Kneser–Haken finiteness for incompressible tori

Apply [F2] to the given finite torus family for the very triangulation in the Statement. Its smooth ambient isotopy retains incompressibility, all cut-component homeomorphism types and the no-product condition. It puts the family into actual normal disks in each parametrized tetrahedron and disjoint arcs in each triangular face, joining distinct edges. This invokes the accepted marked normal-position construction rather than treating a surgery that changes a torus as an isotopy.

In each triangular face group the arcs by the pair of distinct edges they join. Within each of the three types they occur in a parallel stack. Between consecutive arcs of a stack is a rectangle, and collapsing those strips leaves at most one arc of each type, hence at most four complementary regions. Every nonrectangular region survives distinctly in this collapsed picture, so there are at most $4t$ such regions over all faces. The successive finite face-disk cuts and the marked rectangle product maps are supplied by [F2] and [F5]; no unparametrized polygon or missing side matching is being assumed.

Let $m$ be the number of cut components meeting all faces only in rectangles. By [F3] each is an interval bundle. The torus-product possibility is excluded by hypothesis. Thus they have disjoint one-sided compact locally flat cores $Q_1,\dots ,Q_m$, and removing each core leaves its component connected. No ambient triangulation or smoothness of these cores is required.

Form the connected finite dual graph of all torus cuts, allowing loops, and choose a spanning tree. Let $k$ be the number of edges outside that tree. The corresponding $k$ tori, together with all the $Q_j$, have connected complement: remove the cores in the fully cut pieces, which remain connected by step 3.1, and glue back the tree faces. A tree joins all the pieces and cannot create a new disconnection. Pass to $\mathrm{Int}M$ by the finite inward collar push in [F5], chosen disjoint from these compact surfaces; its homotopy fixes them and proves that the interior inclusion induces a homology isomorphism, while their complement remains connected. Now [F4] gives $k+m$ independent mod-two classes, so $k+m\le \beta$, where $\beta ={\dim }_{{\mathbb{F}}_2}{H}_2(M;{\mathbb{F}}_2)$. This number is finite because the supplied triangulation is finite and its simplicial homology computes singular homology.

Write $n=\mathrm{\mid }\mathcal{T}\mathrm{\mid }$ and let $v$ be the number of cut components. The dual graph has $n$ edges, $v$ vertices and $k=n-v+1$ non-tree edges; thus $v=n-k+1$. Each component other than the $m$ rectangular-only ones contains a nonrectangular face region. A physical region belongs to just one cut component: its two tetrahedral sides are glued through its interior, or it is a boundary face with only one incident side. Distinct components therefore use distinct such regions, and step 2.1 gives $v\le m+4t$.

Combining these inequalities yields $n+1\le k+m+4t\le \beta +4t$, which is exactly the stated bound. It also holds for the empty family. If an accumulating cutting process continued past $4t+\beta -1$ cuts while each finite accumulated family retained the stated hypotheses, that stage would contradict the same bound for the fixed original triangulation. Thus no such process continues indefinitely. The proof retains the original countable-choice and smooth-compatible triangulation scope; it does not postulate an arbitrary unrelated TOP marking.