Prove or disprove:
Every continuous map from \(S^2\) to \(S^2\) has a fixed point.
Solution
The statement is false.
1. Define the antipodal map\[ a:S^2\longrightarrow S^2, \qquad a(x)=-x. \]Then \(a\) is continuous.
The map \[ \RR^3\longrightarrow\RR^3, \qquad x\longmapsto -x \] is linear and hence continuous, and it preserves the unit sphere \(S^2\). Its restriction to \(S^2\) is therefore continuous.
2. The map \(a\) has no fixed point.
If \(x\in S^2\) were fixed, then \[ x=a(x)=-x, \] so \(2x=0\) and hence \(x=0\). But \(0\notin S^2\). Thus no point of \(S^2\) is fixed by \(a\).
3. Therefore not every continuous self-map of \(S^2\) has a fixed point.
The antipodal map from 1 is a continuous self-map of \(S^2\) and, by 2, is fixed-point-free.