We now claim that the map f can be lifted to a map f' from P2 to S1 which satisfies p1f' = 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 P2 into the fundamental group of P1. Since the former group is Z2 and the latter is Z, the claim follows since the only homomorphism from Z2 to Z is the trivial one.
We now note that our map f' satisfies
p1f'p2 = fp2 = p1g.Therefore, f'p2 and g are both lifts of f. Therefore, for any x in S2 we have either
g(x) = f'p2(x)or
g(-x)=f'p2(x) = f'p2(-x).In either case, g and f'p2 agree at some x in S2, and so are equal by the unique lifting property of covering maps. However, this is a contradiction since p2, and hence f'p2 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 S2 to S1 with the property that g(-x)=-g(x) for each x in S2.
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 S2 to S1 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 S2, and this contradicts the first claim we made in the proof. QED