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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.

The cellular chain complex for either space has one cell in dimensions $0,2,3,4$, with the only nonzero differential equal to multiplication by $p$ from degree three to degree two. The universal coefficient theorem for cohomology therefore gives, for both $X$ and $Y$,Every product of positive-degree classes vanishes for dimensional reasons, so

Their modulo-$p$ cohomology rings distinguish them. Let $u\in H^2(X;{\mathbb{F}}_p)$ be the class restricting to the standard generator on ${\mathbb{CP}}^2$. The attaching map has degree $p$, so its cellular coboundary vanishes modulo $p$; the classes in degrees two and four restrict isomorphically to those of ${\mathbb{CP}}^2$. Hence $u^2\ne 0$ in $H^4(X;{\mathbb{F}}_p)$. In $Y=M(\mathbb{Z}/p,2)\vee S^4$, the degree-two class comes from the three-dimensional Moore space, so its square is zero; products between distinct wedge summands also vanish. The mod-p cup-square obstruction to a homotopy equivalence now proves

The standard cellular chain complex of the Klein bottle givesThe universal coefficient theorem for cohomology therefore yields $H^0=\mathbb{Z}$, $H^1=\mathbb{Z}$, and $H^2=\mathbb{Z}/2$. The degree-one generator is pulled back from the base circle in the circle-bundle description of $K$, so its square is zero. Thus every positive-degree product vanishes. The same calculation for ${\mathbb{RP}}^2\vee S^1$ gives the integral cohomology ring of the wedge of the real projective plane and a circle, and hence

There is no map $Y\to K$ inducing this isomorphism. The Klein bottle is an aspherical space, and its fundamental group is torsion-free. Any map ${\mathbb{RP}}^2\to K$ therefore induces the trivial homomorphism $\mathbb{Z}/2\to {\pi }_1(K)$ and is null-homotopic. Its pullback on $H^2$ is zero, while the circle summand has no degree-two cohomology. Consequently every map $Y\to K$ is zero on $H^2(K;\mathbb{Z})$ and cannot be a cohomology isomorphism.

Every singular simplex has compact image, and a singular chain is a finite sum of simplices. The image of any chain is therefore compact and lies in some $X_a$. Directedness puts any finite collection of chains into one common $X_b$, soBecause filtered colimits of abelian groups are exact, kernels and images commute with this colimit. Taking homology gives the homology of a directed union:

For an open $U\subseteq {\mathbb{R}}^N$, use the directed family of finite unions of closed rational cubes contained in $U$. Every compact subset of $U$ lies in one such finite polyhedron, and each polyhedron has finitely generated cellular homology. There are only countably many of them, so their direct limit is countable. Thus every $H_i(U;\mathbb{Z})$ is countable.

Cohomology behaves differently because $\mathrm{Hom}$ turns a direct sum into a direct product. The connected open sethas one independent loop around each puncture, so $H_1(U;\mathbb{Z})\cong {⨁}_{n\ge 1}\mathbb{Z}$. Since $H_0(U;\mathbb{Z})=\mathbb{Z}$, the universal coefficient theorem for cohomology giveswhich is uncountable. This is the first cohomology of the countably punctured plane.

Because $S^n$ is simply connected for $n\ge 2$, the homological Serre spectral sequence has constant coefficients and only two nonzero columns:Its only possible nonzero differential isUsing the orientation generator of $H_n(S^n)$ to identify both columns with $H_{\ast }(F)$ defines the Wang homomorphismThe kernel and cokernel descriptions of the two surviving columns splice with the filtration of $H_{\ast }(E)$ to give the Wang sequence over a sphere

Let $F_m$ be the homotopy fiber of a degree-$m$ map $f_m:S^n\to S^n$. Apply this sequence to the fibrationAt the bottom, the map between the two copies of $H_n(S^n)$ is multiplication by $m$. It follows thatThe first torsion group can also be seen from ${\pi }_{n-1}(F_m)\cong \mathbb{Z}/m$ and the Hurewicz theorem; the Wang map then propagates it periodically.

The universal coefficient theorem for cohomology givesEvery product of two positive-degree classes is zero. For $n>2$, this follows immediately because the sum of two degrees of the form $k(n-1)+1$ is not of that form. For $n=2$, degree counting does not suffice, since $H^j(F_m)\cong \mathbb{Z}/m$ for every $j\ge 2$. Use instead the multiplicative cohomological Serre spectral sequence forIts transgression in degree one is multiplication by $m$. Every positive integral cohomology class that survives lies in filtration two, while the product of two such classes lies in filtration four; the base has dimension two, so filtration four is zero. Thus the reduced cohomology is a square-zero ideal, andas a graded ring, with zero multiplication on the second summand.

Repeated Smith normal form puts a chain complex of finitely generated free abelian groups into its elementary decomposition of a finite free chain complex: one-term summands $\mathbb{Z}$ and two-term summands $0\to \mathbb{Z}\stackrel{m}{\to }\mathbb{Z}\to 0$. Applying ${\mathrm{Hom}}_{\mathbb{Z}}(-,\mathbb{Z})$ reverses a two-term summand but keeps the same multiplication by $m$. Reading its homology gives the universal coefficient theorem for cohomologysplit noncanonically. Thus if $H_n(C)\cong {\mathbb{Z}}^{b_n}\oplus T_n$ with $T_n$ finite, then

The universal coefficient theorem for homology with ${\mathbb{F}}_p$ coefficients givesIf the middle group vanishes for every prime $p$, so does the tensor term. Any nonzero free summand survives for every $p$, and any nonzero finite summand survives for a prime dividing its order. Since integral homology is finitely generated, it must vanish in every degree. This is detection of integral acyclicity modulo primes.

For the displayed complex, writeUsing $f_{\#}{d}_C=d_D{f}_{\#}$, one obtainsThus $M$ is the mapping cone of $f_{\#}$ in this sign convention. Its long exact sequence in homology shows that $f_{\ast }$ is an isomorphism exactly when $H_{\ast }(M)=0$. If $f_{\ast }$ is an isomorphism with ${\mathbb{F}}_p$ coefficients for every prime, then $M\otimes {\mathbb{F}}_p$ is acyclic for every $p$. Prime-field detection makes $M$ integrally acyclic, so the integral long exact sequence gives

For a singular $(k+l)$-simplex $\sigma :[v_0,\dots ,v_{k+l}]\to X$, define the cochain-level cup product byIf $\alpha \in C^k(X,A;R)$ and $\sigma$ lies in $A$, its front $k$-face also lies in $A$, so the first factor vanishes. Thus $\alpha ⌣\beta \in C^{k+l}(X,A;R)$.

For the projections from $X\times Y$, the exterior product in cohomology isFor CW pairs over a field $R$, subject to the usual finite-type condition that makes the graded tensor product commute with the relevant direct products, the Künneth theorem says thatMore generally the conclusion holds over a principal ideal domain when one factor has degreewise finitely generated free cohomology. It fails over $\mathbb{Z}$ in general: $H^{\ast }({\mathbb{RP}}^2;\mathbb{Z})$ is $\mathbb{Z}$ in degree zero and $\mathbb{Z}/2$ in degree two, but the integral Künneth and universal coefficient theorem for cohomology calculations giveThe tensor product of the two cohomology groups has no degree-three term, so $\Phi$ is not surjective.

Use the homeomorphismwhich carries the diagonal $\Delta$ to $T^2\times \{0\}$. HenceThe punctured torus deformation retracts onto a wedge of two circles. Let $a,b$ be the degree-one generators from the first torus and $c,d$ those from the punctured torus. Since all cohomology groups are free, the Künneth theorem identifies the integral cohomology ring asThus $ab$, $ac$, $ad$, $bc$, and $bd$ form a basis in degree two; $abc$ and $abd$ form a basis in degree three; and all higher positive degrees vanish.

Write $F=K(\mathbb{Z}/4,1)$, $E=K(\mathbb{Z},2)$, and $B=K(\mathbb{Z},2)$. In total degree at most two, the $E_2$ page of the mod-two Serre spectral sequence hasand $E_2^{1,0}=E_2^{1,1}=0$. A periodic free resolution of the cyclic group gives $H_1(F;\mathbb{Z})=\mathbb{Z}/4$ and $H_2(F;\mathbb{Z})=0$. Hence $H^1(F;{\mathbb{F}}_2)\cong \mathrm{Hom}(\mathbb{Z}/4,{\mathbb{F}}_2)$, while the universal coefficient theorem for cohomology gives $H^2(F;{\mathbb{F}}_2)\cong \mathrm{Ext}(\mathbb{Z}/4,{\mathbb{F}}_2)$; both are one-dimensional.

The edge map $H^2(B;{\mathbb{F}}_2)\to H^2(E;{\mathbb{F}}_2)$ is induced by multiplication by $4$ and is therefore zero modulo two. Consequentlyis an isomorphism. The differential out of $E_2^{0,2}$ is zero, because the total space has a one-dimensional $H^2$ which must survive in filtration zero. Thus for every $r\ge 3$ and $p+q\le 2$,and all other groups in that range vanish.

It follows thatThe surviving filtration-zero class is the restriction of the degree-two class of $E$, so

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.

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.

The first integral singular homology of the Hawaiian earring has a direct summand isomorphic to $\mathbb{Q}$. This consequence of its structural homology theorem suffices, through the universal coefficient theorem for cohomology and nonzero Ext of the rationals with integer coefficients, to show that its integral $H^2$ is nonzero. The precise structural theorem is Theorem 3.1 of www.sas.rochester.edu/mth/sites/doug-ravenel/otherpapers/eda-kawamura2.pdf.

Zero Euler characteristic and $H_3=0$ give free rank $b_1-1$. The universal coefficient theorem for cohomology identifies $H^3(M;\mathbb{Z})=\mathbb{Z}/2$ with $\mathrm{Ext}(H_2(M;\mathbb{Z}),\mathbb{Z})$, so the only torsion in $H_2$ is one copy of $\mathbb{Z}/2$. In particular $b_1\ge 1$.

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.

For a connected closed oriented four-manifold, Poincare duality identifies $H_3$ with $H^1$. The universal coefficient theorem for cohomology identifies $H^1$ with $\mathrm{Hom}(H_1,\mathbb{Z})$ because $H_0=\mathbb{Z}$ has zero Ext. Consequently $H_3$ is torsion-free and its rank equals that of $H_1$.

The universal coefficient theorem relates homology or cohomology with different coefficient groups through tensor and Tor, or Hom and Ext, respectively. The universal coefficient theorem for homology is its tensor-Tor sequence; the universal coefficient theorem for cohomology is its Hom-Ext sequence.

The universal cover of a connected closed manifold of dimension three with finite fundamental group is a closed simply connected three-manifold. It is orientable, since its orientation character has trivial domain. Its first integral homology vanishes, and the universal coefficient theorem for cohomology then gives $H^1=0$. Integral Poincare duality gives $H_2=H^1=0$, while $H_0$ and $H_3$ are $\mathbb{Z}$. Thus it is an integral homology sphere. This proof does not require classifying simply connected three-manifolds.