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Attach two one-handles to a four-dimensional zero-handle, then a two-handle along the commutator of their generators with the product zero framing. The result is $T^2\times D^2$. Its spine is the usual cell structure of the torus: one vertex, two one-cells, and a two-cell attached by $xy{x}^{-1}{y}^{-1}$. An embedded representative of its $H_2$ generator is obtained by taking a once-punctured torus carried by the two one-handles and capping its boundary by the core disk of the two-handle. It meets the relative co-core once and has self-intersection zero. The boundary is $T^3$, whose integral homology has ranks $1,3,3,1$.

A compression body is obtained from a product $F\times I$ by attaching one-handles to $F\times \{1\}$, allowing three-ball components as well. Its negative boundary is the retained copy of $F$ and its positive boundary is the other boundary component. A handlebody has empty negative boundary.

A genus-$g$ handlebody is a three-ball with $g$ one-handles attached. Its boundary is a closed orientable topological surface of genus $g$.

The usual compact closed-manifold hypothesis is needed for the first assertion, or a proper bounded-below Morse exhaustion in an appropriate noncompact version. As literally printed without any such condition, $f(x)=x$ on $\mathbb{R}$ is a Morse function with no critical points, while $b_0=1$. We prove the intended closed-manifold statement, which applies to the surface product later in the question.

By the Morse handle-attachment theorem, crossing an index-$i$ critical point attaches an $i$-handle, and no topological change occurs between critical levels. Retraction to the core disks of handles gives a CW complex model with $c_i$ cells in dimension $i$. Its cellular chain complex has $C_i\cong {\mathbb{Z}}^{c_i}$. Tensor with $\mathbb{Q}$, which computes the ranks $b_i$ of integral homology. Since homology is a quotient of a vector subspace of $C_i\otimes \mathbb{Q}$, $b_i\le c_i$. This argument has not used the Morse inequalities.

The strong Morse inequalities and top-dimensional equality areTheir weak form is $c_i\ge b_i$. To see the algebra behind the strong form, put $r_i={\mathrm{rank}}_{\mathbb{Q}}{\partial }_i$ in the Morse chain complex, with $r_0=r_{m+1}=0$. The kernel-image dimensions give $c_i=b_i+r_i+r_{i+1}$, so the alternating difference at $j$ is exactly $r_{j+1}\ge 0$. Conversely, given the strong inequalities, denote those alternating differences by $S_j$, with $S_{-1}=0$. Then $c_j-b_j=S_j+S_{j-1}\ge 0$, proving that they imply every weak inequality.

They are not equivalent as numerical restrictions. For the Betti numbers $b=(1,0,0,1)$ of $S^3$, the formal counts $c=(2,0,0,2)$ satisfy all weak inequalities and the Euler equality, but at $j=1$ give $c_1-c_0=-2<-1=b_1-b_0$. They therefore cannot be the counts of a Morse function on $S^3$; the strong inequalities detect that additional obstruction.

If all the strong inequalities are equalities, all $r_i$ vanish and $c_i=b_i$. An integer boundary matrix with rational rank zero is the zero matrix. Thus the integral complex itself has zero differential, not merely its rationalization. Consequentlyand higher homology vanishes. In particular there is no homology torsion. This is an integral consequence of a Perfect Morse function, rather than just an assertion about ranks.

For ${\Sigma }_g$, the integral homology is free with Betti numbers $(1,2g,1)$. The Künneth theorem therefore has no torsion terms, and the product's Poincaré polynomial isSumming the weak inequalities yields the lower boundChoose a surface Morse function $h$ with one minimum, $2g$ saddles and one maximum. It can be built from one closed disk, the standard $2g$ one-handles, and a closing closed disk, using the local forms $u^2+v^2$, $-u^2+v^2$, and $-u^2-v^2$ at the handle centres. Regular collars interpolate between the handles without new critical points. Now set $F(x,y)=h(x)+h(y)$. Its critical points are the ordered pairs of those of $h$, and its Hessian matrix is a block direct sum; the indices add. This product of Morse functions has exactly $(2g+2{)}^2$ critical points and realizes every displayed Betti count. Small bumps constant near the critical pairs can separate their values without altering the Hessians or creating new critical points.

The same minimum does not apply to arbitrary smooth functions. Here is an explicit finite-critical-point counterexample, avoiding any assumption that degeneracy leaves the Morse lower bound intact. On $T^2=(\mathbb{R}/\pi \mathbb{Z}{)}^2$, takeIt is well defined because shifting either coordinate by $\pi$ reverses two of its factors. Its derivatives are $\sin v\sin (2u-v)$ and $\sin u\sin (u-2v)$. If either $\sin u$ or $\sin v$ vanishes, both derivatives vanish only at $(0,0)$. Otherwise $2u-v$ and $u-2v$ must be multiples of $\pi$, giving exactly $(\pi /3,2\pi /3)$ and $(2\pi /3,\pi /3)$ in addition to the origin. At the first point the Hessian is $\sqrt{3}\left(\begin{array}{cc}1& -1/2\\ -1/2& 1\end{array}\right)$, positive definite; at the second it is its negative. They are a minimum and maximum. The origin is an isolated degenerate monkey saddle, with leading term $uv(u-v)$. Thus this three-critical-point function on a torus has exactly three critical points.

For $g>1$, perform connected-sum gluing of functions between this torus and a perfect Morse surface of genus $g-1$: delete a small neighbourhood of the torus maximum and of the other surface's minimum, and join their regular boundary circles by a monotone cylinder. Collar coordinates ensure nonzero differential throughout the neck, so precisely those two critical points are lost and none are added. The result on ${\Sigma }_g$ has $3+[2(g-1)+2]-2=2g+1$ critical points, retaining the degenerate point. For $g=1$, simply use $h_0$. Calling this smooth function $\tilde{h}$ and the earlier perfect Morse function $h$, the smooth product $\tilde{h}(x)+h(y)$ hascritical points. This explicitly proves the negative answer to the unrestricted smooth-minimum question; it does not assert that this smaller illustrative count is sharp.

Start with a disk in the zero-handle and attach two untwisted bands passing through the two one-handles. The four passages around its boundary occur in the order $x,y,x^{-1},y^{-1}$, giving an embedded once-punctured torus whose boundary is the attaching circle of the two-handle. Push its interior into the existing four-dimensional handlebody, keeping its boundary on the attaching locus, and cap it with the core disk of a handle. The matching product framing makes the union smooth after rounding its corner. This gives an embedded closed torus $T\subset W$.

The torus uses the core of the two-handle exactly once. Equivalently, it intersects that handle's relative co-core disk transversely in one point. Its cellular two-chain is therefore the generator of $C_2=\mathbb{Z}$, which is a cycle and is not a boundary because there are no three-cells. ThusThis is the zero section in the commutator handlebody of the torus $T^2\times D^2$.

There are no one-handles, so attaching the three two-handles to a four-ball cannot create a fundamental group generator: ${\pi }_1(W)=1$. The handle chain complex has $C_2={\mathbb{Z}}^3$, $C_1=0$ and $C_0=\mathbb{Z}$, givingTo calculate the boundary, orient the two long components down their central strands, and orient the small component clockwise. The six mutual crossings of the long components have negative sign, giving linking number $-3$. Each long component has two positive crossings with the small component, giving linking number $1$. The surgery linking matrix, in the order of framings $4,3,2$, is consequentlyChanging a component orientation changes the corresponding row and column signs, and does not change the resulting homology. The long exact sequence of a pair and Poincare-Lefschetz duality identify $H_2(\partial W)=\ker Q$ and $H_1(\partial W)=\mathrm{coker}Q$. The determinant is nonzero, so the kernel vanishes. The gcd of matrix entries is one; the two-by-two minors include $-6$ and $7$, so their gcd is also one. The Smith normal form therefore has diagonal $1,1,7$, andIn particular the sign of the braid crossings matters: replacing $-3$ by $+3$ while leaving both meridian linkings positive would represent a different link and would give the wrong determinant.

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

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.

A Heegaard splitting of a closed oriented three-manifold is a decomposition $Y=H_0{\cup }_{\Sigma }{H}_1$ into two handlebodies, with their common boundary the Heegaard surface $\Sigma$. With boundary present, the corresponding pieces are compression bodies, whose negative boundaries account for $\partial Y$.

For the given triangulation, take the barycentric subdivision. A regular neighbourhood $H_0$ of the original one-skeleton is a handlebody: thicken vertices to balls and edges to one-handles, then contract a spanning tree. The complementary region is a regular neighbourhood $H_1$ of the dual one-skeleton, whose vertices are tetrahedron centres and whose edges cross triangular faces. It too is a handlebody. Both graphs are connected, and their common boundary supplies the Heegaard splitting. If the original triangulation has $v$ vertices and $e$ edges, its Heegaard surface has genus $e-v+1$; the dual count gives the same number by $v-e+f-t=0$.

Choose an oriented meridian of a knot $\mu$ and the Seifert longitude $\lambda$ supplied by a Seifert surface for the null-homologous knot. For relatively prime integers $p,q$, rational Dehn surgery removes the interior of a tubular neighborhood and attaches a solid torus with its meridian of a solid torus on the unoriented slope $p\mu +q\lambda$. The choices $(p,q)$ and $(-p,-q)$ describe the same slope; $q=0$ is the original meridional filling. The boundary gluing reverses boundary orientation so that the oriented three-manifold extends across the filling.

For integral coefficients $n_i$, attach two-handles to $B^4$ along the components of the framed link, with their indicated Seifert framing shifts. The boundary operation removes $S^1\times D^2$ and inserts $D^2\times S^1$, with meridian $n_i{\mu }_i+{\lambda }_i$. Thus the compact oriented surgery trace satisfiesFor a finite rational coefficient, use a negative continued fractionReplace that component by an integrally framed chain of successive meridians with coefficients $a_0,\dots ,a_k$. Repeated slam-dunk moves give back $p/q$. Perform this replacement for every rationally framed component, leaving meridional fillings out. The resulting integral framed link has the same filled boundary, so its surgery trace proves that every such rational filling bounds a compact oriented four-manifold.

Interpret the paired $A,B,C$ disks as the three one-handles of a genus-three handlebody. Attach two-handles on the two displayed $\beta$ curves. Orient the three disk-crossing generators as $a,b,c$. Reading the signed crossings, starting at the upper-left portion of each attaching curve and changing the start point when needed, givesThe first says $abc=bca$ and the second $abc=cab$. Thus the fundamental group presentation isHere $[u,v]=uv{u}^{-1}{v}^{-1}$. Cyclically changing the starting point, reversing an attaching curve, or changing generator orientations gives equivalent presentations. With $h=abc$, these relations make $h$ central; eliminating $c=b^{-1}{a}^{-1}h$ givesFor the topological identification, thicken the diagram's two nested bands and identify the three paired disk mouths. The complement of those bands is the product region of a pair of pants with a circle; its two compressing curves are exactly $[a,bc]$ and $[ab,c]$. Equivalently, the standard cell decomposition of this product has three one-handles and the two commuting two-handle attachments shown. Thus this is a generalized Heegaard diagram of $P\times S^1$.

The Hopf fibration of $S^3$ has three disjoint regular fibers whose removal leaves $P\times S^1$. Its three fibers are the components of the torus link $T_{3,3}$, as is also apparent from the full three-strand twist in the later link diagram. This identifies $Y$ with that link exterior, using the diagram and product structure rather than just its fundamental group. In particular it is a link exterior in the three-sphere.

A zero-handle in an $n$-dimensional handle decomposition is $D^0\times D^n=D^n$. Its attaching region is empty, so it starts a connected component of the manifold. In dimension four it is a four-ball; adding one-handles and two-handles gives a four-dimensional handlebody with a two-dimensional CW spine.