Let $f:S^{1} \rightarrow S^{1}$ be $z \rightarrow z^{5}$, then prove that $f$ can't be extended to unit disk, i.e. there is no continuous function from the unit disk to a circle which coincides with $f$ on the boundary.
I know that $f$ defines a nonzero element in the fundamental group of circle. If it can be extended to the unit disk. I don't which tool we can resort to to lead a contradiction.
Suppose there is an extension $f:D^1\rightarrow S^1$. Then $f$ is nullhomotopic, since the disk is contractible. This implies the composition $S^1\hookrightarrow D^1\xrightarrow{f}S^1$ is nullhomotopic. But this is the original map, which is not nullhomotopic.