Smooth singular chains compute singular homology

First let $M$ be boundaryless. From any finite list of chains take their finite supports and all iterated faces, identifying equal parametrized maps in each degree. This is a finite face-closed set. Assign each zero-simplex itself and its constant homotopy. Inductively the already assigned face homotopies of a simplex agree on intersections, because the affine face identities produce the same lower-dimensional map. Apply [F3] to give the simplex a smooth replacement and a homotopy with these exact faces. If the simplex was originally smooth, all its faces were smooth and already fixed, so use the fixed clause of [F3]. Only finitely many choices are made at each of finitely many dimensions.

Now let $M$ have boundary and let $K\subseteq M$ be compact. Its boundary part is compact by [F5]. For each eligible boundary chart centred at $q=(z_0,0)$, choose $r>0$ so its closed half-ball of radius $3r$ is inside the chart, and put $\chi (z,s)=1-s_0((\mathrm{\parallel }(z,s)-q{\mathrm{\parallel }}^2-r^2)\mathrm{/}(3r^2))$ with step $s_0$ from [F6]. Extend $\chi$ by zero outside the chart. It is smooth, equals one on the radius-$r$ half-ball, and has compact support in the radius-$2r$ half-ball. The family of all such smaller half-balls covers $K\cap \mathrm{\partial }M$; take a finite subcover. No chart is selected simultaneously for every boundary point.

Replacements respect faces exactly, hence give a chain map $Q$ on the finite graded spans in question. The homotopy prisms give $\mathrm{\partial }P+P\mathrm{\partial }=Q-1$: apply the oriented-prism calculation [F4] to each simplex homotopy; the side terms are the already assigned face prisms, so cancel with $P\mathrm{\partial }$. For a continuous cycle $z$, this gives $Qz-z=\mathrm{\partial }Pz$ with $Qz$ smooth, proving surjectivity. If a smooth cycle $z$ bounds a continuous $b$, use the supports of both $b$ and $z$ in step 1.1. Then $Qz=z$ and $\mathrm{\partial }Qb=Q\mathrm{\partial }b=z$, proving injectivity. Equal nonsmooth faces that cancel in $\mathrm{\partial }b$ have the same replacement, so the equality survives all cancellations. No assertion that a constant unnormalized prism vanishes is needed.

For one chosen chart take $0<\epsilon <r\mathrm{/}2$ and define $P_t(z,s)=(z,s+\epsilon s_0(t)\chi (z,s))$ there, and the identity elsewhere, for every real $t$. The displacement is nonnegative and less than $r\mathrm{/}2$; where it is nonzero the original point is within radius $2r$, so the image remains within radius $5r\mathrm{/}2<3r$. Thus the map is well-defined into $M$. It is jointly smooth across the chart edge because the support is compactly inside the chart. It preserves interior points, equals the identity for $t\le 0$, and at $t=1$ moves inward every boundary point where $\chi >0$. Compose the finitely many maps at the same time to obtain $J_t$ and set $j=J_1$. If an original boundary point of $K$ has not yet moved, all earlier maps have fixed it exactly; eventually its covering bump moves it inward. Once interior, it remains interior. Therefore $j(K)\subseteq M^{\circ }$.

For a smooth simplex with extension $\bar{\tau }:O\to M$, $(x,t)\mapsto J_t(\bar{\tau }(x))$ is smooth into $M$ on $O\times \mathbb{R}$. Hence [F2] makes the prism $P_J$ preserve smooth chains. The ordinary and smooth identities are both $\mathrm{\partial }{P}_J+P_J\mathrm{\partial }=j_{\mathrm{\#}}-1$ by [F4]. If $j\tau$ has image in $M^{\circ }$, restrict its extension to the open inverse image of $M^{\circ }$; it is then a strict smooth simplex into that boundaryless manifold.

For surjectivity let $z$ be a continuous cycle, and let $K$ be the finite union of its simplex images. Step 2.2 gives $j_{\mathrm{\#}}z$ in $M^{\circ }$, homologous to $z$ in $M$ by step 3.1. Boundaryless surjectivity from step 2.1 makes $j_{\mathrm{\#}}z$ homologous in $M^{\circ }$ to a smooth cycle, which is also smooth in $M$. For injectivity let a smooth cycle $z$ satisfy $z=\mathrm{\partial }b$ with $b$ continuous, and include both supports in $K$. Then $j_{\mathrm{\#}}z$ is smooth in $M^{\circ }$ by step 3.1 and bounds $j_{\mathrm{\#}}b$ there. Boundaryless injectivity supplies smooth $c$ in $M^{\circ }$ with $\mathrm{\partial }c=j_{\mathrm{\#}}z$. The smooth prism gives $\mathrm{\partial }(c-P_{J}z)=j_{\mathrm{\#}}z-(j_{\mathrm{\#}}z-z)=z$.

The actual inclusion of complexes commutes with postcomposition by every smooth map, so its induced isomorphism is natural; no naturality of the chosen finite smoothings or pushes is claimed. For an empty compact boundary part take no chart maps and $J=1$. Empty chains have empty support, and all negative chain groups are zero. Degree-zero cycles and one-point manifolds are covered by steps 1.1–2.1; zero-dimensional manifolds have empty boundary. Constant and repeated simplices are retained. All boundary pushes are finite; the sole countable-choice cost is that of [F3].