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Every closed connected orientable three-manifold is obtained by integer Dehn surgery on a finite framed link in the three-sphere. Compare a Heegaard splitting with a standard splitting of the sphere, factor their gluing difference using the Lickorish-Dehn theorem, and realize each twist by unit surgery on a distinct level of a collar of the splitting surface.

The least genus of a Heegaard surface among all Heegaard splittings of a closed orientable three-manifold. Having a splitting of genus $g$ only proves $g_H(M)\le g$; stabilization supplies larger-genus splittings of the same manifold.

The Heegaard surface is the common positive boundary of the two pieces in a Heegaard splitting.

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.

Write $V=M_{S^2}(p^{\ast }/q,\ast )$. Filling one end of $S^1$ times an annulus gives a solid torus, so $V$ is a solid torus. The prescribed algebraic intersection number of curves on an oriented surface convention gives ${\tilde{f}}_2\cdot {\tilde{h}}_2=1$. HenceIn $H_1(V;\mathbb{Z})$, the first Dehn filling imposes $p^{\ast }f-q{h}_1=0$, while $h_1+h_2=0$. ThereforeThe class ${\lambda }_2$ is primitive because $\gcd (p^{\ast },q)=1$, a consequence of $p{p}^{\ast }-q{q}^{\ast }=1$. It is therefore the disk-bounding meridian of a solid torus of $V$. Since ${\mu }_2$ meets it once, Dehn filling $V$ along ${\mu }_2$ glues two solid torus pieces with disk boundaries meeting once. This is the standard genus-one Heegaard splitting of $S^3$, soTo identify the regular fiber geometrically, put it on their common boundary torus. Invert the displayed basis change:Here ${\mu }_2$ bounds a disk in the newly attached solid torus and ${\lambda }_2$ bounds a disk in $V$. Thus the curve has winding numbers $p$ and $q$ in the two complementary solid torus pieces. Both coefficients are positive and $\gcd (p,q)=1$, so it is the positive $(p,q)$ torus knot. This identifies the torus knot directly from the genus-one splitting, without invoking a theorem specifically about torus knots.

A triangulation of a topological space is a homeomorphism from the realization of a simplicial complex to that space. In three-manifold topology, thickening its one-skeleton and the dual one-skeleton constructs a Heegaard splitting.