Let $M$ be a differentiable manifold, $\Delta$ the closed simplex $[p_0, p_1,...,p_k]$. A differential singular $k$-simplex $\sigma$ of $M$ is a smooth mapping $\sigma:\Delta \to M$.

And we construct a chain complex in the same way we construct the chain complex of singular homology, we gain its homology group.

My question is why this homology group equals the singular homology group? I have tried finding in many books but there is no answer of this question.

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    $\begingroup$ Probably I'm missing something, but your question doesn't seem to make sense unless it is taken as "why is singular homology equal to singular homology?" in which case it is not really a question. $\endgroup$ Apr 10 '10 at 3:29
  • $\begingroup$ By a smooth mapping $\sigma:\Delta\to m$ do you mean $\sigma$ is smooth in the interior of $\Delta$? In any case, any smooth map can be approximated by continuous maps and therefore, the inclusion of smooth singular chains into the singular chains is a quasi-isomorphism. Doesn't this prove what you need? $\endgroup$ Apr 10 '10 at 3:43
  • $\begingroup$ By a smooth mapping $\sigma: \Delta -> M $ it means that $\sigma$ is smooth in a open neighborhood of $\Delta$ in $R^{k}$ (R is the set of real numbers). We need the equality of this two homology groups to prove the De Rham theorem: The De Rham cohomology of M is equal to the singular cohomology of M with coefficient $R$. $\endgroup$
    – vu viet
    Apr 10 '10 at 10:08
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    $\begingroup$ You could just give a crude, bare-hands proof... For surjectivity on homology, take a continuous singular cycle. It's homologous to its iterated barycentric subdivisions, whose simplices will eventually map into Euclidean charts. By induction on the number of simplices, and a suitable local smoothing procedure rel boundary, we can find a homology to a nearby smooth cycle. For injectivity, apply the same thing to a bounding chain. (The answers below give much more elegant methods...) $\endgroup$
    – Tim Perutz
    Apr 10 '10 at 15:28
  • $\begingroup$ This was proved by Eilenberg in Singular homology in differentiable manifolds, Ann. of Math. (2) 48 (1947) 670–681. The natural inclusion $S_*^\infty(M) \longrightarrow S_*(M)$, where $M$ is a differentiable manifold and $S_*^\infty$ and $S_*$ are the functors of differentiable and continuous chains, respectively, is a homotopy equivalence. $\endgroup$ Jun 7 at 6:44

This depends on what exactly is a smooth mapping from a simplex to the manifold. The standard definition is that the mapping of a non-open subset $X$ of $\mathbf{R}^n$ to a manifold is smooth iff it can be extended to a smooth mapping of an open neighborhood of $X$. With this definition the comparison theorem is true and a very detailed proof can be found e.g. in the book "Introduction to smooth manifolds" by Lee, chapter 16.


You are asking: can one compute the singular homology using chains which are {\it smooth} mappings rather than just continuous ones.

The answer is that one can. The reason is that there is a chain homotopy between the complex of smooth chains and the complex of continuous chains. In fact, I think a standard smoothing mollification does the trick.

This kind of thing is standard (but rarely written down) for standard situations, e.g. compact smooth manifolds, but is more subtle (and interesting) when one looks at analogous constructions for intersection homology on stratified spaces (cf. the work of Brasselet and collaborators).

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    $\begingroup$ you said "The reason is that there is a chain homotopy between the complex of smooth chains and the complex of continuous chains". Could you explain me the fact more clear or show me a book that prove the fact,please? In Griffith's,in the section "The de Rham theorem",p45, he wrote "by a foundational result from differential topology,..." this two homology group are equal. $\endgroup$
    – vu viet
    Apr 10 '10 at 10:21

For singular and singular differentiable cohomology there's an explanation of why they yield the same result in chapter five of Warner's "Foundations of Differentiable Manifolds and Lie Groups". The reason is that the ('sheafification' of the) two complexes both give you a resolution of the constant sheaf and so the cohomologies must be the same.

  • $\begingroup$ For R is a field, we see that k-th singular homology group (in fact it is a vector space) with coefficient R equals the dual space of k-th singular cohomology group with coefficient R. We have the same conclusion for singular differentiable cohomology. Hence, singular and singular differentiable homology are equal. However, I still want to find a direct answer(use differential topology, not use sheaf theory). $\endgroup$
    – vu viet
    Apr 10 '10 at 13:06
  • $\begingroup$ I second algori's answer. I forgot all about the great book by Lee. Theorem 16.6 on page 417 of the edition I have is probably what you want. The inclusion from differentiable chains to continuous chains gives a maps of complexes and a homotopy is given using the Whitney approximation theorem. "The key to the proof is a systematic construction of a homotopy from each continuous simplex to a smooth one, in a way that respects the restriction to each boundary face of [the standard p-simplex]". As usual Lee gives all the details you'll ever want. $\endgroup$
    – babubba
    Apr 10 '10 at 18:22

ons can find a proof for my quesion in "algebraic topology: a first course" by W. Fulton. The proof bases on the fact that in some simple sample like a ball, these two are identical, and the general case would be implied by applying the Mayer-Vertoris principle (in fact, ones can see a manifold as a set of balls that overlap each other, and that principle allows to compute homology groups by reducing on each ball).


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