MAS61015 Algebraic Topology

24. Borsuk-Ulam and related results

Definition 24.1.

A continuous map f:SnSm is odd (or antipodal) if f(-x)=-f(x) for all xSn.

Example 24.2.

If nm then it is easy to produce examples of odd continuous maps f:SnSm. Most obviously, we can just define

Remark 24.3.

If f:SnSm is odd, then we have a well-defined map f¯:PnPm given by f¯([x])=[f(x)].

Proposition 24.4.

Suppose that f:SnSn is continuous and odd. Then the induced map f*:Hn(Sn;/2)Hn(Sn;/2) is the identity.


As f is odd we have a well-defined map f¯:PnPn given by f¯([x])=[f(x)]. This satisfies pf=f¯p:SnPn, so the right-hand square below commutes:

We claim that the left hand square commutes as well. To see this, define χ:SnSn by χ(x)=-x, so fχ=χf and pχ=p. Consider a continuous map u:ΔkPn. Choose a lift v:ΔkSn. The other lift is then χv, so τ(u)=v+(χv), so


Here fv and χfv are the two lifts of the map pfv=f¯pv=f¯u, so we see that (fv)+(χfv)=τ(f¯#(u)). This shows that f#τ=τf¯#, so the left hand square commutes as claimed.

As the diagram commutes, we see that the maps f* and f¯* are compatible with the maps in the exact sequence obtained in Lemma 23.14. In particular, we have commutative squares

for 1in. It is clear that f¯* gives the identity on H0(Pn;/2), and we have seen that all of the maps Δ are isomorphisms, so it follows inductively that f¯* gives the identity on Hi(Pn;/2) for 1in. We also have a commutative square

We have seen that τ* is an isomorphism and f¯* is the identity so f* is also the identity, as claimed. ∎

Theorem 24.5 (Borsuk-Ulam).

If n>m, then there are no odd continuous maps from Sn to Sm.


Suppose that f:SnSm is odd and continuous. Define i:SmSn by


It is clear that i is odd, so the composite fi:SmSm is odd, so the induced map (fi)*:Hm(Sm;/2)Hm(Sm;/2) must be the identity by Proposition 24.4. In particular, (fi)* is nonzero.

On the other hand, we can define h:[0,1]×SmSn by


(Using the fact that i(x) and em+1 are orthogonal, we see that this does indeed lie in Sn.) This gives a homotopy between i and a constant map, which implies that fi is also homotopic to a constant map, so (fi)*=0. This contradiction shows that no such map f can exist. ∎

Corollary 24.6.

Let g:Snm be a continuous map, with 0<mn. Then there is a point xSn with g(x)=g(-x).


Suppose (for a contradiction) that no such point exists, so g(x)-g(-x) is always nonzero. We can then define f:SnSn-1 by f(x)=(g(x)-g(-x))/g(x)-g(-x). It is easy to check that this is continuous and antipodal, which contradicts Theorem 24.5, as required. ∎

Example 24.7.

There are two opposite points on the Earth’s surface that have the same temperature and also the same atmospheric pressure, as we see by considering the map f:S22 given by

f(a)=( temperature at a, pressure at a).
Theorem 24.8 (Sandwich Slicing Theorem).

Let A1, A2 and A3 be three reasonable subsets of 3. Then there is a plane P3 such that for each set Ai, half of the volume lies on one side of P, and half of the volume lies on the other side.

Example 24.9.

We could have a sandwich, with the top slice of bread filling the set A1, and cheese filling the set A2, and the bottom slice of bread filling A3. The theorem then says that we can make a single straight cut with a knife to share all three components equally.

Example 24.10.

Suppose that the sets Ai are solid balls, with centres ai in general position. Then there is a unique possible choice for P, namely the plane passing through a1, a2 and a3.

We will not be very rigorous about what “reasonable” means, but we will make some comments here and in the body of the proof. To start with, each set Ai should be bounded (which implies that the volume is finite) and the volume should also not be zero.

Proof of Theorem 24.8.

For any unit vector u=(u0,u1,u2,u3)S3 and v=(v1,v2,v3)3 we put m(u,v)=u0+u1v1+u2v2+u3v3, so m(-u,v)=-m(u,v). We then put

P(u) ={v3|m(u,v)=0}
H(u) ={v3|m(u,v)>0},

so P(u) is a plane and H(u) is the half-space on one side of that plane. Note that the plane P(-u) is the same as P(u), and H(-u) is the half-space on the opposite side to H(u).

We now define g:S33 by


It is not too hard to check that this is continuous. By Corollary 24.6, there exists uS3 with g(-u)=g(u), which means that vol(AiH(u))=vol(AiH(-u)) for i=1,2,3. In other words, the plane P(u) bisects each of the sets Ai. ∎