Ordinary barycentric subdivision cannot fix a nonconstant simplicial edge

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.