The Borsuk-Ulam Theorem

I give a proof of the Borsuk-Ulam Theorem which I claim is a simplified version of the proof given in Bredon [1], using chain complexes explicitly rather than homology. Of course this is a matter of taste, and the mathematical content is identical, but in my opinion this proof highlights precisely where and how the contradiction arises. 1 Statements and Preliminaries Theorem 1.1 (Borsuk-Ulam). Let f : S → R be a continuous map for n ∈ N. Then there exists a point x ∈ S such that f(x) = f(−x) ( cf. Theorem 20.2 of Bredon [1]). Lemma 1.2. Let φ : S → S be a continuous map which is equivariant with respect to the antipodal maps on S and S. Then n ≤ m. ( cf. Theorems 20.1, 20.6 of Bredon [1].) Proof of Theorem 1.1 assuming Lemma 1.2. Suppose not. That is suppose that f : S → R such that f(x) 6= f(−x) for all x ∈ S. Then define φ : S → S by: x 7→ f(x)− f(−x) ‖f(x)− f(−x)‖ . This satisfies φ(−x) = −φ(x) i.e. φ is continuous and equivariant with respect to the antipodal maps which contradicts Lemma 1.2 ⇒⇐. Prior to giving a proof of Lemma 1.2, we explain the main ingredient, namely a Z[Z2]-module chain complex of S, which arises by considering it as the universal covering space of RP. Proposition 1.3. There exists a cell decomposition of S with two cells in each dimension 0, . . . , r with Sr−1 ⊂ S embedded as the equator. There is a Z2 ∼= 〈T |T 2 = 1 〉 action corresponding to the antipodal map which transposes the two cells in each dimension. With respect to this decomposition the Z2-equivariant cellular chain complex W [0, r] := C∗(R̃P ;Z) = C∗(RP ;Z[Z2])