from the first factor and $c_j,d_j,{\omega }_2$ from the second. The Künneth theorem gives their graded tensor-product ring, whose additive ranks in degrees $0,1,2,3,4$ areThe top orientation class is $\Omega ={\omega }_1{\omega }_2$. The two fiber surfaces $\{p\}\times {\Sigma }_2$ and ${\Sigma }_2\times \{q\}$ have dual classes ${\omega }_1,{\omega }_2$, intersect once positively and have zero self-intersection. The other sixteen degree-two classes are products of degree-one classes and are dual, up to the chosen orientation convention, to products of handle curves. Their intersections are encoded byThe minus sign in the first mixed formula comes from commuting $c_j$ past $b_k$. These classes together with the fiber classes give the intersection form of a product of two closed oriented surfaces, here nine hyperbolic blocks. Complementary degrees one and three are similarly paired, for example $a_i(b_k{\omega }_2)={\delta }_{ik}\Omega$; geometrically this pairs a three-dimensional product submanifold with a transverse loop. This exhibits the full duality, not just the additive groups.

The Lefschetz fixed-point theorem says that a self-map of a compact polyhedron with nonzero Lefschetz number has a fixed point, where $L(f)={\sum }_q(-1{)}^q\mathrm{tr}(f_{\ast }:H_q(-;\mathbb{Q})\to H_q(-;\mathbb{Q}))$. A map homotopic to the identity has $L(f)=\chi$, and Euler characteristic of a product givesConsequentlyThe product $g_1{g}_2$ itself is not determined by the printed assumptions. For every $h\ge 0$, translation by a nonzero torus point, times the identity on ${\Sigma }_h$, is a fixed-point-free map of ${\Sigma }_1\times {\Sigma }_h$ homotopic to the identity, and its genus product is $h$. Thus the exact conclusion is the criterion for fixed-point-free self-maps homotopic to the identity on products of surfaces, rather than a single value of $g_1{g}_2$.

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.

True: every integer occurs. The simple-connectivity assumptions in parts 1 and 2 are local to those statements and are not imposed here. For a connected closed oriented four-manifold, Poincare duality givessince $b_3=b_1$. This expression has no fixed sign or parity.

Use $\chi ({\mathbb{CP}}^2)=3$, from its three even-dimensional cells, and $\chi (S^1\times S^3)=0$, by Euler characteristic of a product. The Euler characteristic of a connected sum in dimension four isIt follows either by Mayer–Vietoris sequence after deleting balls, or by the intermediate-degree cohomology splitting and the single top and bottom generators. Therefore a connected sum of $r$ projective planes and $s$ copies of $S^1\times S^3$ hasGiven an integer $h$, choose an integer $s\ge \max \{0,\lceil (2-h)/2\rceil \}$ and put $r=h-2+2s\ge 0$. This produces a connected closed oriented four-manifold of Euler characteristic $h$. When $r=s=0$, use $S^4$. ThusIf simple connectivity were added, the possible values would instead satisfy $\chi =2+b_2\ge 2$. This distinction is recorded in Euler characteristics of closed oriented four-manifolds.

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.

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 $\boxed{\chi (X)\in 2\mathbb{Z}}$.