 Solution

Let $\varphi :C\setminus S\to {\mathbb{P}}^n$ be the given algebraic morphism of algebraic varieties. Its value at the generic point supplies homogeneous coordinates $[h_0:\cdots :h_n]$ with $h_i\in k(C)$, not all zero. Thus it defines a rational map of projective varieties on $C$.

Fix a missing point $p\in S$. The local ring of a smooth algebraic curve at $p$ is a discrete valuation ring $A={\mathcal{O}}_{C,p}$, with uniformizer $t$ and valuation $v_p$. PutZero coordinates remain zero. Every $a_i$ is in $A$, and some $a_j$ is a unit because its valuation is zero. Choose a neighborhood $U$ of $p$ on which all $a_i$ are regular and $a_j$ remains nonvanishing; this is possible because membership in the local ring means regularity on a neighborhood. The affine target-chart coordinatesare regular on $U$, so they define a morphism of algebraic varieties $U\to {\mathbb{P}}^n$. At $p$ its homogeneous value is $[a_0(p):\cdots :a_n(p)]$, well-defined because $a_j(p)\ne 0$.

Multiplication of all homogeneous coordinates by the same nonzero rational function does not change their projective point. Thus this local morphism agrees with the original one on the common dense open where their coordinates were chosen. They agree throughout the overlap: the target's diagonal is closed, being defined in homogeneous coordinate pairs by $T_i{U}_j-T_j{U}_i=0$ for all $i,j$. Thus the equalizer is closed and contains that dense open. Apply this construction at each point of the finite set $S$ and glue with the original morphism. The same closed-diagonal argument proves uniqueness. ThereforeThis gives the extension of a rational map from a smooth projective curve by explicit local coordinates. The mechanism is the one-dimensional smooth local ring: it permits removal of the smallest common pole order while leaving at least one unit coordinate.