proof of Borsuk-Ulam theorem

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 f:SnSn has odd degree.

Proof: We go by inductionMathworldPlanetmath on n. Consider the pair (Sn,A) where A is the equatorial sphere. f defines a map


. By cellular approximation, this may be assumed to take the hyperplane at infinity (the n-1-cell of the standard cell structureMathworldPlanetmath on Pn) to itself. Since whether a map lifts to a covering depends only on its homotopy class, f is homotopicMathworldPlanetmath to an odd map taking A to itself. We may assume that f is such a map.

The map f gives us a morphismMathworldPlanetmathPlanetmath of the long exact sequences:


Clearly, the map f|A is odd, so by the induction hypothesis, f|A has odd degree. Note that a map has odd degree if and only if f*:Hn(Sn;2)Hn(Sn,2) is an isomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath. Thus


is an isomorphism. By the commutativity of the diagram, the map


is not trivial. I claim it is an isomorphism. Hn(Sn,A;2) is generated by cycles [R+] and [R-] which are the fundamental classesMathworldPlanetmath of the upper and lower hemispheres, and the antipodal map exchanges these. Both of these map to the fundamental class of A, [A]Hn-1(A;2). By the commutativity of the diagram, (f*([R±]))=f*(([R±]))=f*([A])=[A]. Thus f*([R+])=[R±] and f*([R-])=[R] since f commutes with the antipodal map. Thus f* is an isomorphism on Hn(Sn,A;2). Since Hn(A,2)=0, by the exactness of the sequencePlanetmathPlanetmath i:Hn(Sn;2)Hn(Sn,A;2) is injectivePlanetmathPlanetmath, and so by the commutativity of the diagram (or equivalently by the 5-lemma) f*:Hn(Sn;2)Hn(Sn;2) is an isomorphism. Thus f has odd degree.

The other statement of the Borsuk-Ulam theorem is:

There is no odd map SnSn-1.

Proof: If f where such a map, consider f restricted to the equator A of Sn. This is an odd map from Sn-1 to Sn-1 and thus has odd degree. But the map


factors through Hn-1(Sn)=0, and so must be zero. Thus f|A has degree 0, a contradictionMathworldPlanetmathPlanetmath.

Title proof of Borsuk-Ulam theorem
Canonical name ProofOfBorsukUlamTheorem
Date of creation 2013-03-22 13:10:33
Last modified on 2013-03-22 13:10:33
Owner bwebste (988)
Last modified by bwebste (988)
Numerical id 5
Author bwebste (988)
Entry type Proof
Classification msc 54C99