Borsuk-Ulam theorem

Let $𝑓 : 𝕊 𝑛 \to ℝ 𝑛$ be a continuous function. Then there exists a pair of antipodes $𝑥, - 𝑥 \in 𝕊 𝑛$ such that $𝑓 (𝑥) = 𝑓 (- 𝑥)$ . #m/thm/topology

Proof

#missing/proof

The case $𝑛 = 1$ is easily proven using the Intermediate value theorem.

Corollaries

Antipodal map from a sphere

If $𝑓 : 𝕊 𝑛 \to ℝ 𝑛$ is a continuous odd map, i.e. the following diagram commutes where $𝑎 : 𝑥 \mapsto - 𝑥$ , then there exists $𝑥 0 \in 𝕊 𝑛$ with $𝑓 (𝑥 0) = 0$ . #m/thm/topology

$% https://q.uiver.app/#q=WzAsNCxbMCwwLCJcXG1hdGhiYiBTXm4iXSxbMiwwLCJcXG1hdGhiYiBSXm4iXSxbMCwyLCJcXG1hdGhiYiBTXm4iXSxbMiwyLCJcXG1hdGhiYiBSXm4iXSxbMCwxLCJmIl0sWzIsMywiZiJdLFswLDIsImEiLDEseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJhcnJvd2hlYWQifX19XSxbMSwzLCJhIiwxLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiYXJyb3doZWFkIn19fV1d \begin{tikzcd}[ampersand replacement=\&] {\mathbb S^n} \&\& {\mathbb R^n} \\ \\ {\mathbb S^n} \&\& {\mathbb R^n} \arrow["f", from=1-1, to=1-3] \arrow["f", from=3-1, to=3-3] \arrow["a"{description}, tail reversed, from=1-1, to=3-1] \arrow["a"{description}, tail reversed, from=1-3, to=3-3] \end{tikzcd}$

Proof

By Borsuk-Ulam there exists $𝑥 \in 𝕊 𝑛$ such that $𝑓 𝑥 = 𝑓 𝑎 𝑥$ but by construction $𝑓 𝑎 𝑥 = 𝑎 𝑓 𝑥$ . Hence $𝑎 𝑓 𝑥 = 𝑓 𝑥 = 𝑓 𝑎 𝑥 = 0$ .

Map from a ball antipodal at the boundary

If $𝑓 : 𝔹 𝑛 + 1 \to ℝ 𝑛 + 1$ is a continuous map odd at its boundary $𝜕 𝔹 𝑛 + 1 = 𝕊 𝑛$ , i.e. the following diagram commutes where $𝑎 : 𝑥 \mapsto - 𝑥$ , then there exists $𝑥 0 \in 𝔹 𝑛 + 1$ with $𝑓 (𝑥 0) = 0$ .

$% `calc` is necessary to draw curved arrows. \usetikzlibrary{calc} % `pathmorphing` is necessary to draw squiggly arrows. \usetikzlibrary{decorations.pathmorphing} % A TikZ style for curved arrows of a fixed height, due to AndréC. \tikzset{curve/.style={settings={#1},to path={(\tikztostart) .. controls ($(\tikztostart)!\pv{pos}!(\tikztotarget)!\pv{height}!270:(\tikztotarget)$) and ($(\tikztostart)!1-\pv{pos}!(\tikztotarget)!\pv{height}!270:(\tikztotarget)$) .. (\tikztotarget)\tikztonodes}}, settings/.code={\tikzset{quiver/.cd,#1} \def\pv##1{\pgfkeysvalueof{/tikz/quiver/##1}}}, quiver/.cd,pos/.initial=0.35,height/.initial=0} % TikZ arrowhead/tail styles. \tikzset{tail reversed/.code={\pgfsetarrowsstart{tikzcd to}}} \tikzset{2tail/.code={\pgfsetarrowsstart{Implies[reversed]}}} \tikzset{2tail reversed/.code={\pgfsetarrowsstart{Implies}}} % TikZ arrow styles. \tikzset{no body/.style={/tikz/dash pattern=on 0 off 1mm}} % https://q.uiver.app/#q=WzAsNixbMiwwLCJcXG1hdGhiYiBCXntuKzF9Il0sWzQsMCwiXFxtYXRoYmIgUl57bisxfSJdLFsyLDIsIlxcbWF0aGJiIEJee24rMX0iXSxbNCwyLCJcXG1hdGhiYiBSXntuKzF9Il0sWzAsMCwiXFxtYXRoYmIgU15uIl0sWzAsMiwiXFxtYXRoYmIgU15uIl0sWzAsMSwiZiJdLFsyLDMsImYiXSxbMSwzLCJhIiwxLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiYXJyb3doZWFkIn19fV0sWzQsMCwiXFxpb3RhIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbNSwyLCJcXGlvdGEiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFs0LDUsImEiLDEseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJhcnJvd2hlYWQifX19XV0= \begin{tikzcd}[ampersand replacement=\&] {\mathbb S^n} \&\& {\mathbb B^{n+1}} \&\& {\mathbb R^{n+1}} \\ \\ {\mathbb S^n} \&\& {\mathbb B^{n+1}} \&\& {\mathbb R^{n+1}} \arrow["f", from=1-3, to=1-5] \arrow["f", from=3-3, to=3-5] \arrow["a"{description}, tail reversed, from=1-5, to=3-5] \arrow["\iota", hook, from=1-1, to=1-3] \arrow["\iota", hook, from=3-1, to=3-3] \arrow["a"{description}, tail reversed, from=1-1, to=3-1] \end{tikzcd}$

Proof

The key is to embed the $(𝑛 + 1)$ -ball in the $(𝑛 + 1)$ -sphere via

𝑖 : 𝔹 𝑛 + 1 ↪ 𝕊 𝑛 + 1 𝑥 \mapsto (𝑥, \sqrt 1 - | 𝑥 ‖)

and then define $¯ 𝑓$ to be the unique function so the following diagram commutes

Then $¯ 𝑓$ is an Antipodal map from a sphere and therefore there exists $𝑥' 0, 𝑎 𝑥' 0 \in 𝕊 𝑛 + 1$ so that $𝑓 𝑥' 0 = 𝑓 𝑎 𝑥' 0 = 0$ . By construction either $𝑥' 0 = 𝑖 𝑥 0$ or $𝑎 𝑥' 0 = 𝑖 𝑥 0$ for some $𝑥 0 \in 𝔹 𝑛 + 1$ , and thus $𝑓 𝑥 0 = ¯ 𝑓 𝑖 𝑥 0 = 0$ .

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