Can I define a function $f:S^1\to S^1$ such that $f(e^{2\pi it})=e^{2\pi i(t+\sqrt{2}k)}$ for some $k\in \mathbb{Z}$. I am not sure even if this is well defined. Say this is, then is it a open map?
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Related: Describing continuous functions on $S^1/\sim$. – Andrew D. Hwang Jul 31 '17 at 20:02
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It's a rotation of the circle about its center, through the angle $2\pi \sqrt 2 k$. – DanielWainfleet Aug 01 '17 at 05:03
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$f(e^{2\pi i(t+1)})=e^{2\pi i(t+1+\sqrt2 k)}=e^{2\pi i(t+\sqrt 2k)}$ so $f$ is well defined
Tsemo Aristide
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I am so sorry I did not type the function correctly. I edited it. In the previous for sure it is the identity – Not_a_topologist Jul 31 '17 at 19:55
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write $f_k=e^{2\pi i(t+\sqrt 2k)}$, $f_{-k}$ is the inverse of $f_k$ so $f$ is open. – Tsemo Aristide Jul 31 '17 at 20:07