|
|
|
|
![[parent]](http://images.planetmath.org:8080/images/uparrow.png)
proof of Borsuk-Ulam theorem
|
(Proof)
|
|
|
Proof of the Borsuk-Ulam theorem: I'm going to prove a stronger statement than the one given in the statement of the Borsak-Ulam theorem here, which is:
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.
The map  gives us a morphism of the long exact sequences:
Clearly, the map  is odd, so by the induction hypothesis,  has odd degree. Note that a map has odd degree if and only if
 is an isomorphism. Thus
is an isomorphism. By the commutativity of the diagram, the map
is not trivial. I claim it is an isomorphism.
 is generated by cycles ![$ [R^+]$](http://images.planetmath.org:8080/cache/objects/3621/l2h/img20.png "$ [R^+]$") and ![$ [R^-]$](http://images.planetmath.org:8080/cache/objects/3621/l2h/img21.png "$ [R^-]$") which are the fundamental classes of the upper and lower hemispheres, and the antipodal map exchanges these. Both of these map to the fundamental class of  ,
![$ [A]\in H_{n-1}(A;\mathbb{Z}_2)$](http://images.planetmath.org:8080/cache/objects/3621/l2h/img23.png "$ [A]\in H_{n-1}(A;\mathbb{Z}_2)$") . By the commutativity of the diagram,
![$ \partial(f^*([R^\pm]))=f^*(\partial([R^\pm]))=f^*([A])=[A]$](http://images.planetmath.org:8080/cache/objects/3621/l2h/img24.png "$ \partial(f^*([R^\pm]))=f^*(\partial([R^\pm]))=f^*([A])=[A]$") . Thus
![$ f^*([R^+])=[R^\pm]$](http://images.planetmath.org:8080/cache/objects/3621/l2h/img25.png "$ f^*([R^+])=[R^\pm]$") and
![$ f^*([R^-]) =[R^\mp]$](http://images.planetmath.org:8080/cache/objects/3621/l2h/img26.png "$ f^*([R^-]) =[R^\mp]$") 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.
|
"proof of Borsuk-Ulam theorem" is owned by bwebste.
|
|
(view preamble)
Cross-references: contradiction, factors, restricted, injective, sequence, antipodal map, fundamental classes, cycles, generated by, commutativity, isomorphism, induction hypothesis, exact sequences, morphism, homotopic, class, homotopy, covering, lifts, structure, cell, infinity, hyperplane, approximation, sphere, induction, degree, map, odd, Borsuk-Ulam theorem, proof
This is version 2 of proof of Borsuk-Ulam theorem, born on 2002-11-23, modified 2002-12-05.
Object id is 3621, canonical name is ProofOfBorsukUlamTheorem.
Accessed 5049 times total.
Classification:
| AMS MSC: | 54C99 (General topology :: Maps and general types of spaces defined by maps :: Miscellaneous) |
|
|
|
|
|
|
Pending Errata and Addenda
|
|
|
|
|
|
|
|
|
|
|
