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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.

The possible ranks and the corresponding homeomorphism types areThere is no example for $n=2$ or $n\ge 4$. The uniqueness here is topological, not uniqueness of a Riemannian metric.

To see the restrictions, use the prime decomposition of a closed orientable three-manifold. Its fundamental group is the free product of the prime three-manifold groups. A free product of two nontrivial groups is nonabelian, so at most one summand can have nontrivial fundamental group. The remaining simply connected summands are three-spheres by the Poincaré conjecture. If the nontrivial summand is not irreducible, it is $S^2\times S^1$, giving rank one. If it is irreducible and has infinite fundamental group, the sphere theorem for three-manifolds makes it aspherical. Its fundamental group is then a dimension-three Poincare duality group. But ${\mathbb{Z}}^n$ is a duality group of dimension $n$, as seen from the torus classifying space or its Koszul resolution, so $n=3$.

For rank three, the flat case of the geometrization theorem says that a closed irreducible orientable three-manifold with fundamental group ${\mathbb{Z}}^3$ admits a flat metric. Write it as ${\mathbb{R}}^3/\Gamma$. By the Bieberbach theorem, $\Gamma$ contains a full-rank translation lattice $\Lambda$. If $g=(R,v)\in \Gamma$ commutes with translation by each $\lambda \in \Lambda$, then $R\lambda =\lambda$ for every lattice vector, forcing $R=I$. Since $\Gamma$ itself is abelian, it consists entirely of translations. Its quotient is therefore the three-torus. Rank zero follows directly from the Poincaré conjecture, and rank one was already determined by the prime decomposition of a closed orientable three-manifold. Historically, before the Poincare theorem, the uniqueness conclusions were phrased allowing connected sums with homotopy three-spheres; the modern theorem removes that ambiguity.

The six flat types below mean the six closed orientable flat three-manifold types. There are infinitely many individual flat metrics and there are additional noncompact flat Riemannian manifolds. In each closed case the universal cover is Euclidean space and the deck transformation group is a torsion-free crystallographic group. The holonomy groups of closed orientable flat three-manifolds are $1,C_2,C_3,C_4,C_6,C_2\times C_2$.

The first is a flat torus ${\mathbb{R}}^3/\Lambda$, with trivial holonomy. For each cyclic case choose a planar lattice $L$ admitting the relevant rotation $R_m$, and generate the deck transformation group by translations in $L$ and the screwIts $m$th power is unit translation along the axis. A nontrivial holonomy element has nonzero axial displacement modulo integers, so has no fixed point. All transformations preserve orientation. A compact prism fundamental domain gives compactness. Equivalently, the manifold is a torus mapping torus with finite-order monodromy. In planar lattice bases one may takeThe half-turn uses any planar lattice, the quarter-turn a square lattice, and the third- and sixth-turns a hexagonal lattice. These give respectively the half-turn flat three-manifold, third-turn flat three-manifold, quarter-turn flat three-manifold, and sixth-turn flat three-manifold. Their holonomies have orders $2,3,4,6$, and their first Betti numbers are all one, since the only fixed direction of holonomy is the screw axis.

The sixth type is the Hantzsche-Wendt manifold. A concrete group is generated byTheir linear parts are commuting half-turns about perpendicular axes. Moreover ${\alpha }^2$, ${\beta }^2$ and $(\alpha \beta {)}^2$ are translations by $(1,0,0)$, $(0,1,0)$ and $(0,0,-1)$. Their translation subgroup is the unit cubic lattice and their linear quotient is $C_2\times C_2$. In each nontranslation coset, the displacement along the fixed axis is a half-integer, so the action is free. The quotient is compact and orientable. No vector is fixed by both linear parts, so its first Betti number is zero. Bieberbach classification gives precisely these six affine types; the explicit screw descriptions distinguish them without identifying arbitrary finite quotients with holonomy.

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 $g_H(M)\le g$.

Every closed connected orientable three-manifold is a finite connected sum of prime three-manifolds, unique up to order and the insertion of three-sphere factors. The corresponding fundamental group is the free product of the summand groups. A prime orientable three-manifold is either irreducible or $S^2\times S^1$; this separates aspherical infinite-group cases from the infinite cyclic case.