A controlled nested horn construction embeds a closed three-ball

In a parametrized boundary sphere choose a point outside all cap neighborhoods and use stereographic coordinates. For each cap take a chart $c:D_2^2\to S^2$ whose unit disk is the cap. Such charts may be rescaled so their images are disjoint. A strictly increasing piecewise-linear function of radius, taking $1$ to $\lambda >0$ and fixing $2$, extends by the identity to a sphere homeomorphism and shrinks the cap to $c(D_{\lambda }^2)$. Its inverse uses the inverse radial function. Center the planar chart at zero. Its derivative $L$ is invertible; reverse an input axis if necessary so $\det L>0$. Normalize by $L^{-1}$. For small $r$, the normalized map $g$ has $g(0)=0$, $Dg(0)=I$ and $\mathrm{Lip}(g-\mathrm{i}\mathrm{d})\le ϵ$ on the disk of radius $2r$, by [F2]. With the radial cutoff $\chi =1$ on radius $r$, linear down to zero on radius $2r$, the extension $k=\chi (g-\mathrm{i}\mathrm{d})$, zero elsewhere, has Lipschitz constant at most $3ϵ$. For a segment crossing the support boundary, use its first boundary point to obtain the same estimate. Choose $3ϵ<1\mathrm{/}2$. Then $H=\mathrm{i}\mathrm{d}+k$ is injective since $\mathrm{\mid }H(x)-H(y)\mathrm{\mid }\ge (1-3ϵ)\mathrm{\mid }x-y\mathrm{\mid }$, and onto since $x\mapsto b-k(x)$ has a fixed point in complete ${\mathbb{R}}^2$ for every $b$. Its inverse is Lipschitz. After conjugating by $L$, this gives a homeomorphism supported in the chosen cap neighborhood taking a small ellipse $L(D_{\lambda }^2)$ to the small cap.

Start with a compact standard unknotted torus $X_0$ scaled to diameter at most one. One explicit choice is the image under stereographic projection of $\{\mathrm{\mid }z\mathrm{\mid }\le a\}\subset S^3\subset {\mathbb{C}}^2$, $0<a<1$, with the omitted point chosen in its exterior, followed by a Euclidean dilation. Its disk-times-circle coordinates are those of [F1]. Choose a proper closed core arc and its meridional slice $C_{\mathrm{\varnothing }}$; its relative complement $B_0=X_0\setminus {\mathrm{int}}_{X_0}{C}_{\mathrm{\varnothing }}$ is a cylinder ball meeting it in precisely its two caps. In a prepared $C_s$, insert [F1]'s block using its entire cap-and-side boundary marking, extended through the already parametrized balls as in that lemma. Apply its small-slice preparation with $\eta =2^{-(\mathrm{\mid }s\mathrm{\mid }+1)}$. This gives disjoint child tori $T_{s0},T_{s1}\subset C_s$ and slices $C_{si}\subset T_{si}$ in disjoint closed neighborhoods in $\mathrm{int}{C}_s$, away from incoming caps. Set $P_{si}=T_{si}\setminus {\mathrm{int}}_{T_{si}}{C}_{si}$. Each is a parametrized cylinder ball. The construction fixes all previous pieces and cap maps. Apply [F7] to choose the permitted finite outputs recursively on word length, with lexicographic order within a level.

In the fixed source $D$, let $c_{\mathrm{\varnothing }}=0$, ${\rho }_{\mathrm{\varnothing }}=1$, and recursively $c_{s0}=c_s-({\rho }_s\mathrm{/}2,0),c_{s1}=c_s+({\rho }_s\mathrm{/}2,0),{\rho }_{si}={\rho }_s\mathrm{/}8.$ Let $R_s=\{(u,z):z\ge 0,\mathrm{\mid }u-c_s{\mathrm{\mid }}^2+z^2\le {\rho }_s^2\}$, $Q_s=\{\mathrm{\mid }u-c_s\mathrm{\mid }\le {\rho }_s\}$, $h_s(u)=\sqrt{{\rho }_s^2-\mathrm{\mid }u-c_s{\mathrm{\mid }}^2}$, and $S_s=\{(u,h_s(u)):u\in Q_s\}$. Use $R_{\mathrm{\varnothing }}=D$. The children are disjoint closed half-balls: their centers are distance ${\rho }_s$ apart and the sum of their radii is ${\rho }_s\mathrm{/}4$. They stay below the parent hemisphere since every child point is at distance at most $5{\rho }_s\mathrm{/}8<{\rho }_s$ from the parent center. Thus $U_n={\bigcup }_{\mathrm{\mid }s\mathrm{\mid }=n}{R}_s$ decreases, its constituent diameters are $d_n=2\cdot 8^{-n}$, and ${\bigcap }_n{U}_n$ lies in the flat boundary $z=0$. The upper half-ball $D$ is itself a parametrized closed ball: radial projection from an interior point identifies this compact convex body with a round ball. Along each ray the exit distance is the positive minimum of the plane and sphere exit distances when both exist; these formulas fit continuously where the active exit changes and are bounded away from zero.

An ellipse can itself be changed to a round disk within that neighborhood. Gram–Schmidt on the two columns writes $L=QR$, where $Q$ is a rotation and $R$ is upper triangular with positive diagonal; the latter is a positive diagonal times a shear. Interpolating the rotation angle, positive diagonal entries and shear entry gives a continuous path of invertible positive-determinant matrices from $I$ to $L$. Their norms and inverse norms are bounded on the compact parameter interval by [F3] and [F4]. A sufficiently fine finite subdivision makes every successive quotient matrix $A$ close to $I$. On a fixed small neighborhood, $x\mapsto x+\chi (x)(A-I)x$ then has perturbation Lipschitz constant less than $1\mathrm{/}2$ and is a homeomorphism by the proof of step 1.1. Choose the initial disk sufficiently small that all intermediate ellipses stay where $\chi =1$; the finite composition is exactly $L$ there. Its inverse rounds the ellipse. Together with the inverse of the last map in step 1.1 this rounds the small cap, with all supports missing other caps.

The exact output relations are $T_{s0}\cap T_{s1}=\mathrm{\varnothing }$, $T_{si}\cap \mathrm{\partial }{C}_s$ equal to its one incoming disk, and $P_s\cap C_s$ equal to its two outgoing disks. These are literal relative-interior cuts in the recorded product coordinates, not removal of ambient interiors. Also $\mathrm{diam}{C}_s<2^{-\mathrm{\mid }s\mathrm{\mid }}$ for $s\mathrm{\ne }\mathrm{\varnothing }$, and $T_s\subset C_{s^{-}}$ gives $\mathrm{diam}{T}_s\le 2^{-(\mathrm{\mid }s\mathrm{\mid }-1)}$, with the depth-one bound coming from $X_0$. Define for $n\ge 1$ $B_n=B_0\cup {\bigcup }_{1\le \mathrm{\mid }s\mathrm{\mid }\le n}{P}_s,X_n=B_{n-1}\cup {\bigcup }_{\mathrm{\mid }s\mathrm{\mid }=n}{T}_s\mathrm{.}$ The depth-$n$ tori are disjoint and meet $B_{n-1}$ only in their individual incoming disks. This follows by induction from disjoint siblings and from all subsequent slices lying away from earlier caps. Substitution of $T_s=P_s\cup C_s$ gives $X_n=B_n\cup {\bigcup }_{\mathrm{\mid }s\mathrm{\mid }=n}{C}_s$, $X_{n+1}\subset X_n$, and $B_n\subset B_{n+1}\subset X_{n+1}$. Finite unions are compact by [F3]. Every point of $X_n$ lies in $B_n$ or in one of its attached $C_s$, which has a cap in $B_n$. Therefore $\mathrm{dist}(y,B_n)\le 2^{-n}$ for all $y\in X_n$. No ball-recognition conclusion for $B_n$ has yet been assumed.

Put $A_s=R_s\setminus ({\mathrm{int}}_D{R}_{s0}\cup {\mathrm{int}}_D{R}_{s1})$, including $s=\mathrm{\varnothing }$. Extend the child height functions by zero, and put $g_s=\max (h_{s0},h_{s1})$ on $Q_s$. Then $A_s=\{(u,z):g_s(u)\le z\le h_s(u)\}$ and $j_s(u,z)=(u,g_s(u)+(1-g_s(u)\mathrm{/}{h}_s(u))z)$ for $u\in \mathrm{int}{Q}_s$, with identity at the rim, is a homeomorphism $R_s\to A_s$ fixing $S_s$. Indeed $g_s=0$ near that rim, and $g_s<h_s$ on its compact support by step 1.3. The fiber maps are strictly increasing affine bijections, whose inverses subtract $g_s$ and divide by $1-g_s\mathrm{/}{h}_s$. That denominator is bounded positively away from zero on the support by [F4], and it equals one near the rim; hence both maps are continuous everywhere. Under $j_s^{-1}$ the outgoing hemispheres are the two flat disks $Q_{s0},Q_{s1}$. More generally $A^{\le n}=D\setminus {\bigcup }_{\mathrm{\mid }s\mathrm{\mid }=n+1}{\mathrm{int}}_D{R}_s$ is the union of $A_{\mathrm{\varnothing }}$ and all $A_s$ through depth $n$, with just the parent-child hemisphere intersections. Applying the same vertical formula to the maximum of the depth-$n+1$ heights proves it is a ball.

Finite labelled configurations of disjoint round planar disks can be matched by supported homeomorphisms. For one disk, join its center to a vacant destination by a polygonal path avoiding the protected disks: replace portions of a straight segment inside each protected disk by detours in slightly larger disjoint concentric annuli, and approximate the finitely many circular arcs by polygonal arcs in those annuli. The compact resulting path has positive distance from the protected disks by [F4]. Shrink the moving disk below one tenth of this clearance. For each sufficiently small displacement $v$ along a finite subdivision, use $x\mapsto x+\chi (x)v$, with $\chi =1$ near that disk and supported in a clearance ball. The bound $\mathrm{\mid }v\mathrm{\mid }\mathrm{Lip}\chi <1\mathrm{/}2$ proves invertibility as in step 1.1 and makes the disk move by exactly $v$. Expand it radially at its destination when the intended final disk neighborhood is free. First move every disk to disjoint temporary disks far outside both configurations; then move to the final disks one by one. Shrinking before motion avoids all occupied disks. Thus no coincident occupied destination is assumed free. This proves labelled setwise matching on the sphere after step 2.1.

The matching can have either sign on each boundary circle, with the same global sign. Indeed first match the labelled disks to disks centered on one axis; reflection across that axis preserves every label and reverses each circle, and conjugate it back. Now suppose the whole map on one cap is prescribed. Compare it with a chosen setwise matching, choosing the latter's sign so the required correction $k:D^2\to D^2$ has orientation-preserving boundary map $b$. This means its circle lift has positive increment, as follows directly without a disk classification theorem. Cut domain and target circles at a point and its image. The induced homeomorphism of open intervals is strictly increasing or decreasing by [F5]. In the increasing case its endpoint limits give a lift $\beta :[0,2\pi ]\to \mathbb{R}$ with $\beta (2\pi )=\beta (0)+2\pi$; extend by this translation to all real arguments. In the decreasing case the increment is $-2\pi$, which reflection reverses. For the increasing correction, ${\beta }_s(t)=(1-s)\beta (t)+st$ is strictly increasing, continuous, and obeys ${\beta }_s(t+2\pi )={\beta }_s(t)+2\pi$ for $0\le s\le 1$. Hence it defines a circle homeomorphism. Use $k$ on the unit disk, and $(r,e^{it})\mapsto (r,e^{i{\beta }_{r-1}(t)})$ on its collar $1\le r\le 2$. At radius one the maps agree; at radius two it is the identity. The collar map is a continuous bijection of a compact annulus, so its inverse is continuous by [F3]. Extending by the identity fixes all other caps. Finally a sphere homeomorphism $b_0$ extends through a parametrized ball by $tu\mapsto t{b}_0(u)$ and zero mapping to zero, with the analogous inverse. Thus two- or three-cap balls can be matched with an exact prescribed incoming disk map.

Verify the cap hypotheses of step 4.1 in these particular parametrizations. For $P_s\cong D^2\times [0,1]$, the outgoing caps are its end disks. Extend their disk charts across the rims down the side. The incoming disk is the angular rectangular patch on that side from [F1], away from the removed slice; it has a slightly larger angular chart. A rectangle is parametrized by a disk using an increasing radial adjustment that is the identity near the center, so the chart has nonsingular smooth derivative there. Radial parametrization of the cylinder by a round ball is smooth near the centers of its end and side faces. No smoothness is required at its edges. On the source side, $j_s$ identifies the outgoing caps with disks strictly inside a flat face; the incoming hemisphere has a disk chart smooth near its pole, extended across its rim into the adjacent flat annulus. Such a chart can use polar angular coordinates from the pole and continue its radial parameter past the equator into the flat face. All these extensions miss the other caps. Transport by the already specified abstract ball parametrizations retains these chart properties even when the ambient image is no longer smooth. Consequently choose $h_{\mathrm{\varnothing }}:A_{\mathrm{\varnothing }}\to B_0$ matching the two outgoing caps. Inductively step 4.1 gives $h_s:A_s\to P_s$ extending the exact map already prescribed by its parent's outgoing cap, and matching its two outgoing caps.

Use [F7] for this recursive selection of compatible $h_s$. Finite closed pasting gives $h^{\le n}:A^{\le n}\to B_n$: each finite piece map is continuous, their maps agree on caps, and the only intersections on either side are those same caps. Hence the pasted map is a bijection. Continuity follows because the preimage of a closed set is the finite union of its closed preimages in the closed pieces. Compactness and [F3] make this bijection a homeomorphism. For $n\ge 1$ define $f_n:D\to B_n$ by $h^{\le n-1}$ on $A^{\le n-1}$ and $h_s{j}_s$ on each depth-$n$ half-ball $R_s$. The pieces agree on $S_s$, since $j_s$ fixes it. The same finite-pasting and intersection argument proves $f_n$ a homeomorphism onto $B_n$. One may set $f_0=h_{\mathrm{\varnothing }}{j}_{\mathrm{\varnothing }}$. Every finite embedding has now been proved before taking a limit.

For $m\ge n\ge 1$, $f_m=f_n$ off $U_n$, and $f_m(R_s)\subset T_s$ for every $\mathrm{\mid }s\mathrm{\mid }=n$. Indeed all descendant target pieces and terminal approximations lie within that parent envelope, while earlier source pieces have already fixed maps. Further, $f_n(D\setminus U_n)$ misses all depth-$n$ tori: their intersections with the old body are incoming disks, whose entire preimages $S_s$ already lie in $R_s$. These assertions include every shared cap point. Consequently the diameter bound in step 2.2 gives $\mathrm{\parallel }{f}_m-f_n{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2^{-(n-1)}$. This uses the full torus envelopes, not an unjustified confinement of an unfinished source half-ball to its next slice.

Construct a continuous $q_n:X_n\to D$ agreeing with $(h^{\le n-1}{)}^{-1}$ on $B_{n-1}$ and sending each depth-$n$ torus $T_s$ into $R_s$. The incoming disk of $T_s$ has an extended product neighborhood $D_2^2\times [0,1]$ inside the torus, with cap $D_1^2\times \{0\}$. This comes from the angular side chart and inward radial coordinate of the template torus, transported by all finite maps; the extension past the cap rim misses other attachments. Let $b_s:D_1^2\to S_s$ be the prescribed inverse cap map and $a_s=(c_s,0)$. Put $v(u)=u\mathrm{/}\max (1,\mathrm{\mid }u\mathrm{\mid })$, $w(r)=1$ for $r\le 1$ and $w(r)=2-r$ for $1\le r\le 2$, and $\alpha (u,t)=w(\mathrm{\mid }u\mathrm{\mid })(1-t)$. On this neighborhood set $q_n(u,t)=\alpha (u,t)b_s(v(u))+(1-\alpha (u,t))a_s;$ elsewhere in $T_s$ set $q_n=a_s$. On $\mathrm{\mid }u\mathrm{\mid }=2$ or $t=1$ these formulas agree; at $t=0$, $\mathrm{\mid }u\mathrm{\mid }\le 1$ they give the required inverse cap map. The convexity of $R_s$ keeps the values inside it. Finite closed pasting, with the fixed inverse on $B_{n-1}$, proves continuity on $X_n$. No extension over the entire boundary of a solid torus is being presumed.

For $m\ge n$, step 7.1 and the definitions give $\mathrm{\mid }{q}_n(f_m(x))-x\mathrm{\mid }\le d_n$: it is zero off $U_n$, and otherwise both points lie in the same half-ball $R_s$. Define the closed subset $M_n=\{(u,v)\in X_n^2:\mathrm{\mid }{q}_n(u)-q_n(v)\mathrm{\mid }\ge d_n\}\mathrm{.}$ The set $X_n^2$ is compact as a closed bounded subset of ${\mathbb{R}}^6$, by [F3], so $M_n$ is compact. If it is empty set ${\delta }_n=1$. Otherwise let ${\delta }_n$ be half the minimum of $\mathrm{\mid }u-v\mathrm{\mid }$ on $M_n$, attained by [F4]. This minimum is positive because $u=v$ would force $0\ge d_n>0$. Thus for all $u,v\in X_n$, $\mathrm{\mid }u-v\mathrm{\mid }<{\delta }_n$ implies $\mathrm{\mid }{q}_n(u)-q_n(v)\mathrm{\mid }<d_n$. Combining this with the two error bounds yields the claimed strict inequality $\mathrm{\mid }x-y\mathrm{\mid }<3d_n$ whenever $\mathrm{\mid }{f}_m(x)-f_m(y)\mathrm{\mid }<{\delta }_n$.

Step 7.1 makes $(f_n(x))$ Cauchy for every $x$; [F2] supplies its limit $f(x)$, with uniform error $\mathrm{\parallel }f-f_n{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2^{-(n-1)}$. For any $ϵ>0$, choose $n$ with twice this error below $2ϵ\mathrm{/}3$ and use continuity of $f_n$ at $x$ with error $ϵ\mathrm{/}3$. The triangle inequality proves continuity of $f$. For each fixed $n$, all later images lie in the closed $X_n$, so $f(D)\subset B={\bigcap }_n{X}_n$. By continuity of $q_n$, the estimate $\mathrm{\mid }{q}_n(f_m(x))-x\mathrm{\mid }\le d_n$ passes to the limit. Applying the modulus of $q_n$ just established in step 8.1 directly to $f(x),f(y)$ gives $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }<{\delta }_n\Rightarrow \mathrm{\mid }x-y\mathrm{\mid }<3d_n$. This does not pass a strict inequality through a limit. If $x\mathrm{\ne }y$, choose $n$ with $3d_n<\mathrm{\mid }x-y\mathrm{\mid }$; equality $f(x)=f(y)$ contradicts that implication. Hence $f$ is injective.

If $y\in B$, then step 2.2 gives $\mathrm{dist}(y,B_n)\le 2^{-n}$. Since $f_n$ is onto $B_n$, the uniform error gives $\mathrm{dist}(y,f(D))\le 2^{-n}+2^{-(n-1)}$. More explicitly choose a point of $B_n$ within $2^{-n}+ϵ$ of $y$ and its preimage under $f_n$, then let $ϵ\downarrow 0$. The right side tends to zero. The compact image $f(D)$ is closed by [F3], so $y\in f(D)$. We already have the reverse inclusion, hence $f(D)=B$. By [F3], or directly the modulus in step 9.1, $f$ is a homeomorphism. In particular $B$ is nonempty; no invocation of an empty-set minimum or of an unproved limit-embedding assertion has occurred.

Apply [F6] to $f{\mathrm{\mid }}_{\mathrm{int}D}$, whose domain is open in ${\mathbb{R}}^3$, to obtain $f(\mathrm{int}D)\subset \mathrm{int}B$. Conversely suppose $f(x)\in \mathrm{int}B$ with $x\in \mathrm{\partial }D$. Choose an ambient open ball $V$ around $f(x)$ contained in $B$. The continuous injective inverse $f^{-1}:V\to {\mathbb{R}}^3$ would have open image by [F6], contained in $D$ and containing its boundary point $x$. This is impossible by the defining Euclidean boundary of $D$. Thus $\mathrm{int}B=f(\mathrm{int}D)$, and, since $B$ is closed, $\mathrm{\partial }B=B\setminus \mathrm{int}B=f(\mathrm{\partial }D)\cong S^2$. The radial parametrization in step 1.3 identifies that boundary sphere. This is precisely the inherited AC use of invariance of domain; it was not used to assert the earlier finite maps were embeddings.

Finally check the claimed geometric insertion data throughout this actual recursion. In the extended coordinates of a meridional slice, increasing its disk radius gives an outward side collar outside the old torus. The transported block of [F1] gives an inward collar after replacement. Their annulus parameters agree by the prescribed whole boundary marking, so together they are a two-sided product collar. At each cap end reduce its positive width continuously to avoid the closed cap; the open side is parametrized by the circle and the open core interval. Prepared supports have disjoint closed neighborhoods inside their parent and miss all earlier pieces, so this collar misses the remainder $Y$ of the old body, which meets the slice exactly in its two caps. Process each finite level lexicographically. Old meridians and whiskers in the exterior avoid every remaining closed slice; retain them, and use the transported template meridians and whiskers for new children. The preparation transports those same loops to the future meridional-slice coordinates, fixing the parent boundary. Thus no unrecorded path conjugation or arbitrary replacement picture enters the recursion. If an old exterior is path connected, [F1]'s collared open cover also proves the new exterior is path connected; its algebraic injection still requires the stated free-basis hypothesis, to be checked when used. All scales are positive, $j_s$ fixes every incoming endpoint, and the compression fixes support boundaries. AC was used exactly in steps 1.2 and 6.1 for recursive selections and in step 11.1 through [F6]; the finite matching, inverse formulas and metric estimates add no choice principle. This proves all assertions.