# II.1.21

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This has H = ?, V = ?

### HAPPY

Just parts d,e

Work on the level of presheaves. You clearly have an injection $O \to K$. And there are natural restriction maps $K \to K/O_p$. So there is at least a map $K \to \prod_p K/O_p$

Now let $f$ be a rational function on an open set $U$. Since we're dealing with $\mathbb{P}^1$, after some automorphism we can assume there are no poles at infinity, so write $f = \frac{\prod (x - a_i)}{\prod (x - b_i)}$ the point is that is can poles at only finitely many points $p_1, ..., p_n$ so if $p$ is not one of these points, then $f$ restricts to an element of $O_p$ hence maps to zero in $K/O_p$.

I.e. the map $K \to \prod K/O_p$ is nonzero at only finitely many places, so factors through the direct sum, so there is a factorization on the level of presheaves, so by universal property of sheafification you get $O \to K \to \oplus K/O_p$ and its exact on stalks which gives the result.

For part e, just take a finite number of prescribed principle parts, and sum them, then this will constitute a lift, showing the required map is surjective.