Innermost critical height loops admit strict piecewise linear cap reduction

Choose a saddle plane $h=a$ with critical vertex $q$. The section is a finite graph with every point of degree two except $q$, whose degree is $2r_q$. Following each edge from $q$ until it first returns to $q$ decomposes its nonregular component into simple loops; distinct loops meet only at $q$. Other components are simple circles. A finite collection of such planar loops has an innermost simple loop $C$ bounding a polygonal disk $D$ with $\mathrm{Int}D\cap S=\mathrm{\varnothing }$: choose an innermost bounded complementary face inside a loop, and if its boundary passes $q$, follow its simple loop boundary; the only possible common vertex is $q$. The circle $C$ cuts $S$ into two parametrized PL disks $E_1,E_2$. To check the finite PL parameter assertion, transfer to the supplied standard sphere, choose a tetrahedral vertex off the transferred circle by polygonal chart point pushes, and truncate a small cone disjoint from it. The remaining sphere has a planar triangular disk parameter by mapping its opposite face and three strips to concentric triangles. Apply [F1] to the circle there; the other disk is the surrounding annulus capped by the removed cone, parametrized by cutting a joining strip and using [F1]. Thus ${\mathrm{\Sigma }}_i=E_i\cup D$ are finite PL spheres, with their flat caps explicitly parametrized.

First suppose $q\mathrm{\notin }C$. In each $E_i$ a sufficiently narrow collar of $C$ lies entirely on one side of $h=a$, since $C$ has no critical point and the collar's open connected side has constant height sign. Write $t=\pm (h-a)\ge 0$ with that sign. For small $0<t\le ϵ$, the collar meets its level plane in one polygonal circle $C_t$ approaching $C$. Its bounded planar disk $D_t$ varies in a finite polyhedral layer $Z$ with bottom $D$ and side that collar: subdivide by the finitely many incident triangle planes and use planar crossing parity to select the bounded disk cells. Taking $ϵ$ small keeps every other surface part outside $Z$. Away from $C$ this follows from the positive separation of compact nonincident pieces and $\mathrm{Int}D\cap S=\mathrm{\varnothing }$; along $C$ it follows from the ordinary two-face regular height charts. The layer is a PL manifold with boundary: interior cells have ball charts, side points have half-ball charts, and its bottom rim has quadrant charts from the same planar fans. Its level fibers have local collars at both sides, including its bottom and top. Thus [F3] supplies a height-product disk parameter on all of $Z$. Apply [F4] to replace $D$ and the collar by a cap with one extremum and otherwise regular height. The rest of $E_i$ meets $Z$ only in its top circle. The ball-transfer conclusion of [F4] is therefore applicable. This supplies the required cap replacement for ${\mathrm{\Sigma }}_i$ in the nonpointed case.

Now suppose $q\in C$. For each $i$, the open boundary arc $C\setminus \{q\}$ has a one-sided collar in $E_i$ of constant height sign. Orient $t=\pm (h-a)$ with this sign. Choose one positive ray in each of its two incident face sectors at $q$, very close to the corresponding zero ray of $C$, and call them ${\rho }_1,{\rho }_2$. For small $0<t\le ϵ$, the part of the collar between these rays is one embedded polygonal arc $l_t$ joining $t{\rho }_1$ to $t{\rho }_2$; choose ray generators of height one. This follows by tracing the level segments in the finitely many incident triangles along $C\setminus \{q\}$; none of their other vertices has height between zero and $ϵ$. The two terminal sectors are linear cones. The planar disk $D$ has an empty angular sector at $q$: its link arc lies in the horizontal circle and meets no other surface direction except its endpoints, since its interior is disjoint from the section graph. Just inside the chosen positive side of this arc choose a finite polygonal joining arc at level $ϵ$ from $ϵ{\rho }_1$ to $ϵ{\rho }_2$, disjoint from $S$ except at its ends. Such an arc is obtained by following that empty link sector and projecting radially to the positive level; its compact interior stays in the adjacent open complementary sector. Call this arc $g_{ϵ}$ and put $g_t=(t\mathrm{/}ϵ)g_{ϵ}$. Then $l_t\cup g_t$ is a simple polygon. Its bounded disk $D_t$ contains no other surface piece: away from $q$ use finite positive separations as in step 2.1; in the small conical neighborhood of $q$, the chosen joining arc cuts off exactly the sector adjacent to the empty disk, excluding the remaining link branches. Shrink $ϵ$ once for these finitely many requirements.

Let $Z$ be the layer of these disks for $0<t\le ϵ$, with bottom $D$. It is a compact finite polyhedron: its boundary consists of the bottom disk, the finite surface strip $L=\bigcup l_t$, the cone $G=\bigcup g_t=q\ast g_{ϵ}$, and the top disk $T=D_{ϵ}$; a common plane arrangement with planar parity selects its finitely many solid cells. It is a PL manifold with boundary, also at $q$. The link there has boundary made of the bottom disk arc, two positive surface-sector arcs and the closing arc of $G$, forming one simple polygonal circle on the standard link sphere. The disk on its positive side gives its half-ball chart. The disk assertion follows by the finite tetrahedral planar chart and [F1] as in step 1.1; coning gives the chart. The bottom arc is an ordinary boundary interval of that link disk, so $D$ is locally collared even at $q$. Everywhere at positive height, the side arcs and their endpoints give the ordinary half-space or quadrant height charts; there are no other vertices of the surface in the layer. Therefore [F3] gives a height-product parameter $D\times [0,ϵ]\to Z$. In particular $Z$ is a supplied parametrized PL ball. The old cap part $F=D\cup L$ is a disk with boundary $l_{ϵ}\cup {\rho }_1[0,ϵ]\cup {\rho }_2[0,ϵ]$. Splitting its strip corner $q$ into its two boundary occurrences makes $L$ a finite PL disk with bottom the interval obtained by cutting $C$ at $q$, top $l_{ϵ}$ and two monotone sides ${\rho }_1,{\rho }_2$. Its bottom and positive fibers have local one-sided collars, so [F3] also supplies a height-product interval parameter for this normalized strip. Restoring the two occurrences gives the actual strip in $\mathrm{\partial }Z$.

Supply the convex pointed boundary model required by [F5], rather than assuming it. Let $D_0=[0,2{]}^2\times \{0\}$ with $q_0=(0,0,0)$, and take the convex hull with the top triangle having vertices $(-1,1,ϵ),(1,-1,ϵ),(1,1,ϵ)$. Its face in $x+y=0$ is the triangle $G_0$ from $q_0$ to the first two top vertices, meeting $D_0$ only at $q_0$. The remaining side strip, normalized by splitting $q_0$, has a height-product interval parameter by the same [F3] argument, or directly from its convex faces. Map $D$ to $D_0$, taking $q$ to $q_0$, by [F1]. Match the two normalized strip parameters by their induced bottom interval map; this is height-preserving and matches the two side rays at every height. Map the cone $G$ to $G_0$ by coning its top arc map linearly from $q$ to $q_0$; choose its endpoints to match the strip maps. On the top disk extend the already prescribed boundary map by [F1]. These four maps paste to a finite PL boundary homeomorphism preserving height on $D,L,G,T$. Use the cut in [F5] below a height $\delta <ϵ$. Its replaced boundary disk $A$ contains $D$ and meets $G$ only at $q$; it also misses $T$, whose height is $ϵ$. Since $\mathrm{\partial }Z=D\cup L\cup G\cup T$, these two exclusions give $A\subset D\cup L=F$. In particular $A$ is not all of $F$: its boundary turns through the interior of $L$ before reaching either positive $\rho$-ray. In the convex model this is explicit: take the supporting function $L_0(x,y,t)=x+y$ and cut by $t\le e{L}_0$, with $e>0$ so small that the cut lies below $\delta$. On $D_0\setminus \{q_0\}$, $L_0>0$, whereas on $G_0\setminus \{q_0\}$, $t>0=e{L}_0$; hence the cut meets that triangular side only at $q_0$. The height-preserving leveling map in [F5] retains the prescribed boundary map, so these exclusions pull back to $Z$. Step 3.1 says the remainder of $E_i$ misses $Z$ outside $L$; on the cut ball's old boundary its intersection with $F$ is exactly $A$. Thus the remaining disk of ${\mathrm{\Sigma }}_i$ meets this ball only along $\mathrm{\partial }A$, as required for [F5]'s ball transfer. Replacing $A$ by its new pointed cap meets $t=0$ only at $q$ and introduces no positive-height saddle. The replacement is confined to an arbitrarily short positive height range. On the critical plane it deletes precisely the chosen loop $C$; all other branches at $q$ are outside the layer and remain unchanged.

Perform the respective replacements on both ${\mathrm{\Sigma }}_i$, in a height range containing no other original saddle height, obtaining $S_i^{\mathrm{\prime }}$. In step 2.1 only a new extremum was inserted. In step 5.1 all new positive-height points are regular, and at height $a$ exactly the selected loop has disappeared. Every other original critical-plane loop belongs to one of the two cut disks and is unchanged. If the remaining vertex at height $a$ becomes regular or an extremum in either sphere, that plane no longer contributes to its complexity, which can only decrease the count further. Thus the total contribution at this height is at most the former contribution minus one; at other saddle heights the contributions partition the original ones. Restore generic affine heights by arbitrarily small direction perturbations and compatible subdivisions. This does not add saddles: each new nonextremal vertex has a link circle with exactly two transverse zero crossings, a condition stable under a small perturbation on its finite link edges; the cap extremum has all incident directions of one sign. The other critical link signs and the order of their heights are preserved by choosing the perturbation below the finitely many nonzero sign and height gaps. There are no horizontal surface faces left in the replaced caps. The section graph incidence at each unchanged critical vertex is determined by these signs, so its loop count is preserved. This proves the stated complexity inequality. Finally, supplied balls for $S_i^{\mathrm{\prime }}$ transfer by steps 2.1 or 5.1 to supplied balls for ${\mathrm{\Sigma }}_i$. Apply [F6] to the three disks $E_1,E_2,D$ to obtain the ball bounded by $S$. All constructions and exclusions were finite, so no choice principle was introduced.