Why Can't There Be a Continuous Antipode-Preserving Map from S2 to S1?

[murmillo]

I don't understand the proof of this theorem: There is no continuous antipode-preserving map g: S²→S¹.

The proof is like this: Suppose g: S²→S¹ is continuous and antipode-preserving. Take S¹ to be the equator of S². Then the restriction of g to S¹ is a continuous antipode-preserving map h of S¹ to itself. By a previous theorem, h is not nullhomotopic. The upper hemisphere of S² is homeomorphic to the ball B², and g is a continuous extension of h to the upper hemisphere.

The above makes sense. But I don't understand how that gives a contradiction. The upper hemisphere is contractible, so g is homotopic to a constant map. But why would that imply that h is nullhomotopic? The restriction of the homotopy from g to a constant map to just the equator isn't necessarily a homotopy, right?

 

[mighty2000]

Thank you for bringing up this question about the proof of the theorem. Let me try to explain it in more detail.

First, let's recall the definition of a nullhomotopic map. A map f: X→Y is said to be nullhomotopic if there exists a continuous map F: X×[0,1]→Y such that F(x,0) = f(x) for all x∈X and F(x,1) is a constant map for all x∈X. Intuitively, this means that the map f can be continuously deformed to a constant map.

Now, let's consider the restriction of g to the equator S1 of S2. This restriction is the map h: S1→S1. By the previous theorem mentioned in the proof, we know that h is not nullhomotopic. This means that there is no continuous deformation of h to a constant map. In other words, h cannot be continuously deformed to a point on the equator.

Next, we consider the upper hemisphere of S2, which is homeomorphic to the ball B2. Since g is a continuous extension of h to the upper hemisphere, it follows that g is also not nullhomotopic. This is because if g were nullhomotopic, then h would also be nullhomotopic (since h is just the restriction of g to the equator). But we know that h is not nullhomotopic, so it follows that g cannot be nullhomotopic either.

Now, since the upper hemisphere is contractible, there exists a homotopy H: B2×[0,1]→B2 such that H(x,0) = g(x) for all x∈B2 and H(x,1) is a constant map for all x∈B2. This means that g can be continuously deformed to a constant map on the upper hemisphere.

But this is a contradiction, because we know that g is not nullhomotopic. So, we have shown that there can be no continuous antipode-preserving map g: S2→S1, as it would lead to a contradiction.

I hope this helps to clarify the proof for you. Let me know if you have any further questions or concerns.