Counterexample of Borsuk-Ulam theorem (Discontinuous Functions)

Question

I'm looking to build a function $f:S^2 \to \mathbb R^2$ such that $f(x)\neq f(−x)$ for all $x\in S^2$.

By Borsuk-Ulam Theorem, this function must be discontinuous. I was trying to build a not too complicated function, but I always encountered a problem.

I appreciate any help.

Answer 1

In that case you could try with: $$(x_1,x_2,x_3)\in\mathbb{S}^1\longmapsto f(x_1,x_2,x_3):=\begin{cases}(x_1,x_2) &\text{ if } (x_1,x_2,x_3)\neq (0,0,1) \\ (2,2) &\text{ if } (x_1,x_2,x_3)=(0,0,1) \end{cases}$$

Then obviously $f(x_1,x_2,x_3)=-f(-(x_1,x_2,x_3))$ for all $(x_1,x_2,x_3)\in\mathbb{S}^1\setminus\{(0,0,1),(0,0,-1)\}$. Also, $f(0,0,1)=(2,2)\neq(0,0)=f(0,0,-1)$.

Answer 2

$S^2=\{(\cos u,\sin u \cos v,\sin u \sin v):u\in [0,2\pi)\land |v|\le \pi /2\}.$

Consider the equator $E=\{(\cos u, \sin u,0):u\in [0,2\pi)\}.$

For $u\in [0,2\pi)$ let $g(\cos u, \sin u,0)=u.$

If $\sin u \sin v >0$ let $g(\cos u,\sin u \cos v,\sin u \sin v)=3\pi.$

If $\sin u \sin v <0$ let $g(\cos u,\sin u \cos v,\sin u \sin v)=-3\pi.$

Let $f(x)=(g(x),0)$ for all $x\in S^2.$

$f$ maps $E$ bijectively to $[0,2\pi)\times \{0\}.$ And $x\in E\iff x\ne -x\in E.$

If $x\in S^2$ \ $E$ then $\{f(x),f(-x)\}=\{(3\pi,0),(-3\pi,0)\}$ so $f(x)\ne f(-x).$

Answer 3

Fix two distinct elements $a$ and $b$ in $\Bbb R^2$. Consider an arbitrary $(x, y, z) \in S^2.$

• If $x > 0$, map it to $a$ and if $x < 0$, map it to $b$.
• If $x = 0$, then do the same as above with $y$ instead of $x$.
• If $x = y = 0$, then $z \neq 0$ and thus, we can do the above with $z$ instead of $x$.

Note if $\mathbf x = (x, y, z) \in S^2,$ then both $\bf x$ and $-\bf x$ will follow the rule in the same bullet above and hence, one is mapped to $a$ and the other to $b \neq a$.

Answer 4

Here's an overkill using the Axiom of Choice:

Let $\mathcal A = \{\{\mathbf x, -\mathbf x\} \mid \mathbf x \in S^2\}$ be the collection of all antipodal pairs. From each set in $\mathcal A$, pick exactly one element. Map that to $(1, 0)$ and its antipodal counterpart to $(0, 1)$.