# 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 $f:S^{n}\to S^{n}$ 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:

 $\begin{CD}H_{n}(A;\mathbb{Z}_{2})@>{i}>{}>H_{n}(S^{n};\mathbb{Z}_{2})@>{j}>{}>% H_{n}(S^{n},A;\mathbb{Z}_{2})@>{\partial}>{}>H_{n-1}(A;\mathbb{Z}_{2})@>{i}>{}% >H_{n-1}(S^{n},A;\mathbb{Z}_{2})\\ @V{f^{*}}V{}V@V{f^{*}}V{}V@V{f^{*}}V{}V@V{f^{*}}V{}V@V{f^{*}}V{}V\\ H_{n}(A;\mathbb{Z}_{2})@>{i}>{}>H_{n}(S^{n};\mathbb{Z}_{2})@>{j}>{}>H_{n}(S^{n% },A;\mathbb{Z}_{2})@>{\partial}>{}>H_{n-1}(A;\mathbb{Z}_{2})@>{i}>{}>H_{n-1}(S% ^{n},A;\mathbb{Z}_{2})\\ \end{CD}$

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 $f^{*}:H_{n}(S^{n};\mathbb{Z}_{2})\to H_{n}(S^{n},\mathbb{Z}_{2})$ 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 $S^{n}\to S^{n-1}$.

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.

Title proof of Borsuk-Ulam theorem ProofOfBorsukUlamTheorem 2013-03-22 13:10:33 2013-03-22 13:10:33 bwebste (988) bwebste (988) 5 bwebste (988) Proof msc 54C99