Bijective Morphism $f: C \to \mathbb{P}^1$ affine?

Question

Let $C$ be a integral, proper curve over $k$ such that $h^0(\mathcal{O}_C) := \dim_k H^0(C, \mathcal{O}_C) =1 $ and $h^1(\mathcal{O}_C) := \dim_k H^1(C, \mathcal{O}_C) =0 $ holds

and we have a bijective morphism $f: C \to \mathbb{P}^1$.

My interest is to show that $f$ is an affine morphism, so if $U \subset \mathbb{P}^1$ open affine, then $f^{-1}(U) \subset C$ is also affine.

I suppose that this could be follow from Serre's criterion for affinity:

My attempts: Let $U \subset \mathbb{P}^1$ open affine, $\mathcal{F}$ quasi coherent sheaf then we have by definition $H^0(f^{-1}(U), \mathcal{F}) = H^0(U, f_* \mathcal{F})$.

Here I encounter following problems:

Is $f_*\mathcal{F}$ quasi coherent? Does $H^i(f^{-1}(U), \mathcal{F}) = H^i(U, f_* \mathcal{F})$ hold for $i >0$?

Or has here Serre's criterion to be used in another way?

Answer 1

We have that $f$ is proper by cancellation (namely $C \to k$ is proper and $\mathbb{P}^{1}_{k} \to k$ is separated). Thus the underlying map of $f$ is a homeomorphism (since it is a closed, continuous, bijective map); in particular the restriction $f^{-1}(U) \to U$ is a homeomorphism. Thus $H^{i}(f^{-1}(U),\mathcal{F}) = H^{i}(U,f_{\ast}\mathcal{F})$ for $i \ge 0$ since cohomology is computed as an abelian sheaf on the underlying topological space. The pushforward $f_{\ast}\mathcal{F}$ is quasi-coherent since $f$ is a (trivially) quasi-compact and quasi-separated morphism.

Answer 2

If $C$ is any integral curve over a field and $U$ is any open set with $U\neq C$, then $U$ is affine.