# Proving a nonconstant holomorphic map from $\mathbb{C}_\infty \to Y$ gives a homeomorphism

I have a nonconstant holomorphic map $f : \mathbb{C}_\infty \to Y$, where $\mathbb{C}_\infty$ is the Riemann sphere and $Y$ is compact and connected, and I want to prove that $\mathbb{C}_\infty$ and $Y$ are homeomorphic.

I know that $f$ is surjective, since it's a non-constant map of compact connected Riemann surfaces, so its image is open (open mapping theorem) and closed (image is compact in Hausdorff space), so it's all of $Y$.

Where do I go from here? If the degree of the map is exactly 1, then I'm ok since that means the map is injective and it's already a homeomorphism I think. If it's more than 1, then it's not even injective, so how can I come up with a homeomorphism without knowing more about $f$?

• Such a map need not be a homeomorphism: consider $f(z) = z^2$ on the Riemann sphere. The surfaces will be homeomorphic, but not necessarily via $f$. Look for other tools: show that $Y$ admits a metric of positive curvature? or that its fundamental group is trivial?
– user357151
Apr 29, 2018 at 23:56

I think I've come up with a solution for this.

$Y$ is a Riemann surface, so is orientable. We know it's compact and connected, so this restricts the possibilities for $Y$ to being homeomorphic to the 2-sphere (this is what we want) or a connected sum of $g$ tori.

$$\chi(Y) = 2-2g$$

Riemann-Hurwitz:

$$\chi(X) = (\deg f) \;\chi(Y) - b(f)$$

So:

$$2 = (\deg f) (2-2g) - b(f)$$

$b(f)$ and $\deg f$ are always greater than or equal to zero, so the only $g$ that can possibly work is $g = 0$, which means $Y$ is homeomorphically the Riemann sphere.

This doesn't actually construct the homeomorphism, but that seems like it would require more information.