proof of Borsuk-Ulam theorem

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:

Every odd (that is, antipode-preserving) map $f\mathrm{:}{S}^n\mathrm{\to }{S}^n$ 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

$$\tilde{f}:ℝP^n\to ℝP^n$$

. By cellular approximation, this may be assumed to take the hyperplane at infinity (the $n-1$-cell of the standard cell structure on $ℝP^n$) 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:

$$\begin{array}{ccccccccc}\hfill H_n(A;{\mathbb{Z}}_2)\hfill & \hfill \stackrel{i}{\to }\hfill & \hfill H_n(S^n;{\mathbb{Z}}_2)\hfill & \hfill \stackrel{j}{\to }\hfill & \hfill H_n(S^n,A;{\mathbb{Z}}_2)\hfill & \hfill \stackrel{\partial }{\to }\hfill & \hfill H_{n-1}(A;{\mathbb{Z}}_2)\hfill & \hfill \stackrel{i}{\to }\hfill & \hfill H_{n-1}(S^n,A;{\mathbb{Z}}_2)\hfill \\ \hfill f^{*}\downarrow \hfill & & \hfill f^{*}\downarrow \hfill & & \hfill f^{*}\downarrow \hfill & & \hfill f^{*}\downarrow \hfill & & \hfill f^{*}\downarrow \hfill & & \\ \hfill H_n(A;{\mathbb{Z}}_2)\hfill & \hfill \stackrel{i}{\to }\hfill & \hfill H_n(S^n;{\mathbb{Z}}_2)\hfill & \hfill \stackrel{j}{\to }\hfill & \hfill H_n(S^n,A;{\mathbb{Z}}_2)\hfill & \hfill \stackrel{\partial }{\to }\hfill & \hfill H_{n-1}(A;{\mathbb{Z}}_2)\hfill & \hfill \stackrel{i}{\to }\hfill & \hfill H_{n-1}(S^n,A;{\mathbb{Z}}_2)\hfill \end{array}$$

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 $f^{*}:H_n(S^n;{\mathbb{Z}}_2)\to H_n(S^n,{\mathbb{Z}}_2)$ is an isomorphism. Thus

$$f^{*}:H_{n-1}(A;{\mathbb{Z}}_2)\to H_{n-1}(A;{\mathbb{Z}}_2)$$

is an isomorphism. By the commutativity of the diagram, the map

$$f^{*}:H_n(S^n,A;{\mathbb{Z}}_2)\to H_n(S^n,A;{\mathbb{Z}}_2)$$

is not trivial. I claim it is an isomorphism. $H_n(S^n,A;{\mathbb{Z}}_2)$ 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 $A$, $[A]\in H_{n-1}(A;{\mathbb{Z}}_2)$. 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 $H_n(S^n,A;{\mathbb{Z}}_2)$. Since $H_n(A,{\mathbb{Z}}_2)=0$, by the exactness of the sequence $i:H_n(S^n;{\mathbb{Z}}_2)\to H_n(S^n,A;{\mathbb{Z}}_2)$ is injective, and so by the commutativity of the diagram (or equivalently by the $5$-lemma) $f^{*}:H_n(S^n;{\mathbb{Z}}_2)\to H_n(S^n;{\mathbb{Z}}_2)$ is an isomorphism. Thus $f$ has odd degree.

The other statement of the Borsuk-Ulam theorem is:

There is no odd map $S^n\mathrm{\to }{S}^{n\mathrm{-}\mathrm{1}}$.

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.

Title	proof of Borsuk-Ulam theorem
Canonical name	ProofOfBorsukUlamTheorem
Date of creation	2013-03-22 13:10:33
Last modified on	2013-03-22 13:10:33
Owner	bwebste (988)
Last modified by	bwebste (988)
Numerical id	5
Author	bwebste (988)
Entry type	Proof
Classification	msc 54C99