Smooth singular chain and cochain complexes

For a smooth manifold $M$, possibly with boundary, let $C_k^{\mathrm{\infty }}(M;\mathbb{R})={\mathbb{R}}^{(S_k^{\mathrm{\infty }}(M))}(k\ge 0),C_k^{\mathrm{\infty }}(M;\mathbb{R})=0(k<0),$ using Smooth singular simplex. It is the subspace of Real singular chain complex spanned by the smooth simplices, with the same signed face differential.

This is a subcomplex: an affine face map ${\delta }_i:A^{k-1}\to A^k$ has the open inverse image ${\delta }_i^{-1}(O)$ containing ${\mathrm{\Delta }}^{k-1}$. Composing $\bar{\sigma }$ with this affine map on that inverse image extends $\sigma {\delta }_i$ smoothly into $M$. Thus every face is smooth, and the already proved signed cancellation gives ${\mathrm{\partial }}^2=0$ on this subspace.

Define the smooth singular cochain complex by $C_{\mathrm{\infty }}^k(M;\mathbb{R})={\mathrm{Hom}}_{\mathbb{R}}(C_k^{\mathrm{\infty }}(M;\mathbb{R}),\mathbb{R}),\delta \phi =\phi \mathrm{\partial }\mathrm{.}$ As for Real singular cochain complex, these are arbitrary real functions on the supplied smooth-simplex basis, evaluated by finite sums. Precomposition gives ${\delta }^2\phi =\phi {\mathrm{\partial }}^2=0$. Define $H_k^{\mathrm{\infty }}(M;\mathbb{R})=\ker {\mathrm{\partial }}_k\mathrm{/}\mathrm{im}{\mathrm{\partial }}_{k+1}$ and $H_{\mathrm{\infty }}^k(M;\mathbb{R})=\ker {\delta }^k\mathrm{/}\mathrm{im}{\delta }^{k-1}$; both quotients are licensed by square-zero.

Negative groups vanish and the degree-zero boundary is zero. On a point all simplices are smooth, so the complexes are the ordinary unnormalized point complexes; no constant simplices are discarded. For the empty manifold all groups are zero. The specified subspaces and duals require no choice of extensions or bases.