Differential-character curvature and characteristic class are well-defined

Suppose $\omega$ and ${\omega }^{\mathrm{\prime }}$ both satisfy [F1], and put $\alpha =\omega -{\omega }^{\mathrm{\prime }}$. For each oriented smooth integral $k$-simplex $\sigma$, ${\int }_{\sigma }\alpha \in \mathbb{Z}$. If $k>\dim M$, then $\alpha =0$ by degree, so assume $k\le \dim M$. In a coordinate chart about $x$, choose any ordered $k$ coordinate directions and, for every sufficiently small $t>0$, the oriented affine simplex with vertices $x,x+t{e}_1,\dots ,x+t{e}_k$. Its integral of $\alpha$ is integer-valued, depends continuously on $t$, and tends to $0$ as $t\downarrow 0$; hence it is $0$ for all sufficiently small $t$. Dividing by the simplex volume and taking the limit gives ${\alpha }_x(e_1,\dots ,e_k)=0$. Varying the coordinate directions shows ${\alpha }_x=0$, and $x$ was arbitrary. Thus $\omega ={\omega }^{\mathrm{\prime }}$.

For every smooth integral $k$-cycle $z$, [F1] gives ${\int }_z{\omega }_h\equiv h(0)=0(\mathrm{m}\mathrm{o}\mathrm{d}\mathbb{Z})$, so every period is integral. If $k+1>\dim M$, then $d{\omega }_h=0$ by degree. Otherwise, for every smooth integral $(k+1)$-simplex $\sigma$, Stokes and [F1] give ${\int }_{\sigma }d{\omega }_h={\int }_{\mathrm{\partial }\sigma }{\omega }_h\in \mathbb{Z}$. The oriented-affine-simplex argument of step 1.1, applied in degree $k+1$, shows that every coefficient of $d{\omega }_h$ vanishes. Hence $d{\omega }_h=0$.

The subgroup $Z_{k-1}^{\mathrm{\infty }}$ is free by [F4], so it is projective by [F5]. Lift the homomorphism $h:Z_{k-1}^{\mathrm{\infty }}\to \mathbb{R}\mathrm{/}\mathbb{Z}$ through $\mathbb{R}\to \mathbb{R}\mathrm{/}\mathbb{Z}$ to a homomorphism $\mathrm{\ell }:Z_{k-1}^{\mathrm{\infty }}\to \mathbb{R}$. Since $\mathbb{R}$ is injective, extend $\mathrm{\ell }$ to a real cochain $\tilde{h}$ on $C_{k-1}^{\mathrm{\infty }}$. For each $c\in C_k^{\mathrm{\infty }}$, [F1] implies $\mathrm{Int}({\omega }_h)(c)-\delta \tilde{h}(c)\in \mathbb{Z}$. Stokes and $d{\omega }_h=0$ from step 2.1 show that its coboundary is $0$, so it is an integral cocycle. By [F6], its cohomology class is viewed canonically as an ordinary singular class.

If ${\tilde{h}}^{\mathrm{\prime }}$ is another real extension of a lift of $h$, then $u={\tilde{h}}^{\mathrm{\prime }}-\tilde{h}$ takes integer values on $Z_{k-1}^{\mathrm{\infty }}$. By [F4], $C_{k-1}^{\mathrm{\infty }}=Z_{k-1}^{\mathrm{\infty }}\oplus s(B_{k-2}^{\mathrm{\infty }})$ for some splitting $s$. Extend $u{\mathrm{\mid }}_{Z_{k-1}^{\mathrm{\infty }}}$ to the integer cochain $v(c)=u({\mathrm{pr}}_{Z}c)$. For every $k$-chain $c$, $\mathrm{\partial }c\in Z_{k-1}^{\mathrm{\infty }}$, hence $\delta (u-v)(c)=0$ and $\delta u=\delta v$. Thus the two integral cocycles differ by the integral coboundary $\delta v$, proving independence of $I(h)$.

For a smooth map $f:N\to M$, pullback of chains commutes with boundary and pullback of forms commutes with exterior derivative. The character $f^{\ast }h(z)=h(f_{\ast }z)$ has curvature $f^{\ast }{\omega }_h$. Pulling back a chosen extension $\tilde{h}$ gives an extension for $f^{\ast }h$; its integral cocycle is the pullback of the cocycle in step 3.1. Independence from the extension is step 4.1, so $R$ and $I$ are natural.