A horn replacement block has an injective commutator meridian

For $-1<t<1$ define $\mathrm{\Phi }(\theta ,\varphi ,t)=(\sqrt{(1+t)\mathrm{/}2}{e}^{i\theta },\sqrt{(1-t)\mathrm{/}2}{e}^{i\varphi })\mathrm{.}$ On $T^2\times [-r,r]$, for $0<r<1$, this is a homeomorphism onto its image: the inverse reads the phases of its nonzero coordinates and $t=\mathrm{\mid }z{\mathrm{\mid }}^2-\mathrm{\mid }w{\mathrm{\mid }}^2$. Put $a_0=\sqrt{(1-r)\mathrm{/}2}$, $T_0=\{\mathrm{\mid }z\mathrm{\mid }\le a_0\}$ and $T_1=\{\mathrm{\mid }w\mathrm{\mid }\le a_0\}$. The parametrization $(z,u)\mapsto (z,u\sqrt{1-\mathrm{\mid }z{\mathrm{\mid }}^2})$, with $\mathrm{\mid }z\mathrm{\mid }\le a_0$ and $u\in S^1$, and its inverse identify $T_0$ with a disk times a circle; interchange $z,w$ for $T_1$. The tori are disjoint since $2a_0^2=1-r<1$, and together with the displayed collar they fill $S^3$.

Take also $r<\pi \mathrm{/}4$, let $D$ be the angular square $[-r,r{]}^2$ in the torus, and set $Q=\mathrm{\Phi }(D\times [-r,r])$, $C=S^3\setminus \mathrm{int}Q$. All defining compact sets are compact by [F3]; their parametrizations and inverse formulas show their stated subspace topologies. The interior of $Q$ is exactly the image of the open cube, since $\mathrm{\Phi }$ is a coordinate chart on a neighborhood of the cube. It remains to prove that this particular complement $C$ is a ball; the fact that $Q$ is a ball alone would not suffice.

Here is an explicit straightening of $Q$ for sufficiently small $r$. Put $v=\mathrm{\Phi }(0,0,0)$ and use stereographic coordinates $s(y)=\frac{y-⟨y,v⟩v}{1+⟨y,v⟩}(y\mathrm{\ne }-v)$ in the tangent space $v^{\perp }\cong {\mathbb{R}}^3$. Their inverse is $x\mapsto 2x\mathrm{/}(1+\mathrm{\mid }x{\mathrm{\mid }}^2)+(1-\mathrm{\mid }x{\mathrm{\mid }}^2)v\mathrm{/}(1+\mathrm{\mid }x{\mathrm{\mid }}^2)$. Both formulas are continuous inverses, with $-v$ corresponding to infinity. The three derivatives of $\mathrm{\Phi }$ at zero, in real coordinates, are $(0,1\mathrm{/}\sqrt{2},0,0)$, $(0,0,0,1\mathrm{/}\sqrt{2})$ and $(1\mathrm{/}(2\sqrt{2}),0,-1\mathrm{/}(2\sqrt{2}),0)$, hence independent. The derivative of $s$ there is one half the identity on $v^{\perp }$. Postcompose $s$ with the inverse of the resulting invertible linear map to obtain a chart in which $g=s\mathrm{\Phi }$, with this normalization understood, satisfies $g(0)=0$, $Dg(0)=I$. Its displayed coordinate formulas are continuously differentiable near zero.

On a sufficiently small convex open cube containing $[-2r,2r{]}^3$, derivative continuity and [F2] give $\mathrm{Lip}(g-\mathrm{i}\mathrm{d})\le \epsilon$ and $\mathrm{\mid }g(x)-x\mathrm{\mid }\le \epsilon \mathrm{\mid }x\mathrm{\mid }$. Let ${\chi }_r(x)$ be one for $\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}\le r$, $2-\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}r$ between radii $r,2r$, and zero beyond $2r$. Put $k(x)={\chi }_r(x)(g(x)-x)$ inside that cube and zero outside. Since $\mathrm{Lip}{\chi }_r\le 1\mathrm{/}r$ and $\mathrm{\mid }g(x)-x\mathrm{\mid }\le 2\sqrt{3}\epsilon r$ on its support, the product estimate gives $\mathrm{Lip}k\le (1+2\sqrt{3})\epsilon =q$. For a segment with one endpoint outside the cube, stop at its first boundary point, where $k=0$, to get the same bound; for two outside points both values are zero. Reduce $r$ until $q<1\mathrm{/}2$. Then $H(x)=x+k(x)$ obeys $\mathrm{\mid }H(x)-H(y)\mathrm{\mid }\ge (1-q)\mathrm{\mid }x-y\mathrm{\mid }$. It is injective with Lipschitz inverse on its image. For every $b\in {\mathbb{R}}^3$, $x\mapsto b-k(x)$ is a contraction of complete nonempty ${\mathbb{R}}^3$, so [F1] supplies a solution of $H(x)=b$. Hence $H$ is a homeomorphism, identity off a compact cube, and extends fixing infinity. On $[-r,r{]}^3$ it equals $g$.

The complement of the straight cube, with infinity included, is a closed ball explicitly. For a unit direction $u$ let $\rho (u)=r\mathrm{/}\mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}$. The radial homeomorphism $su\mapsto (s\mathrm{/}\rho (u))u$ carries the cube onto the closed unit ball and the closed exterior of the cube onto the complement of the open unit ball. Its inverse is $tu\mapsto t\rho (u)u$; both extend at zero and infinity because $r\le \rho (u)\le \sqrt{3}r$. Inversion $x\mapsto x\mathrm{/}\mathrm{\mid }x{\mathrm{\mid }}^2$, extended by infinity mapping to zero, takes this last closed exterior onto the closed unit ball. Transport these maps through $H$ and the stereographic chart of step 2.2. This supplies a homeomorphism $C\cong D^3$, rather than invoking Schoenflies.

Since $T_0$ consists of the points on and below $t=-r$, and $T_1$ those on and above $t=r$, their intersections with $\mathrm{\partial }C=\mathrm{\partial }Q$ are exactly $D_0=\mathrm{\Phi }(D\times \{-r\})$ and $D_1=\mathrm{\Phi }(D\times \{r\})$. These are disjoint closed disks. The remaining boundary is $A=\mathrm{\Phi }(\mathrm{\partial }D\times (-r,r))$, and direct subtraction of the defining sets gives $Z=\mathrm{\Phi }((T^2\setminus \mathrm{int}D)\times (-r,r))\mathrm{.}$ The inverse in step 1.1 is the asserted homeomorphism of pairs, after rescaling the interval. The square boundary has an explicit annular collar in the angular chart: use its radial direction and the coordinate $\mathrm{\parallel }(\theta ,\varphi ){\mathrm{\parallel }}_{\mathrm{\infty }}-r$. Together with $t$ this is a two-sided product collar of $A$. It is disjoint from the closed caps since $-r<t<r$; near the ends its width can be decreased continuously if required. Corners do not affect the continuous inverse of these radial coordinates.

The inverse coordinates on $\mathrm{\partial }Q$ give a specified boundary homeomorphism from $\mathrm{\partial }C$ to the boundary of a standard cylinder $D^2\times [-1,1]$, matching caps and side-annulus parameters. This extends across $C$: start with the supplied ball parametrization from step 4.1 and any radial parametrization of the convex cylinder, compare their induced sphere maps with the desired boundary map, and extend the resulting sphere homeomorphism $b$ by $su\mapsto sb(u)$, sending zero to zero. The same formula with $b^{-1}$ is a continuous inverse. Thus the block can be inserted with the entire prescribed cap-and-side marking, not just with unlabelled caps.

Model $F=T^2\setminus \mathrm{int}D$ as the square annulus $P=\{x:r\le \mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}\le \pi \}$, with opposite outer edges identified by a map $q:P\to F$. The homotopy $x\mapsto ((1-t)+t\pi \mathrm{/}\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }})x$ fixes the outer boundary and retracts the annulus onto it. It respects the edge identifications at every $t$. The map $q\times {\mathrm{id}}_I:P\times I\to F\times I$ is a continuous surjection from a compact space to a Hausdorff space: $P\times I$ is a closed bounded Euclidean subset and hence compact by [F3], while $F\times I$ is Hausdorff as a subspace of the torus times $I$. Images of closed subsets of the compact domain are compact and therefore closed by [F3]. Thus $q\times {\mathrm{id}}_I$ is a closed quotient map, and the radial homotopy descends continuously with its time parameter. The outer-edge quotient is a wedge of two circles, so [F4] computes its group freely on the horizontal and vertical edge loops. Choose the inner corner $(-r,-r)$ and the radial path from it to $(-\pi ,-\pi )$. Conjugate the two edge loops by this path to obtain loops $a,b$ at the inner corner. Following the positively oriented inner-square boundary and its radial image reads the four outer edges in order $a,b,a^{-1},b^{-1}$. The radial homotopy with its basepoint track proves that its class is exactly $[a,b]$ with these paths. Explicitly a moving-basepoint homotopy gives this conjugacy by traversing the boundary of its parameter square; the square itself contracts that boundary. Multiplying a whisker by its reverse cancels by linear retracing, so no basepoint-conjugation convention is omitted.

To prepare small future slices, use the product coordinates of step 1.1. In $T_0$ take $K_0=D_{a_0}^2\times [\pi -\nu ,\pi +\nu ]$ in its $w$-phase coordinate, where $0<\nu <\pi \mathrm{/}4$, and use the corresponding slice $K_1$ in $T_1$. They miss the incoming cap patches, whose core angles lie in $[-r,r]$. Enlarge the disk radii to $(1+\delta )a_0$ and angle half-lengths to $(1+\delta )\nu$ for sufficiently small $\delta >0$. These closed coordinate cylinders $N_0,N_1$ are disjoint and lie in $\mathrm{int}C$: their torus-height ranges remain respectively below $-r\mathrm{/}2$ and above $r\mathrm{/}2$, their phases near $\pi$ avoid the deleted box, and their disk radii remain below one. The ambient chart for $N_0$ is $(z,v)\mapsto (z,\sqrt{1-\mathrm{\mid }z{\mathrm{\mid }}^2}{e}^{i(\pi +v)})$, and for $N_1$ interchange coordinates. Each is defined on a neighborhood of its closed cylinder and has the phase-and-disk inverse.

Contracting the interval coordinate of $Z$ to zero retracts the pair onto $(F,\mathrm{\partial }F)$, with the chosen basepoint at height zero. The loop $a$ varies the $z$ phase with $w$ phase $\pi$, hence is a meridian of $T_0$ pushed into the collar; move its height from zero to just above $-r$ for the actual push-off. Similarly $b$ varies the $w$ phase at $z$ phase $\pi$ and is a pushed meridian of $T_1$. These movements transport the same whiskers. The annulus retracts to its circle, whose group is infinite cyclic by [F4]. For every integer $k\mathrm{\ne }0$, the word $[a,b{]}^k$ is reduced of length $4\mathrm{\mid }k\mathrm{\mid }$, so it is nontrivial by [F7]. Thus its map into the free group is injective. Reversing the annulus orientation replaces it by its inverse; a changed path conjugates it and leaves injectivity unchanged, with the conjugation specified by that path.

For the insertion in the Statement, transport all these markings. Write its supplied collar as $A\times (-ϵ,ϵ)$ with normal parameter $u<0$ outside $C^{\mathrm{\prime }}$ and $u>0$ inside; varying widths at the ends may be rescaled to this interval. Define $U=E\cup \{u<ϵ\mathrm{/}3\},V=(\mathrm{int}{C}^{\mathrm{\prime }}\setminus (T_0^{\mathrm{\prime }}\cup T_1^{\mathrm{\prime }}))\cup \{u>-ϵ\mathrm{/}3\},$ where the inequalities refer only to collar points. These are open in $E^{\mathrm{\prime }}$, cover it, and intersect exactly in $A\times (-ϵ\mathrm{/}3,ϵ\mathrm{/}3)$. Points of $E^{\mathrm{\prime }}$ on $\mathrm{\partial }{C}^{\mathrm{\prime }}$ are precisely the annular points, so none is missed. The overlap retracts to $A$. The space $V$ retracts onto the transported $Z$ by replacing its negative normal coordinates by zero; interpolate between $u$ and $\max (u,0)$ and leave $Z$ fixed. Thus $V$ is path connected by step 6.2 and its product model, and $U$ is path connected since it is the old connected exterior with a collar meeting it.

In either cylinder set $\rho (z,v)=\max (\mathrm{\mid }z\mathrm{\mid }\mathrm{/}{a}_0,\mathrm{\mid }v\mathrm{\mid }\mathrm{/}\nu )$. Use radial homothety by $\lambda \in (0,1)$ on $\rho \le 1$, and on $1\le \rho \le 1+\delta$ map radius $\rho$ by the strictly increasing linear function joining $(1,\lambda )$ to $(1+\delta ,1+\delta )$. Outside $N_i$ use the identity. This is a homeomorphism: on each ray its continuous strictly increasing radial function has the displayed piecewise-linear inverse, and at the center both maps are continuous. It takes $K_i$ onto $\lambda K_i$ and fixes the cylinder boundary. Conjugate through the transported copy's map $C\to C^{\mathrm{\prime }}$. The extension by identity is an ambient homeomorphism, since its support is a compact subset of $\mathrm{int}{C}^{\mathrm{\prime }}$; the two supports remain disjoint. Continuity of the transported chart at its center makes the diameter of the image of $\lambda K_i$ tend to zero. Hence a sufficiently small $\lambda$ makes it less than any specified $\eta >0$. Transport the torus product parametrizations and all paths by these same maps.

Inclusion $E\to U$ is a homotopy equivalence by this explicit push. For $u\ge -ϵ\mathrm{/}2$ put $h(u)=-ϵ\mathrm{/}2+(2\mathrm{/}5)(u+ϵ\mathrm{/}2)$, and put $h(u)=u$ below $-ϵ\mathrm{/}2$. Keep points outside this collar portion fixed. On $U$, $u<ϵ\mathrm{/}3$ implies $h(u)<-ϵ\mathrm{/}6$, so the resulting map lands in $E$. Straight interpolation stays in $U$, and for a point already in $E$ stays in $E$. It proves both inverse-homotopy identities. To retain a fixed exterior basepoint choose the collar width smaller if necessary so the basepoint is outside the moving portion. Then both homotopies fix it. A path from that point to the overlap transfers the van Kampen basepoint; on loops the transfer is conjugation by that fixed path, with its inverse provided by the reversed path. This is verified by cancellation of path followed by reverse, as in step 6.2. The overlap generator on the $U$ side is precisely the pushed annular loop $\mu$, and on the $V$ side precisely the marked word of step 7.1.

Apply [F6] to this open cover. Its pushout has the presentation $⟨R,\mu ,a,b\mid \mu =[a,b]⟩\mathrm{.}$ This statement can also be checked directly by the universal property: compatible maps from the free group on $R,\mu$ and the free group on $a,b$ are exactly choices of their images satisfying the one displayed equality. Eliminating $\mu$ gives the free group on $R,a,b$, with inverse maps specified by fixing these letters and replacing $\mu$ by $[a,b]$. By [F5] and step 8.1 this is the actual inclusion-induced map from $E$, not an arbitrary abstract group identification. If a nontrivial reduced word on $R,\mu$ is grouped into alternating nonempty $R$-words and nonzero powers of $\mu$, substitution gives nonempty reduced blocks on disjoint alphabets $R$ and $\{a,b\}$. No letters cancel across their boundaries, and step 7.1 excludes trivial blocks on the second alphabet. Thus the image word is nontrivial. This proves injectivity, including empty $R$ and words with only one block. For a conjugated or inverse marked commutator, every nonzero power is still a nontrivial word in the child free factor, so the same argument applies after its internal reduction.

In those transported coordinates a prepared slice is exactly $D^2\times I_0$ for a proper closed core arc $I_0$. Its relative interior in the torus is $D^2\times {\mathrm{int}}_{S^1}{I}_0$. The complement of that relative interior is $D^2\times \overline{S^1\setminus I_0}$, a cylinder ball, and its intersection with the slice is exactly the two end disks. In particular no lateral annulus is retained in the complementary ball. The incoming cap stays on its side, away from the removed slice. All finite collars and the punctured-torus complement survive under the same ambient homeomorphism. This proves the preparation clause as well as the original block and injection assertions. The bounds require $r>0$, $q<1$ and $\eta >0$; no zero-width pillbox is claimed. The annular loop's zeroth power is the identity, whereas every nonzero power survives by step 7.1. Only finitely many parameters, charts and loops have been instantiated; the contraction iteration and explicit radial maps require no AC.