Bijection between set of morphisms and $K(X)$

Question

Let $X$ be a smooth irreducible curve, and let $f:X\to P^1$ be a morphism. One can easily show that $f$ is either constant or surjective, by using that any regular function on $X$ is constant. Now if $f$ is such that the image of $f$ is not $\{(1:0)\}$ (i.e. $\infty$), then the restriction of $f$ to $X\setminus f^{-1}(\infty)$ defines an element $\tilde f$ of $K(X)$ (rational functions on $X$). I am trying to show that this assignment is a bijection. Injectivity is easy, since we have dense subsets and $P^1$ is Hausdorff. However, I'm struggling with surjectivity. First I thought that if $(U,g)$ is some element of $K(X)$, then one can define an extension to the whole of $X$ by defining $g(x):=\infty$ for all $x\notin U$. However, this map does not seem to be continuous. I have gotten the hint to consider the divisor of $g$, and to cover $X$ by opens $V$ and $W$ such that $f\in O_X(V)$ and $1/f\in O_X(W)$. I have done this, I think, by considering the complements of $\{P\in X:div(f)(P)<0\}$ and $\{P\in X:div(f)>0\}$, but I don't see how that brings me closer to constructing a morphism $X\to P^1$. Any thoughts, hints or help?

Answer 1

You want to cover $X$ by opens $V$ and $W$, where $V$ is the complement of $g$ where $v_P(g) < 0$ and $W$ is the complement of $g$ where $v_P(g)>0$.

So you have a function $g\in \mathcal{O}(V)$ is a regular function on $V$ and $g^{-1} \in \mathcal{O}(W)$ is a regular function on $W$ (this follows, say by the fact that for a normal domain $A$ we have $A = \bigcap_{ht(\mathfrak{p})=1} A_{\mathfrak{p}}$).

This defines two maps $V\rightarrow \mathbb{A}^1$ and $W\rightarrow \mathbb{A}^1$. Cover $\mathbb{P}^1$ (with coordinates $x_0,x_1$) by the standard affine opens $U_0= \{x_0\ne 0\}$ and $U_1 = \{ x_1 \ne 0 \}$, and consider the maps $V\rightarrow \mathbb{A}^1 \simeq U_0 \hookrightarrow \mathbb{P}^1$

and $W \rightarrow \mathbb{A}^1 \simeq U_1 \hookrightarrow \mathbb{P}^1$. We want to show that these two maps agree on the intersection $V\cap W \rightarrow U_0 \cap U_1$, but this is easy to see as the isomorphism $\mathbb{A}^1 - \{0\} \simeq U_0|_{U_0 \cap U_1} = U_1 |_{U_0 \cap U_1} \simeq \mathbb{A}^1$ is given by $k[u,\frac{1}{u}] \rightarrow k[v,\frac{1}{v}]$ sending $u\mapsto \frac{1}{v}$. So this glues to a map to $\mathbb{P}^1$.