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For a singular simplex $\sigma$ of dimension $n$ and a degree-$p$ cochain, use $\sigma ⌢\alpha =\alpha (\sigma |[v_0,\dots ,v_p])\sigma |[v_p,\dots ,v_n]$, setting the result to zero if $p>n$. A continuous map $f$ commutes with both face restrictions, and the pullback cochain evaluates by composing with $f$. Therefore $f_{\ast }(\sigma ⌢f^{\ast }\alpha )=(f\circ \sigma )⌢\alpha$ already on chains. Passing to homology and cohomology proves the displayed cap product identity.

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

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 an $n$-dimensional singular simplex $\sigma$ and a degree-$p$ singular cochain $\alpha$, use the front-face/back-face conventionand set it to zero for $p>n$. Extend linearly over $R$. Passing to homology classes of cycles and cohomology classes of cocycles defines the cap product $H_n(X;R)\times H^p(X;R)\to H_{n-p}(X;R)$.

For a continuous map $f$, the cochain pullback satisfies $(f^{\ast }\alpha )(\tau )=\alpha (f\circ \tau )$, and the induced chain map sends $\sigma$ to $f\circ \sigma$. Since composition commutes with restrictions to faces,The identity holds on chains, so it descends directly to classes and proves the naturality of the cap product: $f_{\ast }(x⌢f^{\ast }\alpha )=f_{\ast }(x)⌢\alpha$. No additional well-definedness verification is required here.

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

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

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

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

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:

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 $R$.

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.