singular homology of a differential manifold - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T04:17:43Z http://mathoverflow.net/feeds/question/20891 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20891/singular-homology-of-a-differential-manifold singular homology of a differential manifold vu viet 2010-04-10T03:15:51Z 2012-06-14T15:35:35Z <p>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 $\sigma:\Delta -> M$. </p> <p>And we construct a chain complex in the same way we construct the chain complex of singular homology, we gain its homology group.</p> <p>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. </p> http://mathoverflow.net/questions/20891/singular-homology-of-a-differential-manifold/20894#20894 Answer by Rafe Mazzeo for singular homology of a differential manifold Rafe Mazzeo 2010-04-10T03:38:10Z 2010-04-10T03:38:10Z <p>You are asking: can one compute the singular homology using chains which are {\it smooth} mappings rather than just continuous ones.</p> <p>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.</p> <p>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).</p> http://mathoverflow.net/questions/20891/singular-homology-of-a-differential-manifold/20916#20916 Answer by babubba for singular homology of a differential manifold babubba 2010-04-10T11:48:11Z 2010-04-10T11:48:11Z <p>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.</p> http://mathoverflow.net/questions/20891/singular-homology-of-a-differential-manifold/20926#20926 Answer by algori for singular homology of a differential manifold algori 2010-04-10T15:18:50Z 2010-04-10T15:18:50Z <p>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.</p> http://mathoverflow.net/questions/20891/singular-homology-of-a-differential-manifold/99618#99618 Answer by vu viet for singular homology of a differential manifold vu viet 2012-06-14T15:35:35Z 2012-06-14T15:35:35Z <p>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).</p>