Simplicial Subdivision and Simplicial Approximation: Examples

The edge has nonempty faces $a,b,ab$, abbreviating braces. Its only length-two chains are $a<ab$ and $b<ab$, so the two subdivided edges meet at $b_{ab}=(a+b)\mathrm{/}2$. There are three singleton chains and one empty chain. Thus its numbers of faces in dimensions $-1,0,1$ are $(1,3,2)$.

The triangle has face labels $a,b,c,ab,ac,bc,abc$. The maximal chains are $a<ab<abc$, $b<ab<abc$, $a<ac<abc$, $c<ac<abc$, $b<bc<abc$, and $c<bc<abc$. The comparable pairs are six vertex-edge pairs, three vertex-triangle pairs, and three edge-triangle pairs, giving twelve edges. The seven singleton chains and one empty chain give face counts $(1,7,12,6)$. Barycenters place these six triangles inside the original triangle and the barycentric homeomorphism identifies their union with it. For a vertex-free complex there is only the empty chain, so the counts reduce to $(1,0,0,0)$.

The augmented cone recursion gives $S[u,v]=[uv,v]-[uv,u]=[u,uv]-[v,uv]$. Hence $S\mathrm{\partial }[a,b,c]=[b,bc]-[c,bc]-[a,ac]+[c,ac]+[a,ab]-[b,ab]$. Prepending $T$ to each term and moving it past the two other vertices changes sign by $(-1{)}^2=1$, giving exactly the six displayed oriented triangles.

For a term $[u,e,T]$, its boundary is $[e,T]-[u,T]+[u,e]$. The first two triangle terms give radial contribution $-[b,T]+[c,T]$ after the $[bc,T]$ terms cancel. The next two give $[a,T]-[c,T]$ after the $[ac,T]$ terms cancel. The last two give $-[a,T]+[b,T]$ after the $[ab,T]$ terms cancel. The three remaining radial contributions sum to zero. The six base terms are exactly the displayed $S\mathrm{\partial }[a,b,c]$, verifying the boundary identity directly, in agreement with the general chain-map lemma.

The first source edge maps to the target vertex $0$ and the second to the target edge, so $g$ is simplicial. The two affine formulas agree at $1\mathrm{/}2$, both giving $0$, and $\mathrm{\mid }g\mathrm{\mid }(0)=0$, $\mathrm{\mid }g\mathrm{\mid }(1)=1$. The target stars are $[0,1)$ and $(0,1]$. The source stars of $0,1\mathrm{/}2,1$ are $[0,1\mathrm{/}2)$, $(0,1)$, $(1\mathrm{/}2,1]$; under $x^2$ they lie respectively in $[0,1)$, $[0,1)$, $(0,1]$. Thus this is even a star approximation to $f$.

The explicit homotopy is $H(x,t)=(1-t)x^2+t\mathrm{\mid }g\mathrm{\mid }(x)$. It is continuous because its piecewise polynomial formulas agree at $x=1\mathrm{/}2$; its values lie in $[0,1]$ as convex combinations. It satisfies $H(x,0)=x^2$, $H(x,1)=\mathrm{\mid }g\mathrm{\mid }(x)$ and $H(0,t)=0,H(1,t)=1$. At the midpoint, $f(1\mathrm{/}2)=1\mathrm{/}4$, $\mathrm{\mid }g\mathrm{\mid }(1\mathrm{/}2)=0$, and $H(1\mathrm{/}2,t)=(1-t)\mathrm{/}4$, so fixing the endpoints does not mean fixing the entire interval. This explicitly realizes the relative theorem for the pair.

The two segments defining $C_0$ are continuous images of compact closed real intervals, so their finite union is compact. Its positive supports are $\{o\}$, $\{v_1\}$, $\{o,v_1\}$ and $\{o,v_2\}$, exactly four open simplices in two edges. Any neighbourhood of $o$ intersects each edge in a relative neighbourhood of its endpoint, hence meets the interior of every edge. Thus $o$ is not locally finite.

Each closed edge meets $Q$ in exactly one point, and the central vertex is not in $Q$. Every subset of $Q$ therefore has finite closed traces on every simplex and is weakly closed. In particular $Q$ is closed and discrete, and its singleton cover has no finite subcover. The numerical parameter $1\mathrm{/}n$ tending to zero does not imply convergence to $o$ in the weak topology: $\mathrm{\mid }K\mathrm{\mid }\setminus Q$ is an open neighbourhood of $o$ missing all these points.

For a compact nonempty $C$, the coordinate $x\mapsto x_{v_n}$ is continuous because it is continuous on every simplex. Let $m_n=\sup \{x_{v_n}:x\in C\}\in [0,1]$. This supremum is attained: otherwise the sets $\{x\in C:x_{v_n}<m_n-1\mathrm{/}k\}$ for positive integers $k$ would cover $C$, and a finite subcover would bound all coordinates by $m_n-1\mathrm{/}{k}_0$ for its largest index $k_0$, contradicting the supremum. If $m_n>0$, there is exactly one point of the entire fan with coordinate $x_{v_n}=m_n$, namely $p_n=(1-m_n)o+m_n{v}_n$, so this point belongs to $C$. These points are canonical and require no countable selections.

If infinitely many $m_n$ are positive, the corresponding $p_n$ form an infinite set with at most one point on each edge, every subset of which is weakly closed just as for $Q$. It is therefore closed in compact $C$, and is compact (adjoin its open complement to any cover to see this directly). Yet it is discrete with an infinite singleton cover, a contradiction. Only finitely many $m_n$ are positive, so $C$ lies in their edges and possibly $o$; if none are positive use the first edge to contain $o$. For empty $C$ no edge is needed.

The map takes $[0,1]$ into itself, fixes both vertices, and is continuous: $\mathrm{\mid }{x}^2-y^2\mathrm{\mid }=\mathrm{\mid }x-y\mathrm{\mid }\mathrm{\mid }x+y\mathrm{\mid }\le 2\mathrm{\mid }x-y\mathrm{\mid }$ on this interval. Any simplicial map whose realization equals $f$ must therefore send $0$ to $0$ and $1$ to $1$. Its affine realization on the single edge must be $(1-x)0+x1=x$.

At $x=1\mathrm{/}2$ this affine map equals $1\mathrm{/}2$, while $f(1\mathrm{/}2)=1\mathrm{/}4$. Hence $f$ is not simplicial in the original triangulation, refuting the assertion. Nevertheless the identity vertex map is a star approximation: $f([0,1))\subseteq [0,1)$ and $f((0,1])\subseteq (0,1]$. The star criterion therefore gives a homotopy to the identity, consistent with finite simplicial approximation. Failure of exact simpliciality is not failure of a simplicial approximation.

The identity is continuous and simplicial on all of $A$. In $\mathrm{sd}K$ the nonempty edge face supplies the midpoint $m=1\mathrm{/}2$ as a vertex. Every subsequent barycentric subdivision retains that geometric point as a singleton-face vertex, so $m$ is a vertex of ${\mathrm{sd}}^{r}K$ for every $r\ge 1$.

A simplicial map to the unchanged target must send $m$ to the target vertex $0$ or $1$. Pointwise agreement on $\mathrm{\mid }A\mathrm{\mid }$ instead requires $g(m)=f(m)=1\mathrm{/}2$, impossible. More generally any source containing a fixed edge mapped identically to a target edge has the same obstruction by restricting to its midpoint. Relative derived subdivision avoids it: for $A=K$, $D_A^{r}K=K$, so the identity itself is simplicial and fixed throughout, as allowed by the relative theorem.