Regularity, compactness, and gluing for decorated knot Floer triangles

Treat rank zero directly. The target and all empty product tori are one point. The fixed-map operator is the zero map from the zero section space to the zero forcing space; it is onto of index zero. The three-marked source has no conformal parameter or automorphism. Hence its unique map is a regular isolated triangle with all marked degrees zero, in its assigned constant-class Spin-c extension. With another prescribed label or a nonzero degree vector the relevant set is empty. A fixed index-one triangle or universal index-minus-one triangle cannot occur. A matching one-parameter family is exactly its compact parameter interval, with its two identical endpoint counts; there are no nonconstant strips, spheres or boundary disks. Thus its differential is zero and its triangle and parameter-boundary identities hold directly. No strip translation is divided out and no empty tuple is claimed to supply a constituent surface point.

Now suppose that $m>0$ and apply [F1] with the marked points and one witness in every original region protected. The resulting representative has no triple torus intersection and has the same followed pair generators. [F2]'s actual collar extension carries the compression disks and Morse data, identifies the actual Spin-c lift and capped classes, and preserves every old strict witness coefficient under the swept-annulus lattice isomorphism. Thus the new triple has the same required strict admissibility and Spin-c indexing. Its polygon spaces represent exactly the transported literal continuous classes, not classes modulo a sphere or fundamental-group action. A constant triangle would lie in all three tori, so none occurs. This explicitly permitted auxiliary curve choice is necessary: at a constant triangle the almost-complex parameter derivative vanishes because $du=0$, so no J-only universal argument can be substituted for it.

Apply [F3] to this actual triple and its restricted pair sectors. The geometric strict-domain theorem first supplies a finite lower strip index and finite nonnegative domains in every upper-index band. Its actual auxiliary all-index regular path and bubble budget then produce the open dense good genuine-collar end data with no nonconstant index at most zero and index-one regularity. Hence the ensuing polygon limits have the required regular strip levels; a claim of index-one regularity alone would not justify that conclusion. The source-local Hilbert perturbations give actual fixed-data and one-parameter universal transversality at every nonconstant triangle, preserving the protected divisors, tori and exact product-angle end charts. There are no constant triangles by step1.2.

A triangle has three marked ends and no conformal modulus or translation action. Its fixed count index is therefore $\mu$, and a one-parameter universal count has index $\mu +1$. The domain bounds in [F2] and actual domain-area inequality in [F3] give one energy bound for each required low-dimensional set, across all its literal classes and all nonnegative degree vectors. Every limiting nonconstant strip has positive index. Every nonconstant disk bubble contributes at least two, every sphere at least $2k$, and every regular triangle vertex has nonnegative count dimension. The unconditional determinant index budget of [F3] thus excludes all bubbles in count dimensions zero and one. The only broken one-dimensional ends have one rigid triangle and one rigid strip. The same calculation with the family parameter counted ONCE gives rigid universal index-minus-one triangles and the stated universal index-zero ends. This retains $k=1$ and general intermediate $Y$; it does not borrow a knot-in-$S^3$ disk theorem outside its scope.

The actual product-chart, correction and exhaustion proof in [F3] supplies exactly one collar for every such compatible broken pair. Its global tail estimate promotes compact-core convergence to the true Sobolev chart, so a rigid total set is compact discrete and finite under Countable Choice. Its explicit Hausdorff collar compactification of a one-dimensional space is a compact one-manifold with finite boundary. Degree vectors and the actual total Spin-c structure are carried by each neck smoothing; restricting either these labels or an actual class gives the corresponding open-and-closed part. Thus neither multiple classes with the same domain nor the Spin-c extension orbit creates a missing or an extra end.

Count the boundary of this actual compact one-manifold over ${\mathbb{F}}_2$. Writing $F$ for the bilinear triangle count, the strip breaks give $dF+F(d\otimes 1)+F(1\otimes d)=0$. The marked vectors add at each actual gluing, so the component weights multiply, with every $w,z$ exponent retained. For a fixed-end one-parameter family its universal one-manifold also has the two parameter endpoints; the universal index-minus-one counts $H$ give $F_1+F_0=dH+H(d\otimes 1)+H(1\otimes d)$. These are finite weighted equations, also after specializing every $w$ color to $U$ and every $z$ color to $V$. If the intermediate input $\mathrm{\Theta }$ is an actual cycle, substituting it removes the $F(1\otimes d)$ term and gives the required one-input chain map and homotopy. Together with the direct point model in step1.1 and the raw convention [F4], this proves exactly the analytic identities. No existence or cycle property of a distinguished top input, elementary cobordism-map construction, or movie-invariance theorem has been assumed to obtain them.