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Is there a conformal diffeomorphism between $\mathbb{R}^3$ and $S^1\times S^2$ minus a point on the $S^2$? I would have thought not, because the fundamental group of $\mathbb{R}^3$ is trivial, but that of $S^1\times S^2$ minus a point on the $S^2$ is not.

user12588
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    It's not clear if you mean $S^1 \times (S^2 \setminus {p})$ or $S^1 \times S^2 \setminus {(p, q)}$. But in both cases the manifolds are not homeomorphic to $\mathbb R^3$. – Arctic Char Mar 06 '23 at 21:13
  • I was thinking more about the first case - thanks. I was almost sure that my argument based on the fundamental groups was correct, but I just wanted to double check. – user12588 Mar 06 '23 at 21:52

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You say yourself that the spaces have different fundamental groups, thus not homotopy equivalent, much less diffeomorphic. So, no.

Igor Rivin
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