singular homology of a differential manifold - MathOverflow

singular homology of a differential manifold

2010-04-10T03:15:51Z

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

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.

Answer by Rafe Mazzeo for singular homology of a differential manifold

2010-04-10T03:38:10Z

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).

Answer by babubba for singular homology of a differential manifold

2010-04-10T11:48:11Z

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.

Answer by algori for singular homology of a differential manifold

2010-04-10T15:18:50Z

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.

Answer by vu viet for singular homology of a differential manifold

2012-06-14T15:35:35Z

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).