= Solution
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> $S'$ and $S''$. 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 $B'$ and $B''$. If both $B'$ and $B''$ 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, $B'$ which is not of this type, and retain $S'$ alone in place of $S$.
The other new piece is $A\cup P\cup B''$. 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 \b[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
$$
\boxed{k+1\leq 4f_2+\dim_{\mathbb F_2}H_1(M;\mathbb F_2).}
$$
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 \b[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 by
$$
r=b_1(X;\mathbb Q)-\sum_P b_1(P;\mathbb Q),
$$
where the sum runs over the common irreducible factors. This uses additivity of first <homology (mathematics)> under <connected sum>, not a claim that the <fundamental group> determines a manifold. Therefore \b[the <orientable> prime decomposition is unique up to order and insertion or deletion of $S^3$ factors].
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