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Let $\phi: S^1 \rightarrow S^1$ homomorphism continouos injective. Then $\phi(x)= x$ or $\phi(x)= \overline{x}$ for all $x \in S^1$, where $\overline{x}$ is conjugate.

Attempt. I proved $\phi$ is bijective function then $\phi$ topological homomorphism. Then we know $\ker \left (\phi \right)={1}$, and $(-1)^2=1$ so

$f(-1)^2=1$ but $-1 \notin \ker \left (\phi \right)$ then $f(-1)=-1$.

Other case is, $i^2=-1$, then $f(i)^2=-1$. Then we have two cases $f(i)=i$ and $f(i)=-i$. In the first case I'll have three fixed points so, I suppose $\phi$ is identity and the other case $\phi$ is the conjugate. But I have problems to proceed with that idea. Am I right with this idea?

Arturo Magidin
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  • What have you tried by using the assumption that $\phi$ is continuous? – Chickenmancer Mar 06 '24 at 17:41
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    See here, here, here for example. – Izaak van Dongen Mar 06 '24 at 18:17
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    The most basic approach in line w/ what OP has already done: focus on the degree 1 case and argue $f$ preserves $2^k$th roots of unity by inducting on $k$. E.g. if $k=2$ and $f(i)=-i$ argue $f$ must have negative degree. The inductive argument is similar [just focus on the std generator being a fixed point] and get the result by continuity [and conjugating $f$ for degree -1 case]. A less intuitive but easier argument would be to do the induction on degree -1 map always sending $2^k$ th roots of unity to their conjugates since an orientation reversing homeomorphism has exactly 2 fixed points. – user8675309 Mar 06 '24 at 20:22

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