**EggMath: The White/Yolk Theorem**

**Proof of the Borsuk-Ulam
Theorem**

Here is an outline of the proof of the
Borsuk-Ulam Theorem; more details can be found in
Section 2.6 of Guillemin and Pollack's book
*Differential Topology*.

As there, we will deal with smooth maps, and
make use of standard results like Sard's
theorem.

Remember that Borsuk-Ulam says that any
*odd* map **f** from **S**^{n}
to itself has odd degree. Here **f** is called
odd if it is equivariant with respect to the
antipodal map: **f(-s)=-f(s)**.

The proof is by induction on **n**. For
**n=1**, lift the map **f** from
**S**^{1} to **S**^{1} to a
map **h** from **R** to **R** with
**h(x+1)=h(x)+deg(d)**. But if **f** is
odd, we find that **h(x+1/2)=h(x)+k/2** for
some odd integer **k**, and it then follows
that **deg(f)=k** is odd.

For the inductive step, let **k** be the
degree of **f**, and let **g** be the
restriction of **f** to the equator
**S**^{n-1}. By Sard's theorem, we can
find a point **a** in the image sphere which is
a regular value for **g** and **f**, meaning
that it is not in the image of **g** and it is
achieved exactly **k** times by **f**.
After a rotation, we can assume that **a** is
exactly the north pole of the sphere.

Because **f** is odd, **k** can be
computed as the number of preimages in the
northern hemisphere of the north or south pole.
Defining **f**^{+} as the restriction
of **f** to the northern hemisphere, composed
with projection to the equatorial plane, we find
that **k** is the number of preimages of
**0** under **f**^{+}.

Now neither north nor south pole is in the
image of **g**, so we can retract **g** onto
the equatorial sphere and get an odd map from
**S**^{n-1} to itself. Applying the
inductive hypothesis, this map has odd degree.

But **g** is the restriction of
**f**^{+} to the boundary. By a
standard lemma in differential topology its degree
is then the number of preimages of the regular
value **0**. (If we restrict
**f**^{+} to small spheres around each
preimage, they have winding number one each.)
Thus we see that **k** must be odd.