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Let $x,y$ be the generators associated with the two pairs of one-handle feet. Following the attaching circle across the identifications marked $a,b$ and $c,d$ gives the word $xy{x}^{-1}{y}^{-1}$, up to reversal and changing generator orientations. Thus the van Kampen theorem givesThe handle decomposition has cellular groups $C_0=\mathbb{Z}$, $C_1={\mathbb{Z}}^2$, $C_2=\mathbb{Z}$, with no higher groups. The boundary of the two-cell is the abelianized attaching word, which is zero. HenceThe zero framing is the product framing of the commutator handlebody of the torus, so the represented manifold is $T^2\times D^2$ and its intersection form is $(0)$. In particular its boundary is $T^2\times S^1=T^3$.

The boundary homology can also be obtained without using the product identification. By Poincare-Lefschetz duality, $H_3(W,\partial W)=H^1(W)={\mathbb{Z}}^2$ and $H_2(W,\partial W)=H^2(W)=\mathbb{Z}$. The map $H_2(W)\to H_2(W,\partial W)$ is its zero intersection form. The long exact sequence of a pair then gives split sequencesThe boundary is connected and oriented, so its zeroth and third homology are both $\mathbb{Z}$. Therefore

Use the boundary-orientation convention in which $-p/q$ surgery on the unknot gives $L(p,q)$. The negative continued fraction isAttach three two-handles to a zero-handle along a chain of unknots, with adjacent linking numbers $+1$ and framings $-3,-2,-2$. Equivalently, take the plumbing of oriented disk bundles over three spheres with Euler numbers $-3,-2,-2$. Successive slam-dunk moves change the boundary surgery coefficient to $-3-1/(-2-1/(-2))=-7/3$, proving that its oriented boundary is $L(7,3)$ in this convention.

The zero sections are a basis of second homology, and the intersection form isIts successive leading principal minors are $-3,5,-7$. Equivalently those of $-Q$ are $3,5,7>0$, so Sylvester's criterion proves that $Q$ is negative definite. This explicitly constructs the required filling.

If lens spaces are instead labeled by positive surgery with its boundary orientation, use the chain with framings $-2,-4$, whose coefficient is $-7/4$; its boundary is orientation-preservingly the positive-surgery $L(7,3)$, since orientation reversal changes $q=4$ to $q=3$. Its negative-definite form is $\left(\begin{array}{cc}-2& 1\\ 1& -4\end{array}\right)$. This is only a convention change in the named boundary, and specifies a valid filling in either labeling system.

For the unknot the three-manifolds are $S^3$, $U_0=S^1\times S^2$, and $U_n=L_n$, where $L_n$ denotes positive $n$-surgery with its surgery orientation. This fixes the orientation independently of competing conventions for the name $L(n,1)$.

Use the grading convention $\mathrm{gr}{\Theta }_{S^3}^{+}=0$ and $\mathrm{gr}{\Theta }_{S^3}^{-}=-2$. Let $R=\mathbb{Z}[U]$ and $\hat{R}=\mathbb{Z}[[U]]$. The ordinary minus groups arewhere $0\le k<n$ andThe lens-space absolute grading can be calculated from the standard torus surgery triple. For $0\le k<n$, the reverse trace structure with evaluation $2k-n$ has a small embedded holomorphic triangle with basepoint multiplicity zero and signed count $\pm 1$, taking $y_k$ to ${\Theta }^{-}$. Its degree shift is ${\Delta }_k=(1-(2k-n{)}^2/n)/4$. Thus $(d_k-2)+{\Delta }_k=-2$, which gives the boxed formula for $d_k$. Additional triangles wind around the surgery region and give the higher powers of $U$ calculated below.

Their infinity groups are obtained by inverting $U$: one Laurent tower for $S^3$ and each lens-space structure, two for the torsion structure on $S^1\times S^2$, and $\mathbb{Z}[U,U^{-1}]/(1-U^{|i|})$ in its non-torsion structures. Their plus groups are respectively ${\mathcal{T}}_0^{+}$, ${\mathcal{T}}_{1/2}^{+}\oplus {\mathcal{T}}_{-1/2}^{+}$, and ${\mathcal{T}}_{d_k}^{+}$. The hat groups are $\mathbb{Z}$ in degree zero, $\mathbb{Z}$ in each degree $\pm 1/2$, and $\mathbb{Z}$ in degree $d_k$. These follow directly by straightening the standard genus-one Heegaard diagrams: the lens space has one generator in each structure and no chain differential, while the two bigons in the torsion $S^1\times S^2$ diagram cancel with coherent signs. The sphere adjunction condition eliminates its non-torsion plus and hat groups, but not its ordinary minus group. The Non-torsion Heegaard Floer homology of the sphere-circle product calculation comes from a two-generator chain complex with chain differential $1-U^{|i|}$; its cokernel is the displayed quotient. For the triangle, complete the free minus towers by replacing $R$ with $\hat{R}$. The non-torsion summands disappear after completion, because $1-U^{|i|}$ has an inverse $1+U^{|i|}+U^{2|i|}+\cdots$ in $\hat{R}$.

For $k=0$, choose signs so that the exact triangle becomesThe zero-surgery trace has $\chi =1$, $\sigma =0$, and $c_1^2=0$ in the torsion structure; hence its absolute degree shift is $-1/2$, taking ${\Theta }^{-}$ to $x_{-}$. The auxiliary-generator map takes $x_{+}$ to $y_0$, so its effective grading shift isFor the underlying cobordism $Z_n$ in its torsion structure, $c_1^2=0$, $\chi =1$, $\sigma =0$, and its actual cobordism shift is $-1/2$. Tensoring with the canonical lens generator adds $d_0=(n-1)/4$ under the connected-sum grading convention, explaining the effective shift. For $n=1$ the auxiliary lens space is $S^3$, and this is simply the other degree-$-1/2$ surgery trace map.

The reverse trace components are explicit in every $k$. For $j=k+n\ell$, put $m_j=2j-n$. Its intersection form is $(-n)$, soIndeed $c_1^2=-m_j^2/n$, $\chi =1$, $\sigma =-1$, and matching source and target degrees gives precisely the displayed exponent. The exponent is nonnegative, and the coefficient is a unit: the infinity trace map is an isomorphism, and restriction to the free minus towers leaves the unique grading-compatible monomial.

With a compatible choice of Coherent orientations in Heegaard Floer homology, the sum $C_k$ is multiplication byFor $k=0$, pair $\ell$ with $1-\ell$: their exponents agree and their signs are opposite, so $P_0=0$, in agreement with the first triangle. For $0<k<n$ the term $\ell =0$ is the unique constant term, and every other exponent is positive. Thus $P_k=1+U(\cdots )$ is a unit in $\hat{R}$. The zero-surgery summand for such a $k$ vanishes, and the exact triangle isIt is isomorphic, as an ungraded completed-module triangle, to one with the final map the identity. The sum generally has no single absolute Heegaard Floer grading shift; each individual trace component has shift ${\Delta }_j$. This distinction and the power series explain why replacing the completed triangle by a polynomial-module triangle is not innocuous.

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.

The geometric argument uses the Mixed Heegaard Floer invariant defined in Question 5(iii). First suppose $g(\Sigma )>0$; the genus-zero qualification is discussed below. Write $r=\Sigma \cdot \Sigma \ge 0$. Blow up $r$ distinct points of the surface. Its proper transform $\tilde{\Sigma }$ in $\tilde{M}=M\#r{\overline{\mathbb{CP}}}^2$ has classChoose the extension of the Spin-c structure withSince $E_j^2=-1$, its exceptional evaluations are $-1$, so the Blowup formula for the mixed Heegaard Floer invariant has no extra power of $U$. Thus ${\Phi }_{\tilde{M},\tilde{\mathfrak{s}}}\ne 0$, andThis calculation is why the self-intersection appears with a plus sign.

It remains to bound the Chern pairing for a square-zero surface $S$. Its oriented normal bundle has Euler number zero and is trivial, so $\partial \nu S=S\times S^1$. Choose an embedded positive-square surface $T$ disjoint from $S$, with positive direction left in its complement, and cut along $N=\partial \nu T$. To see why the choice is possible, impose orthogonality to $[S]$ in the intersection form. A nonzero isotropic class uses at most one positive direction; $b_2^{+}\ge 3$ therefore leaves at least two. Choose an integral positive-square class in that complement, represent it by a surface, and remove algebraically cancelling intersections with $S$ by tubing along $S$. This alters genus but not the class. The neighborhood of $T$ supplies one positive direction and the complement supplies another. Since its normal Euler number is nonzero, $H^1(\nu T)\to H^1(N)$ is onto, proving that the cut is admissible.

Put the outgoing puncture inside $\nu S$, so that the plus Heegaard Floer cobordism map on the side containing $S$ passes through the Floer group of $\partial \nu S$. On the chain level this means splitting the cobordism along that boundary and composing its handle maps; with the fixed determinant class, every intermediate generator has restriction $\mathfrak{t}=\tilde{\mathfrak{s}}{|}_{S\times S^1}$. If the Chern pairing exceeded $2g(S)-2$, the given completed minus group would be zero. As in Question 3, this is the completed convention: for a non-torsion structure, ordinary infinity may retain $(1-U^N)$-torsion, whereas $1-U^N$ is invertible after completion. Such a pairing is positive for $g\ge 1$, so $\mathfrak{t}$ is non-torsion and the completed infinity group ${\mathbf{HF}}^{\infty }(S\times S^1,\mathfrak{t})$ is zero as well. The completed fundamental long exact sequence, whose plus group is the usual plus theory, then makes that plus group zero. The Heegaard Floer cobordism map, and consequently the Mixed Heegaard Floer invariant, would vanish, contradicting nonvanishing. ThereforeSubstituting the preceding blowup calculation provesReversing the orientation of the surface gives the usual absolute-value form of the Adjunction inequality for the mixed Heegaard Floer invariant.

The printed assumptions omit the positive-genus qualification. With the actual Floer groups, the asserted three-dimensional vanishing is false for $S=S^2$: the torsion structure on $S^2\times S^1$ has two nonzero minus towers, although its pairing is $0>-2$. Correspondingly, a small sphere embedded in a ball of a K3 surface has $[S]=0$, $S^2=0$ and Chern pairing zero, while the canonical Mixed Heegaard Floer invariant is nonzero; the desired inequality would read $0\le -2$. Thus the displayed proof establishes the intended positive-genus statement. If the supplied vanishing is instead taken literally as a formal additional axiom for every genus, the same factorization would force the inequality even for a sphere, but that axiom contradicts the actual $S^2\times S^1$ computation.

Remove two disjoint four-balls from $M$, giving $W:S^3\to S^3$. An Admissible cut for a Heegaard Floer mixed map is a separating three-manifold $N$ withThe last condition is the Mayer-Vietoris connecting-map condition. It makes a Spin-c structure on $W$ uniquely determined by its restrictions to the two pieces, avoiding an uncontrolled sum of extensions in the composition law. Such cuts exist when $b_2^{+}(M)>1$: represent an integral positive-square class by an embedded surface whose tubular neighborhood carries one positive direction and whose complement retains another. One can construct an embedded representative by resolving intersections of a representative cycle; the genus may increase without changing its class or square. The nonzero normal Euler number makes the restriction $H^1(\nu T)\to H^1(\partial \nu T)$ surjective, so the connecting-map condition holds. Puncture on either side of the cut.

Write $\mathfrak{t}=\mathfrak{s}{|}_N$ and ${\mathfrak{s}}_i=\mathfrak{s}{|}_{W_i}$. Both infinity Heegaard Floer cobordism maps vanish by positivity. Naturality with respect to the exact sequencegives $\mathrm{im}{F}_{W_1,{\mathfrak{s}}_1}^{-}\subset \ker \iota$ and $F_{W_2,{\mathfrak{s}}_2}^{+}\pi =0$. Exactness identifies $\delta$ with a degree-$-1$ isomorphism from $H{F}^{+}(N)/\mathrm{im}\pi$ onto $\ker \iota$, namely the two versions of Reduced Heegaard Floer homology. Hence the inverse on these reduced groups is defined, and one formsThis is the construction of the Mixed Heegaard Floer invariant, rather than the zero ordinary composite.

Let ${\Theta }^{-}$ be the top minus generator, of degree $-2$, and ${\Theta }^{+}$ the bottom plus generator, of degree zero. For $\zeta \in {\Lambda }^{\ast }(H_1(M)/\mathrm{torsion})$, insert its degree-one homology actions in the Heegaard Floer cobordism maps and defineThis is an integer up to the overall orientation sign. cobordism composition and associativity show independence of the admissible cut and handle choices. Equivalently, fixing the relevant orientation data removes the overall sign ambiguity.

The grading condition can be checked explicitly. Since $\chi (W)=\chi (M)-2$ and $\sigma (W)=\sigma (M)$, the sum of the ordinary cobordism shifts is $d(\mathfrak{s})+1$, where $d(\mathfrak{s})=(c_1(\mathfrak{s}{)}^2-2\chi (M)-3\sigma (M))/4$. The inverse connecting map contributes another $+1$. Therefore the output degree on $U^r{\Theta }^{-}\otimes \zeta$ is $d(\mathfrak{s})-2r-\mathrm{deg}\zeta$, and the displayed invariant vanishes unless $2r+\mathrm{deg}\zeta =d(\mathfrak{s})$.

Finally, two positive pieces are needed, so that both infinity maps vanish and the reduced-group factorization is available. For an admissible cut the connecting-map condition and signature additivity imply that the two positive subspaces inject as orthogonal subspaces of the closed intersection form; hence $b_2^{+}(M)\ge 2$. With only one positive direction no such admissible cut exists, and a chamber-dependent construction requires extra choices. With no positive direction, the ordinary infinity maps need not vanish at all. This explains the stated bound for the invariant defined here.

The map is a homogeneous Laurent polynomial ring homomorphism between the two rank-one infinity towers. Choose homogeneous generators $e_i$ with absolute degrees $q_i$. Its degree isConsequently the general graded-module description isIf the grading does not allow an integer exponent, the map is zero. Negative exponents are allowed in the infinity theory. The notation $F$ here denotes the map called ${\Phi }^{\infty }$ in the question.

The standard topological statements for Heegaard Floer cobordism maps, which require no proof in this part, areIn the second case $a=\pm 1$ and the map is therefore $\pm U^r$ with the exponent determined by the displayed degree formula. Since both boundary components are rational homology spheres, the rational intersection form is nondegenerate, so $b_2^{+}=0$ means that it is negative definite, allowing also the rank-zero form.

The printed hypothesis $b_1(Y_i)=0$ does not imply $b_1(W)=0$. Thus one must not replace the second implication by an unconditional isomorphism assertion for every negative-definite cobordism. A concrete check is $W=(S^1\times S^3)\setminus (B^4\bigsqcup B^4):S^3\to S^3$. It has $b_2^{+}=0$ and $b_1(W)=1$. A one-handle followed by a three-handle gives $e\mapsto e\otimes {\theta }^{+}$ and then kills the ${\theta }^{+}$ summand; hence its infinity map is zero. Its grading shift is one, also incompatible with a nonzero map between the even-graded $S^3$ towers. With only the stated boundary assumption, the general homogeneous Laurent-module description and the two topological implications above are the valid conclusions; determining any remaining coefficient requires the actual cobordism, not just its boundary Betti numbers.

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.

The three handle attachments give $H_2(W;\mathbb{Z})\cong {\mathbb{Z}}^3$: there are no one-handles or three-handles. Use the capped core disks of handles as the homology basis, ordered by the framing labels $-2,-1,+1$. The surgery linking matrix gives the intersection form. Orient the three components so that the outer components have linking number $3$ and the middle component has linking number $1$ with each outer component. There are six crossings between the outer components, and two crossings between the middle component and each outer component; counting their signs gives these linking numbers. Reversing a component changes the corresponding row and column signs and does not change the invariants.

Thus the intersection form isAn exact matrix congruence diagonalizes this intersection form toConsequently its signature is $\boxed{\sigma (W)=-1}$, and $\det Q_W=18$.

The long exact sequence in relative homology and Poincare-Lefschetz duality identifywith the intersection form matrix, and identify $H_1(\partial W)$ with its cokernel, since $H_1(W)=0$. The Smith normal form has first invariant factor $1$, because the entries have greatest common divisor $1$, and second invariant factor $1$, because the upper-left two-by-two matrix minor has determinant $1$. The last invariant factor is $18$. Hence

Choose an oriented Seifert surface $F$ of genus $g$ and an integral basis $a_1,\dots ,a_{2g}$ of $H_1(F;\mathbb{Z})$. Push $a_i$ in the positive normal direction to $F$. The Seifert matrix isIt represents the Seifert form. Its skew-symmetrization $V-V^T$ is the integral intersection form of $F$. The relation to the Alexander polynomial of a knot isThis follows from the presentation of the Alexander module of a knot obtained by cutting the knot exterior along $F$ and stacking the resulting copies; a proof is not needed here.

The printed genus inequality is false. The useful consequence of the determinant relation is the Alexander breadth bound on Seifert genus:Here the breadth of a Laurent polynomial is its largest exponent minus its smallest exponent, so it is unchanged by multiplying by $\pm t^m$. Indeed, for a minimal-genus Seifert surface, the matrix has size $2g_s(K)$ and each entry of $tV-V^T$ has degree at most one. Its nonzero determinant is an ordinary polynomial of degree at most $2g_s(K)$; its breadth is no larger than that degree. Equivalently, the highest exponent of a symmetrically normalized Alexander polynomial of a knot is at most $g_s(K)$. An unnormalized degree is not invariant under Laurent units.

For a counterexample to the printed inequality and the requested example, take the untwisted Whitehead double of a trefoil knot. The picture specifies the two-strand tangle in the zero Seifert framing; the two exterior caps form a Whitehead double clasp. The inset identifies the trefoil knot used as companion.

Its usual genus-one Seifert surface is a zero-framed annulus following the trefoil knot, joined by the clasp band. Choose the annulus core and a curve traversing the clasp band as the basis of its first homology. Zero annulus twisting gives the first self-linking number $0$. The clasp contributes a Hopf band with self-linking number $-1$ in the positive-clasp convention used here. Plumbing the bands contributes one linking in one push-off direction and none in the other. Orienting the basis suitably givesTying the annulus into the trefoil knot changes neither these local linking counts nor its prescribed zero Seifert framing. ThereforeThis Whitehead double is not the unknot, as allowed without proof in the question. Its Seifert genus is consequently at least one and at most one, hence exactly one. Thus $g_s(K)=1$ while every normalized constant Alexander polynomial of a knot has degree zero: a direct counterexample to the printed inequality.

Orient $S^3$ as the boundary of $B^4$. Push the interior of an oriented Seifert surface $F$ slightly into $B^4$, and let $X_F$ be the double branched covering of $B^4$ over this pushed-in surface. Its boundary is the two-fold branched cover of a knot. Define the knot signature byThe signature of a possibly degenerate intersection form means the number of positive eigenvalues minus the number of negative eigenvalues; its radical contributes zero.

Here is why the definition is independent of $F$. Given two choices $F_0,F_1$, glue their branched covers along the common boundary to obtainThis is the double branched covering of $S^4$ over the closed oriented surface obtained by gluing the pushed-in $F_0$ to $-F_1$. Every closed oriented surface in $S^4$ admits a Seifert hypersurface: its complement has an integral meridional class, represent it by a map to the circle, and take a regular value. Near the surface choose the angular map in its trivial normal disk bundle. The closure of a suitable regular fiber is a compact oriented three-manifold bounding the surface. Its trivial normal framing can be chosen to match this construction.

Push the interior of this Seifert hypersurface into $B^5$. The double branched covering of $B^5$ over the resulting properly embedded three-manifold has boundary $X$. The existence of the cover follows from the meridian homomorphism to $\mathbb{Z}/2$; smoothness near the branch set is the local map $(z,u)\mapsto (z^2,u)$. By the hypothesis about boundaries of five-manifolds, $\mathrm{sign}X=0$. Novikov additivity for gluing along an entire closed three-dimensional boundary givesThus the knot signature is independent of the Seifert surface. An isotopy of the knot carries the construction to an equivalent one, so it is also a knot invariant.

To calculate its intersection form, express $F$ as a disk with $2g$ bands. The double branched covering of $B^4$ over the pushed-in disk is again $B^4$. Each band lifts to a two-handle. The capped lifted cores give a basis of $H_2(X_F;\mathbb{Z})$ corresponding to the band-core basis of $H_1(F;\mathbb{Z})$. The framing of the $i$th lifted attaching circle is $2\mathrm{lk}(a_i^{+},a_i)=2V_{ii}$. For distinct bands, the two sheets contribute the two push-off linkings $V_{ij}$ and $V_{ji}$, so the mutual linking number is $V_{ij}+V_{ji}$. HenceA different integral basis changes this matrix by matrix congruence and leaves its signature unchanged.

In Figure 7 the upper knot is a connected sum of knots $T\#T$, while the lower one is $T\#\overline{T}$, where $T$ is the positive trefoil knot. Both knot groups have the same presentationTo see this, the Seifert-van Kampen theorem expresses the knot group of a connected sum of knots as the amalgamated free product of the two summand knot groups, amalgamating their meridians of a knot. A trefoil knot has presentation $⟨a,b\mid aba=bab⟩$ with $a$ a meridian. Reflection gives the same abstract group for the mirror. If it reverses the chosen meridian, send both generators to their inverses; the inverse braid relation is again the same relation. Thus the meridian amalgamation yields the displayed group presentation in both cases.

A band basis for the positive trefoil knot giveswhose symmetrization is negative definite. Therefore $\sigma (T)=-2$ and $\sigma (\overline{T})=2$. Boundary-connected-summing Seifert surfaces makes the Seifert matrix block diagonal, so the knot signature is additive under connected sum of knots. ConsequentlyThe upper knot and lower knot are therefore not isotopic, even though their knot groups are isomorphic. With the opposite global knot signature convention the first value is $+4$, and the distinction is unchanged.

For a CW complex $X$, let $X^k$ be its $k$-skeleton. The cellular chain complex isThe isomorphism uses excision and $H_k(D^k,S^{k-1})\cong \mathbb{Z}$, after choosing an orientation for each cell. In degree zero, take $X^{-1}=\varnothing$ and the free group on the vertices. The differential is the compositeThe long exact sequences of skeleton pairs imply $d_{k-1}{d}_k=0$. The cellular boundary formula describes its coefficient at a $(k-1)$-cell as the mapping degree of the attaching sphere after all other $(k-1)$-cells are collapsed. In degree one this is the signed difference of the two endpoints. The cellular homology theorem identifies the homology of this complex with singular homology: relative homology of successive skeletons is concentrated in their cell dimension, and the exact skeleton sequences leave precisely these kernels modulo images. This is the construction, rather than merely a count of cells.

The filtration ${\mathbb{CP}}^0\subset {\mathbb{CP}}^1\subset {\mathbb{CP}}^2$ gives one cell in each of dimensions zero, two and four, since ${\mathbb{CP}}^k\setminus {\mathbb{CP}}^{k-1}\cong {\mathbb{C}}^k$. There are no odd-dimensional cells, so every cellular differential vanishes. ThusThe complex projective line has one zero-cell and one two-cell. Its product with itself has one zero-cell, two two-cells and one four-cell. Its cellular chain complex again has zero differentials, givingA homeomorphism induces isomorphisms on homology groups. The second groups have different ranks, and therefore

Now orient the Complex projective plane by its complex coordinates. The inclusion of its two-skeleton ${\mathbb{CP}}^1$ identifies the fundamental class of a projective line with a generator $h$ of $H_2({\mathbb{CP}}^2;\mathbb{Z})\cong \mathbb{Z}$: the cellular generator has no incoming three-cell boundary and no outgoing boundary.

Take the projective lines $L_0=\{z_2=0\}$ and $L_1=\{z_1=0\}$. A path of unitary coordinate transformations takes one to the other, so their oriented homology classes are both $h$. They meet in the single point $[1:0:0]$. In its affine chart, let $w=z_1/z_0$ and $v=z_2/z_0$. Their tangent spaces are the complex $w$-axis and complex $v$-axis, so the intersection is a transverse intersection. The ordered real basesconcatenate to the complex orientation of the ambient four-manifold. Thus the local sign is $+1$, andThis computes the intersection form using distinct representatives of the same class, avoiding an ill-defined attempt to count a line intersecting itself as a set. Bilinearity now givesEquivalently, the map $H_2\to \mathrm{Hom}(H_2,\mathbb{Z})$ sending $a$ to $\lambda (a,-)$ is an isomorphism. This is exactly a unimodular intersection pairing, and proves projective lines generate a unimodular intersection form. Reversing the ambient orientation would change the matrix to $(-1)$ and would still be unimodular.

Use the standard complex orientations and the product orientation. Let $a,b\in H^2(S^2\times S^2;\mathbb{Z})$ satisfy $a^2=b^2=0$ and $⟨ab,[S^2\times S^2]⟩=1$, and let $x\in H^2({\mathbb{CP}}^2;\mathbb{Z})$ satisfy $⟨x^2,[{\mathbb{CP}}^2]⟩=1$. These are the cohomology rings of the product of two spheres and the Complex projective plane.

For a map from the Complex projective plane, write $f^{\ast }a=rx$ and $f^{\ast }b=sx$. Naturality of the cup product gives $r^2{x}^2=f^{\ast }(a^2)=0$ and $s^2{x}^2=f^{\ast }(b^2)=0$. Since $x^2$ has infinite order, $r=s=0$. It follows that $f^{\ast }(ab)=0$, so every such map has degree $0$.

In the reverse direction, write $g^{\ast }x=ra+sb$. ThenEvery even mapping degree really occurs. To construct it, identify $S^2$ with ${\mathbb{CP}}^1$ and use the Segre embeddingThe generator $x_3\in H^2({\mathbb{CP}}^3;\mathbb{Z})$ pulls back to $a+b$: restricting to either factor gives a projective line and hence coefficient $1$. By the cellular approximation theorem, $\sigma$ is homotopic to a cellular map. Its four-dimensional domain then maps into the four-skeleton ${\mathbb{CP}}^2$ of ${\mathbb{CP}}^3$. Denote that map by $g_0$; restriction of $x_3$ to ${\mathbb{CP}}^2$ is $x$, so $g_0^{\ast }x=a+b$ and $\mathrm{deg}{g}_0=2$.

For any integer $m$, choose a map $d_m:S^2\to S^2$ of mapping degree $m$. For $m>0$ one may use $z\mapsto z^m$ on the Riemann sphere, for $m<0$ use $z\mapsto {\overline{z}}^{|m|}$, and for $m=0$ use a constant map. Then $g_m=g_0\circ (d_m\times \mathrm{id})$ satisfies $g_m^{\ast }x=ma+b$, so the possible degrees are exactly all even integers, with $g_m$ realizing $2m$. The cellular approximation here deforms this particular map into the skeleton; it does not require a retraction of ${\mathbb{CP}}^3$ onto ${\mathbb{CP}}^2$.

Finally, take the connected sum of oriented manifolds with both summands carrying their standard complex orientations. Its degree-two cohomology has generators $u,v$ withThis follows by choosing the generators supported away from the two balls used to form the connected sum, so mixed cup products vanish while each square gives the common orientation class. WriteIf $\mathrm{deg}h=n$, the three cup product relations sayTaking determinants gives $-(\det P{)}^2=n^2$, which forces $n=0$. Therefore the only possible degree to the connected sum is $0$, realized by a constant map. This is a degree constraint from intersection forms: the indefinite intersection form of the sphere product cannot pull back a definite form with a nonzero degree multiplier.