Geometry Realization of Homology Class - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T00:02:30Z http://mathoverflow.net/feeds/question/110103 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110103/geometry-realization-of-homology-class Geometry Realization of Homology Class Siqi He 2012-10-19T16:31:14Z 2012-10-19T17:12:44Z <p>Hello! My question is about the realization of homology class.</p> <p>The definition of the realizaion of homology class is: for manifold M and a homology class $z\in H_k(M)$, k is an integer. If we find a k-dimensional manifold N and a map $f:N \rightarrow M$ such that $f_* [N]=z$, $[N]$ is the fundamental class, then we call the homology class $z$ can be realized.</p> <p>For this problem, Thom has the following theorem:</p> <p>Thom[1954] For every manifold M, consider a interger coefficient homology class $z\in H_*(M)$, that there exist a interger $l$ and $lz$ can be realized.</p> <p>My Question is Simple: Why we should add this interger $l$?</p> <p>Thom's original paper is written by French and I cann't understand it. Recently, I am reading a paper by A.Gaifullin:Combinatorial Realisation of Cycles and Small Covers and the result is related to Thom's paper.</p> http://mathoverflow.net/questions/110103/geometry-realization-of-homology-class/110106#110106 Answer by Igor Rivin for Geometry Realization of Homology Class Igor Rivin 2012-10-19T17:03:06Z 2012-10-19T17:12:44Z <p>Duplicate of <a href="http://mathoverflow.net/questions/1489/cohomology-and-fundamental-classes" rel="nofollow">this question</a>, which has a very good answer by Eric Wofsey.</p>