Polyhedral two-spheres without height saddles bound parametrized piecewise linear balls

We will use the following finite two-dimensional link construction. A simple polygonal circle $C$ in a standard PL two-sphere bounds two parametrized PL disks. To verify it, replace the sphere by a tetrahedral boundary using its supplied standard parameter and choose a tetrahedral vertex $q$ off $C$. If necessary move a point off $C$ to that vertex by finitely many small polygonal chart point pushes, each obtained by moving the center of a triangle fan in a convex polygon and fixing its boundary. Truncate a sufficiently small cone at $q$ disjoint from $C$. The remainder of the tetrahedral boundary has a finite planar disk parameter: its opposite face maps to a smaller concentric planar triangle, and its three quadrilateral strips to the surrounding triangular annulus, by the disk boundary extension in [F1]. The image of $C$ is a simple polygon inside this triangle. Its inner disk is supplied by [F1]. The region between it and the outer triangle is a PL annulus: cut along a thin strip about a polygonal joining arc, parametrize the resulting polygonal disk by [F1], then restore the strip. Filling the annulus's outer end with the removed cone at $q$ gives the second disk. The same disk boundary-extension result makes their common circle parameters agree. Consequently a cone on either link disk is a half-ball, and coning the two compatible disk parameters gives the standard local pair of a plane and its two half-spaces. This construction applies to the link circle at every point of $S$, including points made vertices by subdivision.

Construct a compact finite polyhedral region with boundary $S$ without assuming a three-dimensional separation theorem. Choose a point $p$ off the finitely many triangle planes and cone each triangle $\tau$ of $S$ to $p$, obtaining finitely many nondegenerate tetrahedra. Regard their sum with coefficients in ${\mathbb{F}}_2$. Its boundary is exactly the triangle sum of $S$: the cone on each sphere edge occurs twice and cancels. Subdivide the union of these tetrahedra by their finitely many face planes and triangulate compatibly by [F2]. Give each full-dimensional cell the parity of the number of tetrahedra containing its interior, and let $Q$ be the union of the closures of odd cells. It is a compact finite polyhedron. At a generic point of a subdivision face not in $S$, the two adjacent parity coefficients agree by the preceding boundary cancellation; at a face in $S$ they differ by one. Thus parity extends as a locally constant function on ${\mathbb{R}}^3\setminus S$ and changes across $S$. To justify extension at subdivision edges or vertices off $S$, join nearby full-dimensional cells by a small polygonal path missing the codimension-two skeleton; each crossed face is outside $S$ and preserves parity. At a point of $S$, use the two half-ball charts of step 1.1; on each side the same small-path argument makes parity constant, and a transverse face crossing shows the two values differ. Therefore $\mathrm{\partial }Q=S$ and $Q$ has a local PL half-ball chart along $S$ and an ordinary interior chart elsewhere.

Both sides of $S$ are connected. The local $Q$ side along $S$ belongs to a single component of $\mathrm{Int}Q$: each surface chart has a connected half-ball side, and paths in the connected sphere propagate that component from one chart to the next. Any additional component would have a nonempty boundary in $S$, since an open nonempty bounded component cannot be closed in ${\mathbb{R}}^3$; a boundary point would contradict the chart description just given. The same reasoning applies to ${\mathbb{R}}^3\setminus Q$: all its local sides along $S$ belong to one component, and any component has a boundary point in $S$ unless it is both open and closed in ${\mathbb{R}}^3$. Hence it is connected; it contains the complement of a large box and is unbounded. These arguments use only elementary polygonal path connectedness of open subsets and connectedness of Euclidean space; a nonempty bounded open component has a boundary point by intersecting a segment from one of its points to a point outside its closure. Thus $Q$ is precisely the closure of the bounded side. Moreover $h$ achieves its minimum and maximum on $Q$ at $S$: an interior extremum of a nonconstant affine coordinate is impossible. They are consequently $h(v_{-})$ and $h(v_{+})$.

At the minimum vertex $v_{-}$, the whole surface link lies strictly above its horizontal plane. The link disk on the $Q$ side supplied by step 1.1 also lies strictly above that plane. Indeed the complementary link component contains every lower direction, which cannot belong to $Q$ by step 3.1; hence the $Q$ link is the other disk, disjoint from the closed lower hemisphere. Choose $\alpha >h(v_{-})$ sufficiently close to the minimum that $Q\cap \{h\le \alpha \}$ lies within the conical star of this vertex. Such a choice exists by compactness of the complement of any smaller vertex neighborhood, whose minimum height is strictly larger. In that star the set is exactly the cone from $v_{-}$ to its planar section $D_{-}=Q\cap \{h=\alpha \}$. This section is a parametrized polygonal disk: radial projection of the upper link disk to the section plane is a homeomorphism, and each cone face cuts it in a finite straight segment. Its boundary is a simple polygon, its interior is its bounded side, and [F1] supplies a finite PL disk parameter. Coning that parameter from $v_{-}$ gives a PL parameter of the cap as a three-ball. Apply the reversed-height argument at $v_{+}$ to obtain $\beta <h(v_{+})$ with a parametrized upper cap and disk section $D_{+}$. Take the caps small enough that $\alpha <\beta$ and neither end level contains a sphere vertex.

Every point of the middle band $P=Q\cap \{\alpha \le h\le \beta \}$ has a local PL height-product chart, with a one-sided chart at the two end levels. At an interior point of $Q$ this is the ordinary linear coordinate chart. At a sphere vertex it follows from the assumed regular link and [F4], using the $Q$ half-space side identified in step 2.1. At an interior point of a sphere triangle its plane is not horizontal because its vertices have distinct heights, so the regular cone condition holds there as well. At an interior point of a sphere edge that edge has nonzero height slope, again by distinct vertex heights; the two incident face half-planes give a link circle with exactly two horizontal crossings, so [F4] applies after subdividing at that point. These charts give precisely the local one-sided fiber collars required by [F5]. Thus [F5] gives a finite PL height-preserving homeomorphism $D_{-}\times [\alpha ,\beta ]\to P$, fixed on the bottom. In particular the whole middle band, not just the individual vertex-free slabs, has a verified finite product parameter.

The product of the disk parameter of $D_{-}$ with the interval identifies $P$ with a convex polygon times an interval, a convex three-ball; compatible prism triangulations in [F2] make this a finite PL parameter. The lower cap meets $P$ in exactly its entire disk $D_{-}$, and the upper cap meets it in exactly $D_{+}$. Both disks lie in the respective boundaries, and the cap interiors are disjoint from the middle band and from one another by their height ranges. Glue the lower cap to $P$ by [F3], obtaining a parametrized three-ball; the remaining upper disk is still a boundary disk, since locally it is the untouched top face of the middle product. Glue the upper cap by [F3] once more. The resulting parametrized ball is exactly $Q$. Its boundary is the original $S$ by step 2.1. All coning, plane arrangements, collar-leveling operations, interval subdivisions and gluings were finite. Thus this no-saddle ball parameter requires neither choice nor an already known general three-dimensional Schoenflies theorem.