Connected sums and prime decomposition

At every $x\in \mathrm{\partial }F$, the inclusion $T_x\mathrm{\partial }F\subset T_{x}F$ and the boundary transversality hypothesis imply $d{i}_x(T_{x}F)+T_{i(x)}S=T_{i(x)}M$. Thus $i$ is transverse to $S$ along $\mathrm{\partial }F$. Transversality is open in the point of the source, and $\mathrm{\partial }F$ is compact, so there is a collar neighbourhood $U$ of $\mathrm{\partial }F$ on which $i$ is transverse to $S$.

Choose a smooth collar coordinate $s$ on $U$ and a smooth function $\gamma :F\to [0,1]$ that vanishes on a smaller boundary collar and equals $1$ outside $U$. Such a function is obtained by composing the collar coordinate with a smooth one-variable cutoff and extending by $1$. Put $\tau ={\gamma }^2$. Then $d\tau =0$ wherever $\tau =0$, and the zero set of $\tau$ lies inside $U$.

Apply [F1] to the boundaryless interior $F^{\circ }$ and the map $i{\mathrm{\mid }}_{F^{\circ }}$, obtaining a smooth family $\mathcal{F}:F^{\circ }\times B\to M$ with $\mathcal{F}(p,0)=i(p)$ and each parameter map a submersion. On $F^{\circ }$ define $G(p,a)=\mathcal{F}(p,\tau (p)a)$, restricting $B$ if necessary so $\tau (p)a\in B$. Where $\tau >0$, the parameter derivative of $G$ is surjective. Where $\tau =0$, $d\tau =0$, so $G(p,a)=i(p)$ and its source derivative is $d{i}_p$; [F5] and step 1.1 make it transverse to $S$. Since $\tau =0$ on a boundary collar, setting $G=i$ on $\mathrm{\partial }F$ extends $G$ smoothly to $F\times B$, fixed near the boundary. Thus the total family is transverse to $S$.

Apply [F2] to $G{\mathrm{\mid }}_{F^{\circ }\times B}$. Its good parameters are dense, because the bad set is null and every open parameter ball has positive measure. For each good parameter $a$, $G_a$ is transverse on $F^{\circ }$; near $\mathrm{\partial }F$ it equals $i$, which is transverse by step 1.1. Hence $G_a$ is transverse on all of $F$.

Choose compatible Riemannian metrics on $F$ and $M$ using [F6]. At each $x\in F$, choose coordinate charts and a linear projection $P_x$ to a two-dimensional tangent plane so that $D(P_x\circ i)$ at $x$ is invertible. By [F5], shrink to a convex coordinate neighbourhood $U_x$ on which this derivative stays within one quarter of the least singular value of its value at $x$. Every sufficiently $C^1$-close map $j$ then has $P_x\circ j$ injective on $U_x$: integrate its derivative along the coordinate segment between two points to obtain a lower bound of one half that least singular value times their coordinate distance. Compactness gives a finite cover by such $U_x$ and a Lebesgue number $\delta >0$. For pairs at source distance at least $\delta$, compactness and injectivity of $i$ give a positive minimum target distance, which persists under sufficiently small $C^0$ perturbations. The differential of $i$ has a positive minimum norm on the compact unit tangent bundle, so immersion also persists under sufficiently small $C^1$ perturbations. Thus all sufficiently $C^1$-close maps are injective immersions; compactness of $F$ and the Hausdorff property of $M$ make them embeddings. Smoothness of $G$ and compactness of $F$ imply $G_b\to i$ in $C^1$ as $b\to 0$, so there is a parameter ball about $0$ for which every $G_b$ is an embedding. By the density in step 4.1 choose a good $a$ in that ball; then $t\mapsto G_{ta}$ is an isotopy through embeddings fixed near $\mathrm{\partial }F$. Apply [F4] to extend it to an ambient isotopy. Its endpoint carries $i(F)$ to a surface transverse to $S$, as required.

Shrink the parametrized ball without collapsing it: $e_t(x)=e(r(t)x)$, where $r$ decreases smoothly from $1$ to a fixed positive $\epsilon$ and is constant near both ends. Every slice is an embedding. For sufficiently small $\epsilon$, its last image lies in one coordinate chart centered at $e(0)$. Write the chart representative of $e$ near $0$ as $f$, with $f(0)=0$ and $L=Df(0)$ invertible. Let $m>0$ satisfy $\mathrm{\mid }Lv\mathrm{\mid }\ge m\mathrm{\mid }v\mathrm{\mid }$. Choose $\epsilon$ so that $\mathrm{\parallel }Df(y)-L\mathrm{\parallel }<m\mathrm{/}2$ for $\mathrm{\mid }y\mathrm{\mid }\le \epsilon$. The interpolation $f_s(x)=\epsilon Lx+(1-s)(f(\epsilon x)-\epsilon Lx)$ has invertible differential and is injective: its difference from $\epsilon L$ is Lipschitz with constant less than $\epsilon m\mathrm{/}2$, so $\mathrm{\mid }{f}_s(x)-f_s(y)\mathrm{\mid }\ge \epsilon m\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}2$. All images lie in the chart after a further shrink. This is an isotopy from the small ball to a linear ellipsoid.

Write $L=QP$ with $Q$ orthogonal and $P$ positive definite. The positive definite path $(1-s)P+s\lambda I$, for small $\lambda >0$, gives an isotopy from that ellipsoid to a coordinate round ball. Reparametrize and concatenate the finite paths smoothly with stationary endpoints. By [F1] the resulting embedding isotopy extends to an ambient isotopy carrying $e(D^n)$ to that round ball. A coordinate ball can be moved along any prescribed finite chain of overlapping charts, shrinking it inside each overlap before the next move; a path between its center and any prescribed target center has such a finite chart subdivision. Within a chart, translation and positive dilation of a sufficiently small round ball are embedding isotopies. Hence any two sufficiently small coordinate balls in the same connected component are ambient isotopic.

Parametrize the given $3$-ball $B$ by $D^3$. Its boundary is thereby the standard $S^2$. Apply steps 1.1–2.1 in dimension two to a parametrization of its disk window $W\subset S^2$, then move and enlarge the resulting small round disk to a hemisphere by latitude disks. The endpoint parametrization in step 2.1 is linear orthogonal followed by a positive scalar; its orthogonal factor is the restriction of a fixed orthogonal symmetry of $S^2$. Consequently the standardization can retain the specified disk parametrization, up to that fixed symmetry, rather than only its image. Thus there is an isotopy $h_t$ of $S^2$, starting at the identity, that carries $W$ to a hemisphere, and, if needed, a fixed orthogonal symmetry makes its endpoint agree with the specified standard disk coordinates. This isotopy extends over $D^3$ explicitly: in a radial annulus use $(r,x)\mapsto (r,h_{\alpha (r)t}(x))$, where $\alpha =0$ near the inner edge and $\alpha =1$ near $r=1$, and use the identity on the smaller ball. Each slice is a diffeomorphism, with inverse $(r,y)\mapsto (r,h_{\alpha (r)t}^{-1}(y))$. It is smooth across the inner edge because it is the identity there; the fixed orthogonal symmetry also extends linearly over $D^3$. This proves standardization of the ball-window pair with its disk coordinates, without extending an arbitrary sphere diffeomorphism.

Standardize the added ball's window using its actual attaching parametrization, as permitted by step 3.1. Its collar map is then the product $(x,t)\mapsto (x,t)$ in the receiving disk coordinates. The standard ball-window pair can be represented as a cap over a flat disk: its remaining boundary is a dome meeting the window at its edge. This representation follows by viewing a round ball as two hemispherical graphs and flattening one graph in its collar; the strictly positive normal-coordinate derivative makes the flattening a diffeomorphism. Making the cap thinner is a positive rescaling of its normal coordinate. Therefore in the collar of a disk $D$ in the receiving boundary the attached ball may be represented as a thin cap, with its attaching map retained. A coordinate neighborhood of the whole closed disk exists: extend its full-rank parametrization across the boundary circle using finitely many boundary charts; the extensions agree with it on the disk, and sufficiently small extensions remain injective by compactness and the inverse function theorem.

In these coordinates the receiving manifold is $t\ge 0$. After rounding the edge, its union with the thin cap is $t\ge \gamma (x)$, where $\gamma \le 0$ is smooth and compactly supported in the disk coordinate neighborhood. Take a smooth cutoff $\rho (t)$ equal to one near the cap and zero beyond the chosen receiving collar. Choose the cap thin enough that $\mathrm{\parallel }\gamma {\rho }^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}<1$. Then $(x,t)\mapsto (x,t-\gamma (x)\rho (t))$ sends $t\ge \gamma (x)$ diffeomorphically to $t\ge 0$: its normal derivative is $1-\gamma (x){\rho }^{\mathrm{\prime }}(t)>0$, its boundary value is zero, and it is the identity at the far end of each collar fiber. It is the identity outside this neighborhood, so it pastes to the identity elsewhere. This proves disk absorption. Applying the inverse construction proves disk removal, with the same support assertion.

The first assertion is established by steps 1.1–2.1, the window assertion by step 3.1, and both attachment assertions by step 5.1. In particular two balls glued along one boundary disk form a ball; and a boundary-attached ball indented into a ball can be removed leaving a ball. Every use of isotopy extension is in dimensions two or three with compact source and stationary endpoints, as required by [F1].

Choose $R>1$ so large that $K\cup f(K)$ lies in the open radius-$R$ ball; this is possible because the continuous image of a compact set is compact, and compact Euclidean sets are bounded. Set $L=R+1$. Define an increasing homeomorphism $\rho :[0,\mathrm{\infty })\to [0,L)$ by $\rho (r)=r$ for $r\le R$ and $\rho (r)=R+(r-R)\mathrm{/}(1+r-R)$ for $r\ge R$. The radial map $y(0)=0$, $y(x)=\rho (\parallel x\parallel )x\mathrm{/}\parallel x\parallel$ for $x\mathrm{\ne }0$, is a homeomorphism from ${\mathbb{R}}^n$ onto the open radius-$L$ ball, and it is the identity on the closed radius-$R$ ball. On the open radius-$L$ ball put $g=y\circ f\circ y^{-1}$. This is a homeomorphism there.

Extend $g$ by the identity on the radius-$L$ sphere. To verify continuity there, put $r=\parallel x\parallel$ and $s=\parallel f(x)\parallel$. The displacement bound gives $s\ge r-C$. For $r>C$, insert $f(x)\mathrm{/}r$ between the two unit vectors to obtain $\parallel \frac{f(x)}{s}-\frac{x}{r}\parallel \le \frac{\mathrm{\mid }s-r\mathrm{\mid }}{r}+\frac{C}{r}\le \frac{2C}{r}\mathrm{.}$ Consequently $\parallel y(f(x))-y(x)\parallel \le \mathrm{\mid }\rho (s)-\rho (r)\mathrm{\mid }+\rho (r)\frac{2C}{r}\le (L-\rho (s))+(L-\rho (r))+\frac{2LC}{r}\mathrm{.}$ The right side tends to zero as $r\to \mathrm{\infty }$, uniformly over directions, since $s\ge r-C$ and $\rho$ is increasing with limit $L$. For an interior point $z=y(x)$ approaching a boundary point $b$, the equality $\parallel z\parallel =\rho (r)$ forces $r\to \mathrm{\infty }$, and $\parallel g(z)-b\parallel \le \parallel g(z)-z\parallel +\parallel z-b\parallel$ tends to zero by this bound. This proves continuity directly. The inverse $f^{-1}$ has the same displacement bound: if $z=f(x)$ then $\parallel f^{-1}(z)-z\parallel =\parallel x-f(x)\parallel \le C$. Applying the same calculation to $f^{-1}$ gives the continuous inverse extension. Extend once more by the identity outside the closed radius-$L$ ball to obtain a homeomorphism $G$ of ${\mathbb{R}}^n$. The formulas and their inverse formulas agree on the joining sphere.

The set $U=\{x:\parallel x\parallel <R,\parallel f(x)\parallel <R\}$ is an open neighborhood of $K$. On $U$ both $y^{-1}(x)=x$ and $y(f(x))=f(x)$, so $G(x)=f(x)$. Thus any piecewise linearity of $f$ on an open neighborhood of $K$ is retained after intersecting that neighborhood with $U$. This is an equality of actual local maps, rather than an approximation statement about their images.

For $0<t\le 1$ define $G_t(x)=tG(x\mathrm{/}t)$ when $\parallel x\parallel \le tL$, and $G_t(x)=x$ when $\parallel x\parallel \ge tL$; put $G_0(x)=x$. The branches agree at $\parallel x\parallel =tL$ because $G$ fixes the radius-$L$ sphere. Each $G_t$ is a homeomorphism, with inverse given by the same formula using $G^{-1}$; it fixes every point outside the radius-$tL$ ball. The formulas give joint continuity for $t>0$. At $t=0$, both $x$ and its image lie in the radius-$tL$ ball whenever a point moves, so $\parallel G_t(x)-x\parallel \le 2tL$. This proves joint continuity at zero, uniformly in $x$. The identical estimate applies to the inverse family, hence $(x,t)\mapsto (G_t(x),t)$ is a homeomorphism with continuous inverse $(x,t)\mapsto (G_t^{-1}(x),t)$. Therefore $G_t$ is an ambient isotopy, supported in the radius-$L$ ball, from the identity to $G$. Together with step 2.1 this proves all conclusions.

If $P$ is empty, take $V$ empty, with the empty triangulation and homotopy; all assertions hold. Otherwise subdivide $K$ sufficiently finely that the closed union of its simplices meeting $P$ lies in $W$. To see this, when $M\setminus W$ is nonempty its compactness and that of $P$ give a positive distance between them in the finite realization metric; take the mesh smaller than that distance. If $W=M$, there is no restriction. After one further barycentric subdivision, the subdivided $P$ is a full subcomplex: vertices of a simplex are a chain of faces of the preceding triangulation, so if all those vertices belong to the subdivided $P$, every face in the chain belongs to $P$, and the whole chain simplex belongs to it. Denote this final triangulation again by $K$. For each vertex $v$ put $a_v=1$ when $v\in P$ and $a_v=0$ otherwise, and extend linearly on simplices to a function $r:M\to [0,1]$. Fullness gives $r^{-1}(1)=P$. Set $V=\{r\ge 1\mathrm{/}2\}$. This is compact, contains the relatively open neighborhood $\{r>1\mathrm{/}2\}$ of $P$, and lies in $W$: every simplex meeting $V$ has a vertex in $P$ and hence belongs to the closed union chosen above.

On a simplex with barycentric coordinates $({\lambda }_v)$ and with $r\ge 1\mathrm{/}2$, define a homotopy by replacing each coordinate at a vertex of $P$ by ${\lambda }_v(r+t(1-r))\mathrm{/}r$ and each other coordinate by $(1-t){\lambda }_v$, for $0\le t\le 1$. The new coordinates are nonnegative and sum to one; their sum at the vertices of $P$ is $r+t(1-r)\ge 1\mathrm{/}2$. The formulas agree on common faces and have denominator bounded below by $1\mathrm{/}2$, so they give a continuous homotopy of $V$ into itself. At time zero it is the identity, at time one it maps into the face spanned by the $P$ vertices, which lies in $P$ by fullness, and it fixes every point of $P$ throughout. This is the required strong deformation retraction. It also gives a path in $V$ from each point to $P$, so $V$ is connected when $P$ is.

The cuts of each simplex by $r=1\mathrm{/}2$ are convex polyhedral cells and agree on common faces. Their finite collection has a compatible triangulation by [F2], including their faces in $P$ and in $\mathrm{\partial }M$. Thus $V$, its level frontier $L=\{r=1\mathrm{/}2\}$, and its boundary trace $A=V\cap \mathrm{\partial }M$ are finite subpolyhedra. It remains to verify the manifold and surface assertions, rather than inferring them merely from this triangulation.

Let $x\in L$, and let $\sigma$ be the unique simplex whose relative interior contains $x$. Since its barycentric coordinates are all positive and their weighted sum is $1\mathrm{/}2$, $\sigma$ has both a vertex of $P$ and one outside $P$. Choose a vector $d$ in its affine span with coordinate sum zero and with $dr(d)=1$, for example the difference between such two vertices. In every coface of $\sigma$, translation a sufficiently small distance along $d$ changes only the coordinates of vertices of $\sigma$; these remain positive near $x$, all other coordinates remain unchanged, and $r$ changes by exactly that distance. The translations agree on shared faces. Consequently the maps $z\mapsto (z-(r(z)-1\mathrm{/}2)d,r(z)-1\mathrm{/}2)$ and $(w,s)\mapsto w+sd$ give inverse local piecewise linear product coordinates with $w\in L$. The simplex-link hypothesis supplies the usual local manifold chart for the remaining coordinates. If $x$ is in the interior of $M$, these coordinates identify $V$ locally with a half-space and $L$ with its boundary plane. If $x\in \mathrm{\partial }M$, the vector $d$ is tangent to the boundary simplex and these coordinates also preserve $\mathrm{\partial }M$; $V$ is locally a quadrant times a line, with its two faces equal to $A$ and $L$.

A quadrant times a line is a piecewise linear half-space: in its two-dimensional factor send the two boundary rays to the positive and negative horizontal rays and send the diagonal ray to the positive vertical ray, extending linearly on the two intervening cones. This is a piecewise linear homeomorphism with piecewise linear inverse, and take its product with the remaining line coordinate. Away from $L$, a point of $V$ has $r>1\mathrm{/}2$ and hence inherits the ordinary manifold or boundary chart of $M$. Together with step 3.1 these charts prove that $V$ is a compact piecewise linear three-manifold, $A$ is a subsurface of $\mathrm{\partial }M$, and $L$ is a proper piecewise linear surface. At their common points the quadrant model identifies their boundaries with the same line, so their common boundary is exactly $A\cap L$. The finite triangulation of step 2.2 and the deformation retraction of step 2.1 finish the proof. All subdivision, triangulation and coordinate selections are finite.

Let $C_{\ast }(K)$ be the simplicial chain complex and $C^{\ast }(K)$ its vector-space dual, with coboundary $\delta$. Every simplicial one-cocycle $a$ is a coboundary. To prove this, construct a twofold simplicial cover from $a$: its vertices are $(v,ϵ)$ for $ϵ\in {\mathbb{F}}_2$, its lifted edge over $vw$ goes from $(v,ϵ)$ to $(w,ϵ+a(vw))$, and its lifted simplex over a simplex with vertices $v_0,\dots ,v_k$ has labels $ϵ+a(v_0{v}_j)$ at $v_j$, with $a(v_0{v}_0)=0$. The cocycle identity on every triangle makes these prescriptions agree on faces and makes the two lifts of each vertex star disjoint copies of that star. Their realization is a genuine twofold covering space of $V$. If it were connected, it would contradict the hypothesis. Otherwise it has exactly two connected components, each projecting homeomorphically to $V$: path lifting connects a chosen fiber point to one point in every other fiber; if the same component contained both points of one fiber, it would contain both points of every fiber and the whole cover would be connected. Choose one component. The label $b(v)$ of its unique vertex over $v$ satisfies $a(vw)=b(v)+b(w)$, so $a=\delta b$. Thus $H^1(C^{\ast }(K))=0$.

Write $d_j:C_j(K)\to C_{j-1}(K)$ for the finite boundary matrices. Gaussian elimination, always taking the first available pivot in the fixed ordering, proves that a matrix and its transpose have the same rank. Consequently $\dim H_1(C_{\ast }(K))=\dim C_1-\mathrm{rank}{d}_1-\mathrm{rank}{d}_2=\dim H^1(C^{\ast }(K))=0.$ Thus $H_1(V;{\mathbb{F}}_2)=0$ by [F1]. No infinite basis or extension of a functional is selected.

For the finite triangulation fixed in step 1.1, there is also a relative dual-cell decomposition of $(V,\mathrm{\partial }V)$. For a $k$-simplex $\sigma$ of $K$, let $B_{\sigma }$ be the closed union of barycentric simplices whose defining face chain begins with $\sigma$, together with all their faces. Its coface poset identifies it piecewise linearly with the cone on the barycentric subdivision of $\mathrm{lk}(\sigma ,K)$. Hence it is a $(3-k)$-ball by the simplex-link hypothesis. For an interior simplex its boundary is the union of the blocks $B_{\tau }$ for proper cofaces $\tau ⊋\sigma$. For a boundary simplex its boundary additionally contains its side in $\mathrm{\partial }V$: in the cone-on-ball model that side is the cone on the boundary of the link ball. The boundary side and the coface blocks are the two parts of its ball boundary. The interiors of the blocks off $\mathrm{\partial }V$ are pairwise disjoint and cover $V\setminus \mathrm{\partial }V$. Indeed, the least face with positive barycentric-chain coordinate determines the unique block interior, while a point whose chain lies wholly in $\mathrm{\partial }V$ belongs to the boundary subcomplex. Starting with a finite triangulation of $\mathrm{\partial }V$ and attaching these blocks in increasing dimension therefore gives a finite relative CW decomposition of $(V,\mathrm{\partial }V)$.

Its relative cellular boundary is $\mathrm{\partial }{B}_{\sigma }={\sum }_{\begin{array}{c}\tau \supset \sigma \\ \dim \tau =k+1\end{array}}{B}_{\tau }(\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{\partial }V)\mathrm{.}$ Each listed block is an embedded codimension-one facet of $B_{\sigma }$ and occurs once. All other boundary faces have smaller dimension or lie in $\mathrm{\partial }V$. Equivalently, in its barycentric triangulation the faces obtained by omitting an intermediate chain member cancel in pairs, and the uncanceled faces are precisely those with the least member replaced by an immediate coface; boundary-side faces are zero in the relative complex. Thus the basis correspondence from the cochain dual to $\sigma$ to $B_{\sigma }$ identifies $C^k(K)$, with coboundary, with the degree-reversed relative dual-cell complex in degree $3-k$. By [F1], this gives $H_2(V,\mathrm{\partial }V;{\mathbb{F}}_2)\cong H^1(C^{\ast }(K))=0.$ This is a finite chain-matrix duality calculation, not an invocation of a general Poincaré duality theorem.

The portion $H_2(V,\mathrm{\partial }V;{\mathbb{F}}_2)\to H_1(\mathrm{\partial }V;{\mathbb{F}}_2)\to H_1(V;{\mathbb{F}}_2)$ of the exact sequence in [F1] now has zero outer groups. Therefore $H_1(\mathrm{\partial }V;{\mathbb{F}}_2)=0$. In particular this holds for each boundary component, since the finite boundary chain complex is the direct sum of its component complexes. Each component is a closed orientable finite PL surface.

For completeness, apply the finite surface procedure [F2] to such a component $S$. Coherent triangle orientations give opposite directions on the two occurrences of every paired polygon side. Cancel adjacent inverse pairs, retaining the terminal sphere bigon. If an uncanceled residue is nonempty and is not that bigon, it has interlaced pairs: otherwise all pairing chords on its boundary circle would be noncrossing, and an innermost chord would give an adjacent inverse pair. Extract an interlaced pair as a commutator block using [F2]'s explicit cut-and-merge identity, reserve that block, and repeat on the residue. Each cancellation deletes a pair, and each extraction removes two pairs from the unprocessed residue, so this terminates with either the sphere bigon or a product of $g\ge 1$ commutator blocks. All cuts and disk identifications are polygonal, and their parametrizations are PL by the disk-extension clause of [F2], so the resulting model is PL homeomorphic to $S$.

A nonempty commutator model has one vertex, $2g$ one-cells, and one two-cell. Its first boundary is zero and its second boundary is zero modulo two, since every side occurs twice. Hence its first homology has dimension $2g>0$ by [F1], contrary to step 4.1. The only remaining model is the sphere bigon, whose quotient is the standard PL two-sphere; one may subdivide its two boundary arcs and realize it as two polygonal disks glued along their common boundary. Thus every component is a PL sphere. All covers, blocks, reductions and pivots used finite data; no arbitrary-index or countable choice was required.

Each double circle has two distinct preimage circles in $D$, and its normal crossing neighborhood is a product of that circle with a planar cross. To check the possible monodromy, orient $D$ and $M$. Each sheet has its induced coorientation, and the ordered pair of sheet coorientations orients their intersection line. If continuation once around the double circle interchanged the sheets, this orientation of the intersection line would be reversed, which is impossible on a circle. Thus the two sheets return individually. The orientation of each sheet and of the intersection line also preserves each of its transverse directions; consequently no pair of opposite arms is interchanged. Trivializing the interval strips in the two sheets therefore gives the product cross. No rotation or reflection monodromy remains.

The two preimage circles of a chosen double circle bound disks $D_1,D_2\subset D$ by [F1]. If they are nested, say $D_1\subset \mathrm{Int}{D}_2$, replace the part $D_2$ by a copy of $D_1$, matching boundary points with the same image under $f$, and retain the outside of $D_2$. If they are disjoint, interchange the maps on $D_1$ and $D_2$, with those same boundary identifications, and retain the rest. The identifications extend over the disks by [F1]. At the joining circle resolve the crossing by connecting adjacent arms of the planar cross instead of opposite arms. The resolution can be made polygonal: in a small square normal section replace the two crossing segments by two disjoint bent polygonal arcs joining the selected adjacent pairs, equal to the old arms near the square boundary. Taking products with successive subintervals of the double circle gives matching polygonal annuli; step 1.1 ensures their end identifications agree. This removes that double circle. The retained domain is a disk, its external boundary map is unchanged, and each retained old sheet piece is used at most once. Outside the crossing neighborhood no new double value is introduced; inside it the replacement sheets are disjoint. Other double curves miss this neighborhood and are either retained or discarded.

Repeat step 2.1 until there are no double circles. The number of double curves strictly decreases at every repetition and was finite initially. Each new disk is still a proper polygonal immersion with at most two preimages per value, regular double arcs, and the original boundary loop. The intermediate disk need not lift to an embedding in any covering space; this proof concerns the immersion in $M$ itself.

Choose a double arc $C$. Its two preimage arcs $a,b$ are disjoint proper arcs in the disk. They cut it into three polygonal disks, in order an outer piece, a middle piece, and the other outer piece. Orient the boundary of $D$, and label its four successive portions by $\alpha ,\beta ,\gamma ,\delta$ so that $\alpha$ and $\gamma$ are the exposed boundary intervals of the outer pieces. The two ways of matching the orientations of $a$ and $b$ give the following boundary words after the two possible crossing resolutions: $\begin{array}{cccc}& \text{original}& \text{first disk}& \text{second disk}\\ \text{same orientation}& \alpha \beta \gamma \delta & \alpha \gamma & \alpha {\beta }^{-1}\gamma {\delta }^{-1}\\ \text{opposite orientations}& \alpha \beta \gamma \delta & \alpha {\gamma }^{-1}& \alpha \delta \gamma \beta \end{array}\mathrm{.}$ Here all boundary paths are their images in $F$; short joining paths inside the endpoint half-disk neighborhoods have been collapsed in the written words. They are homotopic to the actual polygonal boundary paths. To verify the domains and the words, cut along $a,b$, and use the common parametrization of their images $C$ to match the cut shores. Joining the two outer disks along one shore produces the first disk, with the exposed intervals $\alpha ,\gamma$ read in the matching directions. Joining all three disks in the other order along two shores produces the second disk; following its boundary reads the remaining word in the table. Each join is along a full boundary interval, hence gives a disk by [F1]. Each old piece is used at most once in either candidate. This surgery will be applied after the circle eliminations of step 3.1.

Realize each of these two gluings in $M$ by the corresponding adjacent-arm resolution along $C$. In a transverse square use the two disjoint bent arcs described in step 2.1. At an endpoint use its half-space version; the changed boundary arcs stay in $F$, since that endpoint has a neighborhood in the subsurface containing the original boundary germs. Subdivide finitely along $C$ to obtain polygonal product strips, and use a width smaller than their positive distances to all other double curves and to nonincident sheet pieces. Thus the replacement is a proper polygonal immersion with no double value along $C$ and with no newly introduced double curve. Away from the strip it uses a subset of the old sheet pieces, with no repetition. The two-to-one bound is therefore preserved. The remaining double curves are among the old ones other than $C$.

At least one candidate disk still has its boundary class outside $N$. In the first row of the table put $w=\alpha \beta \gamma \delta$, $u=\alpha \gamma$, $v=\alpha {\beta }^{-1}\gamma {\delta }^{-1}$. Cancellation of successive reversed paths gives $w=u{\delta }^{-1}{v}^{-1}u\delta \mathrm{.}$ In the second row put $u=\alpha {\gamma }^{-1}$ and $v=\alpha \delta \gamma \beta$. Then the corresponding cancellation identity is $w=u(\gamma \delta {)}^{-1}{u}^{-1}v(\gamma \delta )\mathrm{.}$ These are identities of composable paths with the indicated common endpoint identifications; inserting basepaths turns them into products of conjugates in ${\pi }_1(F)$. By [F2], if both candidate boundary classes belonged to $N$, so would $w$, contradicting the induction hypothesis. Retain a candidate whose boundary class is outside $N$.

Repeat steps 4.1–6.1. Each repetition removes at least one double arc and introduces none, so after finitely many repetitions there is a proper polygonal immersion of a disk with no double values and with boundary class outside $N$. It is injective. A continuous injection of the compact disk into the Hausdorff manifold $M$ is an embedding; its polygonal local models make this a piecewise linear proper embedding. Its boundary is in $F$, as required. All choices made in the construction are finite; no countable or arbitrary-index choice assumption has been used.

If a cell $C$ meets some level in $[a,b]$, it has vertices strictly below $a$ and strictly above $b$. Indeed the minimum and maximum of the affine function on the convex polytope occur at vertices; a cell meeting the slab cannot have either extremum inside it, by hypothesis. The same holds for every face that meets the slab. Its section $C_t=C\cap {\mathrm{\ell }}^{-1}(t)$ is consequently nonempty for every $t\in [a,b]$ and meets the relative interior of $C$: take an interior point of $C$, and join it to points near the minimum and maximum to obtain interior points below and above $t$, then intersect their segment with the level. In particular no new face begins or ends while the section level traverses this interval.

More precisely, the faces of $C_t$ correspond to the proper faces $G$ of $C$ meeting the slab, together with $C$ itself: intersect each such face with the level and retain the empty face separately. Every face of the section is of this form because an exposing affine functional on its hyperplane extends to the ambient affine span and its maximizing face on the section is cut out by a collection of the supporting facet equations of $C$. The correspondence is injective: $G_t$ meets the relative interior of $G$ by step 1.1, so its smallest containing face of $C$ is $G$. It preserves inclusions and is independent of $t$. The faces of the truncated cell $\widehat{C}=C\cap {\mathrm{\ell }}^{-1}([a,b])$ are its bottom and top sections and the truncated faces $\widehat{G}$ for the same $G$. This follows by adding the two slab inequalities to the facet inequalities of $C$; any proper face not contained in an end hyperplane is obtained from original facet equalities. A face contained in an end hyperplane is a face of that end section. Thus $\widehat{C}$ has exactly the face incidences of the product prism $C_a\times [a,b]$, with end faces distinguished. These incidences agree on intersections of cells of $\mathcal{K}$.

Construct the maps by induction over these common face incidences. On each bottom face use the identity. On the top faces choose compatible PL maps from their corresponding bottom faces as follows: match vertices using the face correspondence, and extend over each higher-dimensional face by coning its already defined boundary map from an interior point to an interior point. Both faces are convex, so the cones cover them with disjoint relative interiors, proving bijectivity and a PL inverse. For every remaining, vertical face pair choose interior points at height $(a+b)\mathrm{/}2$ on both sides; such points exist by step 1.1. The height on the source prism is its second coordinate. Once the boundary map is constructed, triangulate the boundary compatibly and extend over that face by coning between these two interior points. The same convex-cone argument proves that this is a PL homeomorphism on the face. It preserves height: its boundary map already does, its two cone points have the same height, and height is affine on each cone simplex. This induction includes the full truncated cells. Shared faces have only one previously prescribed map, so all maps agree on intersections. Only finitely many cells occur.

Paste the maps to obtain $H$. A point in the relative interior of a source prism cell maps to the relative interior of its corresponding target cell, so the pasted map is a bijection. The cellwise inverses agree on intersections and paste to its inverse. Both are PL on finite compatible subdivisions furnished by [F1], hence continuous. The bottom restriction and height identity follow from step 3.1. A union of original cells is a union of the corresponding prism cells, giving the compatibility assertion. If $P_a$ is empty there is no cell meeting the slab by step 1.1, so both sides are empty. All selections involved finite sets of cells and interior points, and require no countable or arbitrary-index choice.

Its intersection with $z=0$ consists of two distinct rays. Choose a homogeneous PL homeomorphism $g:{\mathbb{R}}^2\to {\mathbb{R}}^2$ taking them to the positive and negative $x$-axis. Explicitly, insert finitely many further rays to divide both plane sectors between the two marked rays into sectors of angle less than $\pi$, do the same for the two target half-planes with corresponding numbers of sectors, and match their cyclically ordered ray generators. Extend linearly on each sector. Each sector map is a linear bijection onto its corresponding convex sector; these maps and their linear inverses agree on rays and paste to inverse PL homeomorphisms. Extend to three-space by $G(u,z)=(g(u),z)$. Replace $S$ temporarily by $G(S)$. Both its horizontal zero rays are now the $x$-axis rays, and $G$ preserves height.

The sections $L_{+}=S\cap \{z=1\}$ and $L_{-}=\{u:(u,-1)\in S\}$ are each simple proper polygonal lines with finitely many bends and exactly two unbounded rays, one tending to each end of the $x$-direction. To see this, radial projection from each open link arc to its level $z=1$ or $z=-1$ is a homeomorphism. A cone triangle meets that level in a segment, ray or empty set; the finitely many triangles therefore give a polygonal line. Near either endpoint of the link arc its last linear cone sector meets the level in a ray parallel to the corresponding horizontal zero ray. These are its only ends; the projection is proper because any compact part of the level has directions bounded away from $z=0$. The same statements remain valid after step 1.1: a ray eventually lies in one sector of the finite homogeneous fan, and its image is a ray parallel to the image of its limiting direction. Thus both lines have horizontal terminal rays directed to opposite ends.

We give an explicit finite PL plane homeomorphism $f$ straightening any such line $L$, with bounded displacement. Take a rectangle $R=[-r,r]\times [-h,h]$ whose interior contains all bends of $L$ and whose two vertical sides meet its horizontal terminal rays once each; enlarge $r,h$ so all finite segments lie in its interior and these are the only boundary intersections. The arc $L\cap R$ divides $R$ into two polygonal disks. Indeed each union of that arc with one of the two boundary arcs between its endpoints is a simple polygon. By the planar separation and disk assertion of [F1], their bounded disks occupy precisely the two sides inside $R$: near the top and bottom edges they occupy the respective sides, their interiors are disjoint by crossing parity, and every point of $R\setminus L$ connects within its polygonal component to one of those sides. Choose on $\mathrm{\partial }R$ an order-preserving PL map fixing both horizontal sides and all corners and taking the two line endpoints to $(-r,0)$ and $(r,0)$. Choose any order-preserving PL parameter of the finite line arc onto the horizontal diameter. On each of the two disks these prescriptions form a boundary homeomorphism onto the corresponding upper or lower half-rectangle. Extend them by [F1]. Their maps agree on the line arc, so they give a PL homeomorphism $f_R:R\to R$ taking that arc to the diameter.

Write the two vertical boundary maps of $f_R$ as $(-r,y)\mapsto (-r,{\gamma }_{-}(y))$ and $(r,y)\mapsto (r,{\gamma }_{+}(y))$. Both ${\gamma }_{\pm }$ are increasing PL homeomorphisms of $[-h,h]$, fix its endpoints, and send the respective terminal-ray height to zero. Outside $R$ extend $f_R$ on the two half-strips by $f(x,y)=(x,{\gamma }_{+}(y))$ for $x\ge r$ and $\mathrm{\mid }y\mathrm{\mid }\le h$, and by the corresponding formula with ${\gamma }_{-}$ for $x\le -r$. Use the identity for $\mathrm{\mid }y\mathrm{\mid }\ge h$. These formulas agree along every seam. Their inverses have the same form with ${\gamma }_{\pm }^{-1}$, so the extension is a PL plane homeomorphism. Its cells consist of finitely many bounded triangles, half-strips and exterior half-planes, all with compatible finite polyhedral subdivisions; their intersections with a bounded box admit finite simplicial subdivisions by [F2]. It sends the entire $L$ to the $x$-axis. Moreover $f(u)-u$ is bounded: on $R$ this follows from compactness, on the half-strips only the second coordinate changes by at most $2h$, and elsewhere it is zero. Its inverse also has bounded displacement, since $f^{-1}(v)-v=u-f(u)$ for $v=f(u)$. Apply this construction independently to $L_{+}$ and $L_{-}$, producing $f_{+}$ and $f_{-}$.

For $z>0$ set $T(u,z)=(z{f}_{+}(u\mathrm{/}z),z)$. For $z<0$ put $s=-z>0$ and set $T(u,z)=(s{f}_{-}(u\mathrm{/}s),z)$. Set $T(u,0)=(u,0)$. These formulas define a finite PL homeomorphism. To check the PL assertion, on a plane cell where $f_{\pm }(v)=Av+b$ its homogenization is $(u,z)\mapsto (Au+zb,z)$ above zero, or $(u,z)\mapsto (Au-zb,z)$ below zero. The homogenized cells are polyhedral cones: an inequality $av\le c$ becomes $au\le cz$ above zero or $au\le -cz$ below zero. There are finitely many such cones. Their intersections with each bounded cube have compatible finite simplicial subdivisions by [F2]; subdividing successive integer cube shells compatibly, always coning from the barycenter of the vertices of each clipped cell, gives the usual locally finite PL interpretation without any countable choice of centers. Bounded displacement implies that their linear formulas match the identity on the zero plane: $\mathrm{\parallel }z{f}_{+}(u\mathrm{/}z)-u\mathrm{\parallel }\le M_{+}z$ and $\mathrm{\parallel }s{f}_{-}(u\mathrm{/}s)-u\mathrm{\parallel }\le M_{-}s$. Hence the formulas extend continuously and agree there. Homogenizing the inverse maps in the same way supplies an inverse with identical continuity and PL properties. In particular $T$ preserves height and fixes the zero plane pointwise.

Every positive-height point of the cone $S$ is uniquely $(zv,z)$ with $v\in L_{+}$, and every negative-height point is uniquely $(sv,-s)$ with $v\in L_{-}$. By step 4.1 their images under $T$ are exactly the positive and negative parts of the vertical plane $y=0$. Its zero-height part is already the full $x$-axis. Therefore $H=T\circ G$, with $G$ from step 1.1 and $T$ applied to the transformed cone, has all the stated properties. For the local consequence take a sufficiently small star at the surface vertex: its finitely many incident triangles agree there with their full cones. Restrict $H$ to a smaller neighborhood of the origin; it gives the required height-preserving local straightening. All fans, polygonal disk extensions, rectangles and affine cells used finite data, so no countable or arbitrary-index choice is needed.

A change of knot representative already comes with an oriented ambient isotopy by [F1]. Applying that isotopy to its ball-arc data carries the cut complement and its boundary collars to the new data, so the standard gluing produces an ambient-isotopic sum. A change of radius in one local straightening chart is the isotopy that rescales the standard ball radially while preserving its oriented diameter. To preserve the entire knot, perform this rescaling by a radial vector field in a larger straightening chart, equal to the required radial field on all swept balls and cut off before the chart boundary. The chart meets the knot only in its diameter line, to which this field is tangent. Its flow therefore preserves the whole knot setwise, including the part outside the moving ball, and identifies the two cut complements and their boundary parametrizations.

For two local oriented ball-arc charts centered at the same point, express them in one smooth coordinate chart. Their transition germ fixes the center, carries the positive tangent line of the knot to itself, and has positive determinant. Shrink to a common ball and use the smooth rescaling homotopy $G_s(x)=s^{-1}(G(sx)-G(0))$ for $s>0$, with $G_0(x)=D{G}_0(x)$. Taylor's formula gives the smooth extension at $s=0$. On a sufficiently small common ball the differentials stay uniformly close to the injective linear map $D{G}_0$, so every $G_s$ is an embedding and preserves the oriented diameter. The subgroup of invertible linear maps preserving that oriented line and the ambient orientation is path-connected: in line-normal coordinates its diagonal blocks lie in $(0,\mathrm{\infty })$ and $G{L}^{+}(2,\mathbb{R})$, and the shear block contracts linearly to zero. Thus the chart germ is isotopic through oriented ball-arc embeddings to the fixed standard chart. Here too ambient extension must preserve the whole knot. On each moving ball the isotopy velocity is tangent to the knot along its diameter. In straightening coordinates $(s,u,v)$ with knot $u=v=0$, its normal components vanish on that axis and have the form $ua+vb$, with smooth coefficients, by integrating their normal derivatives. Extend these coefficients and the tangential component slightly past the moving ball, then cut off inside the larger straightening chart. The resulting extension still has zero normal components along the knot. Finite such extensions and a partition of unity give the velocity-field construction of [F3] tangent to the whole knot, with zero field outside the chart. Uniqueness of its flow keeps the knot setwise fixed while carrying the prescribed balls and their boundary parametrizations.