A continuous map $f:{S}^{n}\to {S}^{m}$ is *odd* (or *antipodal*)
if $f(-x)=-f(x)$ for all $x\in {S}^{n}$.

If $n\le m$ then it is easy to produce examples of odd continuous maps $f:{S}^{n}\to {S}^{m}$. Most obviously, we can just define

$$f({x}_{0},\mathrm{\dots},{x}_{n})=({x}_{0},\mathrm{\dots},{x}_{n},0,\mathrm{\dots},0).$$ |

If $f:{S}^{n}\to {S}^{m}$ is odd, then we have a well-defined map $\overline{f}:\mathbb{R}{P}^{n}\to \mathbb{R}{P}^{m}$ given by $\overline{f}([x])=[f(x)]$.

Suppose that $f\mathrm{:}{S}^{n}\mathrm{\to}{S}^{n}$ is continuous and odd. Then the induced map ${f}_{\mathrm{*}}\mathrm{:}{H}_{n}\mathit{}\mathrm{(}{S}^{n}\mathrm{;}\mathbb{Z}\mathrm{/}\mathrm{2}\mathrm{)}\mathrm{\to}{H}_{n}\mathit{}\mathrm{(}{S}^{n}\mathrm{;}\mathbb{Z}\mathrm{/}\mathrm{2}\mathrm{)}$ is the identity.

As $f$ is odd we have a well-defined map $\overline{f}:\mathbb{R}{P}^{n}\to \mathbb{R}{P}^{n}$ given by $\overline{f}([x])=[f(x)]$. This satisfies $p\circ f=\overline{f}\circ p:{S}^{n}\to \mathbb{R}{P}^{n}$, so the right-hand square below commutes:

We claim that the left hand square commutes as well. To see this, define $\chi :{S}^{n}\to {S}^{n}$ by $\chi (x)=-x$, so $f\circ \chi =\chi \circ f$ and $p\circ \chi =p$. Consider a continuous map $u:{\mathrm{\Delta}}_{k}\to \mathbb{R}{P}^{n}$. Choose a lift $v:{\mathrm{\Delta}}_{k}\to {S}^{n}$. The other lift is then $\chi \circ v$, so $\tau (u)=v+(\chi \circ v)$, so

$${f}_{\mathrm{\#}}(\tau (u))=(f\circ v)+(f\circ \chi \circ v)=(f\circ v)+(\chi \circ f\circ v).$$ |

Here $f\circ v$ and $\chi \circ f\circ v$ are the two lifts of the map $p\circ f\circ v=\overline{f}\circ p\circ v=\overline{f}\circ u$, so we see that $(f\circ v)+(\chi \circ f\circ v)=\tau ({\overline{f}}_{\mathrm{\#}}(u))$. This shows that ${f}_{\mathrm{\#}}\circ \tau =\tau \circ {\overline{f}}_{\mathrm{\#}}$, so the left hand square commutes as claimed.

As the diagram commutes, we see that the maps ${f}_{*}$ and ${\overline{f}}_{*}$ are compatible with the maps in the exact sequence obtained in Lemma 23.14. In particular, we have commutative squares

for $1\le i\le n$. It is clear that ${\overline{f}}_{*}$ gives the identity on ${H}_{0}(\mathbb{R}{P}^{n};\mathbb{Z}/2)$, and we have seen that all of the maps $\mathrm{\Delta}$ are isomorphisms, so it follows inductively that ${\overline{f}}_{*}$ gives the identity on ${H}_{i}(\mathbb{R}{P}^{n};\mathbb{Z}/2)$ for $1\le i\le n$. We also have a commutative square

We have seen that ${\tau}_{*}$ is an isomorphism and ${\overline{f}}_{*}$ is the identity so ${f}_{*}$ is also the identity, as claimed. ∎

If $n\mathrm{>}m$, then there are no odd continuous maps from ${S}^{n}$ to ${S}^{m}$.

Suppose that $f:{S}^{n}\to {S}^{m}$ is odd and continuous. Define $i:{S}^{m}\to {S}^{n}$ by

$$i({x}_{0},\mathrm{\dots},{x}_{m})=({x}_{0},\mathrm{\dots},{x}_{m},0,\mathrm{\dots},0).$$ |

It is clear that $i$ is odd, so the composite $f\circ i:{S}^{m}\to {S}^{m}$ is odd, so the induced map ${(f\circ i)}_{*}:{H}_{m}({S}^{m};\mathbb{Z}/2)\to {H}_{m}({S}^{m};\mathbb{Z}/2)$ must be the identity by Proposition 24.4. In particular, ${(f\circ i)}_{*}$ is nonzero.

On the other hand, we can define $h:[0,1]\times {S}^{m}\to {S}^{n}$ by

$$h(t,x)=\mathrm{cos}(\pi t/2)i(x)+\mathrm{sin}(\pi t/2){e}_{m+1}.$$ |

(Using the fact that $i(x)$ and ${e}_{m+1}$ are orthogonal, we see that this does indeed lie in ${S}^{n}$.) This gives a homotopy between $i$ and a constant map, which implies that $f\circ i$ is also homotopic to a constant map, so ${(f\circ i)}_{*}=0$. This contradiction shows that no such map $f$ can exist. ∎

Let $g\mathrm{:}{S}^{n}\mathrm{\to}{\mathbb{R}}^{m}$ be a continuous map, with $$. Then there is a point $x\mathrm{\in}{S}^{n}$ with $g\mathit{}\mathrm{(}x\mathrm{)}\mathrm{=}g\mathit{}\mathrm{(}\mathrm{-}x\mathrm{)}$.

Suppose (for a contradiction) that no such point exists, so $g(x)-g(-x)$ is always nonzero. We can then define $f:{S}^{n}\to {S}^{n-1}$ by $f(x)=(g(x)-g(-x))/\parallel g(x)-g(-x)\parallel $. It is easy to check that this is continuous and antipodal, which contradicts Theorem 24.5, as required. ∎

There are two opposite points on the Earth’s surface that have the same temperature and also the same atmospheric pressure, as we see by considering the map $f:{S}^{2}\to {\mathbb{R}}^{2}$ given by

$$f(a)=(\text{temperature at}a,\text{pressure at}a).$$ |

Let ${A}_{\mathrm{1}}$, ${A}_{\mathrm{2}}$ and ${A}_{\mathrm{3}}$ be three reasonable subsets of ${\mathbb{R}}^{\mathrm{3}}$. Then there is a plane $P\mathrm{\subset}{\mathbb{R}}^{\mathrm{3}}$ such that for each set ${A}_{i}$, half of the volume lies on one side of $P$, and half of the volume lies on the other side.

We could have a sandwich, with the top slice of bread filling the set ${A}_{1}$, and cheese filling the set ${A}_{2}$, and the bottom slice of bread filling ${A}_{3}$. The theorem then says that we can make a single straight cut with a knife to share all three components equally.

Suppose that the sets ${A}_{i}$ are solid balls, with centres ${a}_{i}$ in general position. Then there is a unique possible choice for $P$, namely the plane passing through ${a}_{1}$, ${a}_{2}$ and ${a}_{3}$.

We will not be very rigorous about what “reasonable” means, but we will make some comments here and in the body of the proof. To start with, each set ${A}_{i}$ should be bounded (which implies that the volume is finite) and the volume should also not be zero.

For any unit vector $u=({u}_{0},{u}_{1},{u}_{2},{u}_{3})\in {S}^{3}$ and $v=({v}_{1},{v}_{2},{v}_{3})\in {\mathbb{R}}^{3}$ we put $m(u,v)={u}_{0}+{u}_{1}{v}_{1}+{u}_{2}{v}_{2}+{u}_{3}{v}_{3}\in \mathbb{R}$, so $m(-u,v)=-m(u,v)$. We then put

$P(u)$ | $=\{v\in {\mathbb{R}}^{3}|m(u,v)=0\}$ | ||

$H(u)$ | $=\{v\in {\mathbb{R}}^{3}|m(u,v)>0\},$ |

so $P(u)$ is a plane and $H(u)$ is the half-space on one side of that plane. Note that the plane $P(-u)$ is the same as $P(u)$, and $H(-u)$ is the half-space on the opposite side to $H(u)$.

We now define $g:{S}^{3}\to {\mathbb{R}}^{3}$ by

$$g{(u)}_{i}=\text{vol}({A}_{i}\cap H(u)).$$ |

It is not too hard to check that this is continuous. By Corollary 24.6, there exists $u\in {S}^{3}$ with $g(-u)=g(u)$, which means that $\text{vol}({A}_{i}\cap H(u))=\text{vol}({A}_{i}\cap H(-u))$ for $i=1,2,3$. In other words, the plane $P(u)$ bisects each of the sets ${A}_{i}$. ∎