TITLE: How to prove the group of automorphisms of $S^1$ as a topological group is $\mathbb Z_2$?
QUESTION [5 upvotes]: The title basically says it all.  How does one prove the group of automorphisms of $S^1$ (the unit circle in $\mathbb C$), as a topological group, is $\mathbb Z_2$?  I was surprised not to find the answer here already, maybe I just didn't use the right search terms but I looked pretty far and wide and didn't find it.  Nor did a general google search turn up the proof.  I should add that this is an exercise in a "basic" topology book so I'm looking for an elementary self-contained proof that does not refer to any major theorems.

REPLY [7 votes]: The elementary, clear way to do this problem is to show that the only continuous group homomorphisms $\mathbf R\rightarrow\mathbf R$ are $x\mapsto \lambda x$ for some $\lambda\in\mathbf R$. Using this and unique path lifting one shows that the only continuous group homomorphisms $\mathbf R\rightarrow S^1$ are $x\mapsto e^{i\lambda x}$ for fixed $\lambda\in\mathbf R$ as before. Finally, $\mathbf R/\mathbf Z\cong S^1$ as top gps, so any continuous gp hom $\mathbf R/\mathbf Z\cong S^1\rightarrow S^1$ is the same as a $\mathbf Z$-periodic continuous gp hom $\mathbf R\rightarrow S^1$; these are just the involutions corresponding to $\lambda=\pm2\pi$. $\blacksquare$