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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.

Let $R=M{U}^{\ast }(\mathrm{pt})$ be the coefficient ring of complex cobordism. We use its standard coefficient theorem: $R\cong \mathbb{Z}[b_1,b_2,\dots ]$ with $|b_j|=-2j$. In particular $R$ is concentrated in nonpositive even degrees. Let $H$ be the quaternionic tautological line bundle on ${\mathbb{HP}}^2$, regarded as a complex rank-two bundle by right multiplication by a fixed imaginary quaternion. SetWe will proveThe argument must establish the multiplication as well as the additive groups, since collapse of a spectral sequence alone does not settle ring extensions.

The quaternionic projective space has one cell in each dimension $0,4,8$, giving integral cohomology $\mathbb{Z}$ in exactly these three degrees. The sphere bundle of $H$ is the quaternionic Hopf fibrationWrite $u=e(H_{\mathbb{R}})=c_2(H)$ for its ordinary degree-four class, fixing its sign by this equality. The Gysin sequence of a sphere bundle gives surjections, in fact isomorphisms,Indeed the corresponding neighboring cohomology groups of $S^{11}$ vanish. Both source and target groups are $\mathbb{Z}$, so these maps make $u$ and $u^2$ generators. There are no higher cohomology groups. This proves the cohomology ring of quaternionic projective space in the form $H^{\ast }({\mathbb{HP}}^2;\mathbb{Z})=\mathbb{Z}[u]/(u^3)$.

Apply the multiplicative Atiyah-Hirzebruch spectral sequence for complex cobordism:The skeletal construction obtains this sequence from the long exact sequences of successive skeleta; the quotients are wedges of spheres, and the first differential is the ordinary cellular differential with coefficient groups $R^t$. Its multiplication on the associated graded group is the ordinary cup product with these coefficients. The finite eight-dimensional skeleton makes the filtration finite, so there is no convergence ambiguity.

Only columns $s=0,4,8$ and even rows $t$ are nonzero. If $r$ is odd, the target column $s+r$ is odd and zero. If $r$ is even, the target coefficient degree $t-r+1$ is odd and zero. The same observations exclude incoming differentials. Consequently $E_{\infty }=E_2$, with associated graded $R$-module freely generated by $1,u,u^2$ in filtrations $0,4,8$.

The canonical map from complex cobordism to integral cohomology preserves the chosen complex orientation and hence sends $q$ to $u$. The class $q$ vanishes on the zero-dimensional skeleton since $R^4=0$, and its leading skeletal term is $u$ in filtration four. Multiplicativity makes the leading term of $q^2$ the generator $u^2$ in filtration eight. Thus $1,q,q^2$ lift the associated graded basis.

This resolves all module extensions explicitly. For a class of total degree $d$, subtract its coefficient $a_0\in R^d$ in filtration zero. From the remainder subtract $a_{1}q$, where $a_1\in R^{d-4}$ is its leading coefficient in filtration four, and then subtract $a_2{q}^2$ with $a_2\in R^{d-8}$. The remaining filtration is beyond dimension eight and is zero. Conversely a relation $a_0+a_{1}q+a_2{q}^2=0$ has zero leading coefficient in each of the three successive filtrations, forcing all three $a_j$ to vanish. ThereforeFinally $q^3$ has degree twelve. Every associated graded contribution in that total degree would have coefficient degree $12-s>0$ for $s=0,4,8$, and all such coefficients vanish. Thus $M{U}^{12}({\mathbb{HP}}^2)=0$ and $q^3=0$. The polynomial map $R[t]/(t^3)\to M{U}^{\ast }({\mathbb{HP}}^2)$ taking $t$ to $q$ is surjective by the basis argument and injective by its uniqueness. This proves the complex cobordism ring of the quaternionic projective plane as a graded ring over the stated coefficient ring, including the absence of any additional coefficient-dependent relation.

Write $k={\mathbb{F}}_p$. The Nottingham group $\mathcal{N}$ is the set of formal series $f(t)=t+{\sum }_{i\ge 2}{a}_i{t}^i$ over $k$, with ordinary composition $f\circ g$ as multiplication and the $t$-adic topology. The identity is $t$. A compositional inverse exists uniquely by solving its coefficients successively, since the linear coefficient is $1$.

Equivalently $\mathcal{N}$ is the group of continuous $k$-automorphisms of $k((t))$ whose action on $(t)/(t^2)$ is the identity. Such an automorphism is determined by the image of $t$, which is $t+O(t^2)$. To identify this with ordinary series composition without reversing multiplication, send $f$ to the automorphism $h(t)\mapsto h(f^{-1}(t))$. Its composite with the automorphism for $g$ is the one for $f\circ g$.

The depth filtration isEach ${\mathcal{N}}_m$ is open and normal, and the coefficient of $t^{m+1}$ identifies ${\mathcal{N}}_m/{\mathcal{N}}_{m+1}$ with $(k,+)$. Thus $[\mathcal{N}:{\mathcal{N}}_m]=p^{m-1}$, and $\mathcal{N}$ is the inverse limit of its finite p-group truncations.

Here is a local-field proof that every finite p-group embeds in $\mathcal{N}$. We first establish the realization step rather than assume the Nottingham embedding itself. Put $F=k((t))$ and $\Gamma =\mathrm{Gal}(F^{\mathrm{sep}}/F)$, and let $I⊴\Gamma$ be the inertia subgroup, the group homomorphism kernel of the action on the algebraic closure of the residue field. We use the standard additive normal basis theorem consequence $H^j(\Gamma ,F^{\mathrm{sep}})=0$ for $j>0$. It follows by taking direct limits of the finite-Galois normal-basis modules, whose additive cohomology vanishes. The Artin-Schreier sequencethen givesThe classes $t^{-m}$ for positive $m$ prime to $p$ are linearly independent: a nonzero finite linear combination has a most negative valuation prime to $p$, whereas a negative valuation of $z^p-z$ is divisible by $p$. Thus there are infinitely many independent continuous characters $\Gamma \to C_p$.

Induct on the order of a finite p-group $E$. Choose a central subgroup $C\cong C_p$ and put $Q=E/C$. By induction there is an onto map $\varphi :\Gamma \to Q$ with $\varphi (I)=Q$, corresponding to a totally ramified $Q$-extension. The central extension $1\to C\to E\to Q\to 1$ is specified by a two-cocycle. Its pullback along $\varphi$ is a coboundary because $H^2(\Gamma ,{\mathbb{F}}_p)=0$. Correcting a set-theoretic section by that coboundary gives a continuous homomorphism ${\rho }_0:\Gamma \to E$ lifting $\varphi$.

We must make the lift onto on inertia, not merely onto on the whole Galois group. Let $A=I\cap \ker \varphi$. On $A$, the lift ${\rho }_0$ takes values in $C$. Characters vanishing on $A$ form a finite-dimensional space: $\Gamma /A$ has a finite subgroup $I/A\cong Q$ and procyclic quotient $\Gamma /I\cong \hat{\mathbb{Z}}$, so a character is determined by finitely many values on generators of these two pieces. The infinite-dimensional character space therefore has infinitely many distinct restrictions to $A$. Choose a character $\chi :\Gamma \to C$ whose restriction to $A$ does not cancel ${\rho }_0{|}_A$, and set $\rho (\gamma )=\chi (\gamma ){\rho }_0(\gamma )$. Centrality of $C$ makes this another homomorphism lifting $\varphi$. Its image on $A$ is nontrivial and hence all of $C$, while its image on $I$ maps onto $Q$. Therefore $\rho (I)=E$. The fixed field of $\ker \rho$ is the required totally ramified finite Galois extension $K/F$ with group $E$. This proves the realization step for every finite p-group.

Since the residue field of $K$ is still $k$, choosing a uniformizer $u$ identifies $K$ with $k((u))$. Indeed successive subtraction of its residue coefficient and division by $u$ gives each element of its valuation ring a unique convergent expansion ${\sum }_{i\ge 0}{b}_i{u}^i$ with $b_i\in k$, and fractions give the Laurent expansions. Each element of $E$ preserves the valuation and fixes $k$, so sends $u$ to $a_{1}u+a_2{u}^2+\cdots$. The linear coefficient is a homomorphism $E\to k^{\times }$. Since $E$ is a $p$-group and $|k^{\times }|=p-1$, that homomorphism is trivial. Thus the faithful Galois action lies in $\mathcal{N}(k)$, completing the embedding proof.

An infinite profinite group is hereditarily just infinite when every open subgroup is infinite and every nontrivial closed normal subgroup of each such open subgroup is open. We sketch the normal-subgroup argument, giving the depth calculation in odd characteristic explicitly. For series of depths $r,s$,where $[f,g]=f^{-1}\circ g^{-1}\circ f\circ g$. The coefficient follows by comparing $f\circ g$ and $g\circ f$ in degree $r+s+1$: their cross terms are $ab(r+1)$ and $ab(s+1)$.

Let $H$ be open and $1\ne K⊴H$ closed. Choose $m$ with ${\mathcal{N}}_m\le H$ and choose $f\in K$ of depth $r$. For every $j\ge m$ with $j\not\equiv r(\mathrm{mod}p)$, commutation with $t+b{t}^{j+1}$ supplies an element of $K$ of depth $r+j$, with any prescribed nonzero leading coefficient by varying $b$. Thus every sufficiently high layer except possibly depths congruent to $2r$ modulo $p$ occurs in $K$.

When $p$ is odd, choose one available depth $u\ge r+m$ with $u\not\equiv 2r,r(\mathrm{mod}p)$. Such a residue exists even for $p=3$. An element $h\in K$ of that depth can be commuted with $t+b{t}^{k-u+1}$ for every sufficiently large missing depth $k\equiv 2r(\mathrm{mod}p)$. The new leading coefficient contains $2u-k\equiv 2(u-r)\ne 0$, so those missing layers occur too. Hence $K$ contains an element with any chosen coefficient in every depth $k\ge M$, for some $M$.

Successive coefficient correction now proves ${\mathcal{N}}_M\le K$: match the coefficient of an arbitrary target in depth $M$ by an element of $K$, correct the error in depth $M+1$, and continue. The partial products converge $t$-adically to the target and remain in $K$, which is closed. Thus $K$ is open. This proves hereditary just infiniteness for odd $p$.

For characteristic two, the leading coefficient alone misses even depths. The standard characteristic-two normal-closure collection lemma supplies the needed extra input: for every $m\ge 1$ and every nonidentity $f\in \mathcal{N}({\mathbb{F}}_2)$, the closed subgroup generated by the conjugates $f^g$ with $g\in {\mathcal{N}}_m$ contains ${\mathcal{N}}_M$ for some $M$. Its proof retains the next coefficients in pairs of group commutators to fill the parity gaps, then uses the same successive coefficient correction. Apply this lemma to $f\in K$ and ${\mathcal{N}}_m\le H$: every such conjugate lies in $K$, so $K$ contains the tail and is open. This states the additional collection result used in the characteristic-two sketch; the odd-prime leading-term argument is not being applied there. Since every open subgroup contains a tail and is infinite, the Nottingham group is hereditarily just infinite for every prime $p$.

For finite-dimensional Complex projective space ${\mathbb{CP}}^r$, use the filtration by its linear projective subspaces. The difference ${\mathbb{CP}}^j\setminus {\mathbb{CP}}^{j-1}$ is ${\mathbb{C}}^j$, so there is one topological cell in every even dimension $0,2,\dots ,2r$, and none in odd dimensions. Every cellular boundary vanishes for degree reasons. The integral cohomology is therefore $\mathbb{Z}$ in those even degrees and zero elsewhere.

For the cohomology ring, let $x\in H^2({\mathbb{CP}}^r;\mathbb{Z})$ be the Poincare dual of a projective hyperplane, using the complex orientation. Generic hyperplanes meet transversely, and a collection of $j$ of them intersects in ${\mathbb{CP}}^{r-j}$. The geometric interpretation of the cup product identifies $x^j$ with the dual of that intersection. Restricting to a linear ${\mathbb{CP}}^j$, the $j$ hyperplanes meet in one point, with positive sign because the transverse normal directions are complex lines. HenceThus $x^j$ is a primitive generator in degree $2j$, not merely a nonzero multiple. There are no classes beyond degree $2r$, soThe point case $r=0$ has ring $\mathbb{Z}$. Taking the union of the finite projective spaces gives $H^{\ast }({\mathbb{CP}}^{\infty };\mathbb{Z})=\mathbb{Z}[x]$, since each fixed cohomological degree stabilizes. This proves the cohomology ring of complex projective space computation.

For the rank bound for a nonsingular complex bilinear map, write $a={\dim }_{\mathbb{C}}U$, $b={\dim }_{\mathbb{C}}V$, $R=\mathrm{im}\varphi$, and $s={\dim }_{\mathbb{C}}R$. Assume $a,b\ge 1$. Injectivity with either nonzero factor fixed implies $\varphi (u\otimes v)\ne 0$ whenever $u,v\ne 0$. Thereforeis well defined and continuous. Scaling either vector only scales the output, and in local projective charts the formula is given by bilinear coordinates with no simultaneous zero. Let $h$ be the target's degree-two generator and $x,y$ the pullbacks of the two factor generators. The Künneth theorem givesFixing $[v]$ or $[u]$ makes $F$ a projective linear embedding induced by an injective linear map. Such an embedding pulls back the hyperplane class to the factor's hyperplane class. Since the two restrictions determine $H^2$ of the product, $F^{\ast }h=x+y$, with the generator of a point factor interpreted as zero.

Put $d=a+b-2$. In the source cohomology ring, every term in $(x+y{)}^d$ vanishes except the one with exponents $a-1,b-1$, andThe binomial coefficient is a positive integer and the final monomial generates the top integral cohomology. If $s\le d$, the class $h^d$ in ${\mathbb{CP}}^{s-1}$ would be zero. This contradiction provesWhen $a=b=1$, the nonzero pure-tensor image already gives $s\ge 1$. A positivity convention is genuinely necessary: if $U=0$, $V={\mathbb{C}}^2$, and $W=0$, all restrictions to the zero tensor subspaces are injective, but the literal bound would read $0\ge 1$. The displayed theorem is therefore the nonzero-factor version. If a nonzero factor is infinite-dimensional, either factor restriction already bounds the image dimension below by that factor dimension; the sum of two positive cardinal dimensions, with at least one infinite, is their maximum, giving the corresponding cardinal bound.

Use cohomology with coefficients in ${\mathbb{F}}_2={\mathbb{Z}}_2$. The cellular homology of real projective space has one cell in degrees zero, one and two, and its boundary coefficients become zero modulo two. Hence there is a one-dimensional cohomology group in each of those degrees. Let $x\in H^1(P;{\mathbb{F}}_2)$ be the Poincare dual of a projective line. Two distinct projective lines intersect transversely in one point, so $⟨x^2,[P{]}_2⟩=1$. ThusThis is the mod-two cohomology ring of real projective space in dimension two. The Künneth theorem over a field makes the cross product an isomorphism of graded rings. Writing $x,y$ for the factor pullbacks givesThere are no additional signs over ${\mathbb{F}}_2$.

The obstruction to a graph avoiding the diagonal is the mod-two diagonal class of the real projective plane. Its dual class $D\in H^2(P\times P;{\mathbb{F}}_2)$ is determined bywhere $i:P\to P\times P$ is the diagonal embedding. Test $z=x^2,xy,y^2$. Each diagonal pullback is $x^2$, with evaluation one. Since $x^2{y}^2$ is the product's top generator, comparison of coefficients givesIf the submanifold were the graph of a smooth map $f:P\to P$, its parametrization would be $j=(\mathrm{id},f)$. For some $\epsilon \in {\mathbb{F}}_2$, $f^{\ast }x=\epsilon x$, and thereforeOn the other hand, the Poincare dual of the diagonal comes from its relative Thom class in $H^2(P\times P,(P\times P)\setminus \Delta ;{\mathbb{F}}_2)$. It restricts to zero on the complement. A graph disjoint from the diagonal would factor through that complement and force $j^{\ast }D=0$, a contradiction. Such a submanifold cannot be the graph of a smooth map $P\to P$. The relative Thom-class argument also applies to continuous self-maps.

False. Take $M={\mathbb{CP}}^2\#{\mathbb{CP}}^2$ and $N=S^2\times S^2$, with their standard product and connected-sum orientations. They are closed simply connected four-manifolds. For the connected sum, removing a four-ball leaves the fundamental group unchanged, and the Seifert-van Kampen theorem across the separating three-sphere preserves simple connectivity. The Mayer–Vietoris sequence computes integral cohomology $\mathbb{Z},{\mathbb{Z}}^2,\mathbb{Z}$ in degrees zero, two and four, and zero in other degrees. The Künneth theorem gives the same groups for $S^2\times S^2$. Thus their Betti numbers agree and all their integral cohomology is torsion-free.

Their cohomology rings nevertheless differ. In $M$, let $u,v$ come from the two projective summands and let $z$ generate degree four. The cohomology ring of the connected sum of two complex projective planes hasThese relations follow by using cycles in the separate summands and the degree-one pinch projections onto them. In $N$, the sphere-factor generators $x,y$ satisfyConsequently every degree-two class of $N$ has squarean even multiple of the top generator, whereas $u^2$ in $M$ is a primitive generator. An integral graded-ring isomorphism cannot change this property, since it sends a generator of $H^4$ to its positive or negative. A homotopy equivalence induces a graded-ring isomorphism, so none exists. Equivalently, their intersection forms are $\mathrm{diag}(1,1)$ and $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, an odd form and an even form respectively.

True for two positive-dimensional factors. In fact degree three can be replaced by any nonzero mapping degree. Suppose $M=X^k\times Y^{m-k}$ with $0<k<m$. The factors are closed and connected. They are also orientable: an orientation-reversing loop in either factor would reverse the product orientation while the other factor stays fixed, contradicting the orientation of $M$.

Let ${\omega }_X,{\omega }_Y$ be their integral top orientation classes, and let $\alpha ,\beta$ be the corresponding projection pullbacks to $M$. The Künneth theorem and the product fundamental class giveFor $f:S^m\to M$, both $f^{\ast }\alpha$ and $f^{\ast }\beta$ vanish, since a sphere has no cohomology in those intermediate positive degrees. Naturality of the cup product therefore givesOn the other hand, the definition of degree three gives $f^{\ast }{\omega }_M=3{\omega }_{S^m}\ne 0$. This contradiction proves the nonzero degree from a sphere obstructs manifold products assertion. The product with a zero-dimensional point factor is the excluded trivial case.

Work with integral cohomology. For a frame $v=(v_1,\dots ,v_k)$ in the complex Stiefel manifold, defineThese fibres form the complement bundle on a complex Stiefel manifold, a complex vector subbundle of the trivial rank-$n$ bundle. Indeedis a smooth orthogonal projector of constant complex rank $n-k$; locally projecting suitable fixed basis vectors and applying Gram-Schmidt process provides bundle frames. The underlying real vector bundle has rank $r=2(n-k)$ and is oriented by its complex structure: a complex change of frame has real determinant equal to the positive number $|{\det }_{\mathbb{C}}A{|}^2$.

A point of its sphere bundle is $(v_1,\dots ,v_k,w)$ with $\Vert w\Vert =1$ and $w$ orthogonal to the frame. Appending $w$ gives a diffeomorphismwhose inverse forgets the last vector while retaining it as the unit vector in the complement. Thus the asserted homotopy equivalence is actually an identification of sphere bundles, for $0\le k<n$.

Start the induction at $V_0({\mathbb{C}}^n)=\{\mathrm{point}\}$, with cohomology the exterior algebra on no generators. SupposeThis is a free abelian group in every degree, and for $k\ge 1$ its smallest positive degree is $2n-2k+1=r+1$. Hence $H^r(V_k;\mathbb{Z})=0$. This vanishing holds at $k=0$ as well, since the base is a point and $r=2n>0$. The Euler class $e(E)$ belongs to this zero group and is therefore zero.

For this oriented rank-$r$ vector bundle, the Gysin sequence of a sphere bundle has segmentBecause $e(E)=0$, this becomesIn particular $H^{\ast }(V_{k+1};\mathbb{Z})$ is torsion-free. At $j=r-1$, choose a class $a\in H^{r-1}(V_{k+1};\mathbb{Z})$ with ${\pi }_{!}a=1$. Its restriction to the sphere fibre is an integral generator. The projection formula for sphere bundle integration, with the conventional fibre-integration sign, isIt can be obtained from the relative cohomology of the disk/sphere pair and the Thom isomorphism theorem. Consequently every cohomology class of $V_{k+1}$ is uniquely expressible as ${\pi }^{\ast }c+{\pi }^{\ast }b⌣a$: subtract the second term using its fibre integral, then use exactness to identify the remaining term; injectivity of ${\pi }^{\ast }$ and the projection formula prove uniqueness. This establishes an additive splitting by the old monomials and the old monomials multiplied by $a$.

To establish the ring, $|a|=r-1$ is odd, so graded commutativity of the cup product gives $a^2=-a^2$, and hence $2a^2=0$. The torsion-freeness just proved now gives $a^2=0$. Graded commutativity also gives the required anticommutation with each old odd generator, and the old relations persist under the injective ring map ${\pi }^{\ast }$. The additive splitting proves that there are no additional relations. This is exactly Gysin ring splitting for an odd-dimensional sphere bundle, and supplies a new exterior generator of degreeInduction therefore proves the full ring calculationOver $\mathbb{Z}$, graded commutativity by itself only kills twice an odd square; torsion-freeness is the essential extra step in this integral cohomology of a complex Stiefel manifold proof.

Finally an ordered orthonormal $n$-frame is a unitary matrix, so $V_n({\mathbb{C}}^n)=U(n)$. Each exterior generator may be included once or omitted from a basis monomial. The rank of degree $j$ is therefore the number of subsets of the generator degrees $1,3,\dots ,2n-1$ summing to $j$. Multiplication of their two-term generating factors givesThe highest exponent is $1+3+\cdots +(2n-1)=n^2$, agreeing with the real dimension of the unitary group, and the total rank is $2^n$.

Use ordinary integer surgery coefficients, so the positive trace has intersection form $(n)$. Let ${\hat{F}}_0\subset K_0$ be the capped Seifert surface. Label the Spin-c structures byOn the trace $X_n:S^3\to K_n$, let ${\hat{F}}_n^2=n$ and label its structures ${\mathfrak{r}}_j$ by $⟨c_1({\mathfrak{r}}_j),[{\hat{F}}_n]⟩=2j-n$. Restriction identifies ${\mathfrak{t}}_k$ with $j\equiv k(\mathrm{mod}n)$. In these compatible labels the Integer-surgery exact triangle in Heegaard Floer homology isHere boldface denotes the U-adic completion of Heegaard Floer homology. Only finitely many zero-surgery structures contribute, by the genus bound on a capped Seifert surface, so the middle direct sum causes no difficulty. The infinite sums of trace maps do require completion. The printed minus notation should be understood in this completed sense; with the polynomial version, a sum over every extension need not even have values in $\mathbb{Z}[U]$. The plus version has the same ordering and needs no completion.

The two ordinary surgery traces give $A_k$ and $C_k$. For $X_0:S^3\to K_0$, $H^2(X_0)=\mathbb{Z}$, and its Spin-c structures ${\mathfrak{q}}_i$ have determinant evaluations $2i$ on the capped Seifert surface. Their restrictions are the unique structure on $S^3$ and ${\mathfrak{s}}_i$ on $K_0$. Thus the component of $A_k$ in the $i$th summand is $F_{X_0,{\mathfrak{q}}_i}^{-}$ for $i\equiv k$. For the orientation-reversed and turned-around trace $V_n:K_n\to S^3$, $H^2(V_n)=\mathbb{Z}$ and ${\hat{F}}_n^2=-n$. Its structures ${\mathfrak{r}}_j$ still have characteristic evaluations $2j-n$, restrict to ${\mathfrak{t}}_{j\mathrm{mod}n}$ on $K_n$ and to the unique structure on $S^3$, andCharacteristic parity explains the allowed evaluations; changing a structure by the generator of $H^2$ changes its evaluation by two, and the boundary restriction reduces the label modulo $n$.

For $n>1$, $B_k$ has an auxiliary lens-space input. More precisely, take the three-boundary surgery cobordism, or equivalently a two-ended cobordism $Z_n:K_0\#L_n\to K_n$, where $L_n$ is positive $n$-surgery on the unknot. Its incoming lens-space factor is fixed to the distinguished structure ${\mathfrak{u}}_0$ and its canonical top Floer generator ${\Theta }_{L_n}$. ThenThis describes the boundary restrictions exactly, including the boundary that must not be suppressed. A concrete Kirby diagram for $Z_n$ starts with the split incoming surgeries of coefficients $0$ and $n$, then attaches a zero-framed circle linking each incoming component once. Its full linking matrix is $\left(\begin{array}{ccc}0& 0& 1\\ 0& n& 1\\ 1& 1& 0\end{array}\right)$. Quotienting cohomology by the two incoming surgery columns gives $H^2(Z_n)=\mathbb{Z}\oplus \mathbb{Z}/n$. Accordingly its Spin-c structures may be labeled $(i,q)$ with incoming restrictions $({\mathfrak{s}}_i,{\mathfrak{u}}_q)$ and outgoing restriction ${\mathfrak{t}}_{i-q}$. In particular, fixing $q=0$ gives exactly one extension for each $i$, and its outgoing class is $i\mathrm{mod}n$. Thus $B_k$ is a Heegaard Floer cobordism map with the fixed auxiliary generator, rather than an unqualified map of a trace from $K_0$ to $K_n$. When $n=1$, $L_n=S^3$ and all three arrows become the familiar sums of ordinary two-ended Heegaard Floer cobordism maps.

Choose orientations on the two sphere factors whose product is the specified orientation of $X$. Give the four factors of $Y=(S^2{)}^4$ the corresponding orientations in the order $1,2,3,4$. The Künneth theorem has no Tor functor terms here and givesLet $u_i\in H^2(Y;\mathbb{Z})$ be the pullback of the positively normalized cohomology class from factor $i$. The cohomology ring iswith $⟨u_1{u}_2{u}_3{u}_4,[Y]⟩=1$. Its generators commute by graded commutativity of the cup product.

Fix a point in each sphere. For each pair $i<j$, let $S_{ij}$ be the four-dimensional submanifold in which factors $i,j$ vary and the remaining factors equal their fixed points. Orient $S_{ij}$ by the ordered factors $i,j$. The six homology classesgive a coordinate sphere intersection basis by the homology cross product description in the Künneth theorem. Moving a fixed point in one of the other two factors produces a disjoint homologous representative of each $S_{ij}$. Hence every basis element has zero self-intersection number.

More precisely, the intersection pairing isFor complementary pairs the representatives intersect transversely at one point. The sign is positive because swapping two-dimensional factors contributes $(-1{)}^{2\cdot 2}=1$. For pairs with a common index, a factor fixed in both representatives can be displaced to make them disjoint. In the paired order $(12,34),(13,24),(14,23)$, the intersection form is three copies of $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$.

For the diagonal map $\delta :X\to Y$, write $a,b\in H^2(X;\mathbb{Z})$ for the two sphere generators. ThenThe coefficient of $[S_{ij}]$ in $[\Delta ]$ is $⟨u_i{u}_j,[\Delta ]⟩$. Pullback and the preceding cup product relations show that these coefficients are one for $12,14,23,34$ and zero for $13,24$. ThereforeThere are exactly two complementary pairings among these four terms, each counted twice in the square. ThusThis computes both the explicit homology class and its intersection pairing, including all orientation signs.

Use the standard orientations on the three spheres and the complex orientation on Complex projective space. Let $h\in H^2({\mathbb{CP}}^3;\mathbb{Z})$ satisfy $⟨h^3,[{\mathbb{CP}}^3]⟩=1$. The cohomology ring of complex projective space is $\mathbb{Z}[h]/(h^4)$. The Künneth theorem givesWrite the induced map on cohomology as $f^{\ast }h=pa+qb+rc$ with integers $p,q,r$. Naturality and graded commutativity of the cup product giveThe mapping degree is therefore $d=6pqr$. This is the factorial degree obstruction for products of two-spheres. Every positive value is a positive multiple of six, so the least possible positive degree is at least $6$.

To attain it, identify each oriented sphere with ${\mathbb{CP}}^1$. Send three projective linear forms to the coefficients of their product:whereNo product of nonzero linear forms is the zero polynomial. Rescaling any factor rescales all coefficients by the same nonzero scalar. The map is therefore well-defined and is a holomorphic map on the products of projective charts.

Choose a homogeneous cubic with three distinct roots, all in an affine chart. Its preimages are exactly the $3!=6$ orderings of its three projective linear factors. In root coordinates, the derivative of the elementary-symmetric coefficient map has determinant equal, up to sign, to the Vandermonde determinant, the product of the three pairwise root differences. This is nonzero for distinct roots, so all six preimages are regular. A nonsingular complex-linear derivative preserves the underlying real orientation, since its real determinant is the squared absolute value of its complex determinant. Each preimage consequently contributes $+1$ to the mapping degree. HenceThis polynomial multiplication map to complex projective space provides the requested example without assuming that a prescribed cohomology homomorphism is realizable by a continuous map.

All homology and cohomology in this solution use ${\mathbb{F}}_2$ coefficients, so no orientation hypothesis is necessary. The Poincare-Lefschetz duality degrees are complementary:The starred notation in the question must be understood with this degree reversal.

Turn the handle decomposition upside down. An absolute $k$-handle becomes a relative $(n-k)$-handle based on $\partial W$. The absolute cellular boundary coefficient counts intersections of an attaching sphere with a belt sphere. In the reversed handle decomposition, the same intersections are counted in the opposite order. Over ${\mathbb{F}}_2$ the two incidence matrices are transposes, with no sign ambiguity. Thus the absolute chain complex is the complementary-degree dual of the relative chain complex. Taking homology and using the universal coefficient theorem for cohomology over a field proves the displayed isomorphism. This is mod-two handle duality.

To obtain a perfect pairing from the assumed closed case, form the double of a manifold $D(W)$ from two copies of $W$. The fold map $D(W)\to W$ is a retraction, so the inclusion of either copy induces an injection on homology. If $a\ne 0$ in $H_k(W)$, its image in $H_k(D(W))$ is nonzero. Closed-manifold mod-two Poincare duality supplies a class $z\in H_{n-k}(D(W))$ with intersection $1$ against that image. Cut a representative of $z$ along the boundary and retain its part in the first copy; it defines $b\in H_{n-k}(W,\partial W)$. Equivalently, apply the quotient map $D(W)\to D(W)/W_{-}\cong W/\partial W$ and excision. Intersections with a representative of $a$ in the interior are unchanged, so $a\cdot b=1$. This proves nondegeneracy in the first variable. Mod-two handle duality gives equal finite dimensions for the two spaces, hence nondegeneracy in the second variable as well.

For the three-dimensional conclusion, setThe long exact sequence in relative homology says $L=\mathrm{im}\partial$, where $\partial :H_2(W,\partial W)\to V$. The closed-surface intersection pairing on $V$ is nondegenerate. A collar calculation gives the adjoint identityIndeed, push $x$ slightly into the collar: intersections with the relative surface representing $b$ correspond to its boundary intersections with $x$. By the absolute-relative perfect pairing already proved, $x$ is orthogonal to every $\partial b$ precisely when $i_{\ast }x=0$. Therefore $L^{\perp }=L$.

For any finite-dimensional space with a perfect pairing, $\dim L+\dim L^{\perp }=\dim V$. Hence the half-lives-half-dies theorem givesBut $H_1({\mathbb{RP}}^2;{\mathbb{F}}_2)\cong {\mathbb{F}}_2$ has dimension one. It cannot be the entire boundary of a compact three-manifold, because the displayed dimension would be $1/2$. This excludes nonorientable three-manifolds as well as orientable ones.

Let $q:{\mathbb{CP}}^8\to Q$ collapse the indicated Complex projective space. Write $x\in H^2({\mathbb{CP}}^8;\mathbb{Z})$ for the positive generator, so the cohomology ring of complex projective space is $\mathbb{Z}[x]/(x^9)$. The inclusion of ${\mathbb{CP}}^3$ sends $x$ to its corresponding generator and is an isomorphism in degrees $0,2,4,6$.

The CW complex structure makes the inclusion a cofibration, so the positive-degree cohomology of $Q$ identifies with the relative cohomology of the pair. Its long exact sequence shows that $q^{\ast }$ is injective in positive degrees, with image the additive subgroup generated by $x^4,x^5,x^6,x^7,x^8$. Equivalently, the surviving cellular cochain complex has one copy of $\mathbb{Z}$ in each of degrees $8,10,12,14,16$ and zero differential. Let $u_i$ denote the class with $q^{\ast }{u}_i=x^i$, for $4\le i\le 8$.

Naturality of the cup product, together with the injectivity of $q^{\ast }$, now determines all multiplication:Every other product of positive-degree basis elements is zero. Indeed, their degrees exceed $16$. In particular, the surviving classes do not form a polynomial algebra on a degree-eight generator: the independent classes in degrees $10,12,14$ must also be retained. This is an instance of the cohomology ring of a collapsed projective subspace.

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.