ProofOfBorsukUlamTheorem 2002-11-23 22:25:14 2002-12-05 21:08:15 Proof Borsuk-Ulam theorem 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amscd} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newcommand{\Z}{\mathbb{Z}} \newcommand{\Zt}{\Z_2} Proof of the Borsuk-Ulam theorem: I'm going to prove a stronger statement than the one given in the statement of the Borsak-Ulam theorem here, which is: {\em Every odd (that is, antipode-preserving) map $f:S^n\to S^n$ has odd degree.} Proof: We go by induction on $n$. Consider the pair $(S^n,A)$ where $A$ is the equatorial sphere. $f$ defines a map $$\tilde{f}:\mathbb{R}P^n\to\mathbb{R}P^n$$. By cellular approximation, this may be assumed to take the hyperplane at infinity (the $n-1$-cell of the standard cell structure on $\mathbb{R}P^n$) to itself. Since whether a map lifts to a covering depends only on its homotopy class, $f$ is homotopic to an odd map taking $A$ to itself. We may assume that $f$ is such a map. The map $f$ gives us a morphism of the long exact sequences: $$\begin{CD} H_n(A;\Zt)@>i>> H_n(S^n;\Zt)@>j>> H_n(S^n,A;\Zt)@>\partial>> H_{n-1}(A;\Zt) @>i>> H_{n-1}(S^n,A;\Zt)\\ @Vf^*VV @Vf^*VV @Vf^*VV @Vf^*VV @Vf^*VV \\ H_n(A;\Zt)@>i>> H_n(S^n;\Zt)@>j>> H_n(S^n,A;\Zt)@>\partial>> H_{n-1}(A;\Zt) @>i>> H_{n-1}(S^n,A;\Zt)\\ \end{CD}$$ Clearly, the map $f|_A$ is odd, so by the induction hypothesis, $f|_A$ has odd degree. Note that a map has odd degree if and only if $f^*:H_n(S^n;\Zt)\to H_n(S^n,\Zt)$ is an isomorphism. Thus $$f^*:H_{n-1}(A;\Zt)\to H_{n-1}(A;\Zt)$$ is an isomorphism. By the commutativity of the diagram, the map $$f^*:H_n(S^n,A;\Zt)\to H_n(S^n,A;\Zt)$$ is not trivial. I claim it is an isomorphism. $H_n(S^n,A;\Zt)$ is generated by cycles $[R^+]$ and $[R^-]$ which are the fundamental classes of the upper and lower hemispheres, and the antipodal map exchanges these. Both of these map to the fundamental class of $A$, $[A]\in H_{n-1}(A;\Zt)$. By the commutativity of the diagram, $\partial(f^*([R^\pm]))=f^*(\partial([R^\pm]))=f^*([A])=[A]$. Thus $f^*([R^+])=[R^\pm]$ and $f^*([R^-]) =[R^\mp]$ since $f$ commutes with the antipodal map. Thus $f^*$ is an isomorphism on $H_n(S^n,A;\Zt)$. Since $H_n(A,\Zt)=0$, by the exactness of the sequence $i:H_n(S^n;\Zt) \to H_n(S^n,A;\Zt)$ is injective, and so by the commutativity of the diagram (or equivalently by the $5$-lemma) $f^*:H_n(S^n;\Zt)\to H_n(S^n;\Zt)$ is an isomorphism. Thus $f$ has odd degree. The other statement of the Borsuk-Ulam theorem is: {\em There is no odd map $S^n\to S^{n-1}$.} Proof: If $f$ where such a map, consider $f$ restricted to the equator $A$ of $S^n$. This is an odd map from $S^{n-1}$ to $S^{n-1}$ and thus has odd degree. But the map $$f^*H_{n-1}(A)\to H_{n-1}(S^{n-1})$$ factors through $H_{n-1}(S^n)=0$, and so must be zero. Thus $f|_A$ has degree 0, a contradiction.