Simplicial Subdivision and Simplicial Approximation

In a finite nonempty vertex set, each simplex is defined by nonnegative coordinates, sum $1$, and zero coordinates outside its face. It is closed and bounded in the ambient finite-dimensional Euclidean space, hence compact. The finite union $\mathrm{\mid }K\mathrm{\mid }$ is also closed and bounded, hence compact by Heine–Borel. If $K$ has no vertices its realization is empty and compact directly.

If $E\subseteq \mathrm{\mid }K\mathrm{\mid }$ is weakly closed, then $E\cap \mathrm{\mid }\sigma \mathrm{\mid }$ is closed in the Euclidean simplex, and thus closed in the ambient Euclidean space since $\mathrm{\mid }\sigma \mathrm{\mid }$ is closed. Their finite union is $E$, so $E$ is Euclidean closed. Conversely, a Euclidean relatively closed $E$ has closed traces on every simplex and is weakly closed. Consequently both topologies agree, and the Euclidean metric and Hausdorff property restrict to $\mathrm{\mid }K\mathrm{\mid }$.

For finite $P\subseteq K$ and closed $E\subseteq \mathrm{\mid }P\mathrm{\mid }$, write $E={\bigcup }_{\sigma \in P}(E\cap \mathrm{\mid }\sigma \mathrm{\mid })$. For any $\tau \in K$, each summand meets $\mathrm{\mid }\tau \mathrm{\mid }$ in a closed subset of the common face $\mathrm{\mid }\sigma \cap \tau \mathrm{\mid }$, hence in a closed subset of $\mathrm{\mid }\tau \mathrm{\mid }$. There are finitely many summands, so $E$ is weakly closed in $\mathrm{\mid }K\mathrm{\mid }$. Conversely an ambient weakly closed set has closed traces on the simplices of $P$. These two implications show that the inclusion induces exactly the topology of $\mathrm{\mid }P\mathrm{\mid }$ and is closed; taking $E=\mathrm{\mid }P\mathrm{\mid }$ proves the closed-image assertion.

Let a point have distinct positive coordinate levels $a_1>\cdots >a_m>0$, put $a_{m+1}=0$, and set $F_j=\{v:x_v\ge a_j\}$. These are nested nonempty faces. With $w_j=\mathrm{\#}{F}_j(a_j-a_{j+1})>0$, a vertex whose coordinate is $a_{\mathrm{\ell }}$ has coordinate ${\sum }_{j=\mathrm{\ell }}^m{w}_j\mathrm{/}\mathrm{\#}{F}_j=a_{\mathrm{\ell }}$ in ${\sum }_j{w}_j{b}_{F_j}$. Also ${\sum }_j{w}_j={\sum }_v{x}_v=1$. Thus every point lies in a chain simplex.

For any strict chain $G_1⊊\cdots ⊊G_s$, the vectors $b_{G_j}$ are linearly independent in barycentric coordinate space: in $\sum c_j{b}_{G_j}=0$, a coordinate in $G_s\setminus G_{s-1}$ gives $c_s\mathrm{/}\mathrm{\#}{G}_s=0$, and descending induction gives every $c_j=0$ (finish with any vertex of $G_1$). Hence they are affinely independent in the original simplex too, by uniqueness of its affine coordinates.

If a point is expressed in one of these chain simplices, delete zero weights. Its coordinate on the successive layers of the remaining chain is strictly decreasing, with consecutive differences equal to the corresponding positive weight divided by the face cardinality. Thus the positive-weight faces are exactly the level sets $F_j$ in the first step and the weights are exactly $w_j$. In two chain representations only common face labels can therefore carry positive weights. Conversely every convex combination of common labels belongs to both simplices. This proves precisely the intersection assertion and hence the triangulation. The empty simplex has no points; a one-vertex simplex has the sole weight $1$.

For every original finite simplex $\sigma$, its chain triangulation gives a bijection $\mathrm{\mid }\mathrm{sd}\sigma \mathrm{\mid }\to \mathrm{\mid }\sigma \mathrm{\mid }$. The inverse formulas agree on common faces because the positive coordinate level sets depend only on the point. Every point of $\mathrm{\mid }K\mathrm{\mid }$ has a finite support face, so these inverses define a single global inverse to $b_K$. This also proves $b_K^{-1}(\mathrm{\mid }A\mathrm{\mid })=\mathrm{\mid }\mathrm{sd}A\mathrm{\mid }$.

On each subdivided simplex $b_K$ is affine into its maximal original simplex and is continuous. The weak topology on $\mathrm{\mid }\mathrm{sd}K\mathrm{\mid }$ therefore implies global continuity: the inverse image of a closed set has closed trace on each simplex. On each original simplex the inverse is affine on finitely many closed chain simplices. A closed set has closed inverse trace on each of these pieces, and their finite union is closed in the original simplex; hence this inverse restriction is continuous. Testing on every original simplex with the weak topology gives continuity of the global inverse. The empty case is the empty homeomorphism, and all formulas restrict identically to subcomplexes.

If the set $J$ of open simplices meeting $C$ were infinite, for each positive integer $n$ let $X_n$ be the nonempty set of ordered $n$-tuples of points of $C$ with distinct support faces. Countable Choice selects one tuple for each $n$. Flatten these finite tuples into a sequence and retain, in their natural order, the first point in each previously unseen support face. There are infinitely many retained points since the tuple of length $n$ supplies $n$ different faces. This gives a sequence $q_j$ in distinct open simplices, using only the stated countable independent choices and least-index deletions.

A closed simplex $\mathrm{\mid }\tau \mathrm{\mid }$ has finitely many faces. Each retained point in it has a distinct support face contained in $\tau$, so it contains only finitely many $q_j$. Every subset of $Q=\{q_j:j\ge 1\}$ consequently has finite closed trace on every simplex and is weakly closed in $\mathrm{\mid }K\mathrm{\mid }$. Thus $Q$ is closed in $C$ and its subspace topology is discrete. Closedness in compact $C$ makes $Q$ compact, whereas its singleton open cover has no finite subcover. This contradiction proves $J$ finite.

Include all faces of the finitely many simplices in $J$ to obtain a finite subcomplex containing $C$; if $C$ is empty use the vertex-free subcomplex. A continuous image of a compact simplex is compact because any open cover pulls back to an open cover with a finite subcover. A finite union of these images is compact, so the preceding conclusion applies to the entire family at once, including an empty family.

Adjoining $a$ stays inside $K$ by the cone hypothesis. Permuting the original vertices changes $[a,v_0,\dots ,v_n]$ by the same sign, so $h$ respects the oriented-chain relations. If $a$ is absent from $s=[v_0,\dots ,v_n]$, expansion gives $\mathrm{\partial }[a,v_0,\dots ,v_n]=s+{\sum }_{i=0}^n(-1{)}^{i+1}[a,v_0,\dots ,{\widehat{v}}_i,\dots ,v_n]=s-h\mathrm{\partial }s$.

If $a=v_j$, then $h(s)=0$. In $h\mathrm{\partial }s$ all terms except deletion of $v_j$ repeat $a$ and vanish. The remaining term is $(-1{)}^j[a,v_0,\dots ,{\widehat{v}}_j,\dots ,v_n]=s$, because moving $a$ back to position $j$ contributes another $(-1{)}^j$. In degree zero this says $\mathrm{\partial }[a,v]+[a]=[v]$ for $v\mathrm{\ne }a$, and $0+[a]=[a]$ for $v=a$. In degree $-1$, $\mathrm{\partial }h(1)=\mathrm{\partial }[a]=1$. Thus the identity holds on all generators and hence all chains.

The identity $1=\mathrm{\partial }h+h\mathrm{\partial }$ is a null homotopy of the identity, hence contractibility. A chain of faces of a nonempty full simplex can always be enlarged by the maximal face; thus its order complex, as defined in Barycentric subdivision of an abstract simplicial complex, is a cone with that face as specified apex, and the same calculation applies.

Set $h_{-1}=0$. For an oriented vertex $s$, $z=f(s)-g(s)$ has augmentation $1-1=0$. It lies in its carrier, so $h_0(s)=c_{s}z$ satisfies $\mathrm{\partial }{h}_0(s)=z$, using $\mathrm{\partial }{c}_s+c_s\mathrm{\partial }=1$. Changing the sign of the generator changes $h_0$ by the same sign.

Suppose $h$ is defined through degree $n-1$ with $\mathrm{\partial }h+h\mathrm{\partial }=f-g$ there. For an oriented $n$-simplex $s$ put $z=f(s)-g(s)-h(\mathrm{\partial }s)$. The nesting of the carriers puts all summands in $C_n(\mathrm{\Phi }(s))$. Since $f,g$ commute with boundary, $\mathrm{\partial }z=(f-g)(\mathrm{\partial }s)-\mathrm{\partial }h(\mathrm{\partial }s)=h{\mathrm{\partial }}^{2}s=0$. Here ${\mathrm{\partial }}^2=0$ is part of the given chain complexes.

Define $h_n(s)=c_{s}z$. The contraction identity gives $\mathrm{\partial }{h}_n(s)=z-c_s\mathrm{\partial }z=z$, so $\mathrm{\partial }{h}_n(s)+h_{n-1}(\mathrm{\partial }s)=f(s)-g(s)$. The formula is alternating in the original oriented representative: the boundary and $f,g$ are alternating, the already defined $h$ is linear, and $c_s$ depends only on the underlying face. It therefore defines a homomorphism without choosing orientations on all simplices. Induction defines all degrees, carried by $\mathrm{\Phi }$, with the asserted identity; in degree $-1$ both maps are the identity on $\mathbb{Z}$, so their difference is zero. This is a chain homotopy by definition.

For a vertex $v$, $\mathrm{\partial }S[v]=\mathrm{\partial }[\{v\}]=1=S\mathrm{\partial }[v]$. The degree $-1$ equation is zero on each side. On an edge $[a,b]$, the formula is $S[a,b]=[ab,b]-[ab,a]=[a,ab]-[b,ab]$, abbreviating singleton face labels by their vertices and $\{a,b\}$ by $ab$. Its boundary is $[b]-[a]=S\mathrm{\partial }[a,b]$.

Assume boundary compatibility through degree $n-1$. In the cone $\mathrm{sd}\sigma$ the contraction identity gives $\mathrm{\partial }S(s)=\mathrm{\partial }{c}_{\sigma }S(\mathrm{\partial }s)=S(\mathrm{\partial }s)-c_{\sigma }\mathrm{\partial }S(\mathrm{\partial }s)=S(\mathrm{\partial }s)-c_{\sigma }S({\mathrm{\partial }}^{2}s)=S(\mathrm{\partial }s)$. For the augmented boundary square at a one-simplex, $\epsilon ([b]-[a])=0$; in higher degrees use the boundary-square theorem. Induction proves the identity in every degree. Geometrically the cone terms on boundary-of-boundary faces cancel, exactly accounting for internal faces. Setting the degree-zero ordinary boundary to zero preserves the identity on ordinary chains.

Each nonempty face is finite, so it has a unique greatest vertex. If ${\sigma }_0⊊\cdots ⊊{\sigma }_q$ is a chain, every $\max {\sigma }_i$ lies in ${\sigma }_q$. Their set is therefore a face of $K$, proving simpliciality. If the entire chain lies over a face $\sigma$, all these vertices belong to $\sigma$, proving the carrier assertion. Repeated selected vertices are permitted for a simplicial map.

The induced chain map sends an oriented chain to its ordered list of selected vertices if distinct, and to zero otherwise. It commutes with the ordinary boundary by the induced-chain-map lemma. In degree zero each vertex is sent to a vertex, so both augmentations equal $1$; extending by the identity in degree $-1$ gives an augmented chain map. The empty complex gives the empty vertex map and identity only in degree $-1$. No existence of a total order on an arbitrary set is inferred: the order is supplied data.

Both $S$ and ${\lambda }_{\mathrm{\#}}$ commute with boundary and augmentation. For a nonempty original face $\sigma$, carry ${\lambda }_{\mathrm{\#}}S$ and $1$ by the full simplex on $\sigma$, a cone with apex its greatest vertex. Indeed $S$ stays over $\sigma$ and $\lambda$ lands in $\sigma$. These carriers are nested under faces. The specified cone contractions and the carried-homotopy lemma give ${\lambda }_{\mathrm{\#}}S\simeq 1$.

For a face chain $\eta =({\sigma }_0<\cdots <{\sigma }_q)$ carry $S{\lambda }_{\mathrm{\#}}$ and $1$ by $\mathrm{sd}{\sigma }_q$, a cone with apex ${\sigma }_q$. The identity lies there; ${\lambda }_{\mathrm{\#}}\eta$ either vanishes or is a face of ${\sigma }_q$, whose subdivision lies there as well. Removing any chain vertex leaves the same or a smaller maximum, so the carriers are nested. The carried-homotopy lemma gives $S{\lambda }_{\mathrm{\#}}\simeq 1$. Its homotopies have $h_{-1}=0$, so restriction also gives ordinary chain homotopies.

Chain-homotopic maps induce equal homology maps, hence $H({\lambda }_{\mathrm{\#}})H(S)=1$ and $H(S)H({\lambda }_{\mathrm{\#}})=1$. These equations hold in every augmented degree and every ordinary degree. If $K$ has no vertices, both augmented degree $-1$ groups are $\mathbb{Z}$ and both maps are the identity. Any other vertex order produces another two-sided inverse $J$ to the same $H(S)$; then $J=JH(S)H({\lambda }_{\mathrm{\#}})=H({\lambda }_{\mathrm{\#}})$, proving independence.

For points $x={\sum }_i{a}_i{v}_i$ and $y={\sum }_j{b}_j{v}_j$ in a simplex, $\mathrm{\parallel }x-y\mathrm{\parallel }\le {\sum }_{i,j}{a}_i{b}_j\mathrm{\parallel }{v}_i-v_j\mathrm{\parallel }\le {\max }_{i,j}\mathrm{\parallel }{v}_i-v_j\mathrm{\parallel }$, since all weights are nonnegative and sum to $1$. The maximum is attained by vertices, so it equals the diameter. A point has diameter zero; the definition of mesh assigns zero to a vertex-free complex without taking a diameter of the empty set.

For nonempty nested faces $F⊊G\subseteq \sigma$, $b_G=(\mathrm{\#}F\mathrm{/}\mathrm{\#}G)b_F+(1-\mathrm{\#}F\mathrm{/}\mathrm{\#}G)b_{G\setminus F}$. Thus $\mathrm{\parallel }{b}_G-b_F\mathrm{\parallel }\le (1-\mathrm{\#}F\mathrm{/}\mathrm{\#}G)\mathrm{diam}\sigma \le n\mathrm{/}(n+1)\mathrm{diam}\sigma$. Vertices of every subdivided simplex form such a chain, so the previous diameter calculation bounds its diameter by this factor. Iteration gives the asserted estimate. Since $0<n\mathrm{/}(n+1)<1$, its powers tend to zero. Zero-dimensional simplices remain points.

For any $x$ in the open or closed vertex star of $v$, some simplex contains both $x$ and $v$, so $\mathrm{\parallel }x-v\mathrm{\parallel }\le m(K)$. For two points $x,y$ in that star, the triangle inequality gives $\mathrm{\parallel }x-y\mathrm{\parallel }\le 2m(K)$, and taking the supremum proves the star bound. Finite weak realization topology agrees with the Euclidean topology, so these estimates use a compatible metric.

If $K$ is vertex-free all assertions concern empty maps. Otherwise enumerate its finitely many vertices. For each positive denominator enumerate, lexicographically, all barycentric rational grid points on every face, allowing repetitions, to obtain a sequence $d_j$. On a face with $s$ vertices, round the first $s-1$ coordinates down to multiples of $1\mathrm{/}N$ and give the last coordinate the remaining mass; every coordinate error is at most $(s-1)\mathrm{/}N$. These points approximate every point of each simplex. Hence the sequence is dense in the finite Euclidean realization, and therefore in the weak realization. The image $C=f(\mathrm{\mid }K\mathrm{\mid })$ is compact: pull an open cover back to compact $\mathrm{\mid }K\mathrm{\mid }$ and take a finite subcover.

Suppose the union of the finite supports of $f(d_j)$ were infinite. Recursively select the least index whose image support is not contained in the finite union of previously selected supports, and call the resulting images $q_n$. Each new point introduces a fresh target vertex. In a fixed closed simplex $\tau$, at most $\mathrm{\#}\tau$ of the points $q_n$ can occur, since each such occurrence introduces a fresh vertex of $\tau$. Every subset of $Q=\{q_n\}$ thus has finite closed trace on each target simplex and is weakly closed. Therefore $Q$ is closed in compact $C$, and every subset is closed in $Q$, making $Q$ discrete. Its singleton cover contradicts compactness. All repeated choices here are least natural indices, not applications of Countable Choice.

The union $W$ of those supports is therefore finite. Let $P$ contain all faces of $L$ whose vertices lie in $W$. This is finite and weakly closed: its intersection with any simplex is a finite union of closed faces. The closed set $f^{-1}(\mathrm{\mid }P\mathrm{\mid })$ contains the dense sequence, so equals $\mathrm{\mid }K\mathrm{\mid }$. By the closed-embedding assertion, $f$ regarded as a map into $\mathrm{\mid }P\mathrm{\mid }$ is continuous for its Euclidean topology.

For any nonempty source face $\sigma$, its barycenter lies in the star of every vertex of $\sigma$. Its image therefore has every $g(v)$, $v\in \sigma$, in its support. That support is a face of $L$, so its subset $g(\sigma )$ is a face. This proves simpliciality. For an arbitrary point $x$ the same reasoning applies to each positive-coordinate vertex of $x$: every corresponding $g(v)$ belongs to $\mathrm{supp}f(x)$. Consequently $\mathrm{\mid }g\mathrm{\mid }(x)={\sum }_v{x}_v{e}_{g(v)}$ lies in that very simplex. In particular the image of $g$ lies in $\mathrm{\mid }P\mathrm{\mid }$.

The realization $\mathrm{\mid }g\mathrm{\mid }$ is affine on each of finitely many closed source simplices; these maps agree on their common faces, so it is continuous into finite Euclidean $\mathrm{\mid }P\mathrm{\mid }$. Hence $H(x,t)=(1-t)f(x)+t\mathrm{\mid }g\mathrm{\mid }(x)$ is jointly continuous into its ambient Euclidean space. The common-carrier conclusion puts its image in $\mathrm{\mid }P\mathrm{\mid }$, so it is continuous into $\mathrm{\mid }P\mathrm{\mid }$ and then $\mathrm{\mid }L\mathrm{\mid }$. At $t=0,1$ it equals $f,\mathrm{\mid }g\mathrm{\mid }$, and if $f(x)=\mathrm{\mid }g\mathrm{\mid }(x)$ it is constant in $t$. If $x\in \mathrm{\mid }A\mathrm{\mid }$ and $f(x)\in \mathrm{\mid }B\mathrm{\mid }$, its carrier is a simplex of $B$, so the entire segment remains in $\mathrm{\mid }B\mathrm{\mid }$.

Use the barycentric homeomorphisms to view every subdivision as a triangulation of the same finite Euclidean polyhedron. The inverse images $f^{-1}({\mathrm{st}}_L(w))$ cover it, since the support of every image point is nonempty. They are open, and the source is compact metric. The Lebesgue-number lemma gives $\delta >0$ such that any nonempty set of diameter less than $\delta$ lies in one of these inverse images. For an empty source use the empty map for every $r$.

Choose $r_0$ with $2m({\mathrm{sd}}^{r}K)<\delta$ for every $r\ge r_0$, using the mesh estimate; in dimension zero $m=0$ already. Each vertex star in that triangulation is nonempty and has diameter at most $2m$. Hence for each of the finitely many vertices $v$ select $g(v)$ with $f(\mathrm{st}(v))\subseteq \mathrm{st}(g(v))$. This is only finite choice. The star criterion gives a simplicial map and a continuous common-carrier straight-line homotopy.

If $\sigma$ is a face of ${\mathrm{sd}}^{r}A$, its barycenter lies in $\mathrm{\mid }A\mathrm{\mid }$, so the support of its image under $f$ is a face of $B$. The star criterion puts all $g(v)$ for $v\in \sigma$ in this support; hence $g(\sigma )\in B$. At every point of $\mathrm{\mid }A\mathrm{\mid }$ the same carrier argument keeps the homotopy inside $\mathrm{\mid }B\mathrm{\mid }$. Pulling back to the abstract subdivided realization gives the stated homotopy of pairs to $f{b}_K^r$, for every $r\ge r_0$.

A simplex of $D_{A}K$ consists of a face $\alpha$ of $A$ and barycenters $b_{{\sigma }_1},\dots ,b_{{\sigma }_s}$ of nested faces outside $A$, strictly containing $\alpha$. The only vertices in $\mathrm{\mid }A\mathrm{\mid }$ are those of $\alpha$: a barycenter of $\sigma \mathrm{\notin }A$ has support $\sigma$, so cannot lie in the subcomplex $A$. Thus the vertices in $A$ span precisely $\alpha$.

For a point in that simplex with a positive coefficient at some $b_{{\sigma }_j}$, take the largest such face ${\sigma }_j$. All its vertex coordinates in the original simplex become positive, with no cancellation, so the original support contains ${\sigma }_j$. Such a point cannot belong to $\mathrm{\mid }A\mathrm{\mid }$, since that would put its support and every subface, including ${\sigma }_j$, in $A$. Conversely every point using only vertices of $\alpha$ lies in $\mathrm{\mid }A\mathrm{\mid }$. Therefore the intersection is exactly $\mathrm{\mid }\alpha \mathrm{\mid }$. The same argument applies with $K$ replaced by each $D_A^{r-1}K$. If $A$ is empty intersections are empty; if $A=K$, every simplex is already in $A$.

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.

Take $h$ and $T_r$ from the neighbourhood-adjustment lemma. Compose its homotopy $H_t$ with $f$ to obtain a homotopy from $f$ to $fh$, fixed on $\mathrm{\mid }A\mathrm{\mid }$. The inverse images under $fh$ of the target vertex stars form an open cover of compact metric $\mathrm{\mid }K\mathrm{\mid }$, so there is a Lebesgue number $\delta >0$. For empty $K$ the empty map already proves the theorem.

Choose $r\ge 3$ so that every star of a $T_r$ vertex in $\mathrm{\mid }{B}_2\mathrm{\mid }$ has diameter less than $\delta$. For each such vertex the Lebesgue-number property gives a target vertex $g(v)$ whose open star contains $fh(\mathrm{st}(v))$. For any remaining vertex $v$, the adjustment lemma gives $a\in A$ with $h(\mathrm{st}(v))\subseteq {\mathrm{st}}_A(a)$. Since $f{\mathrm{\mid }}_A$ is simplicial, a point having positive $a$ coordinate has positive $f(a)$ coordinate in its image, so $f({\mathrm{st}}_A(a))\subseteq {\mathrm{st}}_L(f(a))$; take $g(v)=f(a)$. On $A$ take $a=v$, hence $g(v)=f(v)$. These are finitely many choices.

All vertex stars now satisfy the criterion for $fh$, so $g$ extends simplicially and the criterion supplies $J_t(x)=(1-t)fh(x)+t\mathrm{\mid }g\mathrm{\mid }(x)$. On $\mathrm{\mid }A\mathrm{\mid }$, the simplicial maps $g$ and $f$ agree on vertices and therefore on all affine combinations; also $h$ is the identity there. Thus $J$ fixes $f{\mathrm{\mid }}_A$. Concatenate $f{H}_{2t}$ for $0\le t\le 1\mathrm{/}2$ with $J_{2t-1}$ for $1\mathrm{/}2\le t\le 1$. At the join both equal $fh$, so this is a continuous homotopy from $f$ to $\mathrm{\mid }g\mathrm{\mid }$, fixed on $A$. Its restriction there always lies in the simplicial image $f(A)$, proving the pair assertion.