= Solution
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>. Consequently
$$
\boxed{B_1\cup_D B_2\cong_{mathrm{PL}} B^3.}
$$
The 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: \b[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 $\widetilde S$. Since $S$ is <simply connected>, its lift maps homeomorphically to $S$. By the <Three-dimensional Schoenflies theorem>, $\widetilde S$ 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 $\operatorname{int}B$, 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 \b[every embedded <sphere> in $M$ bounds a ball], and $M$ is an <irreducible three-manifold>.
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