The cone and join metrics and the local product chart of a polyhedral gluing

Truncation. The function $d_{\pi }(x,y)=\min \{\pi ,d_{\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}}(x,y)\}$ is symmetric and vanishes exactly when $d_{\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}}(x,y)=0$, i.e. when $x=y$; it has values in $[0,\pi ]$, so its diameter is at most $\pi$; and for real $a,b\ge 0$ one has $\min \{\pi ,a+b\}\le \min \{\pi ,a\}+\min \{\pi ,b\}$, so the triangle inequality of $d_{\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}}$ passes to $d_{\pi }$. On a pair with $d_{\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}}(x,y)<\pi$ clearly $d_{\pi }=d_{\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}}$.

Cone basics. For points $(r,x),(s,y)\in C(L)$ the definition gives $d_C((r,x),(s,y){)}^2=r^2+s^2-2rs\cos d_{\pi }(x,y)$, so $d_C\ge 0$ and $d_C$ is symmetric, $d_C$ vanishes exactly when $r=s$ and $d_{\pi }(x,y)=0$, and $d_C(o,(r,x))=r$ with $d_C(o,o)=0$; moreover $(r-s{)}^2\le d_C^2\le (r+s{)}^2$, that is $\mathrm{\mid }r-s\mathrm{\mid }\le d_C\le r+s$, and $d_C=r+s$ holds exactly when $\cos d_{\pi }(x,y)=-1$, i.e. $d_{\pi }(x,y)=\pi$. In particular $d_C((t,x),(t^{\mathrm{\prime }},x))=\mathrm{\mid }t-t^{\mathrm{\prime }}\mathrm{\mid }$ for fixed $x$, and $d_C$ is finite whenever $L$ is nonempty, while $C(\mathrm{\varnothing })=\{o\}$ is a point.

The cone triangle inequality, first case. Let $P=(r,x)$, $Q=(t,z)$, $T=(s,y)$ with $\alpha :=d_{\pi }(x,z)$, $\beta :=d_{\pi }(z,y)$, $\gamma :=d_{\pi }(x,y)$ and suppose $\alpha +\beta \le \pi$. In the Euclidean plane with origin $0$ put $A:=r{e}_0$, $B:=t{e}_{\alpha }$, $C:=s{e}_{\alpha +\beta }$ (if $A=0$ or $C=0$ read the point as the origin); then $\mathrm{\mid }A-B\mathrm{\mid }=d_C(P,Q)$ and $\mathrm{\mid }B-C\mathrm{\mid }=d_C(Q,T)$ by the law of cosines [F2], while $\gamma \le \alpha +\beta \le \pi$ and $\cos$ decreasing on $[0,\pi ]$ give $\cos \gamma \ge \cos (\alpha +\beta )$ and hence $d_C(P,T{)}^2=r^2+s^2-2rs\cos \gamma \le r^2+s^2-2rs\cos (\alpha +\beta )=\mathrm{\mid }A-C{\mathrm{\mid }}^2$. So $d_C(P,T)\le \mathrm{\mid }A-C\mathrm{\mid }\le \mathrm{\mid }A-B\mathrm{\mid }+\mathrm{\mid }B-C\mathrm{\mid }=d_C(P,Q)+d_C(Q,T)$.

Development of a minimizing segment. Let $x,y\in L$ have $d_{\pi }(x,y)=\theta <\pi$ and let $\gamma :[0,\theta ]\to L$ be a minimizing segment with $d_{\pi }(\gamma (u),\gamma (w))=\mathrm{\mid }u-w\mathrm{\mid }$. Let $S:=\{t{e}_u\in {\mathbb{R}}^2:t\ge 0,0\le u\le \theta \}$ be the planar sector of angle $\theta$ and define $\mathrm{\Psi }(t{e}_u):=(t,\gamma (u))$ for $t>0$, $\mathrm{\Psi }(0):=o$. Then for $t,t^{\mathrm{\prime }}>0$ the identity $d_C((t,\gamma (u)),(t^{\mathrm{\prime }},\gamma (w)){)}^2=t^2+t^{\mathrm{\prime }2}-2t{t}^{\mathrm{\prime }}\cos \mathrm{\mid }u-w\mathrm{\mid }=\mathrm{\mid }t{e}_u-t^{\mathrm{\prime }}{e}_w{\mathrm{\mid }}^2$ holds, so $\mathrm{\Psi }$ is an isometry of $S$ onto its image and preserves lengths; the straight segment in $S$ from $r{e}_0$ to $s{e}_{\theta }$ has Euclidean length $\sqrt{r^2+s^2-2rs\cos \theta }=d_C((r,x),(s,y))$ and its image is a path in $C(L)$ of the same length joining the two points, hence a minimizing geodesic.

The join: descent and first properties. The relation on $L_1\times L_2\times [0,\pi \mathrm{/}2]$ generated by the stated endpoint identifications is an equivalence relation. Write $E:=\cos \theta \cos {\theta }^{\mathrm{\prime }}\cos d_{\pi }^1(x_1,x_1^{\mathrm{\prime }})+\sin \theta \sin {\theta }^{\mathrm{\prime }}\cos d_{\pi }^2(x_2,x_2^{\mathrm{\prime }})$ and $a:=\cos \theta \cos {\theta }^{\mathrm{\prime }}$, $b:=\sin \theta \sin {\theta }^{\mathrm{\prime }}$; then $a,b\ge 0$ and $a+b=\cos (\theta -{\theta }^{\mathrm{\prime }})\le 1$, so $E\le a+b\le 1$ and $E\ge -a-b\ge -1$ by [F2], that is $E\in [-1,1]$. If $\theta ={\theta }^{\mathrm{\prime }}=0$ then $E=\cos d_{\pi }^1(x_1,x_1^{\mathrm{\prime }})$ depends only on the first coordinates, and if $\theta ={\theta }^{\mathrm{\prime }}=\pi \mathrm{/}2$ then $E=\cos d_{\pi }^2(x_2,x_2^{\mathrm{\prime }})$ depends only on the second, so $E$ is unchanged when a representative is replaced; the formula therefore descends to a symmetric function $G$ on pairs of classes, and $d$ is defined by $\cos d(x,x^{\mathrm{\prime }})=G(x,x^{\mathrm{\prime }})$ with $d(x,x^{\mathrm{\prime }})\in [0,\pi ]$. Moreover $G(x,x^{\mathrm{\prime }})\le 1$ with equality only for $x=x^{\mathrm{\prime }}$: equality forces $\cos (\theta -{\theta }^{\mathrm{\prime }})=1$, that is $\theta ={\theta }^{\mathrm{\prime }}$, and then forces $\cos d_{\pi }^1(x_1,x_1^{\mathrm{\prime }})=1$ when $\cos \theta >0$ and $\cos d_{\pi }^2(x_2,x_2^{\mathrm{\prime }})=1$ when $\sin \theta >0$, so the two classes coincide, and conversely equal classes give $G=1$; hence $d(x,x^{\mathrm{\prime }})=0$ exactly for $x=x^{\mathrm{\prime }}$ and $d$ has diameter at most $\pi$. The classes with $\theta =0$ and $\theta =\pi \mathrm{/}2$ are the images of $L_1$ and $L_2$; when $L_1$ or $L_2$ is empty the quotient of $L_1\times L_2\times [0,\pi \mathrm{/}2]$ is empty, and the stated conventions $L\ast \mathrm{\varnothing }:=L$, $\mathrm{\varnothing }\ast L:=L$, $\mathrm{\varnothing }\ast \mathrm{\varnothing }:=\mathrm{\varnothing }$ together with $C(\mathrm{\varnothing })=\{o\}$ are exactly the cone formula at a one-point link.

Localisation and radial coordinates. Work in the connected component $X_p$ of $p$: chain classes are open and closed unions of cells and are path connected, so this subgluing satisfies (H1)-(H3) and [F4] applies. The union $A$ of all cells not containing $F$ is weakly closed: its trace on any cell is a union of some of that cell's finitely many closed faces, by the intersection condition. It omits $p$, since a face containing a relative interior point of $F$ contains $F$. Thus $B(p,a)\cap A=\mathrm{\varnothing }$ for some $a>0$. There are finitely many incident cells by (H2). In each, the facet inequalities not active at $p$ have positive values at $p$; choosing $b>0$ smaller than all their distances to their supporting hyperplanes ensures that, within Euclidean radius $b$, the cell agrees exactly with $p+T_F{C}_r$. Choose $h<\min \{a,b\}$ and let $Z$ be the gluing of the incident tangent cones, with Euclidean chain metric. On the radial subset $\mathrm{\mid }v\mathrm{\mid }<h$, the map $v\mapsto p_r+v$ into the incident cell, followed by its inclusion in $X$, is well defined and injective: the common-face condition and the gluing cocycle identify exactly the same vectors in $Z$ and points in $X$. For a chain of length $\mathrm{\ell }<h$ starting at $p$, all its vertices lie in $B(p,h)$, hence in incident cells, and radial norms along it are at most $\mathrm{\ell }$ by the Euclidean reverse triangle inequality. Its pullback into $Z$ has the same length. Conversely a radial segment of length $t<h$ maps to a path in $X$ of that length. Taking infima shows $d_X(p,\mathrm{\Phi }(v))=\mathrm{\mid }v\mathrm{\mid }$ for $\mathrm{\mid }v\mathrm{\mid }<h$, and every point of $B(p,h)$ has such coordinates by pulling back a chain from $p$ of length less than $h$.

Triples and antipodal pairs. If a triple has $d_{\pi }(x,y)=\pi$, then $\pi \le d_{\pi }(x,z)+d_{\pi }(z,y)$ by the triangle inequality of step 1.1, so its perimeter is at least $2\pi$; hence a triple of perimeter $<2\pi$ has no side equal to $\pi$, and a fortiori at most one. If no side equals $\pi$ then all three $d_{\pi }$-distances are $<\pi$, hence finite $d_{\mathrm{p}\mathrm{a}\mathrm{t}\mathrm{h}}$-distances, hence all three points lie in one component and the sides are intrinsic path distances, because on a pair at distance $<\pi$ the truncated metric is the componentwise path distance. Finally, a pair of points of the model sphere $S^2$ at round distance $\pi$ is antipodal, since the round distance is $\arccos$ of the inner product.

The cone triangle inequality, second case. With the same notation assume $\alpha +\beta >\pi$; then $\cos \alpha +\cos \beta =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}<0$ because $\frac{\alpha +\beta }{2}\in (\pi \mathrm{/}2,\pi ]$ and $\frac{\mathrm{\mid }\alpha -\beta \mathrm{\mid }}{2}<\pi \mathrm{/}2$. The identity $d_C(P,Q{)}^2=r^2+t^2-2rt\cos \alpha =(r-t\cos \alpha {)}^2+t^2{\sin }^2\alpha$ gives $d_C(P,Q)\ge r-t\cos \alpha$, and similarly $d_C(Q,T)\ge s-t\cos \beta$; hence $d_C(P,Q)+d_C(Q,T)\ge r+s-t(\cos \alpha +\cos \beta )\ge r+s\ge d_C(P,T)$, with strict inequality when $t>0$, where the last inequality is step 1.2.

The product-cone isometry. Let $R:=\sqrt{r^2+s^2}$, $\cos \theta :=r\mathrm{/}R$, $\sin \theta :=s\mathrm{/}R$ for $(r,x),(s,y)\in C(L_1)\times C(L_2)$ with $R>0$, and map $((r,x),(s,y))$ to the point of $C(L_1\ast L_2)$ at distance $R$ from the apex in the direction $(\cos \theta )x+(\sin \theta )y$; this map is a bijection onto $C(L_1\ast L_2)$ (the collapsed cases $r=0$ or $s=0$ giving the points over $L_2$ or $L_1$, and $R=0$ the apex) and it carries the square-sum product distance of two points to the cone distance: expanding with ${\cos }^2+{\sin }^2=1$ gives $d_C((r,x_1),(r^{\mathrm{\prime }},x_1^{\mathrm{\prime }}){)}^2+d_C((s,x_2),(s^{\mathrm{\prime }},x_2^{\mathrm{\prime }}){)}^2=R^2+R^{\mathrm{\prime }2}-2R{R}^{\mathrm{\prime }}(\cos \theta \cos {\theta }^{\mathrm{\prime }}\cos d_{\pi }^1(x_1,x_1^{\mathrm{\prime }})+\sin \theta \sin {\theta }^{\mathrm{\prime }}\cos d_{\pi }^2(x_2,x_2^{\mathrm{\prime }}))$, which is the square of the cone distance of step 1.2 computed with the join formula.

The cone is a metric. Steps 1.2, 1.3 and 2.2 cover all triples: those with a vanishing radius, those with $\alpha +\beta \le \pi$ and those with $\alpha +\beta >\pi$, so $d_C$ satisfies the triangle inequality; symmetry and positivity were noted in step 1.2, and $d_C(P,T)=0$ forces $r=s$ and $d_{\pi }(x,y)=0$, i.e. $P=T$. Hence $d_C$ is a metric on $C(L)$ in all cases, including $d_{\pi }(x,y)=\pi$, different components and $C(\mathrm{\varnothing })=\{o\}$, and it is finite-valued on nonempty $L$.

The apex paths. If $d_{\pi }(x,y)=\pi$ then step 1.2 gives $d_C((r,x),(s,y))=r+s$; the two radial segments from $(r,x)$ to $o$ and from $o$ to $(s,y)$ have length $r$ and $s$ by step 1.2 and concatenate to a path of length $r+s$, so it is minimizing, and for $r=0$ or $s=0$ the same argument with a single radial segment applies. Thus radial segments are geodesics from the apex and the through-apex path realizes the distance whenever $d_{\pi }(x,y)=\pi$.

The join is a metric space. Let $\mathrm{\Phi }:C(L_1)\times C(L_2)\to C(L_1\ast L_2)$ be the bijection of step 2.3 and let $D$ be the cone function on $C(L_1\ast L_2)$ built by the cone formula of step 1.2 from the join function $d$ of step 1.5; the identity of step 2.3 says exactly that $D(X,X^{\mathrm{\prime }})=\rho ({\mathrm{\Phi }}^{-1}X,{\mathrm{\Phi }}^{-1}{X}^{\mathrm{\prime }})$ for all $X,X^{\mathrm{\prime }}\in C(L_1\ast L_2)$, where $\rho$ is the square-sum product metric of $C(L_1)\times C(L_2)$. The cone metrics on $C(L_1)$ and $C(L_2)$ are metrics by step 3.1, so $\rho$ is a metric by [F5] and $D$, the pushforward of $\rho$ along $\mathrm{\Phi }$, is a metric as well; in particular $D(X,X^{\mathrm{\prime }\mathrm{\prime }})\le D(X,X^{\mathrm{\prime }})+D(X^{\mathrm{\prime }},X^{\mathrm{\prime }\mathrm{\prime }})$ for all $X,X^{\mathrm{\prime }},X^{\mathrm{\prime }\mathrm{\prime }}\in C(L_1\ast L_2)$. Now let $x,y,z\in L_1\ast L_2$ and put $A:=d(x,y)$, $B:=d(y,z)$, $C:=d(x,z)$; applying that inequality to the cone points of radii $1,s,1$ over $x,y,z$ gives $\sqrt{2-2\cos C}\le \sqrt{1+s^2-2s\cos A}+\sqrt{1+s^2-2s\cos B}$ for every $s\ge 0$, by the cone formula of step 1.2. If $A+B\ge \pi$ then $C\le \pi \le A+B$ by step 1.5; otherwise let $U:=(1,0)$ and $W:=(\cos (A+B),\sin (A+B))$ in the plane. If $\sin A+\sin B>0$ put $s_{\ast }:=\frac{\sin (A+B)}{\sin A+\sin B}$ and $\lambda :=\frac{\sin A}{\sin A+\sin B}\in [0,1]$; then the point $V:=s_{\ast }(\cos A,\sin A)$ satisfies $(1-\lambda )\cdot 0+\lambda \sin (A+B)=s_{\ast }\sin A$ and $(1-\lambda )\cdot 1+\lambda \cos (A+B)=\frac{\sin B+\sin A\cos (A+B)}{\sin A+\sin B}=\frac{\sin (A+B)\cos A}{\sin A+\sin B}=s_{\ast }\cos A$ — the middle equality being the addition formulas, since $\sin B=\sin (A+B)\cos A-\sin A\cos (A+B)$ — so $V=(1-\lambda )U+\lambda W$ lies on the segment $[U,W]$; if instead $\sin A+\sin B=0$ then $A,B\in \{0,\pi \}$ with $A+B\le \pi$, the only cases being $(A,B)=(0,0),(0,\pi ),(\pi ,0)$, and the choice $s_{\ast }:=1$ gives $V=U$ or $V=W$. Either way $V$ lies on $[U,W]$, and expanding the three squares gives $\mathrm{\mid }U-V\mathrm{\mid }=\sqrt{1+s_{\ast }^2-2s_{\ast }\cos A}$, $\mathrm{\mid }V-W\mathrm{\mid }=\sqrt{1+s_{\ast }^2-2s_{\ast }\cos B}$ and $\mathrm{\mid }U-W\mathrm{\mid }=\sqrt{2-2\cos (A+B)}=2\sin \frac{A+B}{2}$; hence $\sqrt{2-2\cos C}\le \mathrm{\mid }U-V\mathrm{\mid }+\mathrm{\mid }V-W\mathrm{\mid }=\mathrm{\mid }U-W\mathrm{\mid }=2\sin \frac{A+B}{2}$ at $s=s_{\ast }$. Both $\frac{C}{2}$ and $\frac{A+B}{2}$ lie in $[0,\frac{\pi }{2}]$, where $\sin$ is strictly increasing and $\sqrt{2-2\cos t}=2\sin \frac{t}{2}$ for $t\in [0,\pi ]$ by [F2], so $C\le A+B$. Hence the join function satisfies the triangle inequality and, with step 1.5, is a metric on $L_1\ast L_2$ of diameter at most $\pi$.

The cone over a sphere. For $k\ge 1$ the map $C(S^{k-1})\to {\mathbb{R}}^k$, $(t,\xi )\mapsto t\xi$ and $o\mapsto 0$, is a bijection, and for two points it preserves distances because $d_C((t,\xi ),(t^{\mathrm{\prime }},{\xi }^{\mathrm{\prime }}){)}^2=t^2+t^{\mathrm{\prime }2}-2t{t}^{\mathrm{\prime }}\cos d_{\pi }(\xi ,{\xi }^{\mathrm{\prime }})=t^2+t^{\mathrm{\prime }2}-2t{t}^{\mathrm{\prime }}⟨\xi ,{\xi }^{\mathrm{\prime }}⟩=\mathrm{\mid }t\xi -t^{\mathrm{\prime }}{\xi }^{\mathrm{\prime }}{\mathrm{\mid }}^2$; hence $C(S^{k-1})\cong {\mathbb{R}}^k$. For $k=0$ the sphere $S^{-1}$ is empty and $C(\mathrm{\varnothing })=\{o\}={\mathbb{R}}^0$ by the empty convention.

The tangent-cone metric and the cone formula. Put $M={\mathrm{Lk}}_X(p)$ with its cellwise intrinsic angular chain distance. Initially this is an extended pseudometric: reversing and concatenating chains prove symmetry and the triangle inequality, but separation has not yet been proved. The following comparison uses the cosine cone function on the direction quotient $C(M)$ and does not assume separation. For the upper bound, if the truncated angular distance is $\pi$, use the two radial segments through the origin. Otherwise take a link chain of total angle $\mathrm{\Theta }<\pi$ approaching that distance, and develop its successive sectors in a plane: the straight segment between the endpoint radii intersects the intermediate rays and yields a chain in $Z$ of length $\sqrt{t^2+t^{\mathrm{\prime }2}-2t{t}^{\mathrm{\prime }}\cos \mathrm{\Theta }}$. For the lower bound replace a chain in $Z$ by its straight cell segments. If any segment passes through the origin, its total length is at least $t+t^{\mathrm{\prime }}$. Otherwise its radial projections give a link chain; develop the segments with monotonically increasing polar angle, of total angle $\mathrm{\Theta }$ at least the intrinsic angular distance. If $\mathrm{\Theta }<\pi$, the endpoint chord has length at least the cone distance. If $\mathrm{\Theta }\ge \pi$, split at the first crossing of the opposite ray, at radius $s$: the preceding polyline has length at least $t+s$, and the remainder at least $\mathrm{\mid }{t}^{\mathrm{\prime }}-s\mathrm{\mid }$, so the total is at least $t+t^{\mathrm{\prime }}$. These bounds prove the equality, without passing an infinite value to the cosine. Let $\epsilon =h\mathrm{/}4$. Cone chains realizing the upper bound between radii below $\epsilon$ remain at radii below $\epsilon$, so they map to $X$. Conversely chains in $X$ of length below $h-\epsilon$ between points of $B(p,\epsilon )$ lie in $B(p,h)$ and pull back with unchanged length; chains longer than that already exceed the cone distance, which is at most $2\epsilon <h-\epsilon$. Hence $\mathrm{\Phi }$ preserves distances bijectively between these balls. Distinct directions with zero angular distance would give distinct points at the same radius $\epsilon \mathrm{/}2$ with zero distance in $X$, contradicting [F4]. Thus the angular chain distance separates directions; the cone function is a metric by step 3.1, and $\mathrm{\Phi }$ is an isometry $B(o,\epsilon )\subset C(M)\to B(p,\epsilon )$. Scaling tangent vectors scales every Euclidean chain length, so the cone formula and separation also hold on all of $Z$.

Geodesic space. Assume in addition that $L$ is $D_{\pi }$-geodesic. For two points $P=(r,x)$, $T=(s,y)$: if $r=0$ or $s=0$ or $d_{\pi }(x,y)=\pi$ use the radial or the through-apex path of step 4.1; if $d_{\pi }(x,y)<\pi$ the hypothesis supplies a minimizing segment in $L$ and step 1.4 supplies a minimizing geodesic in $C(L)$ joining $P$ to $T$. So every two points of $C(L)$ are joined by a geodesic segment and $C(L)$ is a geodesic metric space.

Associativity, faces and spheres. The unit link of a cone is recovered by $d_{\pi }(u,v)=\arccos ((2-d_C((1,u),(1,v){)}^2)\mathrm{/}2)$, so step 2.3 identifies $L_1\ast L_2$ with the unit link of $C(L_1)\times C(L_2)$; iterating gives $C((L_1\ast L_2)\ast L_3)\cong C(L_1\ast L_2)\times C(L_3)\cong C(L_1)\times C(L_2)\times C(L_3)$, so the two bracketed joins are isometric up to a canonical isometry; when $L_1,L_2$ are round unit spheres the same computation with the round inner product gives $S^{m-1}\ast S^{n-1}\cong S^{m+n-1}$, and the face metrics of the join are the joins of the face metrics because the formula restricts to sub-links.

Containment in the ball. Let $c$ be a minimizing geodesic from $P=(r,x)$ to $T=(s,y)$ in $C(L)$ and let $Q=(t,z)$ be a point of its image with $\alpha :=d_{\pi }(x,z)$, $\beta :=d_{\pi }(z,y)$, $\gamma :=d_{\pi }(x,y)$, so that $d_C(P,Q)+d_C(Q,T)=d_C(P,T)$. If $Q=o$ the claimed radius bound is immediate, so suppose $t>0$. If $\alpha +\beta >\pi$, the strict inequality of step 2.2 contradicts this equality; hence $\alpha +\beta \le \pi$. Then, with $A,B,C$ as in step 1.3, the chain $d_C(P,Q)+d_C(Q,T)=\mathrm{\mid }A-B\mathrm{\mid }+\mathrm{\mid }B-C\mathrm{\mid }\ge \mathrm{\mid }A-C\mathrm{\mid }\ge d_C(P,T)=d_C(P,Q)+d_C(Q,T)$ is an equality throughout, so $A,B,C$ are collinear with $B$ between $A$ and $C$; since the norm is convex along the segment from $A$ to $C$, one has $t=\mathrm{\mid }B\mathrm{\mid }\le \max \{\mathrm{\mid }A\mathrm{\mid },\mathrm{\mid }C\mathrm{\mid }\}=\max \{r,s\}$. This includes $Q=o$ and the endpoints, so every minimizing geodesic joining $P$ to $T$ is contained in the closed ball of radius $\max \{r,s\}$ about $o$.

The face product and its angular link. Let $U=\mathrm{span}(F-p)$. Every active facet normal is perpendicular to $U$, so each tangent cone splits orthogonally as $U\oplus N_F{C}_r$, and the gluing maps respect the common $U$ and the normal sections. Thus $Z$ is the gluing of $U\times N_F{C}_r$. The chain comparison of step 4.4 applied to the normal cones identifies their glued chain distance $d_N$ with the cosine cone function on their angular direction quotient; separation is checked below. The chain metric of $Z$ is the square-sum product metric: each chain has length at least $\sqrt{\mathrm{\mid }u-u^{\mathrm{\prime }}{\mathrm{\mid }}^2+d_N(n,n^{\mathrm{\prime }}{)}^2}$ by the triangle inequality in ${\mathbb{R}}^2$, applied to its nonnegative coordinate lengths; conversely choose a piecewise straight normal path with length approaching $d_N(n,n^{\mathrm{\prime }})$, parametrise it proportionally to length, and move the $U$ coordinate linearly over the same interval. The resulting cellwise path has length $\sqrt{\mathrm{\mid }u-u^{\mathrm{\prime }}{\mathrm{\mid }}^2+{\mathrm{\ell }}_N^2}$, proving the reverse bound on taking infima (constant normal paths cover coincident endpoints; a zero infimum is handled by arbitrarily short chains). Since the chain distance on $Z$ is a metric by step 4.4, the product identity forces $d_N$ to separate points. The cone formula then forces the angular chain distance of distinct normal directions to be positive; reversal and concatenation give its extended metric axioms. Consequently $Z\cong {\mathbb{R}}^k\times C({\mathrm{Lk}}_X(F))$, with the genuine cone metric of step 3.1. Steps 2.3, 4.2 and 4.3 identify this product with $C(S^{k-1}\ast {\mathrm{Lk}}_X(F))$ by a map preserving radial norm. Recovering the truncated angular distance from the distances of unit radial points as in step 5.2 gives $M\cong S^{k-1}\ast {\mathrm{Lk}}_X(F)$ with its angular metric. The ball isometries and all charts preserve intrinsic lengths.