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Consider the same setup with $f$ as mentioned in Continuous bijection which is not a homeomorphism., i.e $f:[0,2\pi)\to S^1$, $\varphi\mapsto (\cos(\varphi), \sin(\varphi))$.

I am confused about the following wrong reasoning that leads me to conclude that $f^{-1}$ is continuous:

To show that $f^{-1}$ is continuous, it suffices to show that $f(U)$ is open for all $U \subseteq [0, 2\pi)$ open. But this seems to be true since we are just mapping it along the circle, such that open sets would map to open sets on the circle. We don't run into any issues with the wrap-around at 0 or $2 \pi$ since we only consider open sets.

What am I getting wrong? Thank you.

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  • $\begingroup$ Do you remember the definition of an open set in $[0,2\pi)$? $\endgroup$
    – Arthur
    Commented Sep 9, 2022 at 4:22
  • $\begingroup$ Is $f([0, \pi))$ open? $\endgroup$
    – Qiaochu Yuan
    Commented Sep 9, 2022 at 4:31
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    $\begingroup$ Oh, I see now... $[0, \pi)$ is in fact open in $[0, 2\pi)$, which maps to something that is neither open nor closed. Thank you! $\endgroup$
    – nerraruzi
    Commented Sep 9, 2022 at 4:33

1 Answer 1

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Perhaps a really elementary mistake as someone who only mostly worked in $\mathbb{R}$ before, and as Arthur and Qiaochu pointed out, we can find an open interval (i.e $[0, \pi)$ that maps to something which is not open.

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