 Solution

Let $A=A(X)=k[x_1,\dots ,x_m]/I(X)$. Polynomial restrictions are regular functions, and a restriction is zero precisely when its class in the coordinate ring is zero. We prove surjectivity.

Using the usual irreducible-variety convention, view a regular function $f$ as an element of the fraction field of $A$. On a neighbourhood of every point it has the form $a/b$, where $a,b\in A$ and $b$ does not vanish at that point. Equality on a nonempty open subset of the irreducible variety is equality of rational functions. DefineThis is an ideal and contains a denominator nonzero at each point. Thus its zero locus is empty. The Hilbert Nullstellensatz implies $J=A$, so $1\in J$ and $f\in A$. ThereforeThe result also holds under a convention allowing reduced reducible affine varieties. One may cover by finitely many principal opens $D(b_i)$, represent $f$ in each localization, and increase powers until the local representatives glue after clearing denominators. Since these opens cover, the ideal generated by those powers is the unit ideal by the Nullstellensatz; a unit linear combination glues their numerators to one element of $A$. Irreducibility is used only for the shorter fraction-field proof, not for the assertion itself.