A continuous map need not be simplicial before subdivision

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.