Unramified holomorphic map is isomorphism

Question

In my course about Riemann surfaces, the professor briefly mentioned the following as a fact that we shall just accept:

If $X$ is a connected, compact Riemann surface and $f:X\to\mathbb C_\infty$ is a holomorphic unramified map, then $f$ is necessarily an isomorphism.

Which ingredients are used in order to prove this result? I'd like to have a go at a proof myself, but I'm not sure if I have the needed knowledge yet.

@MoisheKohan I added it, but I think this is generally part of the definition of a Riemann surface, isn't it? — Zuy, Apr 25 '22 at 15:45

score 1 · Accepted Answer · answered Apr 24 '22 at 08:23

Since $\mathbb{C}_{\infty} = \mathbb{C} \cup \{\infty\}$ is simply connected and $f$ is unramified (i.e. a covering map), $f$ is a homeomorphism and admits a continuous inverse $f^{-1}$. Then the inverse is also holomorphic. The ingredients to understand the first answer would be some covering theory in topology, and I think the course you're taking may cover all the concepts you need to understand second answer.

Unramified holomorphic map is isomorphism

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