Questions: The Borsuk-Ulam Theorem

Question 1

The Borsuk-Ulam theorem for n = 1 says: every continuous map f: S^1 → R has f(x) = f(-x) for some x. This is a consequence of which elementary theorem?

• A. The mean value theorem
• B. The intermediate value theorem, applied to g(x) = f(x) - f(-x)
• C. The Bolzano-Weierstrass theorem
• D. The extreme value theorem

Answer: (B)

Define g(x) = f(x) - f(-x). Then g(-x) = f(-x) - f(x) = -g(x), so g is an odd function on S^1. Since g is continuous and g(-x) = -g(x), if g(x_0) > 0 for some x_0, then g(-x_0) < 0, and by the intermediate value theorem (applied to g along any path from x_0 to -x_0), there exists a point where g = 0, meaning f(x) = f(-x). The higher-dimensional version requires algebraic topology because the intermediate value theorem does not generalize directly.

Question 2

The Borsuk-Ulam theorem implies that no subset of R^n is homeomorphic to S^n.

• T. True
• F. False

Answer: True

If S^n embedded in R^n, the inclusion would give a continuous injection i: S^n → R^n. But Borsuk-Ulam says any continuous f: S^n → R^n must satisfy f(x) = f(-x) for some x ≠ -x. An injection cannot satisfy this (it maps distinct points to distinct values). Therefore no continuous injection S^n → R^n exists. This is a generalization of the invariance of domain: S^n cannot be 'flattened' into R^n without some pair of antipodal points colliding.

Question 3

The ham sandwich theorem states: given n measurable sets in R^n, there exists a single hyperplane that simultaneously bisects all n sets. This follows from Borsuk-Ulam.

• T. True
• F. False

Answer: True

A hyperplane in R^n is determined by a unit normal vector v ∈ S^{n-1} and an offset d ∈ R. For each v, there is a unique offset d_i(v) that bisects the i-th set. Define f: S^{n-1} → R^{n-1} by f(v) = (d_1(v) - d_n(v), ..., d_{n-1}(v) - d_n(v)). By Borsuk-Ulam applied to f: S^{n-1} → R^{n-1}, there exists v with f(v) = f(-v). Since d_i(-v) = -d_i(v) (reflecting the normal reverses the hyperplane), this gives d_i(v) - d_n(v) = -(d_i(v) - d_n(v)) for each i, forcing all differences to be zero. A careful argument then gives a hyperplane bisecting all n sets.

Question 4

State the equivalent 'no equivariant map' formulation of the Borsuk-Ulam theorem and explain why it is equivalent to the antipodal coincidence version.

Model answer: Equivalent formulation: there is no continuous map g: S^n → S^{n-1} satisfying g(-x) = -g(x) (equivariant with respect to the antipodal action). Equivalence: if f: S^n → R^n had f(x) ≠ f(-x) for all x, then g(x) = (f(x) - f(-x))/|f(x) - f(-x)| would be a continuous equivariant map S^n → S^{n-1}. Conversely, an equivariant map g: S^n → S^{n-1} composed with the inclusion S^{n-1} ↪ R^n would give a map with no antipodal coincidence. So the two versions are contrapositives.

The equivariant formulation is often more useful for proofs. The nonexistence of equivariant maps S^n → S^{n-1} can be proved using the homology of real projective spaces (quotienting by the Z/2Z antipodal action) or using degree theory: an equivariant map would induce a map RP^n → RP^{n-1} with specific properties on homology that lead to a contradiction.