## Proof of Theorem

We first show that there is no continuous map *g* from
**S**^{2} to
**S**^{1} with the property that
*g(-x)=-g(x)* for each *x* in
**S**^{2}. We prove this by
contradiction. Suppose that *g* is such a map and let
*f* be the induced map from the projective space
**P**^{2} to the projective space
**P**^{1}. Thus *f* is
defined by *fp*_{2} =
p_{1}g where
*p*_{i} is the canonical
projection from
**S**^{i} to
**P**^{i}.
We now claim that the map *f* can be lifted to a map *f'* from
**P**^{2} to
**S**^{1} which satisfies
*p*_{1}f' = f. To do this,
it suffices to show that the map of fundamental groups induced by *f*
is trivial. But this follows since this induced map is a homomorphism from
the fundamental group of
**P**^{2} into the fundamental
group of
**P**^{1}. Since the former group
is **Z**_{2} and the latter is
**Z**, the claim follows since the only homomorphism from
**Z**_{2} to **Z** is the
trivial one.

We now note that our map *f'* satisfies

*
p*_{1}f'p_{2} =
fp_{2} =
p_{1}g.

Therefore, *f'p*_{2} and *g*
are both lifts of *f*. Therefore, for any *x* in
**S**^{2} we have either
*
g(x) = f'p*_{2}(x)

or
*
g(-x)=f'p*_{2}(x) =
f'p_{2}(-x).

In either case, *g* and *f'p*_{2}
agree at some *x* in
**S**^{2}, and so are equal by
the unique lifting property of covering maps. However, this is a
contradiction since
*p*_{2}, and hence
*f'p*_{2} maps *x* to the
same point as *-x*, whereas we assumed that *g* mapped them to
antipodal points. Therefore, there cannot be a continuous map *g*
from
**S**^{2} to
**S**^{1} with the property that
*g(-x)=-g(x)* for each *x* in
**S**^{2}.
The remainder of the proof of the theorem now follows easily. Assume that
*f* is as in the theorem statement, and that no point *x*
exists as asserted in the theorem. Then we can define a function *g*
from **S**^{2} to
**S**^{1} by

*
g(x)=[f(x)-f(-x)]/|f(x)-f(-x)|
*

where |.| is the Euclidean norm. This function has the property that
*g(x)=g(-x)* for every *x* in
**S**^{2}, and this
contradicts the first claim we made in the proof. QED

*Andrew D. Lewis (andrew at mast.queensu.ca)*