Exercise 4, § 2

Show that there is no homeomorphism of $S_1$ into $\mathbf{R}$. (Observe that the complement in $S_1$ of any point of $S_1$ is connected, and that every connected subset of $\mathbf{R}$ is an interval.)

Deduce that a homeomorphism of $S_1$ onto a subspace of $S_1$ is necessarily a homeomorphism of $S_1$ onto $S_1$ (use Proposition 4 of no. 4).