# Antipode-preserving map

1. Apr 23, 2012

### 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?
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