Homotopy formula for differential characters

For a smooth singular simplex $\sigma :{\mathrm{\Delta }}^r\to M$, triangulate the prism ${\mathrm{\Delta }}^r\times [0,1]$ into the $r+1$ simplices with ordered vertices $(v_0,0),\dots ,(v_j,0),(v_j,1),\dots ,(v_r,1)$ for $0\le j\le r$, orient their sum by the interval-first product orientation $[0,1]\times {\mathrm{\Delta }}^r$, and map it to $M\times J$ by $(x,t)\mapsto (\sigma (x),t)$. Let $P(\sigma )$ be this signed sum. In its boundary each interior face occurs twice with opposite orientations; the two endpoint faces remain as $i_{1\ast }\sigma -i_{0\ast }\sigma$, and the side faces are the negative interval-first prism on the alternating boundary of $\sigma$. Hence $\mathrm{\partial }P(c)=i_{1\ast }c-i_{0\ast }c-P(\mathrm{\partial }c)$ for every finite chain $c$, by additivity.

If $z$ is a smooth integral $(k-1)$-cycle on $M$, then [F2] gives $\mathrm{\partial }P(z)=i_{1\ast }z-i_{0\ast }z$. By [F1], $h_1(z)-h_0(z)=\widehat{h}(\mathrm{\partial }P(z))={\int }_{P(z)}R(\widehat{h})(\mathrm{m}\mathrm{o}\mathrm{d}\mathbb{Z})\mathrm{.}$ The prism triangulation and the product orientation identify the last integral with ${\int }_z{\int }_0^1{\iota }_{{\mathrm{\partial }}_t}R(\widehat{h})dt$. By [F4] this is exactly $a({\int }_{J}R(\widehat{h}))(z)$. Equality on every cycle proves the asserted equality of differential characters.