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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:

Use singular homology with integer coefficients. The Künneth theorem for triangulable spaces gives a split short exact sequence with tensor terms ${⨁}_{i+j=k}{H}_i(X)\otimes H_j(Y)$ and torsion terms ${⨁}_{i+j=k-1}{\mathrm{Tor}}_1(H_i(X),H_j(Y))$. When one factor's homology groups are free abelian, the torsion terms vanish and the tensor map is an isomorphism.

The circle has $H_0(S^1)=H_1(S^1)=\mathbb{Z}$ and all other groups zero. Inductively, the homology groups of a torus are free, andPascal's identity givesGeometrically the generators correspond to choosing $k$ of the $n$ circle factors and taking their product, consistent with the cellular model having one $k$-cell for each such subset and cancelling opposite boundary faces.

The irregularity of an algebraic surface and the geometric genus of an algebraic surface are respectively

If $\pi :X^{\prime }\to X$ is the blowup at a point with exceptional divisor $E$, thenA holomorphic two-form on $X$ pulls back to one on $X^{\prime }$. Conversely, a holomorphic two-form on $X^{\prime }$ descends across $E$: locally it is a form on the punctured smooth surface $X\setminus \{p\}$, and its coefficients extend over the codimension-two point $p$. Equivalently, ${\pi }_{\ast }{K}_{X^{\prime }}=K_X$. Pullback is therefore an isomorphismwhich proves the birational invariance of the geometric genus of a surface in this case.

Let $X=C\times D$, where both smooth projective curves have positive genus. The Künneth theorem gives the irregularity of a product of curvesIf a surface is a hypersurface in projective space, dimension forces it to be a smooth hypersurface $Y\subseteq {\mathbb{P}}^3$. Fromand the intermediate cohomology vanishing for line bundles on projective space, one obtains $H^1(Y,{\mathcal{O}}_Y)=0$. Thus every such hypersurface has irregularity zero, whereas $X$ has positive irregularity. Hence $C\times D$ is not isomorphic to a hypersurface. This is the product of positive-genus curves is not a projective hypersurface obstruction.

The Albanese variety $\mathrm{Alb}(X)$ is the universal abelian variety receiving a pointed morphism from $X$, andLet the smooth image curve of the Albanese morphism be $C$ of genus $g$. Pullback of holomorphic one-forms along the dominant map $X\to C$ is injective, giving $g\le q(X)$. On the other hand, the universal property of the Jacobian extends $C\to \mathrm{Alb}(X)$ to a homomorphismThe Albanese image generates the entire Albanese variety, so this homomorphism is surjective and $q(X)\le \dim \mathrm{Jac}(C)=g$. Therefore the Irregularity from a smooth Albanese curve image is

An orientation of a vector bundle of real rank $d$ over $R$ is a coherent choice of generator of $H^d(E_b,E_b\setminus \{0\};R)$ in every fibre. For a complex bundle of rank $m$, a complex basis gives the real basis $(v_1,i{v}_1,\dots ,v_m,i{v}_m)$. A complex change of basis has positive real determinant $|{\det }_{\mathbb{C}}A{|}^2$, so these local choices agree and give the canonical orientation of a complex vector bundle over $\mathbb{Z}$, hence over every commutative ring $R$.

Let $D(E)$ and $S(E)$ be the disk and sphere bundles of an $R$-oriented rank-$d$ bundle over compact $B$. A Thom class $u_E\in H^d(D(E),S(E);R)$ restricts to the chosen generator on every fibre. The Thom isomorphism theorem states thatThe Euler class of a vector bundle is $e(E)=s^{\ast }{u}_E\in H^d(B;R)$ for the zero section $s$. Substituting the Thom isomorphism into the long exact sequence of the pair $(D(E),S(E))$ gives the Gysin sequence of a sphere bundle

Apply this to the Hopf fibration and induct on $n$. If $x=c_1(\mathcal{O}(1))$ has degree two, the result is the cohomology ring of complex projective space

The Künneth theorem over a principal ideal domain gives a natural short exact sequence with tensor-product and Tor terms; the sequence splits, though not naturally. Here all groups are free, so the Tor term vanishes andIf a graded-ring automorphism sends $x$ to $ax+by$, then $(ax+by{)}^{n+1}=0$. The coefficients of the nonzero mixed monomials force $ab=0$; the same applies to the image of $y$. Invertibility then forces a signed permutation matrix on the basis $x,y$. Conversely, complex conjugation on either factor changes the sign of its degree-two generator, and swapping the factors exchanges $x$ and $y$. Thus the realizable group isthe group of all signed $2$ by $2$ permutation matrices, as described by the cohomology automorphisms of a product of two complex projective spaces.

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.

Let $p:\mathbb{P}(E)\to X$ be the projectivization of a real vector bundle, let $L_E$ be its tautological bundle, and putThe mod-two projective bundle formula says that $H^{\ast }(\mathbb{P}(E);{\mathbb{F}}_2)$ is a free $H^{\ast }(X;{\mathbb{F}}_2)$-module on $1,u,\dots ,u^{d-1}$. The Projective bundle definition of Stiefel–Whitney classes is the unique relation

Apply the splitting principle for real vector bundles. After an injective pullback, writeIf $x_i=w_1(L_i)$, the projective-bundle relation factors assoThe line summands of $E\oplus E^{\prime }$ are the union of the two lists, henceComparing the degree-$k$ components proves the Whitney product formula for Stiefel–Whitney classesInjectivity of the splitting pullback returns the identity to $X$.

For real line bundles, the transition functions take values in $O(1)=\{\pm 1\}$. Tensor product multiplies these signs, while the identification $\{\pm 1\}\cong \mathbb{Z}/2$ turns multiplication into addition. The corresponding degree-one characteristic classes therefore satisfy the First Stiefel–Whitney class of a tensor product of real line bundles formulaEquivalently, this follows from the classification of real line bundles by $H^1(X;{\mathbb{F}}_2)$.

Now take $M={\mathbb{RP}}^n$ and $E=k{\gamma }_{\mathbb{R}}^{1,n+1}$. Sincea line in $E_x$ is the fixed line ${\gamma }_x$ tensored with a line in ${\mathbb{R}}^k$. Thus the projectivization of copies of the real tautological line bundle isLet $x$ and $v$ be the degree-one generators pulled back from the first and second factors. The mod-two cohomology ring of real projective space and the Künneth theorem giveThe tautological line $L_E$ is the tensor product of the two tautological lines, so $w_1(L_E)=x+v$. In the alternative generator $u=w_1(L_E)$, the same ring is

The stable tangent-bundle identitygivesFor the vertical part of the tangent bundle of a projectivized real vector bundle,Each of the $k$ line summands on the right has first Stiefel–Whitney classThe Whitney product formula for Stiefel–Whitney classes therefore yieldsthe Total Stiefel–Whitney class of the projectivization of copies of the tautological line.

Let $a$ and $b$ be the pullbacks of the standard generator of $H^n(S^n;\mathbb{Z})$ from the two factors. The Künneth theorem and graded commutativity of the cup product give

Each homeomorphism acts invertibly on $H^n\cong {\mathbb{Z}}^2$. To respect ordinary composition, send $h$ to $(h^{-1}{)}^{\ast }$; functoriality of induced map on cohomology then defines a homomorphism

For $n=1$, the space is the torus. Every matrix in $GL(2,\mathbb{Z})$ induces a linear homeomorphism ${\mathbb{R}}^2/{\mathbb{Z}}^2\to {\mathbb{R}}^2/{\mathbb{Z}}^2$, so the image is all of $GL(2,\mathbb{Z})$.

For $n=2$, write $h^{\ast }a=pa+qb$. Since $a^2=0$ and $ab=ba$,so $pq=0$; the same argument applies to $h^{\ast }b$. Invertibility then forces the matrix to be a signed permutation matrix. Every such matrix is realized by swapping the two sphere factors and applying an orientation-reversing homeomorphism to either factor. Thus the image consists exactly of the eight signed permutation matrices.

The Schur multiplier of a group $G$ isthe second group homology group with trivial integral coefficients. If $G\cong F/R$ is a free presentation, Hopf's formula states that

Write $I_F=\ker (\mathbb{Z}F\to \mathbb{Z})$ for the augmentation ideal. The presentation relation sequence iswhere $r[R,R]\mapsto 1\otimes (r-1)$. If $F$ is free on a set $S$, then $I_F$ is free as a left $\mathbb{Z}F$-module on the elements $s-1$, so the two modules immediately preceding $\mathbb{Z}$ are free $\mathbb{Z}G$-modules. Resolving the relation module $R/[R,R]$ by free modules and splicing produces a free resolution of $\mathbb{Z}$.

Apply the right-exact functor $\mathbb{Z}{\otimes }_{\mathbb{Z}G}-$ to this partial resolution. Its degree-two homology is the kernel ofThe coinvariant module on the left is $R/[F,R]$. On the right, the map $f-1\mapsto f[F,F]$ identifies the coinvariants with the abelianization $F/[F,F]$. The displayed map is induced by the inclusion $R\hookrightarrow F$, so its kernel isThis proves Hopf's formula.

For an abelian group $A$, the Schur multiplier of an abelian group is $M(A)\cong {\bigwedge }^{2}A$. One way to see the direct-sum rule is the degree-two Künneth theorem:A cyclic group has zero second integral group homology, whileConsequently

The integral Künneth theorem gives a natural short exact sequence containing tensor and Tor terms. Here all cohomology groups are free abelian, so the Tor terms vanish and the external cup product is a ring isomorphism. Thus

A rank-$d$ vector bundle $\pi :E\to B$ is $R$-oriented when its fiber groups $H^d(E_b,E_b\setminus \{0\};R)$ admit a locally coherent choice of generator. Equivalently, it has a Thom class $u_E\in H^d(D(E),S(E);R)$ restricting to that generator on every fiber. The Thom isomorphism theorem states thatThe Euler class of a vector bundle is $e(E)=s^{\ast }{u}_E\in H^d(B;R)$, where $s$ is the zero section. The Gysin sequence of a sphere bundle is

Over ${\mathbb{F}}_2=\mathbb{Z}/2$, the required rings areas in the cohomology ring of complex projective space and the mod-two cohomology ring of real projective space. By the Künneth theorem, the base has ring ${\mathbb{F}}_2[x,y]/(x^{n+1},y^{m+1})$.

Let $L$ be the underlying real plane bundle of the complex tautological bundle on ${\mathbb{CP}}^n$, and let $\lambda$ be the real tautological line bundle on ${\mathbb{RP}}^m$. Under the splitting principle, write the formal Stiefel-Whitney roots of $L$ as $a,b$, so $a+b=w_1(L)=0$ and $ab=w_2(L)=x$. Tensoring with $\lambda$ adds $y=w_1(\lambda )$ to each root. Thus the mod-two Euler class of $L{\otimes }_{\mathbb{R}}\lambda$ is

For $n=m=2$, put $B={\mathbb{CP}}^2\times {\mathbb{RP}}^2$. Multiplication by $e=x+y^2$ onhas ranks $1,1,2,1,1,0,0$ from degrees $0$ through $6$. More explicitly, it is injective through degree three; in degree four its kernel is generated by $x^2+x{y}^2$, and all of degrees five and six lie in its kernel. The Gysin sequence therefore yieldsfor the unit sphere bundle $X=S(L\otimes \lambda )$.

For a space with finite-dimensional homology over a specified field and only finitely many nonzero Betti numbers, this polynomial records their graded dimensions. Its value at one is the total Betti number, and its value at minus one is the Euler characteristic. The Künneth theorem gives $P_{M\times N}(t)=P_M(t)P_N(t)$ over a field. For integer homology ranks, use the rational Betti numbers.

The Cartesian product $S^p\times S^q$ has one cell in each of dimensions $0,p,q,p+q$, counting the two middle cells separately when $p=q$. For positive $p,q$, its integral cohomology ring has generators $a,b$ of degrees $p,q$, with $a^2=b^2=0$ and $ba=(-1{)}^{pq}ab$. The class $ab$ is a product orientation class. This follows from the Künneth theorem.

For a Serre fibration over $S^r$, cover the base by a contractible punctured sphere and a contractible ball. Excision relative to the preimage of the punctured sphere identifies the relative pair with the product of $F$ and $(D^r,S^{r-1})$ up to homotopy. The base pair has a single free homology group in degree $r$, so the Künneth theorem gives the displayed shift without a torsion term. The relative long exact sequence then gives $\cdots \to H_i(F)\to H_i(E)\to H_{i-r}(F)\to H_{i-1}(F)\to \cdots$. The connecting map retains the twisting of the original Serre fibration.