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I was reading this post and was interested in a more detailed topological approach than one of the answers is giving.

I see that if there were a continuous $\varphi : \mathbb{C}\setminus\{0\}\to\mathbb{C}$ such that $\exp(\varphi(z))=z$, then $\exp: \mathbb{C}\to\mathbb{C}\setminus\{0\}$ is a covering projection.

How would I go from here to showing that $\exp$ and $\varphi$ are homeomorphisms, and not just locally? Is this possible?

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  • $\begingroup$ $\exp$ is a covering projection, regardless of the existence of $\varphi$. Assuming that you showed that $\exp$ is a homeomorphism, what would you do next? How would you show $\mathbb C$ and $\mathbb C\smallsetminus 0$ are not homeomorphic? They are both connected, so you might want to look at fundamental groups, but that might be a bit roundabout... $\endgroup$ – Pedro Tamaroff Nov 8 '17 at 16:20
  • $\begingroup$ @PedroTamaroff I thought you might need existenze of such a $\varphi$ so that you know $\exp$ is surjective. The problem I have in mind comes from a complex analysis class where I don't think I can assume that property. That's okay though. I know a little about fundamental groups but I don't get the connection from covering projection to (global) homeomorphism -- only between evenly covered neighbourhoods and their corresponding sheets. $\endgroup$ – laura_b Nov 8 '17 at 22:13

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