Satellite knots and JSJ tori

First apply [F9] to the given surface, fixed pointwise on its boundary. The resulting surface is neat. Incompressibility passes to it, and a boundary-compressing witness passes in either direction; absence of boundary compressions therefore passes as well. Work with this neat representative for the following smooth collar constructions. The boundary-fixing topological ambient extension pulls any resulting boundary-parallel product back to the actual original surface. Thus this preparation does not add a neatness hypothesis to the Statement. Write $\mathrm{\partial }D=\alpha \cup \beta$, where $\alpha$ is the essential proper arc in $A$ and $\beta$ lies in a boundary torus $T$. In an annulus every essential proper arc joins its two boundary circles: cut along a radial spanning arc and use a lift to the strip; an arc returning to the same boundary cuts off a disk there. Consequently both boundary circles of $A$ lie on this same $T$. They are essential on $T$: otherwise choose an innermost disk bounded by one of them in $T$, push it inward off the annulus collar, and obtain a compression of $A$. By [F2] the two circles cut $T$ into two parametrized annuli. The interior of $\beta$ misses them and joins them, so it spans one of these annuli, denoted $A_0$.

Push the interior of $A_0$ a little into its inward collar, keeping its rim on $A$, and round the rim. The union $S=A\cup A_0$ is a torus with a displayed corner before rounding. The curve $\alpha \cup \beta$ on it meets a core of $A$ once algebraically, not merely modulo two: $\alpha$ is a spanning arc of the annulus. It is therefore a nonseparating essential circle. Compress $S$ along $D$, taking two parallel copies of $D$ in its disk neighborhood; the result is a sphere. Irreducibility supplies a parametrized ball bounded by it. The disk neighborhood lies outside this ball: its strip along $\beta$ is joined to $\mathrm{\partial }M$ through the gap left when that strip was removed from $A_0$, without crossing the compressed sphere. A ball in the interior cannot contain that boundary collar. Restoring the disk neighborhood therefore attaches one marked $1$-handle to the ball, rather than removing a possibly knotted handle from it. Its two attaching disk windows have the actual parallel-copy parameters of $D$. They can be put in standard positions by [F3]: standardize one window, shrink the other in the connected complement of the first on the boundary sphere, and move it to the opposite standard disk by the finite supported construction in [F3]'s Proof 1.1–2.1, fixing the first. Thus the restored region $U$ is a smooth solid torus with meridian disk the cocore of this handle. Its boundary is the displayed $S$; all corner changes are restored by the original quadrant collars.

The marking is essential. In the product coordinates $U=D^2\times S^1$ supplied by that ball and handle, let $(\mu ,\lambda )$ denote the meridian-longitude basis of its boundary. Each rim circle of $A$ meets the cocore boundary once algebraically, because it meets $\alpha \cup \beta$ once. Hence its slope is $p\mu +q\lambda$ with $q=\pm 1$. By [F2] normalize the two rim circles to affine circles of that slope. The entire boundary isotopy extends into a radial annulus of $D^2\times S^1$: in the boundary radial coordinate use the same formula $(r,x)\mapsto (r,h_{\eta (r)t}(x))$ as [F3], identity near the inner solid torus. The product diffeomorphism $(z,e^{i\theta })\mapsto (e^{ip\theta }z,e^{iq\theta })$, with inverse $(z,e^{i\phi })\mapsto (e^{-ipq\phi }z,e^{iq\phi })$, carries longitudinal circles to that slope, up to replacing $p$ by its negative according to the chosen inverse. Apply its inverse. The two complementary boundary annuli now correspond to the two arcs between two points on $\mathrm{\partial }{D}^2$, times $S^1$. The disk with one of these marked arcs is a rectangle with its bottom edge marked, with the usual rounded-corner interpretation, by [F3] in dimension two. Taking its product with $S^1$ gives a product between the corresponding annuli: the first is $A\times \{0\}$, and $A\times \{1\}\cup \mathrm{\partial }A\times [0,1]$ is the second annulus $A_0$. Restore the retained quadrant collars to get this product for the actual original $A$ and boundary annulus in $\mathrm{\partial }M$. This proves boundary-parallelity in [F1]'s precise sense. The winding-one condition from $D$ is what justifies this product; complementary annuli in an arbitrary solid torus do not suffice. All selected charts, windows and curve moves were finite, and [F2]–[F3] require no choice.

The solid torus is irreducible. Embed it as a standard unknotted solid torus in $S^3$. A sphere in its interior has two ball sides by [F2]. Its exterior solid torus is connected and misses the sphere, so lies entirely on one side; the other ball is contained in $V$. We will use the following disk-push consequence with its smooth interpretation. If two smooth disks meet just along their common boundary and bound such a ball, their faces admit a common-boundary disk sweep through the ball. To retain the actual faces, take a thin rim collar first. At a transverse rim its normal section is a quadrant; at a previously shared collar first separate the two faces by a small normal graph. Parametrize the two faces by smooth disks with their prescribed rim collars, using [F3]. Round and truncate the rim; the remaining boundary is a smooth sphere with these marked disk and collar portions. The two disk maps and the intervening product annulus give a smooth boundary diffeomorphism to the corresponding standard rounded lens; [F3] extends it over the ball. Restore the quadrant rim using those same product collars. This gives a lens parameter retaining both faces and their rim coordinates. In this parameter use [F4]'s spherical-cap sweep; in the rim section its inward tangent runs between the two face tangents and stays nonzero. The sweep is jointly smooth, including at the disk center, by [F4]'s displayed formulas. It can be continued just past either face in its normal collar and extended ambiently by [F5]. If another surface misses the ball except in one face, the push removes that face's intersections and creates none. The support can be restricted to the ball and arbitrarily thin face and rim collars.

For the classification first replace $F$ by its neat representative using [F9]; both disk-obstruction hypotheses pass to it. Its boundary and abstract surface are unchanged. The topological ambient extension pulls a boundary-parallel product back to the original surface, and no smooth product at an original tangency is asserted. In the rest of the classification denote this neat representative again by $F$. The surface $F$ is compact, since its inclusion is proper and $V$ is compact, and it is orientable, since it is two-sided in oriented $V$. If some component $c$ of $\mathrm{\partial }F$ is nullhomotopic on $\mathrm{\partial }V$, choose one with an innermost boundary disk $Q$ whose interior misses all of $\mathrm{\partial }F$. A short inward push gives a disk with boundary $c$ on $F$ and interior disjoint from $F$. If $F$ is not a disk, $c$ is essential on $F$: a boundary circle of a connected surface can bound a disk in that surface only when that disk is the entire surface, by the local half-disk charts and connectedness. The pushed disk would therefore compress $F$. Thus in this case $F$ is a disk. For the rest of the classification suppose $F$ is not a disk; then every boundary circle is essential on $\mathrm{\partial }V$. Disjoint essential circles on a torus have the same primitive unoriented slope, as the strip reductions of [F4] show. Use those reductions for the finite family and extend into a boundary collar by [F5] to make them parallel affine circles in coordinates $(\mu ,\varphi )\in (\mathbb{R}\mathrm{/}\mathbb{Z}{)}^2$, where $\varphi$ is the longitude coordinate of $V$. If the slope is meridional, choose a standard meridian disk $D_0$ with boundary disjoint from them. Otherwise the affine circles have nonzero $\varphi$ derivative and meet $\mathrm{\partial }{D}_0=\{\varphi =0\}$ transversely.

Finally, if a two-sided torus $T\subset \mathrm{Int}V$ had no compressing disk, [F6] would inject ${\pi }_1(T)\cong {\mathbb{Z}}^2$ into ${\pi }_1(V)\cong \mathbb{Z}$. No such injection exists: for images $a,b\in \mathbb{Z}$ of the two generators, $(b,-a)$ is in the kernel when $(a,b)\mathrm{\ne }(0,0)$, and otherwise the entire group is the kernel. This contradiction yields a smooth compressing disk. This use of [F6] is within its compact smooth scope and does not turn a PL disk into a smooth disk without the supplied bridge. All earlier isotopies are smooth by the explicit cap, graph and finite extension constructions; the full AC assumption covers [F6] and implies the countable choice used elsewhere.

Straighten the neat boundary collars of $F$ in $V$ and double $V$ and $F$ across those collars. The doubled comparison surface is closed in the boundaryless double. The normalized boundary of $D_0$ is transverse to it, or disjoint in the meridional case, by step 2.1. Apply the relative transversality lemma in [F5] to $D_0$, fixing its boundary collar. Choose the perturbation small on the remaining compact disk so that its interior stays in $\mathrm{Int}V$. The motion is stationary on the fixed collar, so the interior extension in [F5] shows that $D_0$ remains ambient isotopic in $V$ to the standard meridian disk. Thus its complement is still a cut ball, and its interior meets $F$ in finitely many transverse curves and arcs. An innermost intersection circle on $D_0$ bounds a disk $D$ with interior missing $F$. Incompressibility makes its boundary bound a disk $D^{\mathrm{\prime }}$ on $F$. The sphere $D\cup D^{\mathrm{\prime }}$ bounds a ball in $V$ by step 1.1, with the corner interpreted there. The connected surface $F\setminus \mathrm{Int}{D}^{\mathrm{\prime }}$ reaches $\mathrm{\partial }V$, misses the sphere off its rim, and hence lies on the exterior side of that ball. The disk push of step 1.1 replaces $D^{\mathrm{\prime }}$ by a disk just past $D$. It removes this circle and any additional circles in $D^{\mathrm{\prime }}$, and creates none. After finitely many pushes no circle remains. The isotopies have interior support and retain the normalized boundary curves.

Suppose an intersection arc remains. An outermost arc $\alpha$ cuts off a disk $D$ in $D_0$, with other boundary arc $\beta \subset \mathrm{\partial }{D}_0$ and interior missing $F$. Boundary-incompressibility implies that $\alpha$ cuts off a disk $D^{\mathrm{\prime }}$ on $F$, with other boundary arc $\gamma$ contained in one component of $\mathrm{\partial }F$. If the slope were meridional there would be no arc endpoints, by the chosen disjoint boundaries. Otherwise $\gamma$ is a nonconstant subarc of an affine boundary circle whose longitude coordinate is strictly monotone on its lift. Both its endpoints have $\varphi =0$, so its lifted longitude displacement is a nonzero integer. But $\beta$ has constant longitude. The union $D\cup D^{\mathrm{\prime }}$, pushed a little along its common edge if necessary, gives a continuous disk cap for the loop $\beta \cup \gamma$ in $V$. Its longitude degree would therefore be zero, contradicting that nonzero integer. Thus there are no arcs. If the slope were nonmeridional, the affine boundary curves would necessarily meet $\mathrm{\partial }{D}_0$, so this case is impossible. We have made $F$ disjoint from $D_0$. Cutting $V$ along $D_0$ leaves a ball. An innermost boundary circle of $F$ on its boundary sphere has a disk there with interior missing all boundary circles of $F$. Push it inward. Unless $F$ is a disk it is a compressing disk, by the boundary-circle argument of step 2.1. This contradicts our supposition that $F$ was not a disk.

Let $D_1,D_2$ be meridian disks with the same boundary $C$. If either disk is not neat, first neatify each separately by [F9], fixing $C$. Apply the following neat-disk argument and then reverse the second embedding path. This gives a smooth embedding isotopy for the broad inputs; their two explicit topological ambient extensions and the middle smooth ambient path give a topological ambient extension. For neat inputs no preliminary change is necessary, and the entire argument below is smooth ambient. Now suppose both disks neat. In coordinates along $C$, use a small normal graph to separate their interior collars while fixing $C$ pointwise: with inward boundary coordinate $q\ge 0$ and a transverse coordinate $y$ on $\mathrm{\partial }V$, the two collars can be arranged as $y=\pm \epsilon q$ on a common smaller strip. First each neat disk is straightened to such a graph by its full-rank collar coordinates; interpolation of the positive inward derivative gives an embedding path on a common compact strip, extended by [F5]. This keeps every boundary point fixed. For general position, truncate a smaller collar of $D_2$; its new boundary misses $D_1$ because the collars have been separated. Straighten the neat collar of $D_1$ and double it across $\mathrm{\partial }V$, obtaining a closed sphere in the boundaryless double. Apply [F5] to the truncated $D_2$ against that sphere, fixing the new boundary collar and making the perturbation sufficiently small to stay in $\mathrm{Int}V$. Extend by the unchanged original collar. The interior-supported extension in [F5] realizes this as an ambient motion in $V$. There are then only finitely many interior intersection circles. An innermost one on $D_1$ and its subdisk on $D_2$ bound a ball by irreducibility. The complement of that subdisk in $D_2$ reaches $C$ outside the ball, so the disk push from step 1.1 removes the circle and any circles inside that subdisk. Repeat until the interiors are disjoint. The resulting disks bound a ball with rim $C$, again by step 1.1 after the rim is rounded. Its common-boundary cap sweep carries one disk to the other, fixing $C$ throughout; at the rim every intermediate tangent is an inward direction between the two collar tangents, so the family remains neat in $V$. Undo the preliminary isotopies of $D_2$ and concatenate with stationary time changes. This gives a jointly smooth isotopy relative to $C$; The velocity vanishes on $C$ throughout this neat path. In a neat boundary chart it is $q$ times a smooth coefficient, as in [F9] and its neatification supplier Proof 5.2. Extend that coefficient and retain the factor $q$; the finite compact flow gives a smooth ambient isotopy fixing all of $\mathrm{\partial }V$, not only $C$.

We have proved that $F$ is a disk. If $\mathrm{\partial }F$ is essential on $\mathrm{\partial }V$, it is a primitive class in the kernel of ${\pi }_1(\mathrm{\partial }V)=\mathbb{Z}\mu \oplus \mathbb{Z}\varphi \to {\pi }_1(V)=\mathbb{Z}\varphi$, since $F$ caps it. It is therefore $\pm \mu$, so $F$ is a meridian disk. If $\mathrm{\partial }F$ is inessential, let $Q$ be its disk on $\mathrm{\partial }V$. The sphere $F\cup Q$, pushed slightly into the interior and rounded, bounds a ball there. Restore the boundary collar as in step 1.1 to get a ball with these two disk faces. Its marked lens parameter gives the boundary-parallel product: extend $Q$ a short distance beyond its rim in $\mathrm{\partial }V$ and retain a smaller collar on $F$; in the normal section the two disk faces are separated by a rectangle, with its outer edge on $\mathrm{\partial }V$. The lens parameter away from that rim and this rectangle near it identify the region with $D^2\times [0,1]$, with $D^2\times \{0\}=F$ and the other disk and lateral annulus in $\mathrm{\partial }V$. Thus $F$ is boundary-parallel in exactly [F1]'s product sense. Conversely a boundary-parallel disk has no essential circle or essential proper arc, so it does satisfy both disk-obstruction hypotheses; this is why it must be retained in the conclusion.

Finally allow the torus $T\subset \mathrm{Int}V$ to be TOP two-sided with its given product neighborhood, without assuming it smooth. Cut along that TOP collar. The cut pieces are compact orientable TOP3-manifolds; there are either two connected pieces with one torus face each, or one connected piece with two torus faces. Indeed every piece meets a face, and each face is connected. If any face map has nontrivial fundamental-group kernel, [F10] gives a proper disk in that cut piece whose boundary has nontrivial class on the face. Its image in $V$ is an embedded TOP compression of $T$ contained in $\mathrm{Int}V$, since its interior misses every cut boundary face and its boundary lies on only the chosen copy of $T$. Suppose therefore all face maps inject. In the two-piece case, [F11] embeds ${\pi }_1(T)$ in their amalgam ${\pi }_1(V)$. In the one-piece case choose a path in the cut piece joining the two face basepoints. Adjoining a thin copy of this path to the collar makes its overlap connected; van Kampen gives the presentation $⟨{\pi }_1(W),t\mid t{i}_{+}(a)t^{-1}=i_{-}(a),a\in {\pi }_1(T)⟩$. One torus-overlap group maps to the collar's torus group and the other to its conjugate by the new path letter, which is exactly the displayed HNN relation. The two injected face groups are its associated subgroups, so HNN normal form [F11] again embeds ${\pi }_1(T)$ in ${\pi }_1(V)$. Both alternatives would inject ${\mathbb{Z}}^2$ into $\mathbb{Z}$, impossible: if the two basis images are $a,b$, then $(b,-a)$ is a nonzero kernel vector when either image is nonzero, and otherwise the whole group is the kernel. Thus some face kernel is nontrivial and the TOP compression exists. This closes the original unqualified torus clause without converting its input or its resulting disk to the smooth category. The preceding smooth classification and meridian-isotopy arguments are unchanged; their original convention/disposition remains owned by the parent assembly.

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.

Each of the two possible bases has nonempty boundary and deformation-retracts to a graph, so ${\pi }_2(B)=0$. The homotopy exact sequence [F2] therefore injects the fiber subgroup ${\pi }_1(S^1)\cong \mathbb{Z}$ into ${\pi }_1(N)$, with image the kernel of $p_{\ast }:{\pi }_1(N)\to {\pi }_1(B)$.

The fundamental group of a pair of pants is free on two generators $a,b$, and its three boundary loops represent $a$, $b$, and $(ab{)}^{-1}$, each of infinite order. The once-punctured Möbius band has free generators $x,c_1$ and boundary loops represented by $c_1$ and $c_1^{-1}{x}^{-2}$; both are nontrivial cyclically reduced words and have infinite order in the free group. Thus every boundary circle of $B$ injects on fundamental groups.

The restriction $T\to \mathrm{\partial }B$ is a circle bundle over a circle, hence induces a surjection ${\pi }_1(T)\to {\pi }_1(\mathrm{\partial }B)$. If $g\in {\pi }_1(T)$ maps trivially to ${\pi }_1(N)$, its projection is a power of the corresponding boundary loop and is trivial in ${\pi }_1(B)$; step 1.2 forces that power to be zero. Therefore $g$ lies in the fiber subgroup, whose inclusion into ${\pi }_1(N)$ is injective by step 1.1, so $g=1$. This proves that ${\pi }_1(T)\to {\pi }_1(N)$ is injective.

The interior of the parametrized PL ball is a contractible boundaryless topological three-manifold. Its second mod-two homology is zero by [F4]. Hence [F1] and [F4] show that its complement of $T$ cannot be connected. The same is true in $B$: push its outer boundary collar slightly inward, fixing the compact interior torus; if $B\setminus T$ were connected, its connected pushed image and the collar homotopy would make the interior complement connected. There are exactly two components. Indeed the two open product-collar sides of connected $T$ are connected, and every complementary component must meet one: otherwise its closure misses $T$ and it is open and closed in connected $B$. More explicitly, at any point of $T$ in the closure of a component, a local flat chart puts some collar-side patch in that component; the connected full side collar then belongs to it. Thus there are at most two components. Their closures $W_1,W_2$ are compact orientable topological three-manifolds, with the corresponding copies of $T$ in their boundaries.

If both maps ${\pi }_1(T)\to {\pi }_1(W_i)$ were injective, choose their coset transversals by [A1]. The given topological product collar gives the connected-overlap van Kampen cover, and [F3] expresses ${\pi }_1(B)$ as their amalgam with ${\pi }_1(T)$ embedded. By [F5] this would embed ${\mathbb{Z}}^2$ into the trivial group of the ball, a contradiction. Thus one side boundary map has nontrivial kernel. This conclusion is topological and has required no triangulation of the cut pieces.

Apply [F2]'s ACTUAL TOP triangulation theorem to that compact orientable side $W_i$, using [A1]. Let $q:\mathrm{\mid }L\mathrm{\mid }\to W_i$ be its resulting finite manifold triangulation. The preimage $q^{-1}(T)$ of the whole boundary component is a connected PL boundary surface subcomplex; no pre-existing compatibility of the original embedded torus with the original ball triangulation was needed. Conjugating by $q$ identifies its inclusion kernel with the nontrivial kernel from step 2.1. Apply finite PL Loop with that entire boundary component and normal subgroup $1$. It supplies a proper embedded PL disk in $\mathrm{\mid }L\mathrm{\mid }$ with boundary representing a nonidentity element of its torus group. Transport it by $q$ and then by the cut-side inclusion into $B$. These maps are homeomorphisms on the disk: the disk meets only this single boundary face and its interior misses all boundary faces. Hence its image $D$ is a TOP embedded disk in $\mathrm{int}B$, with $D\cap T=\mathrm{\partial }D$ and essential boundary on $T$. It is proper in the cut side, as the Statement requires. This completes the unqualified TOP two-sided torus case with the original full AC hypothesis.

Put $X=e(X_P)$ and $Y=E_K$. They are compact orientable smooth three-manifolds with common boundary torus $T_c$. The exterior $Y$ is connected by [F2]. For completeness $X$ is connected as well: in the standard connected solid torus join two points of $X_P$ by a path consisting of finitely many smooth chart arcs. Remove repeated arcs and perturb finitely many joins in small coordinate balls to obtain an embedded arc, fixed near its endpoints; in dimension three finitely many arcs can be made disjoint except at consecutive endpoints by small finite chart perturbations. By [F5] perturb it to avoid the pattern core $P$. On the punctured supplied tube of $P$, radial projection to its boundary, extended by the identity outside that tube, is continuous and sends this path into $X_P$. Consequently all exterior points are connected by paths. The outer torus has no compressing disk by hypothesis, so [F3], with the entire outer boundary component as subsurface and normal subgroup $1$, gives ${\pi }_1(T_c)\hookrightarrow {\pi }_1(X)$. The companion product tube is exactly the supplied input of [F2], so ${\pi }_1(T_c)\hookrightarrow {\pi }_1(Y)$ too. Both applications are smooth; an abstract finite PL disk has not been treated as a smooth disk.

Use finite smooth collars of the common torus to enlarge the two sides to a connected open cover whose intersection retracts to $T_c$. Van Kampen and the amalgam normal form [F4], with transversals supplied by [A1], then make ${\pi }_1(T_c)$ inject into ${\pi }_1(E_{P(K)})$. Any compressing disk would make its essential torus boundary nullhomotopic in the union, contradicting that injection. Thus $T_c$ is incompressible.

The only boundary of the satellite exterior is the inner pattern torus $T_{\mathrm{i}\mathrm{n}}=\mathrm{\partial }e(N(P))$. The two components cut along $T_c$ are the connected pieces $X$ and $Y$ from step 1.1; only $X$ meets $T_{\mathrm{i}\mathrm{n}}$. Suppose there were a boundary-parallel product region $R$ with ends $T_c,T_{\mathrm{i}\mathrm{n}}$. Its interior must lie in $X$: it cannot cross either of its two boundary tori, and its collar at $T_{\mathrm{i}\mathrm{n}}$ is on the $X$ side. Moreover $R$ exhausts $X$. Indeed $\mathrm{Int}R$ is open in $\mathrm{Int}X$ by the smooth full-dimensional product charts, and closed relative to $\mathrm{Int}X$ because $R$ is compact and its boundary lies in $\mathrm{\partial }X$. The interior of the connected manifold $X$ is connected, as is seen by pushing any boundary portions of a path into its finite collar. Hence $\mathrm{Int}R=\mathrm{Int}X$, and taking closures gives $R=X$. Transporting that product by $e^{-1}$ would identify $X_P$ with a product between its two marked ends, contrary to hypothesis. Therefore $T_c$ is not boundary-parallel. For this closed torus boundary-incompressibility is vacuous, so it is essential by [F6].

Let $S$ be a PL sphere in the interior of the selected cut component. Its image in $M$ is a PL sphere missing every cutting torus, so bounds a parametrized PL ball $B$ by [F1]. Each connected cutting torus lies wholly inside or wholly outside that sphere. If a torus lies inside $B$, its existing two-sided neighborhood can be shrunk into $\mathrm{Int}B$, by compactness and a common smaller product width. It remains a PL embedded torus there. [F3] supplies the compatible finite subdivisions needed for [F2], even if the original triangulation of $B$ was different from the one containing the torus.

Such an inside torus would have the PL compressing disk of [F2] entirely in $B\subset M$, contradicting its original incompressibility. Thus $B$ misses every cutting torus. The connected ball interior lies in the same complement component as its collar along $S$, so $B$ lifts to a PL ball in the specified cut component with boundary $S$. Every interior PL sphere there therefore bounds a PL ball. This is precisely PL irreducibility; no smooth-ball or arbitrary-topological-marking claim entered.

The exterior is a compact oriented smooth manifold with torus boundary. To verify the last assertion, orthonormalize the global normal frame of [F2] in the metric used for the tube's disk subbundle. Its chosen tube is then smoothly parametrized by $S^1\times D^2$, so its boundary is $T^2$. The exterior is connected as well: any two distinct points in it are joined by an embedded great-circle arc in $S^3$ (choose a half great circle in the antipodal case). Apply [F6]'s compact relative transversality to this arc and the knot, fixing its endpoint neighborhoods. A transverse one-dimensional arc cannot meet the one-dimensional knot in a three-manifold, so it avoids the knot. There is a continuous retraction $S^3\setminus K\to E_K$, equal to the identity outside the tube and sending a nonzero normal vector of length $r\le ϵ$ radially to the vector of length $ϵ$ in the same direction. It is continuous at the tube boundary and fixes both endpoints. The retracted arc is a path in the exterior. The coincident-endpoint case is constant. Thus $E_K$ is connected, and it is irreducible by [F1]–[F2].

Choose the finite compatible triangulation of [F3] once. If $E_K$ itself is atoroidal or Seifert, the empty family proves existence and is minimal. Otherwise start with the empty family, and whenever a current component $C$ is neither atoroidal nor Seifert, choose an incompressible torus not boundary-parallel in $C$, by [F8]'s definition of failure of atoroidality. It is two-sided: the torus and ambient orientations orient its normal line, and a finite normal-field flow gives a smooth product neighborhood. Every cut component has nonempty torus boundary. Indeed a boundaryless cut component would have no cutting face and no original boundary, hence would give a closed component of the original connected exterior, impossible by step 1.1. Irreducibility of the successive pieces will follow from [F5]. This construction is needed only for a bounded finite number of steps, as established below.

Suppose a finite accumulated family is incompressible in $E_K$. By [F5] its cut pieces inject into $E_K$ and are irreducible. A new torus chosen in $C$ injects into $C$ by [F6], and then into $E_K$ by composition. It is therefore incompressible in $E_K$ by [F6]. The empty family starts this induction. Thus every finite stage consists of incompressible tori in the original exterior and has irreducible cut pieces, and all applications of [F4]–[F5] are within their smooth categories.

No finite stage has a product cut component. Initially the exterior is not good, hence is not $T^2\times I$. If a product first appears after a new cut along $T\subset C$, it must be one of the new components and must contain a boundary copy of $T$. If its other end is an old boundary face of $C$, the displayed product makes $T$ boundary-parallel in $C$, contradicting its choice. If both ends are the two copies of $T$, regluing them makes the entire connected old component $C$ a closed torus bundle; this contradicts the nonempty boundary of $C$ verified in step 2.1. Components unaffected by this cut were already nonproducts. This finite induction proves the no-product condition required by [F4], without assuming that selected tori separate.

Put $B=4t+\beta -1$, a fixed finite integer from [F4]. If no stage had become good after at most $B$ cuts, the construction could be extended through $B+1$ cuts by finite induction: at each of these finitely many stages use step 2.1. Steps 3.1 and 4.1 would then give an incompressible torus family with no product cut component and $B+1$ members, contradicting [F4]. Thus a finite good family exists. This argument selects only a bounded finite list of cuts, rather than postulating an infinite sequence of independent choices. The zero-cut empty family in the nongood initial case also satisfies [F4], so $B\ge 0$ there.

Among the subfamilies of that particular finite good family, take a good subfamily of least cardinality. There is at least one, and its finite power set suffices for the selection. No proper subfamily can be good, so it is inclusion-minimal in exactly [F8]'s sense. This proves existence with all the original hypotheses and no additional condition on the knot.

Finally let two finite inclusion-minimal systems as in the Statement be given. Step 1.1 verifies that their ambient exterior is compact, connected, orientable and irreducible with nonempty torus boundary. Their tori and good cut pieces have exactly [F7]'s smooth hypotheses. Apply that full uniqueness supplier to obtain a jointly smooth ambient isotopy of $E_K$, even fixed near its boundary, carrying one system to the other. The underlying convention is the ordinary smooth ambient isotopy of the knot definitions; no homeomorphism isotopy has replaced it. The conclusion asserts only atoroidal or Seifert pieces and introduces no hyperbolicity assertion.

First apply [F9] to the given surface, fixed pointwise on its boundary. The resulting surface is neat. Incompressibility passes to it, and a boundary-compressing witness passes in either direction; absence of boundary compressions therefore passes as well. Work with this neat representative for the following smooth collar constructions. The boundary-fixing topological ambient extension pulls any resulting boundary-parallel product back to the actual original surface. Thus this preparation does not add a neatness hypothesis to the Statement. If $S$ is boundary-incompressible, the first conclusion holds. Otherwise choose a boundary-compressing disk $D$, with $\mathrm{\partial }D=\alpha \cup \beta$ and $\alpha$ an essential proper arc of $S$. A boundary disk on a torus whose interior misses $\mathrm{\partial }S$ can be pushed inward to compress $S$, unless $S$ is itself a disk: a boundary circle bounding a disk on the connected surface forces the surface to be that disk. An innermost choice among the finitely many inessential boundary circles provides such a boundary disk. But a disk has no essential proper arc. Therefore every boundary circle of $S$ on the torus $T$ containing $\beta$ is essential. By [F2] its complementary regions are annulus strips. The arc $\beta$ lies in one of them. If both endpoints lie on the same boundary copy of that strip, it cuts off a disk there. Gluing this disk to $D$ gives a disk with boundary $\alpha$ and an interval of $\mathrm{\partial }S$; incompressibility of $S$ forces that circle to bound a disk on $S$, so $\alpha$ would cut off a disk on $S$, contrary to essentiality. Thus $\beta$ spans its annulus strip $A_0$.

The two endpoints of $\alpha$ lie on distinct actual boundary circles of $S$. The only alternative after step 1.1 would be that the two ends of $A_0$ are the two copies of one circle $c$, with $c$ the only boundary circle of $S$ on $T$. Two-sidedness rules this out. Along the connected arc $\alpha$, the interior of $D$ lies on one fixed normal side of $S$. Along $c$, that same side restricts to one fixed side of $c$ in $T$, since the proper surface is two-sided. Hence $\beta$ must approach the same side of $c$ at both endpoints; a spanning arc of $T$ cut along $c$ approaches opposite sides. This contradiction proves the assertion. Let $P$ be a small regular neighborhood in $S$ of $\alpha$ and its two incident boundary circles. It is a pair of pants. Its third circle $d$ bounds a disk just off $S$: boundary-compress the annulus strip $A_0$ along $D$ and push the resulting disk a little inward. Its interior misses $S$, because the interior of $A_0$ misses $\mathrm{\partial }S$ and the interior of $D$ misses $S$. Incompressibility makes $d$ bound a disk on $S$. This disk is on the side opposite $P$: if it contained $P$, the two boundary circles of $S$ in $P$ would lie in the interior of that disk, contrary to the surface half-space charts. Gluing the disk to $P$ fills a whole connected component of $S$, so $S$ is an annulus. The original $D$ is its boundary compression. Apply [F3] to get the second conclusion. Every neighborhood and disk surgery used only finitely many explicit surface and quadrant charts; [F2] and [F3] introduce no choice assumption.

First apply [F9] to the given surface, fixed pointwise on its boundary. The resulting surface is neat. Incompressibility passes to it, and a boundary-compressing witness passes in either direction; absence of boundary compressions therefore passes as well. Work with this neat representative for the following smooth collar constructions. The boundary-fixing topological ambient extension pulls any resulting boundary-parallel product back to the actual original surface. Thus this preparation does not add a neatness hypothesis to the Statement. Take a small product neighborhood of $A$ and collars of its incident boundary tori. They exist by a finite coordinate construction: choose the normal side using orientations of $M$ and $A$, combine finitely many adapted chart fields by normalized finite bumps, and use their short flow and the local inverse and compact separation arguments in [F7]. Normalize the two boundary curves on each incident torus by [F3]. Their parallel strip parameters and the product parameter of $A$ supply circle coordinates on the neighborhood, agreeing at the attaching rims. Circle parameter discrepancies have degree $\pm 1$ and are adjusted by interpolation of their increasing, or decreasing, angular lifts; this is finite. Round the quadrant corners in the base coordinate, keeping the circle coordinate. The resulting neighborhood $N_0$ is an ordinary circle bundle with $A$ vertical. If the annulus meets two different original tori, its base is a pair of pants with two original boundary circles and one internal boundary. If it meets one original torus, its band attaches to the same base boundary circle: the two possible band identifications give a pair of pants with two internal boundaries, or a once-punctured Möbius band with one. The total space is orientable, since it is a submanifold of $M$; in the Möbius case the fiber orientation reverses with the base orientation. These are actual finite band-and-collar quotient parameters. The internal frontier consists of at most two tori.

We first fill compressible frontier tori, before addressing boundary-parallel ones. For $N_0$, frontier groups inject into its group by [F2]. Use this injection for the initial iteration. After the only possible intermediate filling, its explicit group calculation below will verify the invariant before another iteration. If a frontier torus is compressible in $M$, choose a compressing disk and put it transverse to the other frontiers by [F3]. Its intersection consists of finitely many circles. Remove any inessential circle on a frontier by an innermost disk replacement: choose its frontier disk with interior disjoint from the compressing disk, replace the latter's subdisk by it and push slightly to the side of the retained disk. The disk parameters in [F3] and quadrant rounding give an embedded smooth disk with the same essential original boundary and fewer intersections. If a remaining intersection is essential on its frontier, an innermost circle on the disk cuts off a compressing disk whose interior misses the entire frontier. If no intersection remains, the original disk already has this property. Thus some frontier $T$ has a clean compressing disk $D$. It cannot lie in $N$, since its boundary class is nonzero on $T$ and ${\pi }_1(T)\to {\pi }_1(N)$ is injective. The connected disk interior consequently lies on the exterior side of $N$.

Compress $T$ along $D$ and push the resulting sphere slightly off the frontier. Irreducibility supplies a ball $B$ bounded by it. The connected $N$ lies outside that ball: it misses the compressed sphere and contains an original boundary torus of $M$, whereas $B$ is an interior ball. The disk neighborhood restored by reversing compression is also on this non-ball side, because its rim strip is joined to $N$ through the gap removed from $T$. Hence reversal attaches a marked $1$-handle to $B$; it does not remove a possibly knotted handle from its interior. Its two disk windows have the parallel-copy parameters of $D$, and [F4] standardizes them. The result is a solid torus $U$ outside $N$, with boundary exactly $T$. It contains no other frontier, since $B$ and the small disk neighborhood miss $N$ except at $T$. The exterior component adjacent to $T$ is exactly $U$: its interior is both open and closed in that component and it has no other boundary. Adjoin it to $N$.

The meridian of $U$ cannot be its attaching fiber slope. If it were, [F3] and its supported boundary collar formula give a smooth meridian disk bounded by that actual fiber. In $N$, choose a base arc from this frontier to the interior of the base arc underlying $A$, avoiding any previous cone point and meeting that arc only at its endpoint. Its inverse image is a vertical annulus meeting $A$ in its one boundary core. Join it to the meridian disk and round the common collar. This is a compression of $A$, a contradiction. In product coordinates $U=D^2\times S^1$, therefore, the primitive regular-fiber class is $a\mu +b\lambda$ with $b\mathrm{\ne }0$. After reversing the fiber orientation take $b>0$. The action $t\cdot (z,w)=(e^{iat}z,e^{ibt}w)$ gives a Seifert model with that boundary slope and one possible multiplicity-$b$ core. Indeed the map $(\zeta ,v)\mapsto (v^a\zeta ,v^b)$ identifies $U$ with the quotient by $(\zeta ,v)\mapsto (e^{-2\pi ia\mathrm{/}b}\zeta ,e^{2\pi i\mathrm{/}b}v)$; since $\gcd (a,b)=1$, choose an inverse generator to obtain exactly the standard fibered-solid-torus cyclic model. Its boundary fibration has the same fiber class as the old one. Use [F5] to match it smoothly and glue the product collars. Thus this filling extends the actual Seifert structure, leaves $A$ vertical and caps that base boundary by a disk with cone order $b$ (regular if $b=1$).

There are initially at most two frontiers. The only filling that can leave another frontier is the pair-of-pants case with just one original boundary torus; its new base is an annulus with at most one cone. Its remaining boundary groups still inject, by this explicit finite calculation. Before filling the orientable circle bundle over a pair of pants has group $F(a,c)\times ⟨h⟩$: lift the two free base generators, and the oriented fiber $h$ is central and injective by [F2]. Filling adds the primitive relation $c^q{h}^p=1$, with $q\ge 1$, by [F2]'s gluing pushout. The quotient by central $h$ is $\mathbb{Z}\ast C_q$. The two boundary base loops have images $a$ and $(ac{)}^{-1}$, both of infinite order by reduced words (for $q=1$, both are $a^{\pm 1}$). Also $h$ has infinite order: the presentation maps to the additive reals by $h\mapsto 1$, $c\mapsto -p\mathrm{/}q$, $a\mapsto 0$. Thus a boundary element projecting trivially must first have zero base exponent, then zero fiber exponent. This proves boundary-torus injection after the intermediate filling. Step 2.1 is therefore applicable again. After finitely many such steps, either no frontier remains or every remaining frontier is incompressible in $M$.

By atoroidality, each remaining frontier $T$ is parallel to an original boundary torus, with a supplied smooth product $P=T^2\times I$. If $P$ is on the exterior side of $N$, it is exactly the adjacent exterior component, so absorb it into $N$, extending the fibration as a product by [F5]. Suppose instead that $N$ lies on the $P$ side. It is then contained in $P$, since the latter occupies that whole component between $T$ and the original boundary. The product contains only one original boundary torus of $M$. Thus this is impossible if $N$ already contains two original boundary tori. If $N$ still has two frontiers, it is the original pair-of-pants bundle with one original boundary; the other frontier $T^{\mathrm{\prime }}$ is an incompressible smooth torus in $P$. By [F6], every cut piece of $P$ along $T^{\mathrm{\prime }}$ injects into ${\pi }_1(P)$. The connected $N$ is one such piece, since its only interior boundary in $P$ is $T^{\mathrm{\prime }}$. But ${\pi }_1(N)$ surjects onto the nonabelian free rank-two base group [F2], while ${\pi }_1(P)={\mathbb{Z}}^2$ is abelian. This is a contradiction. This argument treats $N$ as a possible proper submanifold of the collar; it does not incorrectly infer that nonabelian $N$ cannot sit inside a product without an injection proof.

It remains to exclude the $N$-side collar when there is one frontier and one original boundary. In that case $N=P$, since it has exactly the two boundary tori of $P$ and no interior frontier in it. If its base is the once-punctured Möbius band, [F2]'s free rank-two quotient contradicts the abelian product group. Otherwise this is the intermediate annulus-with-one-possible-cone case. Its central-fiber quotient from step 5.1 is $\mathbb{Z}\ast C_q$, which is nonabelian for $q\ge 2$. Hence $q=1$: the inherited structure is an ordinary circle bundle over an annulus. The vertical $A$ has both ends on the same original base boundary, so its base arc cuts off a disk. That disk has a smooth parameter by [F3] in an annulus cover. Trivialize the bundle over it by radial horizontal transport: combine finitely many local lifts by finite base bumps and solve the radial ODE from its center using [F7]. A fixed point of the initial fiber gives a section disk. Its boundary consists of an essential spanning arc of $A$ and an arc in the original boundary torus. This is a forbidden boundary compression of $A$. Thus the collar can never be on the $N$ side. Every remaining frontier is therefore an exterior product collar and can be absorbed.

Each solid-torus filling or collar absorption removes a frontier, and initially there were at most two. At the end $N$ has no internal frontier; it is nonempty, closed and relatively open in connected $M$, hence equals $M$. Its fibration is the smooth Seifert structure constructed in steps 4.1 and 6.1. The neat representative of the annulus was retained throughout. The conclusion is a smooth Seifert structure on the unchanged ambient manifold, so it proves the original choice-free Statement without asserting that an original tangential annulus is smoothly ambient isotopic to a vertical one. The only product-group comparison used the explicit flat-product compression and cut-injection bridge [F6]; no arbitrary smooth/PL compatibility or general topological triangulation assertion entered. All choices and modifications were finite.

By [F2], $T_c$ is incompressible. The hypothesis on the two cut components and [F3] show that the singleton family $\{T_c\}$ is a JSJ system. The empty family is not a JSJ system because its sole cut component is $M$, which by hypothesis is neither atoroidal nor Seifert-fibered. Thus $\{T_c\}$ is inclusion-minimal.

Apply [F4] to the minimal system $\{T_c\}$. It follows that the unique minimal system consists of one torus ambient isotopic to $T_c$. The decomposition of the satellite exterior in [F1] shows that the side inside the companion solid torus is the pattern exterior and the outside side is exactly $E_K$. The ambient isotopy supplied by [F4] is jointly smooth and starts at the identity. Apply its final diffeomorphism to the cut decomposition: it carries the original outside component $E_K$ to the corresponding outside component for the isotopic torus. The boundary of $M$ remains on the pattern side, so the component labels are unambiguous.