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A fundamental normal surface has a nonzero admissible coordinate vector which is not the sum of two nonzero admissible vectors. Fixing the allowed quadrilateral type in every tetrahedron reduces the problem to the Hilbert basis of a rational cone. There are finitely many choices of quadrilateral types and finitely many indecomposable vectors for each choice.

The Haken sum of compatible normal surfaces is obtained by resolving their intersection curves in the way that preserves the normal disk types. It adds normal surface coordinates, Euler characteristic and mod-two homology classes. The alternative local resolution can create returning face arcs and need not be normal.

Relative to a triangulation, a normal surface meets every tetrahedron in disjoint normal disks: four types of triangles cutting off vertices and three types of quadrilaterals separating opposite pairs of edges. A normal disk meets any edge at most once; its face arcs join distinct edges. Two different quadrilateral types cannot coexist in a tetrahedron.

For $t$ tetrahedra, count the seven disk types in each tetrahedron. Across each glued face the counts of each of its three normal arc types must agree. These are homogeneous linear equations with integer coefficients. An admissible solution also has at most one nonzero quadrilateral type in each tetrahedron. Matching the arcs in their boundary order reconstructs the normal surface up to an isotopy preserving the tetrahedra.

The weight of a normal surface counts its intersections with the edges of the triangulation. It is a positive linear function of its coordinates: each corner of a normal disk contributes the reciprocal of the number of tetrahedral sectors around its edge. Thus a nontrivial decomposition into compatible coordinate vectors gives both summands strictly smaller weight.

A normal surface meets each tetrahedron in disjoint disks of the four triangular and three quadrilateral types. Its arcs in a face join distinct edges, each disk meets an edge at most once, and at most one quadrilateral type occurs in a tetrahedron. These last requirements exclude, for example, a disk with an octagonal boundary.

We first establish the surgery operation needed for the sphere system. Suppose a disk $D$ disjoint from the other spheres compresses one sphere $S$. It produces spheres $S^{\prime }$ and $S^{\prime \prime }$. Small parallel copies of the three spheres bound a three-times punctured three-sphere $P$. Denote the old piece on the uncompressed side by $A$, and the two pieces on the compressed side by $B^{\prime }$ and $B^{\prime \prime }$. If both $B^{\prime }$ and $B^{\prime \prime }$ were punctured three-spheres, then $P$ glued to them would also be a punctured three-sphere, contrary to the hypothesis on the original cut manifold. Choose, say, $B^{\prime }$ which is not of this type, and retain $S^{\prime }$ alone in place of $S$.

The other new piece is $A\cup P\cup B^{\prime \prime }$. It too cannot be a punctured three-sphere. If it were, cap its external boundary spheres to make $S^3$. All the spheres separating $A$ from the attached pieces then bound three-balls by the Three-dimensional Schoenflies theorem; the component containing $A$ would be a punctured three-sphere, contradicting the hypothesis on $A$. Thus the sphere surgery preserving nontrivial complementary pieces keeps exactly $k$ spheres and keeps every complementary piece nontrivial. The assumption that all spheres separate ensures that the old and new pieces used here really are distinct sides.

Apply general position to the whole sphere system, avoiding the vertices and making it transverse to edges and faces. We describe a terminating normalization, allowing the replacement operation just proved as well as isotopy. If some component of its intersection with a tetrahedron is not a disk, choose a boundary circle innermost among circles not bounding such disk components. A disk just inside the tetrahedral boundary gives a compression; if intervening disk components occur, first push them into collars of their boundary disks. Perform the compression and retain one new sphere as above. This replaces a non-disk tetrahedral piece by disk pieces without increasing edge intersections. Repetition is finite: every compression divides the original sphere along a separating circle, and the sum of $1-\chi$ over the non-disk planar tetrahedral pieces decreases; disk components can then be put in boundary collars. A closed sphere component wholly in a tetrahedron is impossible, because it would cut off a three-ball, contrary to the complementary-piece condition.

Now each tetrahedral piece is a disk. A face circle innermost in a face allows its incident disk to be moved through that face disk, removing the circle. This reduces the number of face intersections without increasing the edge weight. A returning face arc similarly gives an edge bigon across which the surface can be pushed, reducing the number of edge intersections by two. Restore tetrahedral disk pieces by the preceding operations whenever necessary.

It remains to exclude multiple intersections of a disk with an edge, which the absence of returning face arcs alone does not exclude. Suppose a tetrahedral disk meets an edge twice. Among such disks choose a pair of intersections consecutive on the edge and with no other surface intersection between them. Such a pair exists: an intervening disk crosses the edge interval an even number of times, so passing to an innermost intervening pair eventually gives consecutive points. Join them by an arc in the disk. The disk pieces divide the tetrahedron into three-balls, and this arc together with the edge interval lies in the boundary of one of these three-balls. It therefore bounds a bigon with interior disjoint from the surface. Push across the bigon to reduce edge weight by two. At fixed weight the previous moves reduce face and non-disk complexity; the edge weight itself is a nonnegative integer. Hence the whole process terminates. A final tetrahedral disk meets each edge at most once, and its boundary separates either one vertex from three or two vertices from two. It is respectively a normal triangle or quadrilateral. Thus a normal system of exactly $k$ spheres with the required complementary-piece property exists.

For the finite decomposition, we need a bound on such normal systems, not merely an assertion that they have been normalized. Let $f_2$ be the number of faces of the fixed triangulation. In a triangular face cut by normal arcs, all but at most four complementary regions are rectangles between adjacent parallel arcs. Thus at most $4f_2$ complementary three-dimensional components contain a nonrectangular face region. Every remaining component is assembled from product regions between parallel normal disks, so is an interval bundle whose boundary components are spheres. Its connected base is either $S^2$ or ${\mathbb{RP}}^2$. The product $S^2\times I$ is a punctured three-sphere, and is excluded. The other possibility is the twisted interval bundle over ${\mathbb{RP}}^2$, namely ${\mathbb{RP}}^3$ with a three-ball removed. Each such piece contributes a distinct ${\mathbb{F}}_2$ summand to $H_1(M;{\mathbb{F}}_2)$: gluing along spheres does not identify first-homology classes, by the Mayer–Vietoris sequence. Since $k$ separating spheres give $k+1$ components, we obtain the explicit finite bound

For an arbitrary closed connected $X$, first cut out nonseparating spheres. A neighbourhood of one such sphere together with an arc joining its two sides is a punctured $S^2$-bundle over $S^1$; the bundle is the product or the twisted bundle according to the orientation of the attaching identification. Its complement is connected, so this extracts a connected sum factor. The mod-two first Betti number of the remaining capped manifold decreases by one. Hence only finitely many such extractions are possible. The remaining manifolds have only separating spheres.

Whenever a remaining factor is not a prime three-manifold, split it along a sphere into two factors neither homeomorphic to $S^3$. Viewed in the original remaining manifold, all these successive splitting spheres form a disjoint system whose pieces are not punctured three-spheres. The normalization and bound just proved prohibit infinitely many such splittings. This proves existence of a finite prime decomposition, including for $X$ without assuming orientability.

For completeness, both sphere bundles extracted above are prime three-manifolds. Their fundamental group is $\mathbb{Z}$, so a separating sphere and the van Kampen theorem express $\mathbb{Z}$ as a free product in which one capped factor has trivial fundamental group. Its punctured piece lifts homeomorphically to the universal cover $S^2\times \mathbb{R}$, identified with ${\mathbb{R}}^3\setminus \{0\}$. That compact lifted piece has one spherical boundary and must be the bounded side of this sphere in ${\mathbb{R}}^3$. It misses the origin, so the Three-dimensional Schoenflies theorem makes it a three-ball. The corresponding capped factor is $S^3$, as required.

Suppose now that $X$ is orientable. A prime factor which is not an irreducible three-manifold contains a nonseparating sphere: an essential separating sphere would already split it nontrivially. Extracting the corresponding sphere bundle shows that the factor must be $S^2\times S^1$. Thus the other nontrivial primes are irreducible.

Consider two prime decompositions. Represent each by a sphere system, including one nonseparating sphere in each $S^2\times S^1$ factor. The capped pieces are now the irreducible factors and copies of $S^3$. Make the two sphere systems transverse. To remove their intersections, choose a circle innermost on a sphere of the second system and surger the first system along its disk. Retain both surgery spheres temporarily. The old sphere and the new pair bound a three-times punctured sphere; adding the pair and then deleting the old sphere only inserts or removes spherical punctures.

This operation cannot alter an irreducible capped piece: every new sphere in that piece bounds a three-ball, so it can only split off a punctured $S^3$ piece. The other pieces were already punctured $S^3$, where the same conclusion follows from the Three-dimensional Schoenflies theorem. After small displacements the surgery removes the chosen intersection circle and introduces none. Repeat until the sphere systems are disjoint. Their union is a common refinement, and its non-$S^3$ capped irreducible pieces are precisely those of each original decomposition. Hence these factors agree, up to order and orientation-preserving homeomorphism.

Finally, the number $r$ of $S^2\times S^1$ factors is determined bywhere the sum runs over the common irreducible factors. This uses additivity of first homology under connected sum, not a claim that the fundamental group determines a manifold. Therefore the orientable prime decomposition is unique up to order and insertion or deletion of $S^3$ factors.

For $t$ tetrahedra, a normal surface has a vector $x\in {\mathbb{Z}}_{\ge 0}^{7t}$ counting four triangle types and three quadrilateral types in each tetrahedron. On every glued face, the numbers of arcs of each of its three types must agree on the two sides. These give homogeneous integer linear equationsFor closed surfaces the coordinates of disks meeting an unglued boundary face must also vanish. The nonnegativity and matching equations are not the whole admissibility condition: in each tetrahedron at most one of its three quadrilateral coordinates can be nonzero. Conversely, for an admissible vector, arrange the prescribed parallel disks and glue their face arcs in order. This reconstructs an embedded normal surface, unique up to normal isotopy.

A fundamental normal surface corresponds to a nonzero admissible vector which cannot be written as $y+z$ with both $y$ and $z$ nonzero admissible vectors. These indecomposable vectors form a fundamental set: every admissible vector is their finite sum. To find them, fix which quadrilateral type is allowed in each tetrahedron. This leaves the integer points of a rational polyhedral cone $Ax=0,x\ge 0$, with some coordinates set to zero. Compute its Hilbert basis of a rational cone, for example by subdividing into rational simplicial cones and enumerating the lattice points in their fundamental parallelepipeds along with their integral ray generators, then removing decomposable generators. Repeat the finite list of quadrilateral choices, remove duplicates, and remove vectors decomposable in their compatible sector. Finiteness follows because each of these pointed rational cones has a finite Hilbert basis.

Let $E$ be the link exterior. By Alexander duality, for a two-component linkAn embedded connected closed surface separates $S^3$; its mod-two class is nonzero exactly when the two link components lie on different sides. The connectedness qualification matters: two parallel splitting spheres form a disconnected surface with zero total class. We use the criterion only for connected surfaces.

Normalize a splitting sphere by isotopy, disk compressions and removal of components lying inside tetrahedra. A compression replaces a sphere by two spheres whose mod-two classes add to the old class; at least one is still a splitting sphere. The discarded tetrahedral spheres have zero class. Thus there is a normal splitting sphere. Choose one, $S$, of least normal surface weight.

Suppose its vector is decomposable. Realize $S=A+B$ as a Haken sum, choosing a decomposition with the least possible number of intersection curves. Two elementary exchange observations are useful. First, $A$ and $B$ can be taken connected. If their total union has more than two components, perform regular exchanges connecting components until exactly two embedded groups remain, resolving any intersections within each group. Since the final sum is connected, this can be done before the last joining exchange; it uses up at least one of the old intersection curves. The resulting two normal surfaces still have total vector $x(S)$, contradicting minimality. Second, no intersection curve separates both $A$ and $B$. If it did, resolve that curve first: the two separated sides give two groups. Resolve any remaining self-intersections within each group and continue exchanges between groups only while the result remains disconnected. Just before the exchange making the final connected surface, there are again exactly two embedded normal groups with fewer intersection curves. Both assertions follow from the local four-bank resolution at an intersection circle; internal crossings are resolved before regarding a group as one embedded summand.

Every closed embedded surface in $S^3$ is orientable. Additivity of Euler characteristic in the Haken sum givesFor connected orientable closed surfaces the positive Euler characteristic is $2$, and the only zero Euler characteristic is that of a torus. Thus one summand, say $A$, is a sphere and the other $B$ is a torus. Both have smaller weight than $S$. The sphere $A$ cannot split the link by minimality, soEach intersection circle separates the sphere $A$, so the second exchange observation makes it nonseparating on $B$. In particular, these circles are essential parallel curves on the torus.

Cut $A$ along the intersection circles. There are at least two innermost disk patches. Choose one of smallest weight, $D$; then $2w(D)\le w(A)$. Its interior misses $B$, and its boundary is essential on $B$. Compress the torus along two slightly separated copies of $D$. The result is a sphere $C$, and the compression takes place in the exterior, soThus $C$ is a splitting sphere.

This alone is not quite a contradiction if the last weight inequality is equality: $C$ need not yet be normal. At the compression seam it uses the alternative, rather than regular, exchange between the disk patch and the torus. An intersection circle of normal disk sheets must cross a tetrahedral face; it cannot be a closed curve entirely in two disk interiors in one tetrahedron, which can be put in standard position with only intersection arcs. In a face, parallel arcs of the same type do not cross. At a crossing of different normal arc types, the alternative exchange joins two endpoints on the same edge, producing a returning arc. Pushing across its edge bigon strictly reduces weight. Normalize the resulting surface, retaining a nonzero-class sphere at each compression as before. We obtain a normal splitting sphere of weight strictly less than $w(S)$, a contradiction. Therefore

The algorithm is now finite. Triangulate the exterior from a link diagram, enumerate its fundamental admissible vectors with the closed-surface boundary condition, reconstruct their surfaces, and inspect the connected components and Euler characteristics. A fundamental vector yields a connected surface, since otherwise its component vectors decompose it. A connected closed surface in $S^3$ with Euler characteristic two is a sphere. Compute its mod-two class in the finite simplicial chain complex, or equivalently check which link boundary tori lie on either side. Answer yes exactly when such a sphere has nonzero class. The argument proves that every split link is detected, and every detected sphere witnesses splitting. For more than two components, $H_2(E;{\mathbb{F}}_2)\cong {\mathbb{F}}_2^{\ell -1}$ and a nonzero sphere class separates a nonempty proper subset of the components; the same least-weight argument and finite enumeration apply.