Barycentric subdivision and prism preserve smooth singular chains

If $a:{\mathrm{\Delta }}^j\to {\mathrm{\Delta }}^k$ is affine, it extends to an affine map $\bar{a}:A^j\to A^k$. The open inverse image ${\bar{a}}^{-1}(O)$ contains ${\mathrm{\Delta }}^j$, and $\bar{\sigma }\bar{a}$ extends $\sigma a$ smoothly into $M$. Each simplex in $S\sigma$ and $T\sigma$ has this form: coning affine simplices appends a fixed barycenter vertex and so remains affine, at every stage of the finite recursion [F2]. Therefore both operators preserve smooth chains. Their affine images stay in ${\mathrm{\Delta }}^k$, so they also preserve any specified image-containing subset of $M$.

Let $B$ be a smooth target-valued extension of $(x,t)\mapsto H(\sigma (x),t)$ to an open neighbourhood $W$ of ${\mathrm{\Delta }}^k\times [0,1]$. Each prism simplex ${\lambda }_i:{\mathrm{\Delta }}^{k+1}\to {\mathrm{\Delta }}^k\times [0,1]$ is affine. Its affine extension has open inverse image of $W$ containing ${\mathrm{\Delta }}^{k+1}$, and $B{\lambda }_i$ is the required extension into the target. Thus every term of the signed prism is smooth. The equality [F3] is an equality in this subcomplex because every term is now in it.

For a smooth homotopy $H:M\times [0,1]\to N$, with smoothness interpreted by local coordinate extensions also at the time endpoints, replace it by $\widehat{H}(p,t)=H(p,s(t))$ for all real $t$. This is smooth: near an endpoint use a local coordinate extension of $H$ and compose with $(p,t)\mapsto (p,s(t))$; near an interior time ordinary smooth composition suffices. It is target-valued for every real $t$ because $s(t)\in [0,1]$. Then $(x,t)\mapsto \widehat{H}(\bar{\sigma }(x),t)$ is smooth on $O\times \mathbb{R}$ and satisfies step 2.1. Its endpoint maps are exactly those of $H$, so [F3] gives the same difference of induced chain maps.

The qualification in step 2.1 is necessary for an unmodified prism with a boundary target. Take $M$ a point and $H(t)=t$ into $N=[0,\mathrm{\infty })$. This is a smooth homotopy on the interval, but its prism is the path $t\mapsto t$, which has no smooth target-valued extension across $0$: any such nonnegative extension has a local minimum at $0$ and derivative zero, whereas its right derivative would be one. Time flattening avoids this obstruction. In degree zero $S=1$ and $T=0$; the flattened homotopy prism is still a smooth path. Empty chains and empty domains give zero operators, repeated affine vertices are allowed, and all formulas are finite and choice-free.