The closed-form exact sequence of a differential extension

A closed form $\eta \in {\mathrm{\Omega }}_{\mathrm{c}\mathrm{l}}^{n-1}(M;V)$ determines a de Rham class $[\eta ]$. If ${\eta }^{\mathrm{\prime }}=\eta +d\zeta$ is another closed representative, then its quotient class in ${\mathrm{\Omega }}^{n-1}\mathrm{/}d{\mathrm{\Omega }}^{n-2}$ equals that of $\eta$, so $a([{\eta }^{\mathrm{\prime }}])=a([\eta ])$. Thus $a_{\mathrm{c}\mathrm{l}}$ is well defined. Exactness of [F1] at the form quotient says its kernel consists of ${\mathrm{ch}}_{\mathrm{\Omega }}(E^{n-1})$; passing to de Rham classes identifies this with $\mathrm{ch}(E^{n-1})$. Hence the sequence is exact at $H_{\mathrm{d}\mathrm{R}}^{n-1}$.

If $y\in \ker (R,I)$, then $I(y)=0$, so [F1] gives $y=a([\eta ])$ for some form class $[\eta ]$. The curvature identity gives $d\eta =R(y)=0$, hence $\eta$ is closed and $y=a_{\mathrm{c}\mathrm{l}}([\eta ])$. Conversely, each closed form class maps to an element with $I=0$ and $R=0$, so $\ker (R,I)=\mathrm{im}(a_{\mathrm{c}\mathrm{l}})$. This proves exactness at ${\widehat{E}}^n(M)$.

Let $(\omega ,x)\in P^n(M)$. Surjectivity of $I$ in [F1] gives $y\in {\widehat{E}}^n(M)$ with $I(y)=x$. By commutativity of the differential-cohomology square, $[R(y){]}_{\mathrm{d}\mathrm{R}}=\mathrm{ch}(x)=[\omega {]}_{\mathrm{d}\mathrm{R}}$. Thus $\omega -R(y)=d\eta$ for some $\eta \in {\mathrm{\Omega }}^{n-1}(M;V)$. Then $y+a([\eta ])$ has underlying class $x$ and curvature $R(y)+d\eta =\omega$. Therefore $(R,I)$ is onto $P^n(M)$; the zero group at the right is exact by surjectivity.

If $M=\mathrm{\varnothing }$, the differential forms, topological cohomology groups and differential-extension groups on the empty object are zero, so the displayed sequence is the zero sequence and exact. For nonempty $M$, steps 1.1–2.1 prove all exactness assertions.