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I think I read somewhere that all continuous homomorphisms from $\mathbb{S}^{1}$ to itself are of the form $z \mapsto z^{n}$ for some $n \in \mathbb{Z}$. Is this result true? I am not able to prove it. Nor am I able to find any counterexample.

What I have done so far is:

If for some $\theta \in [0,2\pi)$, $e^{\iota\theta}$ goes to $e^{\iota\phi}$ for some $\phi \in [0,2\pi)$, then $e^{\iota n\theta}$ goes to $e^{\iota n \phi}$ for all $n \in \mathbb{Z}$. I thought I could take $n$ to be any rational here and then extend to reals by continuity; but that is not true as a multiple of $2\pi$ may creep in.

Next I tried something using $n^{th}$ roots of unity; but again apart from a number of cases, I could not go very far.

Could someone point me in the right direction?

Vishal Gupta
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1 Answers1

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Hint If $p : \mathbb{R} \to \mathbb{R}$ is a continuous homomorphism then $p$ is of the form $x \mapsto m x$ for some $m \in \mathbb{R}$.

Continuous homomorphisms $\mathbb{R} \to S^1$ are of the form $e^{i m x}$ for $m\in \mathbb{R}$.(using covering map and lifting property)

If $p : S^1 \to S^1$ and $q : \mathbb{R} \to S^1$ are continuous homomorphisms then so is $p \circ q : \mathbb{R} \to S^1$. We have $1 = p (q (0))$. We also know $q$ maps $0$ to $1$ hence $q (0) = e^{i 2 \pi k}$ for some $k \in \mathbb{Z}$. but $1 = e^{i 2 \pi n}$ for some $n \in \mathbb{Z}$. Hence $p(z) = z^m$ for some $m \in \mathbb{Z}$.

Myshkin
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