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For a torsion Spin-c structure, the relative grading of Heegaard Floer homology lifts canonically to a rational absolute grading, normalized by the bottom plus generator of $S^3$ having degree zero. Multiplication by $U$ has degree $-2$, and a Heegaard Floer cobordism map has degree $(c_1^2-2\chi -3\sigma )/4$. The correction term $d(Y,\mathfrak{s})$ is the lowest grading of the image of $H{F}^{\infty }$ in $H{F}^{+}$ for a rational homology sphere. A pure tower then has bottom plus degree $d$ and top minus degree $d-2$ in the $i<0$ minus convention. Keeping that shift explicit prevents errors of two in surgery and mixed-invariant calculations.

Heegaard Floer homology assigns groups $H{F}^{\circ }(Y,\mathfrak{s})$ to a closed oriented three-manifold with a Spin-c structure. A pointed Heegaard diagram of genus $g$ gives product Lagrangian tori ${\mathbb{T}}_{\alpha },{\mathbb{T}}_{\beta }\subset {\mathrm{Sym}}^g(\Sigma )$; the chain complex is generated by their intersection points and its differential counts Whitney disk classes represented by holomorphic disks. The minus, infinity, plus and hat versions use different powers or quotients of the variable $U$. The Heegaard Floer cobordism maps and the Mixed Heegaard Floer invariant relate these groups to smooth four-dimensional topology.

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.

Completing the minus chain complex in powers of $U$ replaces $\mathbb{Z}[U]$ by the formal power series ring $\mathbb{Z}[[U]]$; its homology is written ${\mathbf{HF}}^{-}$. This permits infinite sums of Heegaard Floer cobordism maps whose exponents tend to infinity, as occur in surgery formulas. A series with constant coefficient $\pm 1$ is a unit: its inverse is found recursively, or by the geometric series for its positive-order part. Such a series need not have a polynomial inverse, which is why exact triangles involving sums over all Spin-c extensions require a completion in the minus theory. Individual homogeneous cobordism maps are still defined on the ordinary minus groups.