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Two piecewise linear balls of dimension three glued along a common boundary disk form another piecewise linear ball. The Schoenflies theorem identifies each boundary pair with a sphere and a hemisphere. Coning the resulting boundary homeomorphisms extends them over the balls, which then become the two halves of a standard ball. The same argument shows that removing a ball attached along one boundary disk leaves a ball.

The two-dimensional Schoenflies theorem says that a simple closed piecewise linear curve in $S^2$ separates it into two components whose closures are piecewise linear balls of dimension two. In particular, a homeomorphism of the curve to the equator extends to a piecewise linear homeomorphism of the whole sphere.

A piecewise linear ball is a space piecewise linearly homeomorphic to a standard simplex. Write $D=B_1\cap B_2$. The complementary disk to $D$ in each $\partial B_j$ is a disk by the two-dimensional Schoenflies theorem. Thus each pair $(\partial B_j,D)$ is piecewise linearly homeomorphic to a sphere with a chosen hemisphere. Choose these boundary maps to agree on $D$. Extend them to the interiors by coning from an interior point, after compatible subdivisions. The resulting maps identify $B_1$ and $B_2$ with the two halves of a standard ball, glued along its equatorial disk. ConsequentlyThe same boundary parametrization proves a useful companion fact: if a ball is indented by removing a smaller ball attached along a boundary disk, the closure of what remains is again a ball. In the standard boundary-disk model, move the indented disk through a collar neighbourhood to a flat disk.

A handle decomposition is a filtration obtained by attaching handles $D^r\times D^{n-r}$ along $\partial D^r\times D^{n-r}$ in the boundary of the preceding stage. The number $r$ is the handle index; index-zero handles are disjoint balls, and index-$n$ handles close the remaining spherical boundary components. Starting with a triangulation of $M$, thicken its vertices, then its edges, and successively its higher-dimensional simplexes. In a barycentric subdivision these thickenings have precisely the indicated handle and attaching-region product structures. Adding collars between successive attachments gives a collared handle decomposition.

For a critical level handle embedding, assign successive distinct heights $t_j$ to the handles. At height $t_j$ the horizontal slice contains the $j$th handle and its attaching region. Between handle heights, the embedded boundary of the previously attached handles travels along its product collar; there is no topological change there. The lower boundary collar meets the attaching region of the next handle, and the upper collar leaves its new boundary. The stated ambient isotopy therefore separates all topology changes into horizontal handles and all intervening motion into collars. For an embedded surface, the local changes are the appearance of a disk, attachment of a band, and disappearance of a disk.

Here is the three-dimensional conclusion, with the distinction between the bounded and unbounded regions made explicit: a piecewise linear embedded $S^2$ in ${\mathbb{R}}^3$ bounds a piecewise linear three-ball; in the one-point compactification $S^3$, both complementary closures are three-balls. The unbounded closure in ${\mathbb{R}}^3$ itself is not a compact ball.

Put the embedded sphere in critical level handle embedding position. At a regular horizontal level its intersection with the plane is a finite disjoint union of simple closed curves. An innermost such curve bounds a planar disk whose interior misses the sphere. Cut the sphere along this curve and insert two slightly displaced copies of the disk. This replaces one sphere by two spheres. Repeat at regular levels between the critical heights, using innermost curves first. The number of curves at these chosen levels decreases at each operation, so the process stops. Each resulting sphere then lies in a slab containing at most one horizontal handle.

These terminal spheres bound balls directly. With no handle the surface is a disk transported through a collar and capped at its ends, the boundary of $D^2\times I$. A minimum or maximum merely supplies one of those caps. At a band level there are two planar boundary circles on one side and one on the other. Cap the two circles, straighten each planar disk using the two-dimensional Schoenflies theorem, and straighten the joining band in its planar collar. If the circles are not nested, this gives two balls joined along a boundary disk; if they are nested, it gives a ball with one boundary-attached ball removed. The gluing three-balls along a boundary disk argument and its indentation version show in both cases that the terminal sphere bounds a ball. This also explains why the planar position of the band matters: no assertion about an arbitrary knotted band is being made.

Reverse the disk surgeries. Suppose the two resulting spheres bound balls $A$ and $B$. Their disjoint boundaries imply that the balls are disjoint or one contains the other. Shrink the small displacement of the two surgery caps back to the original planar disk. In the disjoint case, the pre-surgery sphere bounds the union of two balls along that disk. In the nested case, it bounds the larger ball with the smaller boundary-attached ball removed. Both are balls by the preceding elementary facts. Induction through the finitely many surgeries proves the Three-dimensional Schoenflies theorem. To treat the other region in $S^3$, remove a point inside the ball just constructed and apply the same argument in the resulting copy of ${\mathbb{R}}^3$.

A 3-manifold is an irreducible three-manifold when every piecewise linear embedded two-sphere bounds a ball. Suppose $p:{\mathbb{R}}^3\to M$ is a universal cover, and lift an embedded sphere $S$ to $\tilde{S}$. Since $S$ is simply connected, its lift maps homeomorphically to $S$. By the Three-dimensional Schoenflies theorem, $\tilde{S}$ bounds a compact ball. The inverse image of $S$ is locally finite, so only finitely many other lifted spheres lie inside this ball. Choose an innermost one, bounding a ball $B$ whose interior meets no lift of $S$.

The interior of $B$ maps to one component $U$ of $M\setminus S$. This restriction is a covering map onto $U$: a path in $U$ starting in the image lifts into $\mathrm{int}B$, since it cannot cross a lifted boundary sphere. The covering is proper, because a compact subset of $U$ has preimage closed in the compact ball and cannot meet $\partial B$. Near $S$, the boundary map $\partial B\to S$ is one-to-one, so this connected covering has one sheet. The map on $B$ is therefore a homeomorphism onto $U\cup S$. Thus every embedded sphere in $M$ bounds a ball, and $M$ is an irreducible three-manifold.

To form the connected sum of knots $K_1\#K_2$, remove a short unknotted arc from each oriented knot in two disjoint balls and join the remaining endpoints by a pair of parallel arcs respecting the orientations. Equivalently, a sphere meeting a knot in two points cuts it into two one-string tangles; close each by an arc in the sphere to recover its summands. Sliding the small joining balls along the knots shows that the isotopy type does not depend on these choices. The unknot is the identity for this operation. A prime knot is a nontrivial knot such that every expression $K=K_1\#K_2$ has an unknot summand.

We first prove additivity of Seifert genus, which also supplies a terminating measure for decomposition. Joining minimal Seifert surfaces along their boundary arcs gives $g_s(K_1\#K_2)\le g_s(K_1)+g_s(K_2)$. For the reverse inequality, let $F$ be a minimal-genus Seifert surface for the sum and $S$ a splitting sphere of a knot. Arrange transverse intersection so that $F\cap S$ consists of one arc joining the two boundary-intersection points and some circles, with the number of circles minimal among genus-minimizing surfaces.

Each circle separates the sphere into two disks; since it is disjoint from the intersection arc, one disk contains neither marked endpoint. Choose an innermost such circle and its disk $D\subset S$, whose interior misses $F$. If its boundary were essential in $F$, compressing $F$ along $D$ would lower the genus of its component with boundary K: for a nonseparating curve genus drops by one, and for a separating essential curve the removed boundaryless side has positive genus. This contradicts minimal genus. Its boundary therefore bounds a disk in $F$. Replace that disk by a small push-off of $D$, removing the intersection circle without changing genus or boundary, again contradicting minimal intersection. Thus no circles remain.

Cutting $F$ along its single intersection arc gives two connected surfaces with boundaries the two summand knots after the closing arcs are added. Euler characteristic shows that their genera add to $g(F)$. Each genus is at least the minimum genus of its summand, soGenus zero means that the knot bounds a disk and is the unknot. Every nontrivial knot therefore has positive genus. Induct on this nonnegative integer: a nontrivial nonprime knot splits into two nontrivial summands, each with strictly smaller genus by the equality above, and induction decomposes both into primes. This proves every knot is a finite connected sum of prime knots, with the unknot the empty sum. Uniqueness is not needed here.

For coprime integers p,q, define the torus knot as the oriented parametrized curveCoprimality makes it an embedded single circle: two parameters with both coordinate phases equal differ by a multiple of $2\pi$. It lies on the standard unknotted torus separating the two solid tori. Negative exponents change the slope/orientation convention; they are allowed. If an exponent is zero, coprimality forces the other to be plus or minus one, giving an unknot.

Assume now $|p|,|q|\ge 2$. We give a group-theoretic proof that torus knots are prime, making the peripheral ingredient explicit. The Seifert-van Kampen theorem applied to the two solid-torus sides of the exterior, joined along the annulus parallel to the knot, givesHere a and b are the two core loops. The element $z=a^p=b^q$ is central. Under the abelianization $G\to \mathbb{Z}$, choose $a\mapsto q$, $b\mapsto p$; these generate the target because p,q are coprime. Thus $z\mapsto pq\ne 0$, so z is nontrivial. A meridian has, up to conjugacy, the form $m=a^u{b}^v$ with $qu+pv=1$. This follows by tracing the meridional disk across the two torus sides, using a dual primitive slope intersecting the knot slope once. Different Bezout solutions change u by a multiple of p and v by the opposite multiple of q, leaving the same word because z is central.

Quotienting by z gives $G/⟨z⟩=C_{|p|}\ast C_{|q|}$. Neither u nor v is zero modulo the corresponding cyclic order, by $qu+pv=1$. Hence the image of m is a cyclically reduced word of two syllables. All its nonzero powers remain reduced, so it has infinite order. In particular G is nonabelian and the knot is not an unknot.

We use the Loop theorem in its usual form: a noninjective boundary fundamental-group map of a three-manifold supplies an embedded compression disk. For a nontrivial knot, its boundary torus is incompressible. To see why, every sphere in its exterior bounds a ball on the side not containing the knot, by the Schoenflies theorem. A compression of the boundary torus would produce a sphere and make the exterior a solid torus. The compression disk has zero homology class in the exterior, so its boundary is the preferred longitude; attaching the annulus from that longitude to the knot gives a spanning disk and an unknot. This proves that nontrivial knot exteriors have incompressible boundary. Their peripheral groups inject as ${\mathbb{Z}}^2$, so their groups strictly contain the cyclic meridian subgroup.

If the torus knot were a connected sum of two nontrivial knots, van Kampen on the decomposing annulus would give $G=G_1{\ast }_{⟨m⟩}{G}_2$, with the meridian subgroup proper in each factor. The center of a proper amalgam lies in the amalgamated subgroup. One proof uses its Bass-Serre tree: a central elliptic element fixes a vertex orbit and its convex hull, which is the whole tree; a central translation would force the minimal tree to be its invariant axis, but a nontrivial vertex reflection reverses that translation. Thus a central element fixes every edge and lies in the edge subgroup.

The central element z would therefore equal $m^k$. Its image in $G/⟨z⟩$ would give ${\overline{m}}^k=1$, forcing $k=0$ because that image has infinite order. This contradicts $z\ne 1$. Consequently