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I wanted find all possible Homeomorphism between $S^1\to S^1$

I know that $f:S^1\to S^1$ defined as $f(e^{it})=e^{ait}$ for $a=\{1,-1\}$ are some possible Homeomorphism. also Rotation are also possible i.e. $g:S^1\to S^1$ defined as $g(e^{it})=e^{(a+t)i}$ for $a\in [0,2\pi)$

But What are the all possible? I do not know how answer such question

Please give me Hint I wanted to solve this problem.

Any Help will be appreciated

Curious student
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    I don't know that all possible homeomorphisms can be described that easily. There are many of them. Also, your $f$ is only a homeomorphism for $a=\pm1$. Otherwise it's not injective. – Arthur Sep 05 '19 at 04:49
  • Yes .Sorry I missed that – Curious student Sep 05 '19 at 04:52
  • Between the question statement and with claim about $f(z) = z^a$ for all $a \in \Bbb Z$, I wonder if instead of "homeomorphism" OP wants to know about "group homomorphisms". – Travis Willse Sep 05 '19 at 04:53
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    Actually I was doing excerise of Hatcher There is question that we have to write all group homemorphism between $\pi(S^1)\to \pi(S^1)$ in terms of Homomorphism $S^1\to S^1$ But I know that $\pi(S^1)=\mathbb Z$ all set of homomorphism between $Z\to Z$ are of form $x\to ax$ .SO I wanted to know what are possible – Curious student Sep 05 '19 at 04:59
  • For a homomorphism $\pi_1(S^1)\to\pi_1(S^1)$ given by $x\mapsto ax$ the corresponding map $S^1\to S^1$ is simply $z\mapsto z^a$ (complex numbers multiplication). Which are not homeomorphisms except for $a=-1,1$. And in fact for $a\neq -1,1$ the map $x\mapsto ax$ does not arise from a homeomorphism. – freakish Sep 05 '19 at 05:00
  • Sir In Hatcher sec 1.1 problem 10 question is that SHow that for every homomorphism between $\pi(S^1)\to\pi(S^1)$ can be realised as induced homeomorphism between $S^1\to S^1$.This is excersise question – Curious student Sep 05 '19 at 05:05
  • @SRJ You mean "Algebraic Topology"? What exactly page is it? I can't find it. – freakish Sep 05 '19 at 05:08
  • Page 39 question 12. – Curious student Sep 05 '19 at 05:09
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    @SRJ Right. So read question 12 again and very carefuly. There is no word "homeomorphism" there. :) It's unfortunate that so different objects differ only by "e" in their name. Anyway he simply states that every homomorphism is an induced homomorphism (which is true due to special nature of $S^1$, generally it is not true for any $X$). – freakish Sep 05 '19 at 05:10
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    ohh Sir Extermly sorry I had not seen that properly I think. – Curious student Sep 05 '19 at 05:17
  • @reuns Did you mean to tag OP instead of me? My only previous comment here was just a speculation that OP meant 'homomorphisms' and not 'homeomorphisms' (which turned out to be the case). – Travis Willse Sep 05 '19 at 05:53
  • I wanted to tag Arthur sorry – reuns Sep 05 '19 at 05:54
  • You might also find this interesting, re: your actual question. – Simone Ramello Sep 05 '19 at 08:02

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