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:Sn→Sn has odd degree.
Proof: We go by induction on n. Consider the pair (Sn,A) where A is the equatorial sphere.
f defines a map
˜f:ℝPn→ℝPn |
. By cellular approximation, this may be
assumed to take the hyperplane at infinity (the n-1-cell of the standard cell structure on
ℝPn) to itself. Since whether a map lifts to a covering depends only on its homotopy
class, f is homotopic
to an odd map taking A to itself. We may assume that f is such a map.
The map f gives us a morphism of the long exact sequences:
Hn(A;ℤ2)i→Hn(Sn;ℤ2)j→Hn(Sn,A;ℤ2)∂→Hn-1(A;ℤ2)i→Hn-1(Sn,A;ℤ2)f*↓f*↓f*↓f*↓f*↓Hn(A;ℤ2)i→Hn(Sn;ℤ2)j→Hn(Sn,A;ℤ2)∂→Hn-1(A;ℤ2)i→Hn-1(Sn,A;ℤ2) |
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;ℤ2)→Hn(Sn,ℤ2) is an
isomorphism. Thus
f*:Hn-1(A;ℤ2)→Hn-1(A;ℤ2) |
is an isomorphism. By the commutativity of the diagram, the map
f*:Hn(Sn,A;ℤ2)→Hn(Sn,A;ℤ2) |
is
not trivial. I claim it is an isomorphism. Hn(Sn,A;ℤ2) is generated by cycles [R+] and
[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,
[A]∈Hn-1(A;ℤ2). 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;ℤ2). Since Hn(A,ℤ2)=0, by the exactness of the sequence
i:Hn(Sn;ℤ2)→Hn(Sn,A;ℤ2) is injective
, and so by the commutativity of the diagram (or equivalently
by the 5-lemma) f*:Hn(Sn;ℤ2)→Hn(Sn;ℤ2) is an isomorphism. Thus
f has odd degree.
The other statement of the Borsuk-Ulam theorem is:
There is no odd map Sn→Sn-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
f*Hn-1(A)→Hn-1(Sn-1) |
factors through Hn-1(Sn)=0, and so must be zero. Thus f|A has degree 0, a
contradiction.
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 |