One fixed number of barycentric subdivisions makes every singular simplex cover small

For any proposed $m\ge 0$, define the smooth path ${\sigma }_m(t)=\sin (2\pi 2^{m}t)$, $0\le t\le 1$. In dimension one, the cone recursion gives the two half-interval parametrizations, one forward and one backward, with coefficient equal to their orientation sign. Inducting on subdivisions gives one affine parametrization of each dyadic interval $[j{2}^{-m},(j+1)2^{-m}]$, again with its orientation sign as coefficient: subdivision bisects each interval and the two new signs multiply its previous sign.

On a forward dyadic parametrization, ${\sigma }_m$ becomes $p(t)=\sin (2\pi t)$; on a backward one it becomes $q(t)=-\sin (2\pi t)$. Therefore $S^m{\sigma }_m=A[p]-B[q]$, where $A,B$ count forward and backward pieces and $A+B=2^m>0$. The two maps are distinct because $p(1\mathrm{/}4)=1$ and $q(1\mathrm{/}4)=-1$, so no cancellation between them occurs in the free chain group. Each has image $[-1,1]$, which lies in neither $U$ nor $V$. At least one nonzero basis coefficient therefore belongs to a non-small simplex, and the chain is not cover-small.

This proves $\mathrm{\forall }m\mathrm{\exists }{\sigma }_m$ failing the fixed cover's proposed bound; it does not assert that one simplex fails all bounds. For $m=0$ the witness is $p$ itself. Each witness has equal endpoints zero but is a nonconstant smooth simplex; repeated endpoint values cause no cancellation as step 2.1 checks. A constant simplex is already small. The empty chain is always small, and all formulas and witnesses are explicit without choice.