3

I want to prove the Borsuk-Ulam Theorem:

For every continuous map $f:S^2\rightarrow \mathbb{R}^2$ there exist a pair of antipodal points $x$ and $-x$ in $S^2$ with $f(x)=f(-x)$

I did google search and I got the following hint:enter image description here

My Efforts

I know there is quotient map $q:S^2\rightarrow RP^2$ and by property of quotient maps any continuous function from $S^2$ will factor through $q$, so I get a map from $\tilde{f}:\mathbb{R}P^2 \rightarrow \mathbb{R}^{2}$

I have inclusion map, $i:\mathbb{R}^2-\{0\} \rightarrow \mathbb{R}^2$. I have quotient map, $q': \mathbb{R}^2-\{0\}\rightarrow RP^1$. So I get a map $\tilde{i}:RP^1\rightarrow \mathbb{R}^2$

How should I proceed from here ? I am not able to construct a map from $RP^2$ to $RP^1$

For the next step, I think I have to use the fact that $\pi_1(RP^2)=\mathbb{Z}/2\mathbb{Z}$ and $\pi_1(RP^1)=\mathbb{Z}$

Community
  • 1
Tensor_Product
  • 1,483
  • It's not true that any continuous map factors through the quotient map. A map $f$ factors through the quotient if and only if it is constant on the fibers of the quotient map. – Alex Ortiz Sep 05 '17 at 04:06
  • @AOrtiz Am I going on right track? I am not able to see how a continuous map $f$ from $S^2$ to $\mathbb{R}^2$ induces a map from $RP^2$ to $RP^1$. – Tensor_Product Sep 05 '17 at 04:15

2 Answers2

3

I think that the following should work:

If there were some $f:S^2 \to \mathbb R^2$, $f(x) \neq f(-x)$, then $f(x)-f(-x) \neq 0$ for any $x \in S^2$.

This gives a function $\phi:S^2 \to S^1$ given by $x \mapsto \frac{f(x)-f(-x)}{\|f(x)-f(-x)\|}$.

Notice that $\phi(x) = - \phi(-x)$ for all $x \in S^2$.

Compose this map with the double covering self-map $\rho : S^1 \to S^1$ which makes the identification $\rho(z)=\rho(-z)$ for all $z \in S^1$.

Now, $\rho \circ \phi(x)=\rho \circ \phi(-x)$.

By the universal property of the quotient map, this induces a map $$\widetilde{\rho \circ \phi}: \mathbb RP^2 \to (S^1 \cong \mathbb RP^1) $$ To get the final contradiction, this map induces a nontrivial homomorphism from the group $\mathbb Z / 2 \mathbb Z = \pi_1(\mathbb RP^2)$ to the group $\mathbb Z = \pi_1(\mathbb S^1)$, because a closed path in $\mathbb RP^2$ representing a generator of $\pi_1(\mathbb RP^2)$ lifts to a path in $S^2$ with antipodal endpoints, which maps to a path in $S^1$ with antipodal endpoints, which projects via the map $\rho$ to a closed path in $S^1$ representing an odd element of $\pi_1(S^1) = \mathbb Z$.

Lee Mosher
  • 135,265
Andres Mejia
  • 21,467
1

Theorem: If $m > 1$, there exists no antipodal map $g:\mathbb{S}^m \rightarrow \mathbb{S}^1$ (page 124).

If no such x exists, define $g:\mathbb{S}^2 \rightarrow \mathbb{S}^1$ by
$$g(x) = \frac{f(x) - f( -x)}{\|{f(x) - f( -x) \|}}$$ Clearly $g$ is an antipodal map, and this contradicts the theorem.

Mojtaba
  • 690