What are the continuous automorphisms of $\Bbb T$?

Question

I wanted to check my reasoning on this problem. From standard Pontrjagin duality arguments, it's not hard to see that the continuous homomorphisms of the torus (to itself) are nothing more than the maps $z\mapsto z^n$ where $n\in\Bbb Z$. If $n\neq \pm 1$, these maps are clearly not bijective since we have multiple $n$th roots. If $n=0$, this gives rise to the trivial homomorphism which is definitely not bijective. This leaves us with $n=\pm 1$. If $n=1$, then this is clearly an automorphism. It's also not hard to see that $n=-1$ corresponds to an automorphism. Is my logic sound? Is the automorphism group of $\Bbb T$ composed of nothing more than the functions $z\mapsto z$ and $z\mapsto z^{-1}$?

Answer 1

The continuous automorphisms are simply $z\mapsto z$ and $z\mapsto z^{-1}$ as stated in the original post.

Answer 2

Ah I just asked the same question because I didn't find this one until it was pointed out to me. But I was looking for an elementary proof. So I posted one there, if you care the link is here: How to prove the group of automorphisms of $S^1$ as a topological group is $\mathbb Z_2$?