Relative subdivision neighbourhood adjustment

By fullness, in $T_1$ the fixed complex $A$ is induced on its vertices. Let $B_1$ be the induced subcomplex on the other vertices. They are disjoint. Form $T^{+}=D_{A\cup B_1}{T}_1$: it subdivides just the mixed simplices of $T_1$. Its new vertices are barycenters $b_{\sigma }$ of mixed faces. The face $\sigma \cap A$ is nonempty by mixedness and is a face by fullness; choose one of its vertices $a_{\sigma }$. Define $h$ to fix the vertices of $A\cup B_1$ and send $b_{\sigma }$ to $a_{\sigma }$. All choices are finite.

For a simplex of $T^{+}$, its base face belongs to $A\cup B_1$ and its additional vertices are barycenters of a nested chain of mixed faces of $T_1$. Every image vertex lies in the largest face of this chain, or in the base face if the chain is empty. Hence the images span a simplex of $T_1$, so $h$ extends simplicially from $T^{+}$ to $T_1$. Moreover each $h(v)$ belongs to the positive $T_1$ support of $v$: this is immediate for a fixed old vertex, and a mixed barycenter is positive at every vertex of its face. If $x\in {\mathrm{st}}_{T^{+}}(v)$, its positive coefficient at $v$ therefore makes its $h(v)$ coordinate positive in $T_1$. Thus ${\mathrm{st}}_{T^{+}}(v)\subseteq {\mathrm{st}}_{T_1}(h(v))$, which verifies the star hypothesis for the identity map and the vertex assignment $h$. For any point $x$ in such a simplex, $x$ and $h(x)$ lie in the same original $T_1$ simplex. Thus $H_t(x)=(1-t)x+th(x)$ remains there. Finite simplexwise affine formulas agree on faces, so $h$ and $H$ are continuous in finite Euclidean realization, and $H_0=1,H_1=h$, with $H_t{\mathrm{\mid }}_A=1$.

If a simplex of $T^{+}$ contains a vertex $a\in A$, its base face is in $A$, and every other vertex is a mixed-face barycenter mapped to $A$. Its image simplex lies in $A$, since $A$ is full in $T_1$. Moreover a positive coefficient of $a$ remains positive at $a$ in the image. Therefore $h({\mathrm{st}}_{T^{+}}(a))\subseteq {\mathrm{st}}_A(a)$. The union of these open stars is an open neighbourhood of $\mathrm{\mid }A\mathrm{\mid }$ mapped into $\mathrm{\mid }A\mathrm{\mid }$.

The full relative subdivision $T_2$ refines $T^{+}$: it additionally subdivides $B_1$ and uses the same barycenters on the mixed faces, coning their refined boundaries. It retains the vertices of $A$ and their positive barycentric coordinates, so ${\mathrm{st}}_{T_2}(a)\subseteq {\mathrm{st}}_{T^{+}}(a)$. If $v$ is a vertex of any further $T_r$ outside $\mathrm{\mid }{B}_2\mathrm{\mid }$, its minimal carrier face in $T_2$ has an $A$-vertex $a$ with positive coordinate at $v$. Every point of ${\mathrm{st}}_{T_r}(v)$ has positive coefficient at $v$ in some refined simplex; the $T_2$ coordinate of $a$, affine and nonnegative on that simplex, is then positive. Thus this star is contained in ${\mathrm{st}}_{T_2}(a)$ and its $h$-image in ${\mathrm{st}}_A(a)$. For $v\in A$ its carrier is the vertex itself.

Let $B_3$ be the induced subcomplex of $T_3$ on vertices outside $A$. No simplex of $T_3$ containing an $A$-vertex can meet $\mathrm{\mid }{B}_2\mathrm{\mid }$: its relative face-chain description has an $A$ base face and all outside barycenters lie on $T_2$ faces containing that base. Every point of such a simplex has a positive coordinate at some vertex of that base in $T_2$, whereas points of $\mathrm{\mid }{B}_2\mathrm{\mid }$ have all those coordinates zero. Consequently every $T_3$ simplex meeting $\mathrm{\mid }{B}_2\mathrm{\mid }$ is in $B_3$. For $r\ge 3$ and a vertex $v\in \mathrm{\mid }{B}_2\mathrm{\mid }$, every $T_r$ simplex containing $v$ lies in a $T_3$ simplex meeting $\mathrm{\mid }{B}_2\mathrm{\mid }$, hence in $B_3$. Since $B_3$ is disjoint from $A$, its further relative subdivisions are ordinary barycentric subdivisions. Each such star has diameter at most $2m({\mathrm{sd}}^{r-3}{B}_3)$, which tends to zero by the mesh estimate.

If $A$ is vertex-free then $B_1=T_1$, $T^{+}=T_1$ and $h=1$; the neighbourhood can be empty and the mesh estimate handles all stars. If $A=K$, then $T^{+}=K$, $h=1$, and $B_2$ is vertex-free, so only the near clause occurs. These constructions also cover a vertex-free $K$, with empty maps.