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I try to show $\mathbb{S}^1\cong[0,1)$, by the map $f(x) = (\cos2\pi x,\sin2\pi x)$, for $x\in[0,1)$. It's clear that $f$ is continuous and bijective. But I don't know how to show the inverse map $f^{-1}$ is continuous also. Any ideas?

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    $\begingroup$ It isn't.${{}}$ $\endgroup$ – Lord Shark the Unknown Apr 11 at 19:34
  • $\begingroup$ So what the isomorphic mean here? Just need $f$ to be continuous and bijective? $\endgroup$ – QD666 Apr 11 at 19:36
  • $\begingroup$ Show that $f^{-1}$ is discontinuous by considering what $f^{-1}$ does near the point $(1,0)$. $\endgroup$ – kccu Apr 11 at 19:39
  • $\begingroup$ We can't say what "isomorphic" means here - you introduced the word in your question. Where did you get it from? $\endgroup$ – Rob Arthan Apr 11 at 20:06
  • $\begingroup$ it's from an introduction to manifold class. The professor never actually mentioned about what "isomorphic" he refers to? I think I should probably ask him. Thanks anyways $\endgroup$ – QD666 Apr 11 at 20:10

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