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A finite collection of disjoint splitting spheres of a knot encodes a connected sum of knots. Cut along every sphere and fill each spherical boundary with a three-ball containing a straight arc joining its two marked endpoints. The resulting oriented knots are the factors. The system is complete when its nontrivial factors are prime knots.

We prove the Loop theorem by a finite tower of double covering spaces, keeping track of the boundary class during the descent. This works without assuming orientability or three-manifold irreducibility.

Let $F$ be the boundary component containing the basepoint. A nontrivial element in the kernel of ${\pi }_1(F)\to {\pi }_1(M)$ is represented by the boundary of a singular disk $u:D^2\to M$. Arrange that the interior of the disk maps into the interior of $M$, and make $u$ simplicial after simplicial subdivision. Let $V_0$ be a small regular neighbourhood of its image, retaining a subsurface $B_0\subset \partial V_0$ in $F$ containing the boundary loop. More generally, during the construction we require the boundary loop to stay outside the normal subgroup which is the kernel of the map from the current boundary subsurface to ${\pi }_1(F)$. This records exactly the information needed at the end.

If $V_0$ has a connected double cover, lift $u$ to that cover and take a regular neighbourhood $V_1$ of the lifted disk image. Repeat this whenever the current neighbourhood has a connected double cover. This procedure is finite. Indeed, let $C_i$ be the simplicial image of the disk at stage $i$. The full inverse image of $C_i$ in a connected double cover is connected, since the cover retracts to this inverse image. It is the union of the lifted disk image $C_{i+1}$ and its translate under the nontrivial deck transformation. These two images must meet. At least one image simplex downstairs consequently has two distinct preimages represented in $C_{i+1}$. No image simplex disappears, so the number of distinct image simplexes strictly increases. This number is at most the number of simplexes of the fixed domain disk. Thus the tower proof of the loop theorem terminates.

At the top, $V_r$ has no connected double cover. Equivalently, $H^1(V_r;{\mathbb{F}}_2)=0$. Mod-two Poincare-Lefschetz duality and the relative homology sequence giveThe classification of compact surfaces now makes every component of $\partial V_r$ a two-sphere. The lifted boundary subsurface $B_r$ is planar. Its fundamental group is normally generated by its boundary circles. Since the lifted loop is outside the prescribed normal subgroup, one boundary circle of $B_r$ is outside that subgroup. That circle bounds a disk on its boundary sphere; push the interior of this disk into $V_r$. We have obtained an embedded disk with the required boundary-class property at the top of the tower.

We next descend one double cover. Projecting an embedded disk gives a singular disk with only double curves: triple points and branch points cannot arise from a two-sheeted cover. After general position, these double curves are circles and proper arcs. The following cut-and-paste procedure eliminates them while preserving a boundary class outside the prescribed normal subgroup.

First eliminate the circular double curves. Take a double circle innermost on one of its preimage sheets. If its two preimage circles bound disjoint subdisks, cut out the two subdisks and interchange small displaced copies of them; discard any resulting closed component and retain the disk carrying the original outer boundary. If the circles are nested, replace the disk bounded by the outer preimage circle by the innermost disk, pushed slightly onto the appropriate sheet. Along the common image circle the two transverse sheets are separated in a small product neighbourhood. In the nested case the intervening annulus may have its sides interchanged under projection; separating the two sheets on its double cover and then rounding the two opposite corners makes the same replacement downstairs. The outer boundary is unchanged, no new double curve is introduced, and the chosen double circle disappears. Repeating leaves only double arcs.

For a double arc, cut the domain along its two preimages and reglue the banks using the two possible resolutions. There are two disk choices. At least one has boundary outside the prescribed normal subgroup. Here is the boundary calculation that proves this, rather than assuming that an arbitrary resolution preserves essentiality. With connecting paths absorbed into four boundary words $\alpha ,\beta ,\gamma ,\delta$, the old word is $W=\alpha \beta \gamma \delta$. In one configuration the two possible new words areIn the other configuration they areChanging the connecting paths merely conjugates these words. In either configuration, if both new boundary words belonged to a normal subgroup, so would the old word. Choose a disk whose word stays outside the subgroup. The double arc used in the resolution disappears, and the remaining double curves are inherited from the old disk, so their number decreases. Remove any circular double curves again, and repeat. The resulting disk is embedded.

Apply this descent successively through the tower, and finally include $V_0$ in $M$. Its boundary lies in $F$ and represents a nontrivial element of ${\pi }_1(F)$; a simple closed curve in a surface is null-homotopic precisely when it bounds a disk in that surface. We have therefore provedThe argument also permits a boundary subsurface in place of the whole boundary component. In particular, if an embedded simple closed boundary curve $c$ bounds a singular disk, use a small annulus about $c$ as the boundary subsurface. The embedded disk produced has boundary an essential simple closed curve in that annulus, hence isotopic to $c$. Extend that isotopy through a boundary collar. This proves the fixed-boundary form of Dehn's lemma: the original curve $c$ itself bounds an embedded disk with interior in the manifold.

Now consider the knot map. Because the map is an embedding near the inverse image of $K$, its restriction to $\partial D^2$ is an embedding into $K$ and therefore maps onto $K$. No interior point can map to $K$, since it would duplicate a boundary image. Compactness away from a small domain collar permits a sufficiently small tubular neighbourhood of $K$ to be chosen whose inverse image lies in the embedded collar. Its intersection with this collar cuts off an annulus. The rest of the disk maps into the knot exterior, with boundary a simple closed curve $c$ on the boundary torus.

The annulus shows that $c$ runs once along $K$. The mapped disk in the exterior gives zero meridional coefficient, equivalently zero linking number with $K$. Thus $c$ is a longitude. Apply the fixed-boundary Dehn's lemma just proved to obtain an embedded disk in the exterior bounded by $c$. Attach the embedded collar annulus between $c$ and $K$. This gives an embedded spanning disk for $K$. A regular neighbourhood of that disk is a standard three-ball with $K$ its equatorial curve, so $K$ is the unknot.

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.

Apply the boundary-degree argument to each source cut component. It maps to one target cut component and is a homeomorphism on each of its boundary components onto distinct target boundary components. Its relative degree must be $\pm 1$, so it meets the whole boundary of that target component. Since the global cut boundary map is one-to-one, two source components cannot map to the same target component; since it is onto, every target component is accounted for. Thus the cut components correspond bijectively, and each paired map satisfies the original hypotheses.

Induct on the remaining number of cuts in the chosen target three-manifold hierarchy, keeping fixed all boundary pieces already straightened. At the final stage the target components are three-balls. Each corresponding source component is irreducible with a single spherical boundary, and is therefore a three-ball by the inward-pushed-boundary argument above. Extend its prescribed boundary homeomorphism over the three-ball by coning. The original three-ball map and this homeomorphism are homotopic relative to the boundary: after identifying the target with a convex three-ball, take their straight-line homotopy. This establishes the induction base.

Reverse the cuts. On both copies of every cutting surface the two constructed maps agree with the fixed product homeomorphism of its collar, so the component homeomorphisms glue. Their homotopies also glue, because all cut boundaries have stayed fixed. Repeating through the three-manifold hierarchy givesThe proof derives the three-dimensional rigidity from the three-manifold hierarchy and the permitted surface-map straightening; it does not use the desired three-dimensional homeomorphism assertion as a quoted theorem.

Choose a connected proper incompressible surface $S$ with nonempty boundary as the first cut in a three-manifold hierarchy of $N$. Homotope $f$ relative to $\partial M$ into transverse position and let $F=f^{-1}(S)$.

We first make $F$ incompressible. If $D$ is a compression disk for $F$, its boundary maps to a loop in $S$ which is null-homotopic in $N$. Since $S$ is fundamental-group-injective, this loop is null-homotopic in $S$. Fill it in $S$ and alter $f$ in a small neighbourhood of $D$ so that its inverse image is obtained by compressing $F$ along $D$. The alteration is a homotopy: the filling and the old disk give a sphere map in $N$, which is null-homotopic by the asphericity proved above. A product collar of $S$ specifies the two sides of the modification. The integer ${\sum }_j\max (0,1-\chi (F_j))$ decreases at every essential compression; sphere components can be removed separately. A resulting sphere component of $F$ bounds a three-ball in $M$ by three-manifold irreducibility; move the map of that three-ball to one side of $S$ to remove it. The required extension and homotopy exist because the higher homotopy groups of $N$ vanish. Continuing leaves an incompressible $F$ with no spherical components.

For each component $F_j$, the map on fundamental groups to $S$ is injective. Indeed, its composition with the injection ${\pi }_1(S)\to {\pi }_1(N)$ equals the composition of ${\pi }_1(F_j)\to {\pi }_1(M)$ with the isomorphism $f_{\ast }$. Both maps in the latter composition are injective. A closed nonspherical orientable $F_j$ cannot occur: its fundamental group would inject into ${\pi }_1(S)$, a free group, whereas every subgroup of a free group is free and a closed positive-genus surface group is not free. Thus every component of $F$ has boundary.

Apply the permitted surface-map homotopy result to $F_j\to S$, fixing its boundary. A fundamental-group-injective map of compact surfaces is homotopic either to a covering or, in the annulus case here, to a map into one target boundary circle. The exceptional annulus case is impossible: its two distinct boundary circles would map onto the same circle of $\partial S$, contradicting the injectivity of the original boundary map. If $S$ is a disk, injectivity makes each $F_j$ a disk, and the unique boundary circle directly gives a single disk homeomorphism. Otherwise each component is homotopic to a cover of $S$.

Every point of $\partial S$ has exactly one preimage on $\partial F$, since $f{|}_{\partial M}$ is a homeomorphism. Every cover component contributes its positive degree to the number of preimages on every target boundary circle. Thus there is exactly one component and its covering degree is one. We have proved that $F\to S$ can be homotoped to a homeomorphism, relative to its boundary.

Extend this surface homotopy to $M$, and use the two-sided collars to make $f$ a product homeomorphism on a neighbourhood of $F$. The transverse normal direction is a degree-one local crossing, so the collars on its two sides match the two sides of $S$; any superfluous folds can be removed in the normal interval. We may now cut $M$ along $F$ and $N$ along $S$, obtaining a map whose restriction to the entire cut boundary is a homeomorphism. The other hypotheses also survive: the cut manifolds are irreducible, have nonempty boundary, and the inclusion of every cut component into its uncut manifold is fundamental-group-injective by the normal form of the van Kampen theorem. A loop killed by the cut map is consequently killed by $f_{\ast }$ and hence is trivial in its source cut component.

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.