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Two nonseparating compressing disks on opposite sides with the same boundary form an embedded sphere. Cutting each handlebody along its disk leaves it connected, so the sphere is nonseparating in the glued three-manifold. Capping the two resulting sphere boundaries gives a splitting of one lower genus of the remaining summand; the removed summand is $S^1\times S^2$.

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 generalized Heegaard diagram permits compression bodies and a nonempty negative boundary. Equivalently, a suitable handlebody with two-handles attached along only part of a compressing system can retain torus boundary components. Disk crossings supply generators and attaching-curve words supply relators for its fundamental group.

A Heegaard splitting divides a closed three-manifold into two handlebodies with common boundary. For a three-manifold with boundary, one uses compression bodies, which can retain negative boundary components.

For the example, take a once-punctured torus $P$, an essential nonseparating simple closed curve $c\subset P$, andThe inclusion of the annulus $A$ induces the injection $⟨c⟩\hookrightarrow {\pi }_1(P)$, so $A$ is an incompressible surface. Cutting gives $(P\text{ cut along }c)\times I$, a pair of pants times an interval. A compact connected surface with boundary times an interval is a handlebody of genus $1-\chi (P)$: cut the surface to a disk by proper arcs and reverse those cuts as one-handle attachments. Both the punctured torus and the pair of pants have Euler characteristic $-1$. ThereforeThis self-reproducing incompressible annulus cut also shows why arbitrary incompressible cutting choices need not terminate.

A three-manifold hierarchy is a finite sequence of cuts along properly embedded two-sided incompressible surfaces, with no spherical components, whose final manifold is a disjoint union of three-balls. Disks are permitted. In an orientable ambient manifold two-sided surfaces are orientable; for an ambient manifold which is not orientable we permit two-sided surfaces without assuming their orientability as well.

For $S^1\times F$, first cut along $\{p\}\times F$, obtaining $I\times F$. This surface is two-sided and its fundamental group injects as a direct factor. Next choose essential two-sided curves in $F$ until the cut surface has boundary, and cut along their products with $I$. For orientable $F$ one nonseparating curve suffices. For $F$ which is not orientable and has genus at least two, the boundary of an embedded Möbius band is a two-sided essential curve; neither side is a disk, so its fundamental group injects. Cutting this curve leaves pieces with boundary, including the Möbius band. For each resulting compact surface choose a finite system of proper arcs cutting it into disks. The corresponding vertical disks give the remaining three-manifold hierarchy cuts. Essential annuli have injective fundamental group, and disks have trivial fundamental group, so all cuts are incompressible. The terminal products are $D^2\times I$, hence three-balls. This construction covers the possibilities without orientability in the stated range as well as the orientable ones.

Here is a terminating construction for a compact connected orientable irreducible three-manifold $M$ with nonempty boundary. We use the permitted general results in the following precise forms: a boundary-incompressible non-ball component has a connected two-sided nonseparating incompressible surface with nonempty boundary; and a sufficiently large collection of disjoint closed incompressible surfaces has two parallel components with no intervening member. These are respectively the existence and finiteness results for incompressible surfaces.

First perform essential boundary disk compressions as long as possible. Their number is finite: each nonseparating compression decreases boundary genus, while a separating compression divides a boundary component into two of smaller genus. More explicitly, the sum of $\max (0,2g-1)$ over the closed boundary components decreases at every essential compression. Each cut preserves three-manifold irreducibility. To see this for any incompressible cutter, a sphere in the cut manifold bounds a three-ball in the uncut manifold. A closed nonspherical incompressible component of the cutter cannot lie in this three-ball, since its fundamental group could not inject into the trivial group. A component with boundary cannot lie in an interior three-ball either. Thus the three-ball misses the cutter and survives in the cut manifold. Fundamental-group injectivity of two-sided incompressible surfaces follows from the Loop theorem, applied to their two boundary copies after cutting.

Push the boundary of each non-ball residual component slightly into the original $M$ to make a closed frontier $F_1$, omitting spherical components. This frontier is incompressible on its residual side, since all boundary compressions there have been exhausted. It is also incompressible on the side already cut off. Indeed, a hypothetical compression disk on that side can be made transverse to the earlier cutting surfaces. An innermost intersection circle bounds a disk on its cutting surface by incompressibility; replacing the corresponding subdisk removes the circle, without introducing intersections with earlier cutters. Repeat in the order of the cuts. The disk would finally lie in a product collar of the frontier, where its boundary is fundamental-group-injective, a contradiction. Thus $F_1$ is incompressible in $M$.

If a residual component is not a three-ball, choose the permitted nonseparating incompressible surface with boundary, cut it, and compress the new boundary fully again. Push that boundary in to obtain a new closed frontier $F_2$ disjoint from $F_1$. Continue. Three-manifold irreducibility persists, and the same innermost-circle proof makes every nonspherical frontier incompressible in the original $M$.

We explain why the parallel frontiers forced by the finiteness theorem cannot occur in this construction. The required product fact is: a connected nonseparating incompressible surface in $G\times I$ cannot be proper and disjoint from $G\times \{1\}$. To prove it, intersect the surface with vertical annuli $c\times I$ for essential curves $c$ in a closed $G$. Inessential intersection circles can be removed using incompressibility and the three-balls supplied by three-manifold irreducibility. An outermost intersection arc has both endpoints on the lower boundary; boundary compression along its annular bigon reduces the intersections. At least one resulting component stays nonseparating, since the mod-two relative class of the original surface is the sum of the resulting classes. Essential intersection circles can likewise be removed by surgery along the annulus running to its lower boundary. Retain a nonseparating incompressible component after each surgery. These operations reduce the intersection number and cut the base surface to one with boundary. Now intersect with the vertical disks of a proper arc system cutting that base to a disk. The same innermost-circle and outermost-arc operations make the surface disjoint from these disks. It then lies in a three-ball $D^2\times I$, where any proper two-sided surface separates. This is the contradiction proving the product fact.

If two frontiers in the construction were parallel with no other frontier between, their product region would contain one of the intervening nonseparating cutters, disjoint from the later frontier. The product fact forbids this. Consequently the closed-frontier construction has only finitely many stages. Once a residual irreducible component has only spherical boundary, push one boundary sphere inward. The three-ball it bounds cannot be on the collar side, which contains a boundary component of the manifold. The three-ball and the collar therefore exhaust that component, making it a three-ball. We have constructed a finite three-manifold hierarchy, and can choose all its cutting surfaces to have nonempty boundary.

We will use one further consequence in the map argument: these manifolds are aspherical. Work backwards through the three-manifold hierarchy. A three-ball is contractible. At one gluing stage, the van Kampen theorem normal form identifies the lifts of the cut pieces and cutting surface in the universal cover with a tree of spaces. The lifted pieces are contractible by induction, and the lifted cutting surfaces are contractible because the surfaces have boundary and are fundamental-group-injective. A finite subtree is contractible by removing terminal vertices; every sphere map meets only finitely many pieces. All higher homotopy groups therefore vanish at this stage, and induction proves asphericity from a three-manifold hierarchy. Apply this tree argument separately at each gluing stage.

The following three entries complete a single proof of the relative-boundary homeomorphism assertion.

An $N$-dimensional stratified pseudomanifold has a closed filtration $\varnothing =X_{-1}\subseteq X_0\subseteq \cdots \subseteq X_N=X$ with each nonempty $X_s\setminus X_{s-1}$ an $s$-dimensional manifold, dense top stratum, and $X_{N-1}=X_{N-2}$. Each point of an $s$-dimensional stratum has a stratum-preserving conical neighbourhood ${\mathbb{R}}^s\times cL$, where $L$ is a compact $(N-s-1)$-dimensional stratified pseudomanifold. The definition is inductive, starting with zero-dimensional spaces. There are no singular codimension-one strata; connectedness of links is not required. One may work in the usual triangulable category used for Witt bordism.

A rational Witt space is such a stratified pseudomanifold for which every stratum of odd codimension $2r+1$ has link $L$ satisfying $I{H}_r^{\bar{m}}(L;\mathbb{Q})=0$. No extra condition is imposed at even codimension. A Witt space with collared boundary has Witt interior and Witt boundary, and a stratum-preserving collar $\partial X\times [0,\epsilon )$ with the product stratification. An orientation means an orientation of the regular stratum, inducing the boundary orientation on its collar.

A three-pronged graph, with its central vertex and open edges as strata, is a stratified space but not a one-dimensional pseudomanifold: the central vertex is a singular codimension-one stratum, and deleting it leaves three local branches instead of the two branches of a one-manifold. For a non-Witt pseudomanifold, take the suspension of $T^2$. Its two vertices have codimension three and link $T^2$, whose middle group is $I{H}_1(T^2;\mathbb{Q})=H_1(T^2;\mathbb{Q})={\mathbb{Q}}^2$. It violates the Witt space condition.

The rational Witt bordism classification in dimensions divisible by four states that, for every $k\ge 1$, the middle intersection form induces an isomorphismHere $W(\mathbb{Q})$ is the Witt group of a field, so a nonsingular symmetric form represents zero precisely when it is metabolic, that is, when it has a half-dimensional totally isotropic subspace. The Complex projective plane gives a nonzero example: its middle group is generated by the degree-two class $h$, with $⟨h^2,[{\mathbb{CP}}^2]⟩=1$. Its form is $⟨1⟩$, of signature one; a metabolic form has signature zero. Thus $[{\mathbb{CP}}^2]\ne 0$ in ${\Omega }_4^{\mathrm{Witt}}$.

For a closed oriented Witt space $X$ of dimension $2k+1$, take the closed coneAway from its vertex it is a stratified product with an interval, so all old Witt space conditions persist. At the new vertex the codimension is $2k+2$, which is even; hence its odd-dimensional link $X$ creates no new Witt condition. The collar $X\times (1-\epsilon ,1]$ is explicit, and the product orientation may be chosen to induce the given orientation on $X$. This is a compact oriented Witt space with collared boundary, proving

Now let $\dim X=2$. Its singular strata are zero-dimensional. Their links are compact one-dimensional pseudomanifolds, which have no singular strata and are therefore nonempty finite disjoint unions of circles. A single circle is allowed, since a chosen stratum need not be intrinsically singular. The Witt condition is automatic in dimension two, because singular strata have even codimension.

Construct the normalization of a two-dimensional pseudomanifold concretely. Remove disjoint small conical neighbourhoods of the finitely many zero-stratum points. The remainder is a compact oriented surface with boundary, with one boundary circle for each local branch. Cap every such circle with a separate disk. The result $\tilde{X}$ is a closed compact oriented topological surface. Map each new disk homeomorphically onto its branch of the original cone, sending all the disk centres belonging to one old singular point to that point. Along the disk boundaries this agrees with the identification with the remainder, so it defines a continuous surjection $\pi :\tilde{X}\to X$. It is a homeomorphism off these finitely many centres, in particular over the prescribed open stratum. Every fibre is a point except over a zero-stratum point, where it is the finite set of its branch centres. Thus every fibre is zero-dimensional, as required.

Each component of $\tilde{X}$ bounds an oriented handlebody: a surface of genus $g$ is the boundary of a three-ball with $g$ one-handles attached. Let $H$ be the disjoint union of these handlebodies, oriented so that $\partial H=\tilde{X}$. Formgluing $x\in \partial H$ to $(\pi (x),0)$. The quotient is compact and Hausdorff, and its only remaining boundary is $X\times \{1\}$, with an explicit product collar. Near regular points of the gluing locus, two half-ball neighbourhoods glue to a three-ball; the induced orientations match after choosing the collar orientation in the usual boundary-gluing convention.

It remains to check the singular links, rather than merely assert that this quotient is a Witt space. If an old point $p$ has $r$ circle branches, the new one-dimensional stratum $\{p\}\times (0,1)$ has link the disjoint union of those $r$ circles. This has codimension two and imposes no Witt condition. At $(p,0)$, each branch supplies a half-three-ball from $H$ and another from the collar. The halves glue to a full three-ball; the different branches share only the positive collar axis. Intersecting with a small link therefore gives $r$ copies of $S^2$ identified at one common point. This is a two-dimensional pseudomanifold with zero middle intersection homology: an allowable one-cycle avoids the common point, so decomposes into cycles in the $r$ punctured spheres, each homeomorphic to ${\mathbb{R}}^2$; each such cycle bounds there. Consequently $I{H}_1({\bigvee }_{a=1}^r{S}^2;\mathbb{Q})=0$. This is exactly the Witt condition at the new codimension-three stratum. The boundary has the required product stratification. Hence $Z$ is a three-dimensional Witt space with collared boundary with $\partial Z\cong X$, proving

For the final product assertion, “product” means a nontrivial product with at least two positive-dimensional factors; a single four-dimensional factor would allow the preceding ${\mathbb{CP}}^2$ counterexample. State the standard rational intersection-homology product fact used here: products of Witt spaces are Witt, their intersection cohomology satisfies the Künneth theorem, and the complementary-degree intersection pairing on elementary tensors is the product of the factor pairings, with the sign $(-1{)}^{\mathrm{deg}\beta \mathrm{deg}{\alpha }^{\prime }}$ for $\alpha \otimes \beta$ and ${\alpha }^{\prime }\otimes {\beta }^{\prime }$. Terms whose degrees in a factor do not add to that factor's dimension pair to zero. Regrouping factors reduces total dimension four to cases $1+3$ or $2+2$.

In the $1+3$ case, write $X=A\times B$. The middle intersection cohomology isEach summand is totally isotropic, because the degrees in $A$ are respectively $0+0$ and $1+1$, neither equal to one. They pair perfectly with one another by factor Poincare duality. Thus either summand has half the total dimension and is a Lagrangian for the middle symmetric form.

For dimensions $2+2$, the middle group is the direct sumThe outside summands are isotropic and pair perfectly with one another. Each factor's degree-one pairing is a nonsingular skew-symmetric form, so choose a Lagrangian subspace $L_A\subset I{H}^1(A)$. Inside the central summand, $L_A\otimes I{H}^1(B)$ is isotropic and has half its dimension. It is orthogonal to both outside summands by their degrees. Thereforeis an isotropic subspace of half the full middle dimension. In both cases the form is metabolic. The Witt bordism classification now proves the desired $[X]=0$ in ${\Omega }_4^{\mathrm{Witt}}$. This also proves the assertion for more factors, using the regrouping just described.

Work in the smooth or piecewise-linear category; embedded spheres and disks are locally flat. A closed connected orientable three-manifold is an irreducible three-manifold when every embedded $S^2$ bounds an embedded three-ball. It is a prime three-manifold when $M\cong P\#Q$ forces $P\cong S^3$ or $Q\cong S^3$. A separating sphere gives such a connected sum after capping its two sides with balls. Irreducibility therefore implies primeness, but a prime manifold can still have a nonseparating sphere.

A geometrizable three-manifold is one whose prime summands, after sphere splitting and capping, can be cut along disjoint essential tori into pieces whose interiors admit complete geometric structures modeled on the eight Thurston geometries:Here a geometric structure is a locally homogeneous quotient structure modeled on the stated simply connected space. The original manifold need not itself carry one geometry before decomposition.

A Heegaard splitting is $M=H_1{\cup }_{\varphi }{H}_2$, where $H_1,H_2$ are handlebodies with their boundaries identified. Its genus is the common boundary genus. The Heegaard genus $g_H(M)$ is the least possible genus of such a splitting. Thus “genus $g$” as a minimum requires both a genus-$g$ splitting and the nonexistence of one of smaller genus; an arbitrary genus-$g$ splitting only proves $g_H(M)\le g$.

Let $D_{\alpha },D_{\beta }$ be the disjoint complete meridian-disk system of $H_1$, and let $D_{\gamma },D_{\delta }$ be that of $H_2$. The identification $\varphi (\alpha )=\gamma$ joins $D_{\alpha }$ and $D_{\gamma }$ into an embedded sphere $S$ in $M$. Each disk is nonseparating in its handlebody: cutting along it leaves a genus-one handlebody, and cutting along the other then gives a ball. After cutting $M$ along $S$, the two cut handlebodies are connected and are still joined along the complement of the common boundary circle. Hence $S$ is nonseparating.

Cap the two new sphere boundaries with balls. The resulting closed manifold $N$ inherits a genus-one Heegaard splitting: its two sides are the solid tori obtained by cutting the original handlebodies along their common disks. Their meridians are the surviving curves $\beta$ and $\delta$. The hypothesis that $\varphi (\beta )$ avoids $\gamma$ makes it descend to the compressed torus, and its single intersection with $\delta$ remains a single intersection there.

A genus-one splitting whose two meridians intersect once is $S^3$. To see this directly, choose the target torus basis $(\mu ,\lambda )$ with $\mu$ its meridian. The other meridian is $a\mu \pm \lambda$. A meridional Dehn twist, extending across the target solid torus, removes the integer $a$, leaving the longitude. Gluing a second solid torus with meridian equal to that longitude is the standard splitting of $S^3\subset {\mathbb{C}}^2$ into the neighborhoods of its two coordinate circles. Therefore $N\cong S^3$.

By the common nonseparating disks in a Heegaard splitting construction and part (b), reversing the cut and caps givesThe product metric on $S^2\times \mathbb{R}$ is homogeneous, and quotienting by an integer translation in the second factor gives $S^2\times S^1$. Thus $M$ carries this Thurston geometry and is a geometrizable three-manifold.

There is a wording qualification: if “Heegaard genus two” means the minimum defined in part (a), the disk data contradict it. The product splits as $S^1\times D^2$ on each hemisphere of $S^2$, giving a genus-one splitting, while genus zero gives only $S^3$, so $g_H(S^1\times S^2)=1$. The genuine construction is a genus-two splitting of a genus-one manifold; the geometrizability conclusion remains valid. The proof used the complete disk systems, the common-sphere cut-and-cap construction, and the explicitly justified genus-one meridian-intersection criterion.

The Lickorish-Dehn theorem says that every orientation-preserving homeomorphism of a compact connected orientable surface, fixing its boundary pointwise, is isotopic relative to the boundary to a finite product of Dehn twists and their inverses about simple closed curves. Boundary-parallel curves are permitted. For a closed surface of genus $g\ge 1$, a finite collection of curves suffices to generate its mapping class group. If boundary components are allowed to be permuted, twists alone do not express those permutations; the boundary condition is part of the precise statement.

The fundamental theorem of Dehn surgery asserts that every closed connected orientable three-manifold is integer surgery on a finite framed link in $S^3$. To prove it, first choose a Heegaard splitting of $M$ of genus $g$. Existence of such splittings is a standard background result: a triangulation, with spanning trees collapsed in its primal and dual one-skeletons, gives complementary regular neighborhoods of finite graphs, hence two handlebodies. Choose a standard genus-$g$ splitting of $S^3$, formed by stabilizing its genus-zero splitting. Identify the two pairs of handlebodies. If $f_0$ is the sphere's gluing map and $f$ is the gluing map for $M$, then $f_0^{-1}f$ is an orientation-preserving self-homeomorphism of the common surface. The Lickorish-Dehn theorem writes it, up to isotopy, as a product ${\tau }_{c_1}^{{ϵ}_1}\cdots {\tau }_{c_m}^{{ϵ}_m}$, ${ϵ}_i=\pm 1$.

Here is the local surgery mechanism. Place $c$ on an intermediate level of a product collar of the surface. Its tubular neighborhood has longitude $\lambda$ given by that surface and meridian $\mu$ around the normal disk. Remove the neighborhood and fill with meridian slope $\mu \pm \lambda$. Cut the remaining annular neighborhood of $c$ along a transverse arc. Going around the new meridian now makes one extra positive or negative turn in the longitude direction. On gluing the cut annulus back, the two collar levels are therefore identified by $(\theta ,s)\mapsto (\theta \pm \rho (s),s)$, precisely the Dehn twist from part (a). Outside this annulus the identification is unchanged. This proves unit surface-framed surgery realizes a Dehn twist; its surgery coefficient is $+1$ or $-1$ relative to the surface framing, with sign fixed by the chosen normal convention.

Put the curves $c_i$ on distinct collar levels and order those levels to realize the indicated composition. They are disjoint as curves in the three-manifold even if their surface projections intersect. Performing the prescribed surgeries replaces $f_0$ by $f$, up to isotopy. An isotopy of a gluing map extends across a collar, so the filled manifold is homeomorphic to $M$.

Finally each surface framing differs from the preferred longitude framing in $S^3$ by an integer number of twists. Thus its coefficient $\pm 1$ becomes an integer coefficient in the usual knot longitude convention. These finitely many disjoint curves form the required framed link $L$:The ingredients used were Heegaard-splitting existence, the stated twist-generation theorem, and the explicit local unit-surgery construction.

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.

Label the four crossings $A,B,C,D$ in top, left-middle, right-middle, bottom order. Label the bounded regions $d$ (left), $e$ (right), $a$ (upper middle), $b$ (central bigon), $c$ (lower middle), and the unbounded region $o$. The drawing preserves the crossings and the orientation of the supplied figure-eight knot diagram. Its overpassing strands are the outer-left/upper-right strand at $A$, the upper-left/lower-central strand at $B$, the upper-central/lower-right strand at $C$, and the lower-left/outer-right strand at $D$.

Here is the Dehn presentation Heegaard diagram specified by this drawing. Replace the underlying four-valent planar graph $G$ by a small three-dimensional regular neighborhood $N(G)$ and put $\Sigma =\partial N(G)$. Since $G$ has four vertices and eight edges, $\Sigma$ has genus $8-4+1=5$. On $\Sigma$, take the five $\alpha$ curves bordering the bounded faces, pushed slightly onto the boundary of $N(G)$. They bound a complete disk system in the complementary handlebody $S^3\setminus \mathrm{int}N(G)$; their labels are ${\alpha }_d,{\alpha }_e,{\alpha }_a,{\alpha }_b,{\alpha }_c$.

The four $\beta$ curves bound the disks dual to the four short crossing tunnels in $N(G)\setminus \mathrm{int}\nu K_8$. Lift the crossing circles shown in red onto the upper and lower sheets of $\Sigma$, routing them along the corresponding face boundaries; the crossing's overpassing strand determines the lift. Orient the $\alpha$ disk normals and the $\beta$ curves so that their signed successive intersections with the $\alpha$ curves areAt $A,D$ the missing fourth corner is the unbounded region $o$, whose generator is omitted. This sheet-lifting prescription, together with the projected graph, specifies the curves on the Heegaard surface; the red circles alone are only their local projections.

To recover the knot exterior from $\Sigma \times I$, attach five two-handles along the $\alpha$ curves on one side and cap the resulting sphere with a three-handle. On the other side attach four two-handles along the $\beta$ curves. Compressing that side leaves the boundary torus of the knot exterior, which is retained. In particular, the counts are five $\alpha$ compressions and four $\beta$ compressions, rather than two complete genus-five disk systems.

Let $P$ be a once-punctured torus and let $c$ be a nonseparating essential curve. The annulus $c\times [0,1]$ is an incompressible surface in $P\times [0,1]$. Cutting gives a pair of pants times an interval. Both products are genus-two handlebodies, since a compact connected surface with boundary times an interval is a handlebody of genus $1-\chi (P)$.

Every odd-dimensional closed oriented Witt space bounds its closed cone: the new vertex has odd-dimensional link, so introduces no Witt obstruction. For a closed oriented two-dimensional pseudomanifold, normalize to a surface, fill each surface component by a handlebody, and attach a collar of the original space using the finite normalization map. Along the singular collar arcs the links are unions of circles; at their interior starting points the links are wedges of two-spheres. These links have zero middle intersection homology, so the construction is a three-dimensional Witt null-bordism.