Borsuk-Ulam Theorem

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

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

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

$$\begin{aligned} \tau:C_n(X;\mathbb{Z}_2)&\rightarrow C_n(\tilde{X};\mathbb{Z}_2)\\ \sigma&\mapsto\tilde{\sigma}_1+\tilde{\sigma}_2 \end{aligned}$$

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

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

$$\bar{f}:\mathbb{R}P^n\rightarrow\mathbb{R}P^n$$

Consider the following diagram of chain group:

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

$$\widetilde{\bar{f}\circ\sigma_1}+\widetilde{\bar{f}\circ\sigma_2}=f\circ\tilde{\sigma}_1+f\circ\tilde{\sigma}_2.$$

Hence there is a long exact sequence of homology groups:

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

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

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

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

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

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