Algebraic topological proof

Assume that ${\displaystyle h:S^{n}\to S^{n-1}}$ is an odd continuous function with ${\displaystyle n>2}$ (the case ${\displaystyle n=1}$ is treated above, the case ${\displaystyle n=2}$ can be handled using basic covering theory). By passing to orbits under the antipodal action, we then get an induced continuous function ${\displaystyle h':\mathbb {RP} ^{n}\to \mathbb {RP} ^{n-1}}$ between real projective spaces, which induces an isomorphism on fundamental groups. By the Hurewicz theorem, the induced ring homomorphism on cohomology with ${\displaystyle \mathbb {F} _{2}}$ coefficients [where ${\displaystyle \mathbb {F} _{2}}$ denotes the field with two elements],

${\displaystyle \mathbb {F} _{2}[a]/a^{n+1}=H^{*}\left(\mathbb {RP} ^{n};\mathbb {F} _{2}\right)\leftarrow H^{*}\left(\mathbb {RP} ^{n-1};\mathbb {F} _{2}\right)=\mathbb {F} _{2}[b]/b^{n},}$

sends ${\displaystyle b}$ to ${\displaystyle a}$. But then we get that ${\displaystyle b^{n}=0}$ is sent to ${\displaystyle a^{n}\neq 0}$, a contradiction.^[3]

One can also show the stronger statement that any odd map ${\displaystyle S^{n-1}\to S^{n-1}}$ has odd degree and then deduce the theorem from this result.

Combinatorial proof

The Borsuk–Ulam theorem can be proved from Tucker's lemma.^[2][4][5]

Let ${\displaystyle g:S^{n}\to \mathbb {R} ^{n}}$ be a continuous odd function. Because g is continuous on a compact domain, it is uniformly continuous. Therefore, for every ${\displaystyle \epsilon >0}$, there is a ${\displaystyle \delta >0}$ such that, for every two points of ${\displaystyle S_{n}}$ which are within ${\displaystyle \delta }$ of each other, their images under g are within ${\displaystyle \epsilon }$ of each other.

Define a triangulation of ${\displaystyle S_{n}}$ with edges of length at most ${\displaystyle \delta }$. Label each vertex ${\displaystyle v}$ of the triangulation with a label ${\displaystyle l(v)\in {\pm 1,\pm 2,\ldots ,\pm n}}$ in the following way:

• The absolute value of the label is the index of the coordinate with the highest absolute value of g: ${\displaystyle |l(v)|=\arg \max _{k}(|g(v)_{k}|)}$.
• The sign of the label is the sign of g, so that: ${\displaystyle l(v)=\operatorname {sgn}(g(v))|l(v)|}$.

Because g is odd, the labeling is also odd: ${\displaystyle l(-v)=-l(v)}$. Hence, by Tucker's lemma, there are two adjacent vertices ${\displaystyle u,v}$ with opposite labels. Assume w.l.o.g. that the labels are ${\displaystyle l(u)=1,l(v)=-1}$. By the definition of l, this means that in both ${\displaystyle g(u)}$ and ${\displaystyle g(v)}$, coordinate #1 is the largest coordinate: in ${\displaystyle g(u)}$ this coordinate is positive while in ${\displaystyle g(v)}$ it is negative. By the construction of the triangulation, the distance between ${\displaystyle g(u)}$ and ${\displaystyle g(v)}$ is at most ${\displaystyle \epsilon }$, so in particular ${\displaystyle |g(u)_{1}-g(v)_{1}|=|g(u)_{1}|+|g(v)_{1}|\leq \epsilon }$ (since ${\displaystyle g(u)_{1}}$ and ${\displaystyle g(v)_{1}}$ have opposite signs) and so ${\displaystyle |g(u)_{1}|\leq \epsilon }$. But since the largest coordinate of ${\displaystyle g(u)}$ is coordinate #1, this means that ${\displaystyle |g(u)_{k}|\leq \epsilon }$ for each ${\displaystyle 1\leq k\leq n}$. So ${\displaystyle |g(u)|\leq c_{n}\epsilon }$, where ${\displaystyle c_{n}}$ is some constant depending on ${\displaystyle n}$ and the norm ${\displaystyle |\cdot |}$ which you have chosen.

The above is true for every ${\displaystyle \epsilon >0}$; since ${\displaystyle S_{n}}$ is compact there must hence be a point u in which ${\displaystyle |g(u)|=0}$.