Borsuk-Ulam Theorem

This post is mainly based on Chapter 2B of Algebraic Topology written by Allen Hatcher.

Prop. An odd map f:SnSnf:S^n\rightarrow S^n, satisfying f(x)=f(x)f(x)=-f(-x) for all xSnx\in S^n, must have odd degree.

Proof. Consider a two-fold covering space p:X~Xp:\tilde{X}\rightarrow X. Define

τ:Cn(X;Z2)Cn(X~;Z2)σσ~1+σ~2\begin{aligned} \tau:C_n(X;\mathbb{Z}_2)&\rightarrow C_n(\tilde{X};\mathbb{Z}_2)\\ \sigma&\mapsto\tilde{\sigma}_1+\tilde{\sigma}_2 \end{aligned}

where σ~1\tilde{\sigma}_1 and σ~2\tilde{\sigma}_2 are two lifting of σ\sigma. Then there is an exact sequence of homology

Since p:SnRPnp:S^n\rightarrow\mathbb{R}P^n is a two-fold covering map, then we can apply the above transfer diagram to SnS^n and RPn\mathbb{R}P^n. Since f:SnSnf:S^n\rightarrow S^n is odd, then ff induces a well-defined map


Consider the following diagram of chain group:

Here the right-hand square is commutative, as pf=fˉpp\circ f=\bar{f}\circ p, and the left-hand square is commutative, as


Hence there is a long exact sequence of homology groups:

Therefore, by induction of ii, fˉ\bar{f} induces an isomorphism of homology groups on each 1in11\leq i\leq n-1, namely

fˉ:Hi(RPn;Z2)Hi(RPn;Z2)\bar{f}_*:H_i(\mathbb{R}P^n;\mathbb{Z}_2)\rightarrow H_i(\mathbb{R}P^n;\mathbb{Z}_2)

is an isomorphism. As for the nn-th homology, notice that the following diagram:

Therefore, the induced homomorphism f:Hn(Sn;Z2)Hn(Sn;Z2)f_*:H_n(S^n;\mathbb{Z}_2)\rightarrow H_n(S^n;\mathbb{Z}_2) is an isomorphism, which is a the multiplication of degf(mod2)\deg f\pmod{2}. Hence degf\deg f is odd. \square

Borsuk-Ulam theorem appears as a corollary of the above result.

Cor. (Borsuk-Ulam) For every map f:SnRnf:S^n\rightarrow\mathbb{R}^n, there exists a point xSnx\in S^n, such that f(x)=f(x)f(x)=f(-x).