proof of Borsuk-Ulam theorem
Every odd (that is, antipode-preserving) map has odd degree.
Proof: We go by induction on . Consider the pair where is the equatorial sphere. defines a map
. By cellular approximation, this may be assumed to take the hyperplane at infinity (the -cell of the standard cell structure on ) to itself. Since whether a map lifts to a covering depends only on its homotopy class, is homotopic to an odd map taking to itself. We may assume that is such a map.
is an isomorphism. By the commutativity of the diagram, the map
is not trivial. I claim it is an isomorphism. is generated by cycles and which are the fundamental classes of the upper and lower hemispheres, and the antipodal map exchanges these. Both of these map to the fundamental class of , . By the commutativity of the diagram, . Thus and since commutes with the antipodal map. Thus is an isomorphism on . Since , by the exactness of the sequence is injective, and so by the commutativity of the diagram (or equivalently by the -lemma) is an isomorphism. Thus has odd degree.
The other statement of the Borsuk-Ulam theorem is:
There is no odd map .
Proof: If where such a map, consider restricted to the equator of . This is an odd map from to and thus has odd degree. But the map
factors through , and so must be zero. Thus has degree 0, a contradiction.
|Title||proof of Borsuk-Ulam theorem|
|Date of creation||2013-03-22 13:10:33|
|Last modified on||2013-03-22 13:10:33|
|Last modified by||bwebste (988)|